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Intermediate Algebra 2e

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Contents

  1. Preface
  2. Foundations
    1. Introduction
    2. Use the Language of Algebra
    3. Integers
    4. Fractions
    5. Decimals
    6. Properties of Real Numbers
  3. Solving Linear Equations
    1. Introduction
    2. Use a General Strategy to Solve Linear Equations
    3. Use a Problem Solving Strategy
    4. Solve a Formula for a Specific Variable
    5. Solve Mixture and Uniform Motion Applications
    6. Solve Linear Inequalities
    7. Solve Compound Inequalities
    8. Solve Absolute Value Inequalities
  4. Graphs and Functions
    1. Introduction
    2. Graph Linear Equations in Two Variables
    3. Slope of a Line
    4. Find the Equation of a Line
    5. Graph Linear Inequalities in Two Variables
    6. Relations and Functions
    7. Graphs of Functions
  5. Systems of Linear Equations
    1. Introduction
    2. Solve Systems of Linear Equations with Two Variables
    3. Solve Applications with Systems of Equations
    4. Solve Mixture Applications with Systems of Equations
    5. Solve Systems of Equations with Three Variables
    6. Solve Systems of Equations Using Matrices
    7. Solve Systems of Equations Using Determinants
    8. Graphing Systems of Linear Inequalities
  6. Polynomials and Polynomial Functions
    1. Introduction
    2. Add and Subtract Polynomials
    3. Properties of Exponents and Scientific Notation
    4. Multiply Polynomials
    5. Dividing Polynomials
  7. Factoring
    1. Introduction to Factoring
    2. Greatest Common Factor and Factor by Grouping
    3. Factor Trinomials
    4. Factor Special Products
    5. General Strategy for Factoring Polynomials
    6. Polynomial Equations
  8. Rational Expressions and Functions
    1. Introduction
    2. Multiply and Divide Rational Expressions
    3. Add and Subtract Rational Expressions
    4. Simplify Complex Rational Expressions
    5. Solve Rational Equations
    6. Solve Applications with Rational Equations
    7. Solve Rational Inequalities
  9. Roots and Radicals
    1. Introduction
    2. Simplify Expressions with Roots
    3. Simplify Radical Expressions
    4. Simplify Rational Exponents
    5. Add, Subtract, and Multiply Radical Expressions
    6. Divide Radical Expressions
    7. Solve Radical Equations
    8. Use Radicals in Functions
    9. Use the Complex Number System
  10. Quadratic Equations and Functions
    1. Introduction
    2. Solve Quadratic Equations Using the Square Root Property
    3. Solve Quadratic Equations by Completing the Square
    4. Solve Quadratic Equations Using the Quadratic Formula
    5. Solve Equations in Quadratic Form
    6. Solve Applications of Quadratic Equations
    7. Graph Quadratic Functions Using Properties
    8. Graph Quadratic Functions Using Transformations
    9. Solve Quadratic Inequalities
  11. Exponential and Logarithmic Functions
    1. Introduction
    2. Finding Composite and Inverse Functions
    3. Evaluate and Graph Exponential Functions
    4. Evaluate and Graph Logarithmic Functions
    5. Use the Properties of Logarithms
    6. Solve Exponential and Logarithmic Equations
  12. Conics
    1. Introduction
    2. Distance and Midpoint Formulas; Circles
    3. Parabolas
    4. Ellipses
    5. Hyperbolas
    6. Solve Systems of Nonlinear Equations
  13. Sequences, Series and Binomial Theorem
    1. Introduction
    2. Sequences
    3. Arithmetic Sequences
    4. Geometric Sequences and Series
    5. Binomial Theorem

Preface

Welcome to Intermediate Algebra 2e, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.

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Errata

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Format

You can access this textbook for free in web view or PDF through openstax.org, and for a low cost in print.

About Intermediate Algebra 2e

Intermediate Algebra 2e is designed to meet the scope and sequence requirements of a one-semester Intermediate Algebra course. The book’s organization makes it easy to adapt to a variety of course syllabi. The text expands on the fundamental concepts of algebra while addressing the needs of students with diverse backgrounds and learning styles. Each topic builds upon previously developed material to demonstrate the cohesiveness and structure of mathematics.

Coverage and Scope

Intermediate Algebra 2e continues the philosophies and pedagogical features of Prealgebra and Elementary Algebra, by Lynn Marecek and MaryAnne Anthony-Smith. By introducing the concepts and vocabulary of algebra in a nurturing, non-threatening environment while also addressing the needs of students with diverse backgrounds and learning styles, the book helps students gain confidence in their ability to succeed in the course and become successful college students.

The material is presented as a sequence of small, and clear steps to conceptual understanding. The order of topics was carefully planned to emphasize the logical progression throughout the course and to facilitate a thorough understanding of each concept. As new ideas are presented, they are explicitly related to previous topics.

  • Chapter 1: Foundations
    Chapter 1 reviews arithmetic operations with whole numbers, integers, fractions, decimals and real numbers, to give the student a solid base that will support their study of algebra.
  • Chapter 2: Solving Linear Equations and Inequalities
    In Chapter 2, students learn to solve linear equations using the Properties of Equality and a general strategy. They use a problem-solving strategy to solve number, percent, mixture and uniform motion applications. Solving a formula for a specific variable, and also solving both linear and compound inequalities is presented.
  • Chapter 3: Graphs and Functions
    Chapter 3 covers the rectangular coordinate system where students learn to plot graph linear equations in two variables, graph with intercepts, understand slope of a line, use the slope-intercept form of an equation of a line, find the equation of a line, and create graphs of linear inequalities. The chapter also introduces relations and functions as well as graphing of functions.
  • Chapter 4: Systems of Linear Equations
    Chapter 4 covers solving systems of equations by graphing, substitution, and elimination; solving applications with systems of equations, solving mixture applications with systems of equations, and graphing systems of linear inequalities. Systems of equations are also solved using matrices and determinants.
  • Chapter 5: Polynomials and Polynomial Functions
    In Chapter 5, students learn how to add and subtract polynomials, use multiplication properties of exponents, multiply polynomials, use special products, divide monomials and polynomials, and understand integer exponents and scientific notation.
  • Chapter 6: Factoring
    In Chapter 6, students learn the process of factoring expressions and see how factoring is used to solve quadratic equations.
  • Chapter 7: Rational Expressions and Functions
    In Chapter 7, students work with rational expressions, solve rational equations and use them to solve problems in a variety of applications, and solve rational inequalities.
  • Chapter 8: Roots and Radical
    In Chapter 8, students simplify radical expressions, rational exponents, perform operations on radical expressions, and solve radical equations. Radical functions and the complex number system are introduced
  • Chapter 9: Quadratic Equations
    In Chapter 9, students use various methods to solve quadratic equations and equations in quadratic form and learn how to use them in applications. Students will graph quadratic functions using their properties and by transformations.
  • Chapter 10: Exponential and Logarithmic Functions
    In Chapter 10, students find composite and inverse functions, evaluate, graph, and solve both exponential and logarithmic functions.
  • Chapter 11: Conics
    In Chapter 11, the properties and graphs of circles, parabolas, ellipses and hyperbolas are presented. Students also solve applications using the conics and solve systems of nonlinear equations.
  • Chapter 12: Sequences, Series and the Binomial Theorem
    In Chapter 12, students are introduced to sequences, arithmetic sequences, geometric sequences and series and the binomial theorem.

All chapters are broken down into multiple sections, the titles of which can be viewed in the Table of Contents.

Changes to the Second Edition

The Intermediate Algebra 2e revision focused on mathematical clarity and accuracy. Every Example, Try-It, Section Exercise, Review Exercise, and Practice Test item was reviewed by multiple faculty experts, and then verified by authors. This intensive effort resulted in hundreds of changes to the text, problem language, answers, instructor solutions, and graphics.

However, OpenStax and our authors are aware of the difficulties posed by shifting problem and exercise numbers when textbooks are revised. In an effort to make the transition to the 2nd edition as seamless as possible, we have minimized any shifting of exercise numbers. For example, instead of deleting or adding problems where necessary, we replaced problems in order to keep the numbering intact. As a result, in nearly all chapters, there will be no shifting of exercise numbers; in the chapters where shifting does occur, it will be minor. Faculty and course coordinators should be able to use the new edition in a straightforward manner.

Also, to increase convenience, answers to the Be Prepared Exercises will now appear in the regular solutions manuals, rather than as a separate resource.

A detailed transition guide is available as an instructor resource at openstax.org.

Pedagogical Foundation and Features

Examples

Each learning objective is supported by one or more worked examples, which demonstrate the problem-solving approaches that students must master. Typically, we include multiple examples for each learning objective to model different approaches to the same type of problem, or to introduce similar problems of increasing complexity.

All examples follow a simple two- or three-part format. First, we pose a problem or question. Next, we demonstrate the solution, spelling out the steps along the way. Finally (for select examples), we show students how to check the solution. Most examples are written in a two-column format, with explanation on the left and math on the right to mimic the way that instructors “talk through” examples as they write on the board in class.

Be Prepared!

Each section, beginning with Section 2.1, starts with a few “Be Prepared!” exercises so that students can determine if they have mastered the prerequisite skills for the section. Reference is made to specific Examples from previous sections so students who need further review can easily find explanations. Answers to these exercises can be found in the supplemental resources that accompany this title.

Try It

A dark grey right-pointing chevron symbol is centered within a white square, framed by a soft, light blue gradient border. This icon typically represents 'next,' 'forward,' or 'expand'. Try it The Try It feature includes a pair of exercises that immediately follow an Example, providing the student with an immediate opportunity to solve a similar problem with an easy reference to the example. In the PDF and the Web View version of the text, answers to the Try It exercises are located in the Answer Key

How To

An avatar of a person with a neutral expression, next to a speech bubble containing three ellipses, suggesting thought, a pending response, or a message being typed. How To Examples use a three column format to demonstrate how to solve an example with a certain procedure. The first column states the formal step, the second column is in words as the teacher would explain the process, and then the third column is the actual math. A How To procedure box follows each of these How To examples and summarizes the series of steps from the example. These procedure boxes provide an easy reference for students.

Media

A teal-colored play button icon, depicted as a right-pointing triangle, centered within a white square with a soft, blurred teal border. It signifies the start of video or audio playback. Media The “Media” icon appears at the conclusion of each section, just prior to the Self Check. This icon marks a list of links to online videos that reinforce the concepts and skills introduced in the section.

Disclaimer: While we have selected videos that closely align to our learning objectives, we did not produce these tutorials, nor were they specifically produced or tailored to accompany Intermediate Algebra 2e.

Self Check

The Self Check includes the learning objectives for the section so that students can self-assess their mastery and make concrete plans to improve.

Art Program

Intermediate Algebra 2e contains many figures and illustrations. Art throughout the text adheres to a clear, understated style, drawing the eye to the most important information in each figure while minimizing visual distractions.

This figure shows three x y-coordinate planes. The first plane shows two lines which intersect at one point. Under the graph it says, “The lines intersect. Intersecting lines have one point in common. There is one solution to this system.” The second x y-coordinate plane shows two parallel lines. Under the graph it says, “The lines are parallel. Parallel lines have no points in common. There is no solution to this system.” The third x y-coordinate plane shows one line. Under the graph it says, “Both equations give the same line. Because we have just one line, there are infinitely many solutions.”

Section Exercises

Each section of every chapter concludes with a well-rounded set of exercises that can be assigned as homework or used selectively for guided practice. Exercise sets are named Practice Makes Perfect to encourage completion of homework assignments.

  • Exercises correlate to the learning objectives. This facilitates assignment of personalized study plans based on individual student needs.
  • Exercises are carefully sequenced to promote building of skills.
  • Values for constants and coefficients were chosen to practice and reinforce arithmetic facts.
  • Even and odd-numbered exercises are paired.
  • Exercises parallel and extend the text examples and use the same instructions as the examples to help students easily recognize the connection.
  • Applications are drawn from many everyday experiences, as well as those traditionally found in college math texts.
  • Everyday Math highlights practical situations using the concepts from that particular section
  • Writing Exercises are included in every exercise set to encourage conceptual understanding, critical thinking, and literacy.

Chapter Review Features

Each chapter concludes with a review of the most important takeaways, as well as additional practice problems that students can use to prepare for exams.

  • Key Terms provide a formal definition for each bold-faced term in the chapter.
  • Key Concepts summarize the most important ideas introduced in each section, linking back to the relevant Example(s) in case students need to review.
  • Chapter Review Exercises include practice problems that recall the most important concepts from each section.
  • Practice Test includes additional problems assessing the most important learning objectives from the chapter.
  • Answer Key includes the answers to all Try It exercises and every other exercise from the Section Exercises, Chapter Review Exercises, and Practice Test.

Answers to Questions in the Book

Answers to Examples and Be Prepared! questions are provided just below the question in the book. All Try It answers are provided in the Answer Key. Odd-numbered Section Exercises, Review Exercises, and Practice Test questions are provided to students in the Answer Key. Even-numbered answers are provided only to instructors in the Instructor Answer Guide via the Instructor Resources page.

Additional Resources

Student and Instructor Resources

We’ve compiled additional resources for both students and instructors, including Getting Started Guides, manipulative mathematics worksheets, and an answer guide to the section review exercises. Instructor resources require a verified instructor account, which can be requested on your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book.

Partner Resources

OpenStax partners are our allies in the mission to make high-quality learning materials affordable and accessible to students and instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partner resources for your text, visit your book page on openstax.org.

About the Authors

Senior Contributing Authors

Lynn Marecek, Santa Ana College

Andrea Honeycutt Mathis, Northeast Mississippi Community College

Reviewers

Shaun Ault, Valdosta State University
Brandie Biddy, Cecil College
Kimberlyn Brooks, Cuyahoga Community College
Michael Cohen, Hofstra University
Robert Diaz, Fullerton College
Dianne Hendrickson, Becker College
Linda Hunt, Shawnee State University
Stephanie Krehl, Mid-South Community College
Yixia Lu, South Suburban College
Teresa Richards, Butte-Glenn College
Christian Roldán- Johnson, College of Lake County Community College
Yvonne Sandoval, El Camino College
Gowribalan Vamadeva, University of Cincinnati Blue Ash College
Kim Watts, North Lake college
Libby Watts, Tidewater Community College
Matthew Watts, Tidewater Community College

Introduction

A photo of a prosthetic hand created by a 3D printer.
This hand may change someone’s life. Amazingly, it was created using a special kind of printer known as a 3D printer. (credit: U.S. Food and Drug Administration/Wikimedia Commons)

For years, doctors and engineers have worked to make artificial limbs, such as this hand for people who need them. This particular product is different, however, because it was developed using a 3D printer. As a result, it can be printed much like you print words on a sheet of paper. This makes producing the limb less expensive and faster than conventional methods.

Biomedical engineers are working to develop organs that may one day save lives. Scientists at NASA are designing ways to use 3D printers to build on the moon or Mars. Already, animals are benefitting from 3D-printed parts, including a tortoise shell and a dog leg. Builders have even constructed entire buildings using a 3D printer.

The technology and use of 3D printers depend on the ability to understand the language of algebra. Engineers must be able to translate observations and needs in the natural world to complex mathematical commands that can provide directions to a printer. In this chapter, you will review the language of algebra and take your first steps toward working with algebraic concepts.

Use the Language of Algebra

Learning Objectives

By the end of this section, you will be able to:

  • Find factors, prime factorizations, and least common multiples
  • Use variables and algebraic symbols
  • Simplify expressions using the order of operations
  • Evaluate an expression
  • Identify and combine like terms
  • Translate an English phrase to an algebraic expression

This chapter is intended to be a brief review of concepts that will be needed in an Intermediate Algebra course. A more thorough introduction to the topics covered in this chapter can be found in the Elementary Algebra 2e chapter, Foundations.

In algebra, we use a letter of the alphabet to represent a number whose value may change or is unknown. Commonly used symbols are a, b, c, m, n, x, and y. Further discussion of constants and variables appears later in this section.

Find Factors, Prime Factorizations, and Least Common Multiples

The numbers 2, 4, 6, 8, 10, 12 are called multiples of 2. A multiple of 2 can be written as the product of 2 and a counting number.

Multiples of 2: 2 times 1 is 2, 2 times 2 is 4, 2 times 3 is 6, 2 times 4 is 8, 2 times 5 is 10, 2 times 6 is 12 and so on.

Similarly, a multiple of 3 would be the product of a counting number and 3.

Multiples of 3: 3 times 1 is 3, 3 times 2 is 6, 3 times 3 is 9, 3 times 4 is 12, 3 times 5 is 15, 3 times 6 is 18 and so on.

We could find the multiples of any number by continuing this process.

Counting Number 1 2 3 4 5 6 7 8 9 10 11 12
Multiples of 2 2 4 6 8 10 12 14 16 18 20 22 24
Multiples of 3 3 6 9 12 15 18 21 24 27 30 33 36
Multiples of 4 4 8 12 16 20 24 28 32 36 40 44 48
Multiples of 5 5 10 15 20 25 30 35 40 45 50 55 60
Multiples of 6 6 12 18 24 30 36 42 48 54 60 66 72
Multiples of 7 7 14 21 28 35 42 49 56 63 70 77 84
Multiples of 8 8 16 24 32 40 48 56 64 72 80 88 96
Multiples of 9 9 18 27 36 45 54 63 72 81 90 99 108

Multiple of a Number

A number is a multiple of n if it is the product of a counting number and n.

Another way to say that 15 is a multiple of 3 is to say that 15 is divisible by 3. That means that when we divide 15 by 3, we get a counting number. In fact, 15÷3 is 5, so 15 is 5·3.

Divisible by a Number

If a number m is a multiple of n, then m is divisible by n.

If we were to look for patterns in the multiples of the numbers 2 through 9, we would discover the following divisibility tests:

Divisibility Tests

A number is divisible by:

   2 if the last digit is 0, 2, 4, 6, or 8.

   3 if the sum of the digits is divisible by 3.

   5 if the last digit is 5 or 0.

   6 if it is divisible by both 2 and 3.

   10 if it ends with 0.

Is 5,625 divisible by ⓐ 2? ⓑ 3? ⓒ 5 or 10? ⓓ 6?

Solution
ⓐ
This table illustrates a divisibility test for the number 5,625 by 2, presenting the question, the condition checked, and the final result.
Is 5,625 divisible by 2?
Does it end in 0, 2, 4, 6 or 8? No.
5,625 is not divisible by 2.
ⓑ
Demonstrating the divisibility rule for 3 with 5,625.
Is 5,625 divisible by 3?
What is the sum of the digits? 5+6+2+5=18
Is the sum divisible by 3? Yes.
5,625 is divisible by 3.
ⓒ
This table demonstrates how to determine if 5,625 is divisible by 5 or 10 by examining its last digit.
Is 5,625 divisible by 5 or 10?
What is the last digit? It is 5. 5,625 is divisible by 5 but not by 10.
ⓓ
This table illustrates the divisibility test for the number 5,625 by 6, explaining why it is not divisible.
Is 5,625 divisible by 6?
Is it divisible by both 2 and 3? No, 5,625 is not divisible by 2, so 5,625 is not divisible by 6.

Is 4,962 divisible by ⓐ 2? ⓑ 3? ⓒ 5? ⓓ 6? ⓔ 10?

Solution

ⓐ yes ⓑ yes ⓒ no ⓓ yes
ⓔ no

Is 3,765 divisible by ⓐ 2? ⓑ 3? ⓒ 5? ⓓ 6? ⓔ 10?

Solution

ⓐ no ⓑ yes ⓒ yes ⓓ no
ⓔ no

In mathematics, there are often several ways to talk about the same ideas. So far, we’ve seen that if m is a multiple of n, we can say that m is divisible by n. For example, since 72 is a multiple of 8, we say 72 is divisible by 8. Since 72 is a multiple of 9, we say 72 is divisible by 9. We can express this still another way.

Since 8·9=72, we say that 8 and 9 are factors of 72. When we write 72=8·9, we say we have factored 72.

8 times 9 is 72. 8 and 9 are factors. 72 is the product.

Other ways to factor 72 are 1·72,2·36,3·24,4·18, and 6·12. The number 72 has many factors: 1,2,3,4,6,8,9,12,18,24,36, and 72.

Factors

In the expression a·b, both a and b are called factors. If a·b=m, and both a and b are integers, then a and b are factors of m.

Some numbers, such as 72, have many factors. Other numbers have only two factors. A prime number is a counting number greater than 1 whose only factors are 1 and itself.

Prime number and Composite number

A prime number is a counting number greater than 1 whose only factors are 1 and the number itself.

A composite number is a counting number greater than 1 that is not prime. A composite number has factors other than 1 and the number itself.

The counting numbers from 2 to 20 are listed in the table with their factors. Make sure to agree with the “prime” or “composite” label for each!

This table has three columns, 19 rows and a header row. The header row labels each column: number, factors and prime or composite. The values in each row are as follows: number 2, factors 1, 2, prime; number 3, factors 1, 3, prime; number 4, factors 1, 2, 4, composite; number 5, factors, 1, 5, prime; number 6, factors 1, 2, 3, 6, composite; number 7, factors 1, 7, prime; number 8, factors 1, 2, 4, 8, composite; number 9, factors 1, 3, 9, composite; number 10, factors 1, 2, 5, 10, composite; number 11, factors 1, 11, prime; number 12, factors 1, 2, 3, 4, 6, 12, composite; number 13, factors 1, 13, prime; number 14, factors 1, 2, 7, 14, composite; number 15, factors 1, 3, 5, 15, composite; number 16, factors 1, 2, 4, 8, 16, composite; number 17, factors 1, 17, prime; number 18, factors 1, 2, 3, 6, 9, 18, composite; number 19, factors 1, 19, prime; number 20, factors 1, 2, 4, 5, 10, 20, composite.

The prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Notice that the only even prime number is 2.

A composite number can be written as a unique product of primes. This is called the prime factorization of the number. Finding the prime factorization of a composite number will be useful in many topics in this course.

Prime Factorization

The prime factorization of a number is the product of prime numbers that equals the number. These prime numbers are called the prime factors.

To find the prime factorization of a composite number, find any two factors of the number and use them to create two branches. If a factor is prime, that branch is complete. Circle that prime. Otherwise it is easy to lose track of the prime numbers.

If the factor is not prime, find two factors of the number and continue the process. Once all the branches have circled primes at the end, the factorization is complete. The composite number can now be written as a product of prime numbers.

How to Find the Prime Factorization of a Composite Number

Factor 48.

Solution
Step 1 is to find two factors whose product is 48 and use these numbers to create two branches. The two branches originating from 48 are formed by the factors 2 and 24. Step 2 is to circle the prime factor. This completes that branch. In this case, 2 is circled as it is prime. Step 3 is to treat the composite factor as a product, break it into two more factors and continue the process. 24 is not prime. It is broken into 4 and 6. 4 and 6 are not prime. 4 is broken into its factors 2 and 2, both of which are circled. 6 is not prime. It is broken into factors 2 and 3, both of which are circled. Step 4 is to write the original composite number as the product of all the circled primes. 48 is 2 into 2 into 2 into 2 into 3.


We say 2·2·2·2·3 is the prime factorization of 48. We generally write the primes in ascending order. Be sure to multiply the factors to verify your answer.

If we first factored 48 in a different way, for example as 6·8, the result would still be the same. Finish the prime factorization and verify this for yourself.

Find the prime factorization of 80.

Solution

2·2·2·2·5

Find the prime factorization of 60.

Solution

2·2·3·5

Find the prime factorization of a composite number.

  1. Find two factors whose product is the given number, and use these numbers to create two branches.
  2. If a factor is prime, that branch is complete. Circle the prime, like a leaf on the tree.
  3. If a factor is not prime, write it as the product of two factors and continue the process.
  4. Write the composite number as the product of all the circled primes.

One of the reasons we look at primes is to use these techniques to find the least common multiple of two numbers. This will be useful when we add and subtract fractions with different denominators.

Least Common Multiple

The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both numbers.

To find the least common multiple of two numbers we will use the Prime Factors Method. Let’s find the LCM of 12 and 18 using their prime factors.

How to Find the Least Common Multiple Using the Prime Factors Method

Find the least common multiple (LCM) of 12 and 18 using the prime factors method.

Solution
Step 1 is to write each number as a product of primes. The number 12 is written as a product of 2, 2 and 3. The number 18 is written as a product of 2, 3 and 3. Step 2 is to list the primes of each number such that primes are vertically matched when possible. The factors of 12 are listed as 2, 2 and 3. The factors of 18 are written below this. The first 2 at the top lines up with the first two at the bottom. The second 2 at the top does not line up with anything. The 3 at the top lines up with a 3 at the bottom. The last 3 at the bottom does not line up with anything. Hence, four columns are made. Step 3 is to bring down the number from each column. When a column has the same number at the top and the bottom, that number is brought down. When a column has only one number that number is brought down. The numbers brought down are 2, 2, 3 and 3. Step 4 is to multiply the factors. The numbers brought down are multiplied with each other to get the LCM. The LCM is 2 into 2 into 3 into 3 equal to 36.

Notice that the prime factors of 12 (2·2·3) and the prime factors of 18 (2·3·3) are included in the LCM (2·2·3·3). So 36 is the least common multiple of 12 and 18.

By matching up the common primes, each common prime factor is used only once. This way you are sure that 36 is the least common multiple.

Find the LCM of 9 and 12 using the Prime Factors Method.

Solution

36

Find the LCM of 18 and 24 using the Prime Factors Method.

Solution

72

Find the least common multiple using the Prime Factors Method.

  1. Write each number as a product of primes.
  2. List the primes of each number. Match primes vertically when possible.
  3. Bring down the columns.
  4. Multiply the factors.

Use Variables and Algebraic Symbols

In algebra, we use a letter of the alphabet to represent a number whose value may change. We call this a variable and letters commonly used for variables are x,y,a,b,c.

Variable

A variable is a letter that represents a number whose value may change.

A number whose value always remains the same is called a constant.

Constant

A constant is a number whose value always stays the same.

To write algebraically, we need some operation symbols as well as numbers and variables. There are several types of symbols we will be using. There are four basic arithmetic operations: addition, subtraction, multiplication, and division. We’ll list the symbols used to indicate these operations below.

Operation Symbols

Operation Notation Say: The result is…
Addition a+b a plus b the sum of a and b
Subtraction a−b a minus b the difference of a and b
Multiplication a·b,ab,(a)(b), (a)b,a(b) a times b the product of a and b
Division a÷b,a/b,ab,ba a divided by b the quotient of a and b;
a is called the dividend, and b is called the divisor

When two quantities have the same value, we say they are equal and connect them with an equal sign.

Equality Symbol

a=b is read “a is equal to b.”

The symbol “=” is called the equal sign.

On the number line, the numbers get larger as they go from left to right. The number line can be used to explain the symbols “<” and “>”.

Inequality

For a less than b, a is to the left of b on the number line. For a greater than b, a is to the right of b on the number line.

The expressions a<b or a>b can be read from left to right or right to left, though in English we usually read from left to right. In general,

a<bis equivalent tob>a.For example,7<11is equivalent to11>7.a>bis equivalent tob<a.For example,17>4is equivalent to4<17.

Inequality Symbols

Inequality Symbols Words
a≠b a is not equal to b.
a<b a is less than b.
a≤b a is less than or equal to b.
a>b a is greater than b.
a≥b a is greater than or equal to b.

Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in English. They help identify an expression, which can be made up of number, a variable, or a combination of numbers and variables using operation symbols. We will introduce three types of grouping symbols now.

Grouping Symbols

Parentheses()Brackets[]Braces{}

Here are some examples of expressions that include grouping symbols. We will simplify expressions like these later in this section.

8(14−8)21−3[2+4(9−8)]24÷{13−2[1(6−5)+4]}

What is the difference in English between a phrase and a sentence? A phrase expresses a single thought that is incomplete by itself, but a sentence makes a complete statement. A sentence has a subject and a verb. In algebra, we have expressions and equations.

Expression

An expression is a number, a variable, or a combination of numbers and variables using operation symbols.

ExpressionWordsEnglish Phrase3+53 plus 5the sum of three and fiven−1nminus onethe difference ofnand one6·76 times 7the product of six and sevenxyxdivided byythe quotient ofxandy

Notice that the English phrases do not form a complete sentence because the phrase does not have a verb.

An equation is two expressions linked by an equal sign. When you read the words the symbols represent in an equation, you have a complete sentence in English. The equal sign gives the verb.

Equation

An equation is two expressions connected by an equal sign.

EquationEnglish Sentence3+5=8The sum of three and five is equal to eight.n−1=14nminus one equals fourteen.6·7=42The product of six and seven is equal to forty-two.x=53xis equal to fifty-three.y+9=2y−3yplus nine is equal to twoyminus three.

Suppose we need to multiply 2 nine times. We could write this as 2·2·2·2·2·2·2·2·2. This is tedious and it can be hard to keep track of all those 2s, so we use exponents. We write 2·2·2 as 23 and 2·2·2·2·2·2·2·2·2 as 29. In expressions such as 23, the 2 is called the base and the 3 is called the exponent. The exponent tells us how many times we need to multiply the base.

The expression shows the number 2, with the number 3 written to its top right. 2 is labeled base and 3 is labeled exponent. This means multiply 2 by itself, three times, as in 2 times 2 times 2.

Exponential Notation

We say 23 is in exponential notation and 2·2·2 is in expanded notation.

an means multiply a by itself, n times.

The expression shown is a to the nth power. Here a is the base and n is the exponent. This is equal to a times a times a and so on, repeated n times. This has n factors.

The expression an is read a to the nth power.

While we read an as “a to the nth power”, we usually read:

a2“asquared”a3“acubed”

We’ll see later why a2 and a3 have special names.

Table 8 shows how we read some expressions with exponents.

Expression In Words
72 7 to the second power or 7 squared
53 5 to the third power or 5 cubed
94 9 to the fourth power
125 12 to the fifth power

Simplify Expressions Using the Order of Operations

To simplify an expression means to do all the math possible. For example, to simplify 4·2+1 we would first multiply 4·2 to get 8 and then add the 1 to get 9. A good habit to develop is to work down the page, writing each step of the process below the previous step. The example just described would look like this:

4·2+18+19

By not using an equal sign when you simplify an expression, you may avoid confusing expressions with equations.

Simplify an Expression

To simplify an expression, do all operations in the expression.

We’ve introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values.

For example, consider the expression 4+3·7. Some students simplify this getting 49, by adding 4+3 and then multiplying that result by 7. Others get 25, by multiplying 3·7 first and then adding 4.

The same expression should give the same result. So mathematicians established some guidelines that are called the order of operations.

Use the order of operations.

  1. Parentheses and Other Grouping Symbols
    • Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
  2. Exponents
    • Simplify all expressions with exponents.
  3. Multiplication and Division
    • Perform all multiplication and division in order from left to right. These operations have equal priority.
  4. Addition and Subtraction
    • Perform all addition and subtraction in order from left to right. These operations have equal priority.

Students often ask, “How will I remember the order?” Here is a way to help you remember: Take the first letter of each key word and substitute the silly phrase “Please Excuse My Dear Aunt Sally”.

ParenthesesPleaseExponentsExcuseMultiplicationDivisionMyDearAdditionSubtractionAuntSally

It’s good that “My Dear” goes together, as this reminds us that multiplication and division have equal priority. We do not always do multiplication before division or always do division before multiplication. We do them in order from left to right.

Similarly, “Aunt Sally” goes together and so reminds us that addition and subtraction also have equal priority and we do them in order from left to right.

Simplify: 18÷6+4(5−2).

Solution
A mathematical expression reads 18 divided by 6 plus 4 times the quantity 5 minus 2.
Parentheses? Yes, subtract first. A mathematical expression is displayed, showing '18 ÷ 6 + 4(3)', which requires application of the order of operations to solve.
Exponents? No.
Multiplication or division? Yes.
Divide first because we multiply and divide left to right. The image shows a mathematical expression
Any other multiplication or division? Yes.
Multiply. A mathematical expression shows the numbers 3 and 12 being added together, appearing in a reddish-brown hue against a white background.
Any other multiplication of division? No.
Any addition or subtraction? Yes.
Add. The number 15 is prominently displayed in black text against a plain white background, standing out with clear visibility.

Simplify: 30÷5+10(3−2).

Solution

16

Simplify: 70÷10+4(6−2).

Solution

23

When there are multiple grouping symbols, we simplify the innermost parentheses first and work outward.

Simplify: 5+23+3[6−3(4−2)].

Solution
A mathematical expression featuring addition, exponentiation, multiplication, and subtraction within parentheses and brackets: 5 + 2^3 + 3[6 - 3(4 - 2)].
Are there any parentheses (or other
grouping symbols)? Yes.
The image shows the mathematical expression 5 + 2^3 + 3[6 - 3(4 - 2)].
Focus on the parentheses that are inside the
brackets. Subtract.
A mathematical expression featuring a combination of integers, exponents, multiplication, subtraction, and brackets: 5 + 2^3 + 3[6 - 3(2)]. The '3(2)' part is highlighted in red.
Continue inside the brackets and multiply. A mathematical expression reads 5 plus 2 to the power of 3 plus 3 times the quantity 6 minus 6, where the second 6 is highlighted in red.
Continue inside the brackets and subtract. A mathematical expression reads 5 + 2^3 + 3[0] on a white background.
The expression inside the brackets requires
no further simplification.
Are there any exponents? Yes. Simplify exponents. A mathematical expression showing 5 + 8 + 3[0], where the '3[0]' term is highlighted in red.
Is there any multiplication or division? Yes.
Multiply. The mathematical equation '5 + 8 + 0' is clearly displayed, showing three digits being added together.
Is there any addition of subtraction? Yes.
Add. The number 13 plus 0 is displayed in red text on a plain white background.
Add. The image features the number '13' in a dark gray font, centered on a plain white background.

Simplify: 9+53−[4(9+3)].

Solution

86

Simplify: 72−2[4(5+1)].

Solution

1

Evaluate an Expression

In the last few examples, we simplified expressions using the order of operations. Now we’ll evaluate some expressions—again following the order of operations. To evaluate an expression means to find the value of the expression when the variable is replaced by a given number.

Evaluate an Expression

To evaluate an expression means to find the value of the expression when the variable is replaced by a given number.

To evaluate an expression, substitute that number for the variable in the expression and then simplify the expression.

Evaluate when x=4: ⓐ x2 ⓑ 3x ⓒ 2x2+3x+8.

Solution
ⓐ
The mathematical expression 'X squared' or 'X^2' is displayed, representing a variable raised to the power of two.
The text 'Replace x with 4.' is shown in a black, sans-serif font, with the number '4' in red. The number 4 with a superscript 2, representing 4 squared, is displayed on a white background.
Use definition of exponent. The mathematical expression '4 x 4' is prominently displayed, representing a simple multiplication problem where four is multiplied by itself.
Simplify. The number '16' is displayed in a dark gray font against a plain white background.

ⓑ
A mathematical expression shows the number 3 raised to the power of x, written as 3^x, against a plain white background.
The text reads 'Replace x with 4.', indicating a mathematical or programming instruction to substitute the variable 'x' with the number 4. The number 3 raised to the power of 4, or 3^4, is displayed against a white background.
Use definition of exponent. The number 3 multiplied by itself four times, represented as 3 • 3 • 3 • 3.
Simplify. The number '81' is displayed in a sans-serif font on a white background.

ⓒ
A mathematical expression reads '2x² + 3x + 8' against a white background.
The image displays the instruction 'Replace x with 4.' in black text on a white background, with the number 4 highlighted in red. A mathematical expression reads as two multiplied by four squared, plus three multiplied by four, plus eight. The number four is highlighted in red in both instances.
Follow the order of operations. A mathematical expression displaying the sum of products and a single digit: 2 multiplied by 16, plus 3 multiplied by 4, plus 8. The expression is 2(16) + 3(4) + 8.
A mathematical expression displays the sum of three numbers: thirty-two plus twelve plus eight.
The number 52 is displayed in a plain, gray font against a clean white background, appearing simple and clear.

Evaluate when x=3, ⓐ x2 ⓑ 4x ⓒ 3x2+4x+1.

Solution

ⓐ 9 ⓑ 64 ⓒ 40

Evaluate when x=6, ⓐ x3 ⓑ 2x ⓒ 6x2−4x−7.

Solution

ⓐ 216 ⓑ 64 ⓒ 185

Identify and Combine Like Terms

Algebraic expressions are made up of terms. A term is a constant, or the product of a constant and one or more variables.

Term

A term is a constant or the product of a constant and one or more variables.

Examples of terms are 7,y,5x2,9a, and b5.

The constant that multiplies the variable is called the coefficient.

Coefficient

The coefficient of a term is the constant that multiplies the variable in a term.

Think of the coefficient as the number in front of the variable. The coefficient of the term 3x is 3. When we write x, the coefficient is 1, since x=1·x.

Some terms share common traits. When two terms are constants or have the same variable and exponent, we say they are like terms.

Look at the following 6 terms. Which ones seem to have traits in common?

5x7n243x9n2

We say,

7 and 4 are like terms.

5x and 3x are like terms.

n2 and 9n2 are like terms.

Like Terms

Terms that are either constants or have the same variables raised to the same powers are called like terms.

If there are like terms in an expression, you can simplify the expression by combining the like terms. We add the coefficients and keep the same variable.

Simplify.4x+7x+xAdd the coefficients.12x

How To Combine Like Terms

Simplify: 2x2+3x+7+x2+4x+5.

Solution
Step 1 is to identify the like terms in 2 x squared plus 3 x plus 7 plus x squared plus 4 x plus 5. The like terms are 2 x squared and x squared, then 3 x and 4 x, then 7 and 5. Step 2 is to rearrange the expression so the like terms are together. Hence, we have 2 x squared plus x squared plus 3 x plus 4 x plus 7 plus 5. Step 3 is to combine the like terms to get 3 x squared plus 7 x plus 12.

Simplify: 3x2+7x+9+7x2+9x+8.

Solution

10x2+16x+17

Simplify: 4y2+5y+2+8y2+4y+5.

Solution

12y2+9y+7

Combine like terms.

  1. Identify like terms.
  2. Rearrange the expression so like terms are together.
  3. Add or subtract the coefficients and keep the same variable for each group of like terms.

Translate an English Phrase to an Algebraic Expression

We listed many operation symbols that are used in algebra. Now, we will use them to translate English phrases into algebraic expressions. The symbols and variables we’ve talked about will help us do that. Table 14 summarizes them.

Operation Phrase Expression
Addition a plus b
the sum of a and b
a increased by b
b more than a
the total of a and b
b added to a
a+b
Subtraction a minus b
the difference of a and b
a decreased by b
b less than a
b subtracted from a
a−b
Multiplication a times b
the product of a and b
twice a
a·b,ab,a(b),(a)(b)

2a
Division a divided by b
the quotient of a and b
the ratio of a and b
b divided into a
a÷b,a/b,ab,ba

Look closely at these phrases using the four operations:

The sum of a and b, the difference of a and b, the product of a and b, the quotient of a and b.

Each phrase tells us to operate on two numbers. Look for the words of and and to find the numbers.

Translate each English phrase into an algebraic expression:

ⓐ the difference of 14x and 9 ⓑ the quotient of 8y2 and 3 ⓒ twelve more than y ⓓ seven less than 49x2

Solution

ⓐ The key word is difference, which tells us the operation is subtraction. Look for the words of and and to find the numbers to subtract.

The difference of 14 x and 9, 14 x minus 9.

ⓑ The key word is quotient, which tells us the operation is division.

The quotient of 8 y squared and 3, divide 8 y squared by 3, 8 y squared divided by 3. This can also be written as 8 y squared slash 3 or 8 y squared upon 3.

ⓒ The key words are more than. They tell us the operation is addition. More than means “added to.”

twelve more thanytwelve added toyy+12

ⓓ The key words are less than. They tell us to subtract. Less than means “subtracted from.”

seven less than49x2seven subtracted from49x249x2−7

Translate the English phrase into an algebraic expression:

ⓐ the difference of 14x2 and 13 ⓑ the quotient of 12x and 2 ⓒ 13 more than z
ⓓ 18 less than 8x

Solution

ⓐ 14x2−13 ⓑ 12x÷2
ⓒ z+13 ⓓ 8x−18

Translate the English phrase into an algebraic expression:

ⓐ the sum of 17y2 and 19 ⓑ the product of 7 and y ⓒ Eleven more than x ⓓ Fourteen less than 11a

Solution

ⓐ 17y2+19 ⓑ 7y
ⓒ x+11 ⓓ 11a−14

We look carefully at the words to help us distinguish between multiplying a sum and adding a product.

Translate the English phrase into an algebraic expression:

ⓐ eight times the sum of x and y ⓑ the sum of eight times x and y

Solution

There are two operation words—times tells us to multiply and sum tells us to add.

ⓐ Because we are multiplying 8 times the sum, we need parentheses around the sum of x and y, (x+y). This forces us to determine the sum first. (Remember the order of operations.)

eight times the sum ofxandy8(x+y)

ⓑ To take a sum, we look for the words of and and to see what is being added. Here we are taking the sum of eight times x and y.

The sum of 8 times x and y is 8 x plus y.

Translate the English phrase into an algebraic expression:

ⓐ four times the sum of p and q
ⓑ the sum of four times p and q

Solution

ⓐ 4(p+q) ⓑ 4p+q

Translate the English phrase into an algebraic expression:

ⓐ the difference of two times x and 8
ⓑ two times the difference of x and 8

Solution

ⓐ 2x−8 ⓑ 2(x−8)

Later in this course, we’ll apply our skills in algebra to solving applications. The first step will be to translate an English phrase to an algebraic expression. We’ll see how to do this in the next two examples.

The width of a rectangle is 14 less than the length. Let l represent the length of the rectangle. Write an expression for the width of the rectangle.

Solution
This table demonstrates the step-by-step process of translating a verbal phrase describing a rectangle's length into its corresponding algebraic expression.
Write a phrase about the width of the rectangle. 14 less than the length
Substitute l for “the length.” l
Rewrite less than as subtracted from. 14 subtracted from l
Translate the phrase into algebra. l − 14

The length of a rectangle is 7 less than the width. Let w represent the width of the rectangle. Write an expression for the length of the rectangle.

Solution

w−7

The width of a rectangle is 6 less than the length. Let l represent the length of the rectangle. Write an expression for the width of the rectangle.

Solution

l−6

The expressions in the next example will be used in the typical coin mixture problems we will see soon.

June has dimes and quarters in her purse. The number of dimes is seven less than four times the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes.

Solution
Step-by-step process for translating a word problem describing the number of dimes into an algebraic expression.
Write a phrase about the number of dimes. seven less than four times the number of quarters
Substitute q for the number of quarters. 7 less than 4 times q
Translate 4 times q. 7 less than 4q
Translate the phrase into algebra. 4q − 7

Geoffrey has dimes and quarters in his pocket. The number of dimes is eight less than four times the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes.

Solution

4q−8

Lauren has dimes and nickels in her purse. The number of dimes is three more than seven times the number of nickels. Let n represent the number of nickels. Write an expression for the number of dimes.

Solution

7n+3

Key Concepts

  • Divisibility Tests
    A number is divisible by:
      2 if the last digit is 0, 2, 4, 6, or 8.
      3 if the sum of the digits is divisible by 3.
      5 if the last digit is 5 or 0.
      6 if it is divisible by both 2 and 3.
      10 if it ends with 0.
  • How to find the prime factorization of a composite number.
    1. Find two factors whose product is the given number, and use these numbers to create two branches.
    2. If a factor is prime, that branch is complete. Circle the prime, like a bud on the tree.
    3. If a factor is not prime, write it as the product of two factors and continue the process.
    4. Write the composite number as the product of all the circled primes.
  • How To Find the least common multiple using the prime factors method.
    1. Write each number as a product of primes.
    2. List the primes of each number. Match primes vertically when possible.
    3. Bring down the columns.
    4. Multiply the factors.
  • Equality Symbol
    a=b is read “a is equal to b.”
    The symbol “=” is called the equal sign.
  • Inequality
    For a less than b, a is to the left of b on the number line. For a greater than b, a is to the right of b on the number line.
  • Inequality Symbols
    Inequality Symbols Words
    a≠b a is not equal to b.
    a<b a is less than b.
    a≤b a is less than or equal to b.
    a>b a is greater than b.
    a≥b a is greater than or equal to b.
  • Grouping Symbols
    Parentheses()Brackets[]Braces{}
  • Exponential Notation
    an means multiply a by itself, n times.
    The expression an is read a to the nth power.
  • Simplify an Expression
    To simplify an expression, do all operations in the expression.
  • How to use the order of operations.
    1. Parentheses and Other Grouping Symbols
      • Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
    2. Exponents
      • Simplify all expressions with exponents.
    3. Multiplication and Division
      • Perform all multiplication and division in order from left to right. These operations have equal priority.
    4. Addition and Subtraction
      • Perform all addition and subtraction in order from left to right. These operations have equal priority.
  • How to combine like terms.
    1. Identify like terms.
    2. Rearrange the expression so like terms are together.
    3. Add or subtract the coefficients and keep the same variable for each group of like terms.
    Operation Phrase Expression
    Addition a plus b
    the sum of a and b
    a increased by b
    b more than a
    the total of a and b
    b added to a
    a+b
    Subtraction a minus b
    the difference of a and b
    a decreased by b
    b less than a
    b subtracted from a
    a−b
    Multiplication a times b
    the product of a and b
    twice a
    a·b,ab,a(b),(a)(b)


    2a
    Division a divided by b

    the quotient of a and b
    the ratio of a and b
    b divided into a
    a÷b,a/b,ab,ba

Practice Makes Perfect

Identify Multiples and Factors

In the following exercises, use the divisibility tests to determine whether each number is divisible by 2, by 3, by 5, by 6, and by 10.

84

Solution

Divisible by 2, 3, 6

96

896

Solution

Divisible by 2

942

22,335

Solution

Divisible by 3, 5

39,075

Find Prime Factorizations and Least Common Multiples

In the following exercises, find the prime factorization.

86

Solution

2·43

78

455

Solution

5·7·13

400

432

Solution

2·2·2·2·3·3·3

627

In the following exercises, find the least common multiple of each pair of numbers using the prime factors method.

8, 12

Solution

24

12, 16

28, 40

Solution

280

84, 90

55, 88

Solution

440

60, 72

Simplify Expressions Using the Order of Operations

In the following exercises, simplify each expression.

23−12÷(9−5)

Solution

5

32−18÷(11−5)

2+8(6+1)

Solution

58

4+6(3+6)

20÷4+6(5−1)

Solution

29

33÷3+4(7−2)

3(1+9·6)−42

Solution

149

5(2+8·4)−72

2[1+3(10−2)]

Solution

50

5[2+4(3−2)]

8+2[7−2(5−3)]−32

Solution

5

10+3[6−2(4−2)]−24

Evaluate an Expression

In the following exercises, evaluate the following expressions.

When x=2,
ⓐ x6
ⓑ 4x
ⓒ 2x2+3x−7

Solution

ⓐ 64 ⓑ 16 ⓒ 7

When x=3,
ⓐ x5
ⓑ 5x
ⓒ 3x2−4x−8

When x=4,y=1
x2+3xy−7y2

Solution

21

When x=3,y=2
6x2+3xy−9y2

When x=10,y=7
(x−y)2

Solution

9

When a=3,b=8
a2+b2

Simplify Expressions by Combining Like Terms

In the following exercises, simplify the following expressions by combining like terms.

7x+2+3x+4

Solution

10x+6

8y+5+2y−4

10a+7+5a−2+7a−4

Solution

22a+1

7c+4+6c−3+9c−1

3x2+12x+11+14x2+8x+5

Solution

17x2+20x+16

5b2+9b+10+2b2+3b−4

Translate an English Phrase to an Algebraic Expression

In the following exercises, translate the phrases into algebraic expressions.


ⓐ the difference of 5x2 and 6xy
ⓑ the quotient of 6y2 and 5x
ⓒ Twenty-one more than y2
ⓓ 6x less than 81x2

Solution

ⓐ 5x2−6xy ⓑ 6y25x
ⓒ y2+21 ⓓ 81x2−6x


ⓐ the difference of 17x2 and 5xy
ⓑ the quotient of 8y3 and 3x
ⓒ Eighteen more than a2;
ⓓ 11b less than 100b2


ⓐ the sum of 4ab2 and 3a2b
ⓑ the product of 4y2 and 5x
ⓒ Fifteen more than m
ⓓ 9x less than 121x2

Solution

ⓐ 4ab2+3a2b ⓑ 20xy2
ⓒ m+15 ⓓ 121x2−9x


ⓐ the sum of 3x2y and 7xy2
ⓑ the product of 6xy2 and 4z
ⓒ Twelve more than 3x2
ⓓ 7x2 less than 63x3


ⓐ eight times the difference of y and nine
ⓑ the difference of eight times y and 9

Solution

ⓐ 8(y−9) ⓑ 8y−9


ⓐ seven times the difference of y and one
ⓑ the difference of seven times y and 1


ⓐ five times the sum of 3x and y
ⓑ the sum of five times 3x and y

Solution

ⓐ 5(3x+y) ⓑ 15x+y


ⓐ eleven times the sum of 4x2 and 5x
ⓑ the sum of eleven times 4x2 and 5x

Eric has rock and country songs on his playlist. The number of rock songs is 14 more than twice the number of country songs. Let c represent the number of country songs. Write an expression for the number of rock songs.

Solution

2c+14

The number of women in a Statistics class is 8 more than twice the number of men. Let m represent the number of men. Write an expression for the number of women.

Greg has nickels and pennies in his pocket. The number of pennies is seven less than three times the number of nickels. Let n represent the number of nickels. Write an expression for the number of pennies.

Solution

3n−7

Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.

Writing Exercises

Explain in your own words how to find the prime factorization of a composite number.

Solution

Answers will vary.

Why is it important to use the order of operations to simplify an expression?

Explain how you identify the like terms in the expression 8a2+4a+9−a2−1.

Solution

Answers will vary.

Explain the difference between the phrases “4 times the sum of x and y” and “the sum of 4 times x and y”.

Self Check

ⓐ Use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 7 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: identify multiples and apply divisibility tests, find prime factorizations and least common multiples, use variables and algebraic symbols, simplify expressions using the order of operations, evaluate an expression, identify and combine like terms, translate English phrases to algebraic expressions. The remaining columns are blank.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

coefficient
The coefficient of a term is the constant that multiplies the variable in a term.
composite number
A composite number is a counting number that is not prime. It has factors other than 1 and the number itself.
constant
A constant is a number whose value always stays the same.
divisible by a number
If a number m is a multiple of n, then m is divisible by n.
equation
An equation is two expressions connected by an equal sign.
evaluate an expression
To evaluate an expression means to find the value of the expression when the variables are replaced by given numbers.
expression
An expression is a number, a variable, or a combination of numbers and variables using operation symbols.
factors
If a·b=m, then a and b are factors of m.
least common multiple
The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both numbers.
like terms
Terms that are either constants or have the same variables raised to the same powers are called like terms.
multiple of a number
A number is a multiple of n if it is the product of a counting number and n.
order of operations
The order of operations are established guidelines for simplifying an expression.
prime factorization
The prime factorization of a number is the product of prime numbers that equals the number.
prime number
A prime number is a counting number greater than 1 whose only factors are 1 and the number itself.
simplify an expression
To simplify an expression means to do all the math possible.
term
A term is a constant, or the product of a constant and one or more variables.
variable
A variable is a letter that represents a number whose value may change.

Integers

Learning Objectives

By the end of this section, you will be able to:

  • Simplify expressions with absolute value
  • Add and subtract integers
  • Multiply and divide integers
  • Simplify expressions with integers
  • Evaluate variable expressions with integers
  • Translate phrases to expressions with integers
  • Use integers in applications

A more thorough introduction to the topics covered in this section can be found in the Elementary Algebra 2e chapter, Foundations.

Simplify Expressions with Absolute Value

A negative number is a number less than 0. The negative numbers are to the left of zero on the number line. See Figure 1.

Figure shows a horizontal line marked with numbers at equal distances. At the center of the line is 0. To the right of this, starting from the number closest to 0 are 1, 2, 3 and 4. These are labeled positive numbers. To the left of 0, starting from the number closest to 0 are minus 1, minus 2, minus 3 and minus 4. These are labeled negative numbers.
The number line shows the location of positive and negative numbers.

You may have noticed that, on the number line, the negative numbers are a mirror image of the positive numbers, with zero in the middle. Because the numbers 2 and −2 are the same distance from zero, each one is called the opposite of the other. The opposite of 2 is −2, and the opposite of −2 is 2.

Opposite

The opposite of a number is the number that is the same distance from zero on the number line but on the opposite side of zero.

Figure 2 illustrates the definition.

Figure shows a number line with the numbers 3 and minus 3 highlighted. These are equidistant from 0, both being 3 numbers away from 0.
The opposite of 3 is −3.

Opposite Notation

−ameans the opposite of the numbera The notation−ais read as “the opposite ofa.”

We saw that numbers such as 3 and −3 are opposites because they are the same distance from 0 on the number line. They are both three units from 0. The distance between 0 and any number on the number line is called the absolute value of that number.

Absolute Value

The absolute value of a number is its distance from 0 on the number line.

The absolute value of a number n is written as |n| and |n|≥0 for all numbers.

Absolute values are always greater than or equal to zero.

For example,

−5is 5 units away from 0, so|−5|=5. 5 is 5 units away from 0, so|5|=5.

Figure 3 illustrates this idea.

Figure shows a number line showing the numbers 0, 5 and minus 5. 5 and minus 5 are equidistant from 0, both being 5 units away from 0.
The numbers 5 and −5 are 5 units away from 0.

The absolute value of a number is never negative because distance cannot be negative. The only number with absolute value equal to zero is the number zero itself because the distance from 0 to 0 on the number line is zero units.

In the next example, we’ll order expressions with absolute values.

Fill in <,>, or = for each of the following pairs of numbers:

ⓐ |−5|__−|−5| ⓑ 8__−|−8| ⓒ −9__−|−9| ⓓ −(−16)__|−16|.

Solution
ⓐ
Steps demonstrating the comparison of absolute values, specifically |−5| and −|−5|, by simplification and ordering.
|−5|__−|−5|
Simplify. 5__−5
Order. 5>−5
|−5|>−|−5|
ⓑ
Steps for simplifying and comparing mathematical expressions, including those with absolute values.
8__−|−8|
Simplify. 8__−8
Order. 8>−8
8>−|−8|
ⓒ
Steps showing the simplification of -|-9| and its comparison with -9, demonstrating that both are equal.
−9__−|−9|
Simplify. −9__−9
Order. −9=−9
−9=−|−9|
ⓓ
Steps to simplify and compare the absolute value and negative of a negative integer, demonstrating their equality.
−(−16)__|−16|
Simplify. 16__16
Order. 16=16
−(−16)=|−16|

Fill in <,>, or = for each of the following pairs of numbers:

ⓐ −9__−|−9| ⓑ 2__−|−2| ⓒ −8__|−8| ⓓ −(−9)__|−9|.

Solution

ⓐ = ⓑ > ⓒ <
ⓓ =

Fill in <,>, or = for each of the following pairs of numbers:

ⓐ 7__−|−7| ⓑ −(−10)__|−10| ⓒ |−4|__−|−4| ⓓ −1__|−1|.

Solution

ⓐ > ⓑ = ⓒ >
ⓓ <

We now add absolute value bars to our list of grouping symbols. When we use the order of operations, first we simplify inside the absolute value bars as much as possible, then we take the absolute value of the resulting number.

Grouping Symbols

Parentheses()Braces{} Brackets[]Absolute value||

In the next example, we simplify the expressions inside absolute value bars first just as we do with parentheses.

Simplify: 24−|19−3(6−2)|.

Solution
Step-by-step evaluation of a mathematical expression following the order of operations.
24−|19−3(6−2)|
Work inside parentheses first:
subtract 2 from 6. 24−|19−3(4)|
Multiply 3(4). 24−|19−12|
Subtract inside the absolute value bars. 24−|7|
Take the absolute value. 24−7
Subtract. 17

Simplify:19−|11−4(3−1)|.

Solution

16

Simplify: 9−|8−4(7−5)|.

Solution

9

Add and Subtract Integers

So far, we have only used the counting numbers and the whole numbers.

Counting numbers1,2,3… Whole numbers0,1,2,3….

Our work with opposites gives us a way to define the integers. The whole numbers and their opposites are called the integers. The integers are the numbers …−3,−2,−1,0,1,2,3…

Integers

The whole numbers and their opposites are called the integers.

The integers are the numbers

…−3,−2,−1,0,1,2,3…,

Most students are comfortable with the addition and subtraction facts for positive numbers. But doing addition or subtraction with both positive and negative numbers may be more challenging.

We will use two color counters to model addition and subtraction of negatives so that you can visualize the procedures instead of memorizing the rules.

We let one color (blue) represent positive. The other color (red) will represent the negatives.

Figure show two circles labeled positive blue and negative red.

If we have one positive counter and one negative counter, the value of the pair is zero. They form a neutral pair. The value of this neutral pair is zero.

Figure shows a blue circle and a red circle encircled in a larger shape. This is labeled 1 plus minus 1 equals 0.

We will use the counters to show how to add:

5+3−5+(−3)−5+35+(−3)

The first example, 5+3, adds 5 positives and 3 positives—both positives.

The second example, −5+(−3), adds 5 negatives and 3 negatives—both negatives.

When the signs are the same, the counters are all the same color, and so we add them. In each case we get 8—either 8 positives or 8 negatives.

Figure on the left is labeled 5 plus 3. It shows 8 blue circles. 5 plus 3 equals 8. Figure on the right is labeled minus 5 plus open parentheses minus 3 close parentheses. It shows 8 blue circles labeled 8 negatives. Minus 5 plus open parentheses minus 3 close parentheses equals minus 8.

So what happens when the signs are different? Let’s add −5+3 and 5+(−3).

When we use counters to model addition of positive and negative integers, it is easy to see whether there are more positive or more negative counters. So we know whether the sum will be positive or negative.

Figure on the left is labeled minus 5 plus 3. It has 5 red circles and 3 blue circles. Three pairs of red and blue circles are formed. More negatives means the sum is negative. The figure on the right is labeled 5 plus minus 3. It has 5 blue and 3 red circles. Three pairs of red and blue circles are formed. More positives means the sum is positive.

Add: ⓐ −1+(−4) ⓑ −1+5 ⓒ 1+(−5).

Solution
ⓐ
A mathematical expression showing the addition of negative one and negative four, written as -1 + (-4).
A row of six identical reddish-orange circular objects with darker outlines, possibly beads or pills, arranged horizontally on a white background.
1 negative plus 4 negatives is 5 negatives The number -5, representing a negative integer.
ⓑ
The image displays the mathematical expression '-1 + 5' in black text on a white background, representing a simple arithmetic addition problem.
A purple oval groups a light blue circle above a light red circle. To their right, four more light blue circles are arranged horizontally, all against a white background.
There are more positives, so the sum is positive. A black number 4 is visible against a bright white background.
ⓒ
The image shows the mathematical expression '1 + (-5)' in a black font against a white background.
A horizontal row of seven circles, with the first two enclosed in a purple oval. One of the enclosed circles is light blue, while the other six circles, including one inside the oval, are peach-colored.
There are more negatives, so the sum is negative. -4

Add: ⓐ −2+(−4) ⓑ −2+4 ⓒ 2+(−4).

Solution

ⓐ −6 ⓑ 2 ⓒ −2

Add: ⓐ −2+(−5) ⓑ −2+5 ⓒ 2+(−5).

Solution

ⓐ −7 ⓑ 3 ⓒ −3

We will continue to use counters to model the subtraction. Perhaps when you were younger, you read “5−3” as “5 take away 3.” When you use counters, you can think of subtraction the same way!

We will use the counters to show to subtract:

5−3 −5−(−3) −5−3 5−(−3)

The first example, 5−3, we subtract 3 positives from 5 positives and end up with 2 positives.

In the second example, −5−(−3), we subtract 3 negatives from 5 negatives and end up with 2 negatives.

Each example used counters of only one color, and the “take away” model of subtraction was easy to apply.

Figure on the left is labeled 5 minus 3 equals 2. There are 5 blue circles. Three of these are encircled and an arrow indicates that they are taken away. The figure on the right is labeled minus 5 minus open parentheses minus 3 close parentheses equals minus 2. There are 5 red circles. Three of these are encircled and an arrow indicates that they are taken away.

What happens when we have to subtract one positive and one negative number? We’ll need to use both blue and red counters as well as some neutral pairs. If we don’t have the number of counters needed to take away, we add neutral pairs. Adding a neutral pair does not change the value. It is like changing quarters to nickels—the value is the same, but it looks different.

Let’s look at −5−3 and 5−(−3).

The image displays the mathematical expression '-5 - 3' on a white background, representing a subtraction operation involving two negative integers or a negative integer and a positive integer being subtracted. A mathematical expression reads '5 - (-3)' against a white background.
Model the first number. Five light peach-colored circles with red outlines are arranged horizontally on a white background. Five identical light blue circles are arranged in a horizontal row across a white background, suggesting a visual representation of a rating system or a step-by-step process.
We now add the needed neutral pairs. Nine red circles and three blue circles are arranged on a white background, with six red circles in a line and three red circles above three blue circles. Four light blue circles are arranged in a row on the left. To their right, three more light blue circles are in a row, with three red circles directly below them.
We remove the number of counters modeled by the second number. Seven red circles in a row, with three light blue circles highlighted below by a purple oval and an arrow, suggesting a counting or subtraction concept. A horizontal row of seven light blue circles is shown, with three red circles below them encircled by a purple oval and a left-pointing arrow.
Count what is left. Eight light peach-colored circles with red borders are arranged in a horizontal row, with a small gap separating the fifth and sixth circles. A horizontal row of eight light blue circles with a thin blue outline, spaced evenly apart, with a larger gap between the fifth and sixth circles.
A basic arithmetic equation is displayed, showing '-5 - 3 = -8' in black text against a white background, demonstrating the subtraction of two negative numbers resulting in a larger negative number. A mathematical equation is displayed on a white background, reading '5 - (-3) = 8' in black font.
A black, elongated figure-eight symbol with a horizontal bar across its center, set against a plain white background. The number 8.

Subtract: ⓐ 3−1 ⓑ −3−(−1) ⓒ −3−1 ⓓ 3−(−1).

Solution
ⓐ
A magenta arrow shows a loop from the second light blue circle, encircling the first, and returning to the second, with three light blue circles aligned horizontally. The image displays the numeric text '3-1' in a simple, clear font against a white background.
Take 1 negative from 3 negatives and get 2 negatives positives. 2
ⓑ
A purple looping arrow encircles the leftmost of three horizontally aligned orange circles. The arrow's tail starts near the circle, makes a full loop, and points to the left. A mathematical expression showing the subtraction of negative one from negative three: -3 - (-1).
Take 1 positive from 3 negatives and get 2 negatives. A stark, clear image displays the number '-2' in a bold, sans-serif font against a plain white background, appearing as if a mathematical value or a counter.
ⓒ
A minimalist graphic features five light red circles with dark red outlines. Four circles are closely spaced on the left, followed by a gap, and then a single circle on the right, all against a white background. The image displays the mathematical expression '-3 -1' against a plain white background, suggesting an operation that would result in -4.
Take 1 positive from the one added neutral pair. A light blue circle is encircled by a purple oval, with a purple arrow indicating a clockwise rotational path around the bottom curve of the oval. The number -4 is prominently displayed in the center of a white background.
ⓓ
Four light blue circles with darker blue outlines, arranged in a row with a gap after the third circle against a white background. The image shows the mathematical expression 3 - (-1), which represents subtracting negative one from three. This operation is equivalent to adding one to three.
Take 1 negative from the one added neutral pair. A reddish-orange circle is enclosed within a purple oval, with a purple arrow indicating a counter-clockwise rotation or orbit around the circle. The digit 4 is clearly displayed on a plain white background.

Subtract: ⓐ 6−4 ⓑ −6−(−4) ⓒ −6−4 ⓓ 6−(−4).

Solution

ⓐ 2 ⓑ −2 ⓒ −10 ⓓ 10

Subtract: ⓐ 7−4 ⓑ −7−(−4) ⓒ −7−4 ⓓ 7−(−4).

Solution

ⓐ 3 ⓑ −3 ⓒ −11 ⓓ 11

Have you noticed that subtraction of signed numbers can be done by adding the opposite? In the last example, −3−1 is the same as −3+(−1) and 3−(−1) is the same as 3+1. You will often see this idea, the Subtraction Property, written as follows:

Subtraction Property

a−b=a+(−b)

Subtracting a number is the same as adding its opposite.

Simplify: ⓐ 13−8 and 13+(−8) ⓑ −17−9 and −17+(−9) ⓒ 9−(−15) and 9+15 ⓓ −7−(−4) and −7+4.

Solution
ⓐ
This table illustrates that subtracting a positive integer is equivalent to adding its negative counterpart, using 13 minus 8 as an example.
Subtract. 13−8and13+(−8)55
ⓑ
This table demonstrates integer subtraction, illustrating the equivalence of a - b and a + (-b) with a specific example.
Subtract. −17−9and−17+(−9) −26−26
ⓒ
This table illustrates the equivalence of subtracting a negative number and adding a positive number with an example calculation.
Subtract. 9−(−15)and9+15 2424
ⓓ
This table demonstrates integer subtraction, showing the equivalence between subtracting a negative number and adding a positive number.
Subtract. −7−(−4)and−7+4 −3−3

Simplify: ⓐ 21−13 and 21+(−13) ⓑ −11−7 and −11+(−7) ⓒ 6−(−13) and 6+13 ⓓ −5−(−1) and −5+1.

Solution

ⓐ 8,8 ⓑ −18, −18
ⓒ 19,19 ⓓ −4, −4

Simplify: ⓐ 15−7 and 15+(−7) ⓑ −14−8 and −14+(−8) ⓒ 4−(−19) and 4+19 ⓓ −4−(−7) and −4+7.

Solution

ⓐ 8,8 ⓑ −22, −22
ⓒ 23,23 ⓓ 3,3

What happens when there are more than three integers? We just use the order of operations as usual.

Simplify: 7−(−4−3)−9.

Solution
Step-by-step simplification of the mathematical expression 7 - (-4 - 3) - 9.
7−(−4−3)−9
Simplify inside the parentheses first. 7−(−7)−9
Subtract left to right. 14−9
Subtract. 5

Simplify: 8−(−3−1)−9.

Solution

3

Simplify: 12−(−9−6)−14.

Solution

13

Multiply and Divide Integers

Since multiplication is mathematical shorthand for repeated addition, our model can easily be applied to show multiplication of integers. Let’s look at this concrete model to see what patterns we notice. We will use the same examples that we used for addition and subtraction. Here, we are using the model just to help us discover the pattern.

We remember that a·b means add a, b times.

The figure on the left is labeled 5 dot 3. Here, we need to add 5, 3 times. Three rows of five blue counters each are shown. This makes 15 positives. Hence, 5 times 3 is 15. The figure on the right is labeled minus 5 open parentheses 3 close parentheses. Here we need to add minus 5, 3 times. Three rows of five red counters each are shown. This makes 15 negatives. Hence, minus 5 times 3 is minus 15.

The next two examples are more interesting. What does it mean to multiply 5 by −3? It means subtract 5,3 times. Looking at subtraction as “taking away”, it means to take away 5, 3 times. But there is nothing to take away, so we start by adding neutral pairs on the workspace.

The figure on the left is labeled 5 open parentheses minus 3 close parentheses. We need to take away 5, three times. Three rows of five positive counters each and three rows of five negative counters each are shown. What is left is 15 negatives. Hence, 5 times minus 3 is minus 15. The figure on the right is labeled open parentheses minus 5 close parentheses open parentheses minus 3 close parentheses. We need to take away minus 5, three times. Three rows of five positive counters each and three rows of five negative counters each are shown. What is left is 15 positives. Hence, minus 5 times minus 3 is 15.

In summary:

5·3=15−5(3)=−15 5(−3)=−15(−5)(−3)=15

Notice that for multiplication of two signed numbers, when the

signs are thesame, the product ispositive. signs aredifferent, the product isnegative.

What about division? Division is the inverse operation of multiplication. So, 15÷3=5 because 5·3=15. In words, this expression says that 15 can be divided into 3 groups of 5 each because adding five three times gives 15. If you look at some examples of multiplying integers, you might figure out the rules for dividing integers.

5·3=15so15÷3=5−5(3)=−15so−15÷3=−5 (−5)(−3)=15so15÷(−3)=−5 5(−3)=−15so−15÷(−3)=5

Division follows the same rules as multiplication with regard to signs.

Multiplication and Division of Signed Numbers

For multiplication and division of two signed numbers:

Same signs Result
•  Two positives Positive
•  Two negatives Positive

If the signs are the same, the result is positive.

Different signs Result
•  Positive and negative Negative
•  Negative and positive Negative

If the signs are different, the result is negative.

Multiply or divide: ⓐ −100÷(−4) ⓑ 7·6 ⓒ 4(−8) ⓓ −27÷3.

Solution
ⓐ
This table demonstrates the division of two negative numbers, showing the mathematical operation, explanation of the sign rule, and the final positive quotient.
−100÷(−4)
Divide, with signs that are the same the quotient is positive. 25
ⓑ
This table demonstrates an example of multiplication with same signs, showing the problem description and its calculated result.
7·6
Multiply, with same signs. 42
ⓒ
This table illustrates a multiplication problem involving integers with different signs and its corresponding result.
4(−8)
Multiply, with different signs. −32
ⓓ
Illustrates the rule for dividing numbers with different signs, demonstrating that -27 ÷ 3 yields a negative quotient (-9).
−27÷3
Divide, with different signs, the quotient is negative. −9

Multiply or divide: ⓐ −115÷(−5) ⓑ 5·12 ⓒ 9(−7) ⓓ −63÷7.

Solution

ⓐ 23 ⓑ 60 ⓒ −63 ⓓ −9

Multiply or divide: ⓐ −117÷(−3) ⓑ 3·13 ⓒ 7(−4) ⓓ −42÷6.

Solution

ⓐ 39 ⓑ 39 ⓒ −28 ⓓ −7

When we multiply a number by 1, the result is the same number. Each time we multiply a number by −1, we get its opposite!

Multiplication by −1

−1a=−a

Multiplying a number by −1 gives its opposite.

Simplify Expressions with Integers

What happens when there are more than two numbers in an expression? The order of operations still applies when negatives are included. Remember Please Excuse My Dear Aunt Sally?

Let’s try some examples. We’ll simplify expressions that use all four operations with integers—addition, subtraction, multiplication, and division. Remember to follow the order of operations.

Simplify: ⓐ (−2)4 ⓑ −24.

Solution

Notice the difference in parts (a) and (b). In part (a), the exponent means to raise what is in the parentheses, the −2 to the 4th power. In part (b), the exponent means to raise just the 2 to the 4th power and then take the opposite.

ⓐ
Step-by-step calculation of (-2) to the power of 4, showing expansion and multiplication.
(−2)4
Write in expanded form. (−2)(−2)(−2)(−2)
Multiply. 4(−2)(−2)
Multiply. −8(−2)
Multiply. 16
ⓑ
Demonstrates the step-by-step evaluation of the expression -2^4, showing expansion, sequential multiplication, and the final numeric result.
−24
Write in expanded form. −(2·2·2·2)
We are asked to find the opposite of 24.
Multiply. −(4·2·2)
Multiply. −(8·2)
Multiply. −16

Simplify: ⓐ (−3)4 ⓑ −34.

Solution

ⓐ 81 ⓑ −81

Simplify: ⓐ (−7)2 ⓑ −72.

Solution

ⓐ 49 ⓑ −49

The last example showed us the difference between (−2)4 and −24. This distinction is important to prevent future errors. The next example reminds us to multiply and divide in order left to right.

Simplify: ⓐ 8(−9)÷(−2)3 ⓑ −30÷2+(−3)(−7).

Solution
ⓐ
Step-by-step evaluation of the expression 8(-9) ÷ (-2)^3, demonstrating the order of operations.
8(−9)÷(−2)3
Exponents first. 8(−9)÷(−8)
Multiply. −72÷(−8)
Divide. 9
ⓑ
This table demonstrates the step-by-step solution of a mathematical expression following the order of operations.
−30÷2+(−3)(−7)
Multiply and divide left to right, so divide first. −15+(−3)(−7)
Multiply. −15+21
Add. 6

Simplify: ⓐ 12(−9)÷(−3)3 ⓑ −27÷3+(−5)(−6).

Solution

ⓐ 4 ⓑ 21

Simplify: ⓐ 18(−4)÷(−2)3 ⓑ −32÷4+(−2)(−7).

Solution

ⓐ 9 ⓑ 6

Evaluate Variable Expressions with Integers

Remember that to evaluate an expression means to substitute a number for the variable in the expression. Now we can use negative numbers as well as positive numbers.

Evaluate 4x2−2xy+3y2 when x=2,y=−1.

Solution
The image displays the mathematical expression 4x^2 - 2xy + 3y^2.
The text reads: 'Substitute x = 2, y = -1. Use parentheses to show multiplication.' The image displays the mathematical expression 4(2)^2 - 2(2)(-1) + 3(-1)^2, with the number 2 highlighted in red and -1 in light blue, indicating potential substitutions or a specific calculation.
Simplify exponents. A mathematical expression showing operations of multiplication, subtraction, and addition: 4 multiplied by 4, minus 2 multiplied by 2 multiplied by -1, plus 3 multiplied by 1.
Multiply. A mathematical expression showing the calculation 16 - (-4) + 3.
Subtract. A simple mathematical equation '20 + 3' is displayed in black text against a plain white background.
Add. The number 23 is displayed in a dark grey, sans-serif font against a plain white background.

Evaluate: 3x2−2xy+6y2 when x=1,y=−2.

Solution

31

Evaluate: 4x2−xy+5y2 when x=−2,y=3.

Solution

67

Translate Phrases to Expressions with Integers

Our earlier work translating English to algebra also applies to phrases that include both positive and negative numbers.

Translate and simplify: the sum of 8 and −12, increased by 3.

Solution
This table demonstrates the step-by-step process of translating a verbal mathematical phrase into an expression and then simplifying it to the final numerical result.
thesumof–8and–––−12increased by 3
Translate. [8+(−12)]+3
Simplify. Be careful not to confuse the brackets with an absolute value sign. (−4)+3
Add. −1

Translate and simplify the sum of 9 and −16, increased by 4.

Solution

(9+(−16))+4;−3

Translate and simplify the sum of −8 and −12, increased by 7.

Solution

(−8+(−12))+7;−13

Use Integers in Applications

We’ll outline a plan to solve applications. It’s hard to find something if we don’t know what we’re looking for or what to call it! So when we solve an application, we first need to determine what the problem is asking us to find. Then we’ll write a phrase that gives the information to find it. We’ll translate the phrase into an expression and then simplify the expression to get the answer. Finally, we summarize the answer in a sentence to make sure it makes sense.

How to Solve Application Problems Using Integers

In the morning, the temperature in Kendallville, Indiana was 11 degrees. By mid-afternoon, the temperature had dropped to −9 degrees. What was the difference in the morning and afternoon temperatures?

Figure shows a glass thermometer, with temperature markings ranging from minus 10 to 30. Two markings are highlighted, minus 9 degrees C and 11 degrees C.
Solution
Step 1 is to read the problem and make sure all the words and ideas are understood. Step 2 is to identify what we are asked to find. Here, we need to find the difference of the morning and afternoon temperatures. Step 3 is to write a phrase that gives the information to find it. In this case, the phrase is the difference of 11 and minus 9. Step 4 is to translate the phrase to an expression. Here, we write 11 minus open parentheses minus 9 close parentheses. In step 5, we simplify the expression to get 20. Step 6 is to answer the question with a complete sentence: The difference in temperatures was 20 degrees.

In the morning, the temperature in Anchorage, Alaska was 15 degrees. By mid-afternoon the temperature had dropped to 30 degrees below zero. What was the difference in the morning and afternoon temperatures?

Solution

The difference in temperatures was 45 degrees.

The temperature in Denver was −6 degrees at lunchtime. By sunset the temperature had dropped to −15 degrees. What was the difference in the lunchtime and sunset temperatures?

Solution

The difference in temperatures was 9 degrees.

Use Integers in Applications.

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are asked to find.
  3. Write a phrase that gives the information to find it.
  4. Translate the phrase to an expression.
  5. Simplify the expression.
  6. Answer the question with a complete sentence.

Access this online resource for additional instruction and practice with integers.

  • Subtracting Integers with Counters

Key Concepts

  • Opposite Notation
    −ameans the opposite of the numbera The notation−ais read as “the opposite ofa.”
  • Absolute Value
    The absolute value of a number is its distance from 0 on the number line.
    The absolute value of a number n is written as |n| and |n|≥0 for all numbers.
    Absolute values are always greater than or equal to zero.
  • Grouping Symbols
    Parentheses()Braces{} Brackets[]Absolute value||
  • Subtraction Property
    a−b=a+(−b)
    Subtracting a number is the same as adding its opposite.
  • Multiplication and Division of Signed Numbers
    For multiplication and division of two signed numbers:
    Same signs Result
    •  Two positives Positive
    •  Two negatives Positive

    If the signs are the same, the result is positive.
    Different signs Result
    •  Positive and negative Negative
    •  Negative and positive Negative

    If the signs are different, the result is negative.
  • Multiplication by −1
    −1a=−a
    Multiplying a number by −1 gives its opposite.
  • How to Use Integers in Applications.
    1. Read the problem. Make sure all the words and ideas are understood
    2. Identify what we are asked to find.
    3. Write a phrase that gives the information to find it.
    4. Translate the phrase to an expression.
    5. Simplify the expression.
    6. Answer the question with a complete sentence.

Practice Makes Perfect

Simplify Expressions with Absolute Value

In the following exercises, fill in <,>, or = for each of the following pairs of numbers.


ⓐ |−7|___−|−7|
ⓑ 6___−|−6|
ⓒ |−11|___−11
ⓓ −(−13)___−|−13|

Solution

ⓐ > ⓑ > ⓒ > ⓓ >


ⓐ −|−9|___|−9|
ⓑ −8___|−8|
ⓒ |−1|___−1
ⓓ −(−14)___−|−14|


ⓐ −|2|___−|−2|
ⓑ −12___−|−12|
ⓒ |−3|___−3
ⓓ |−19|___−(−19)

Solution

ⓐ = ⓑ = ⓒ > ⓓ =


ⓐ −|−4|___−|4|
ⓑ 5___−|−5|
ⓒ −|−10|___−10
ⓓ −|−0|___−(−0)

In the following exercises, simplify.

|15−7|−|14−6|

Solution

0

|17−8|−|13−4|

18−|2(8−3)|

Solution

8

15−|3(8−5)|

18−|12−4(4−1)+3|

Solution

15

27−|19+4(3−1)−7|

10−3|9−3(3−1)|

Solution

1

13−2|11−2(5−2)|

Add and Subtract Integers

In the following exercises, simplify each expression.


ⓐ −7+(−4)
ⓑ −7+4
ⓒ 7+(−4).

Solution

ⓐ −11 ⓑ −3 ⓒ 3


ⓐ −5+(−9)
ⓑ −5+9
ⓒ 5+(−9)

48+(−16)

Solution

32

34+(−19)

−14+(−12)+4

Solution

−22

−17+(−18)+6

19+2(−3+8)

Solution

29

24+3(−5+9)


ⓐ 13−7
ⓑ −13−(−7)
ⓒ −13−7
ⓓ 13−(−7)

Solution

ⓐ 6 ⓑ −6 ⓒ −20 ⓓ 20


ⓐ 15−8
ⓑ −15−(−8)
ⓒ −15−8
ⓓ 15−(−8)

−17−42

Solution

−59

−58−(−67)

−14−(−27)+9

Solution

22

64+(−17)−9

ⓐ 44−28 ⓑ 44+(−28)

Solution

ⓐ 16 ⓑ 16

ⓐ 35−16 ⓑ 35+(−16)

ⓐ 27−(−18) ⓑ 27+18

Solution

ⓐ 45 ⓑ 45

ⓐ 46−(−37) ⓑ 46+37

(2−7)−(3−8)

Solution

0

(1−8)−(2−9)

−(6−8)−(2−4)

Solution

4

−(4−5)−(7−8)

25−[10−(3−12)]

Solution

6

32−[5−(15−20)]

Multiply and Divide Integers

In the following exercises, multiply or divide.


ⓐ −4·8
ⓑ 13(−5)
ⓒ −24÷6
ⓓ −52÷(−4)

Solution

ⓐ −32 ⓑ −65 ⓒ −4
ⓓ 13


ⓐ −3·9
ⓑ 9(−7)
ⓒ 35÷(−7)
ⓓ −84÷(−6)


ⓐ −28÷7
ⓑ −180÷15
ⓒ 3(−13)
ⓓ −1(−14)

Solution

ⓐ −4 ⓑ −12 ⓒ −39
ⓓ 14


ⓐ −36÷4
ⓑ −192÷12
ⓒ 9(−7)
ⓓ −1(−19)

Simplify and Evaluate Expressions with Integers

In the following exercises, simplify each expression.

ⓐ (−2)6 ⓑ −26

Solution

ⓐ 64 ⓑ −64

ⓐ (−3)5 ⓑ −35

5(−6)+7(−2)−3

Solution

−47

8(−4)+5(−4)−6

−3(−5)(6)

Solution

90

−4(−6)(3)

(8−11)(9−12)

Solution

9

(6−11)(8−13)

26−3(2−7)

Solution

41

23−2(4−6)

65÷(−5)+(−28)÷(−7)

Solution

−9

52÷(−4)+(−32)÷(−8)

9−2[3−8(−2)]

Solution

−29

11−3[7−4(−2)]

8−|2−4(4−1)+3|

Solution

1

7−|5−3(4−1)−6|

9−3|2(2−6)−(3−7)|

Solution

−3

5−2|2(1−4)−(2−5)|

(−3)2−24÷(8−2)

Solution

5

(−4)2−32÷(12−4)

In the following exercises, evaluate each expression.

y+(−14) when
ⓐ y=−33 ⓑ y=30

Solution

ⓐ −47 ⓑ 16

x+(−21) when
ⓐ x=−27 ⓑ x=44

(x+y)2 when
x=−3,y=14

Solution

121

(y+z)2 when
y=−3,z=15

9a−2b−8 when
a=−6 and b=−3

Solution

−56

7m−4n−2 when
m=−4 and n=−9

3x2−4xy+2y2 when
x=−2,y=−3

Solution

6

4x2−xy+3y2 when
x=−3,y=−2

Translate English Phrases to Algebraic Expressions

In the following exercises, translate to an algebraic expression and simplify if possible.

the sum of 3 and −15, increased by 7

Solution

(3+(−15))+7;−5

the sum of −8 and −9, increased by 23


ⓐ the difference of 10 and −18
ⓑ subtract 11 from −25

Solution

ⓐ 10−(−18);28
ⓑ −25−11;−36


ⓐ the difference of −5 and −30
ⓑ subtract −6 from −13

the quotient of −6 and the sum of a and b

Solution

−6a+b

the product of −13 and the difference of c and d

Use Integers in Applications

In the following exercises, solve.

Temperature On January 15, the high temperature in Anaheim, California, was 84°. That same day, the high temperature in Embarrass, Minnesota, was −12°. What was the difference between the temperature in Anaheim and the temperature in Embarrass?

Solution

96°

Temperature On January 21, the high temperature in Palm Springs, California, was 89°, and the high temperature in Whitefield, New Hampshire, was −31°. What was the difference between the temperature in Palm Springs and the temperature in Whitefield?

Football On the first down, the Chargers had the ball on their 25-yard line. They lost 6 yards on the first-down play, gained 10 yards on the second-down play, and lost 8 yards on the third-down play. What was the yard line at the end of the third-down play?

Solution

21 yards

Football On first down, the Steelers had the ball on their 30-yard line. They gained 9 yards on the first-down play, lost 14 yards on the second-down play, and lost 2 yards on the third-down play. What was the yard line at the end of the third-down play?

Checking Account Mayra has $124 in her checking account. She writes a check for $152. What is the new balance in her checking account?

Solution

−$28

Checking Account Reymonte has a balance of −$49 in his checking account. He deposits $281 to the account. What is the new balance?

Writing Exercises

Explain why the sum of −8 and 2 is negative, but the sum of 8 and −2 is positive.

Solution

Answers will vary.

Give an example from your life experience of adding two negative numbers.

In your own words, state the rules for multiplying and dividing integers.

Solution

Answers will vary.

Why is −43=(−4)3?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 6 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: simplify expressions with absolute value, add and subtract integers, multiply and divide integers, simplify and evaluate expressions with integers, translate English phrases to algebraic expressions, use integers in applications. The remaining columns are blank.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

absolute value
The absolute value of a number is its distance from 0 on the number line.
integers
The whole numbers and their opposites are called the integers.
negative numbers
Numbers less than 0 are negative numbers.
opposite
The opposite of a number is the number that is the same distance from zero on the number line but on the opposite side of zero.

Fractions

Learning Objectives

By the end of this section, you will be able to:

  • Simplify fractions
  • Multiply and divide fractions
  • Add and subtract fractions
  • Use the order of operations to simplify fractions
  • Evaluate variable expressions with fractions

A more thorough introduction to the topics covered in this section can be found in the Elementary Algebra 2e chapter, Foundations.

Simplify Fractions

A fraction is a way to represent parts of a whole. The fraction 23 represents two of three equal parts. See Figure 1. In the fraction 23, the 2 is called the numerator and the 3 is called the denominator. The line is called the fraction bar.

Figure shows a circle divided in three equal parts. 2 of these are shaded.
In the circle, 23 of the circle is shaded—2 of the 3 equal parts.

Fraction

A fraction is written ab, where b≠0 and

a is the numerator and b is the denominator.

A fraction represents parts of a whole. The denominator b is the number of equal parts the whole has been divided into, and the numerator a indicates how many parts are included.

Fractions that have the same value are equivalent fractions. The Equivalent Fractions

Property allows us to find equivalent fractions and also simplify fractions.

Equivalent Fractions Property

If a, b, and c are numbers where b≠0,c≠0,

then ab=a·cb·c and a·cb·c=ab.

A fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator.

For example,

  23 is simplified because there are no common factors of 2 and 3.

  1015 is not simplified because 5 is a common factor of 10 and 15.

We simplify, or reduce, a fraction by removing the common factors of the numerator and denominator. A fraction is not simplified until all common factors have been removed. If an expression has fractions, it is not completely simplified until the fractions are simplified.

Sometimes it may not be easy to find common factors of the numerator and denominator. When this happens, a good idea is to factor the numerator and the denominator into prime numbers. Then divide out the common factors using the Equivalent Fractions Property.

How To Simplify a Fraction

Simplify: −315770.

Solution
Step 1 is to rewrite the numerator and denominator to show the common factors. If needed, use a factor tree. Here, we rewrite 315 and 770 as the product of the primes. Starting with minus 315 divided by 770, we get, minus 3 times 3 time 5 times 7 divided by 2 times 5 times 7 times 11. Step 2 is to simplify using the Equivalent Fractions Property by dividing out common factors. We first mark out the common factors 5 and 7 and then divide them out. This leaves minus 3 times 3 divided by 2 times 11. Step 3 is to multiply the remaining factors, if necessary. We get minus 9 by 22.

Simplify: −69120.

Solution

−2340

Simplify: −120192.

Solution

−58

We now summarize the steps you should follow to simplify fractions.

Simplify a fraction.

  1. Rewrite the numerator and denominator to show the common factors.
    If needed, factor the numerator and denominator into prime numbers first.
  2. Simplify using the Equivalent Fractions Property by dividing out common factors.
  3. Multiply any remaining factors.

Multiply and Divide Fractions

Many people find multiplying and dividing fractions easier than adding and subtracting fractions.

To multiply fractions, we multiply the numerators and multiply the denominators.

Fraction Multiplication

If a, b, c, and d are numbers where b≠0, and d≠0, then

ab·cd=acbd

To multiply fractions, multiply the numerators and multiply the denominators.

When multiplying fractions, the properties of positive and negative numbers still apply, of course. It is a good idea to determine the sign of the product as the first step. In Example 2, we will multiply a negative by a negative, so the product will be positive.

When multiplying a fraction by an integer, it may be helpful to write the integer as a fraction. Any integer, a, can be written as a1. So, for example, 3=31.

Multiply: −125(−20x).

Solution

The first step is to find the sign of the product. Since the signs are the same, the product is positive.

A mathematical expression showing the product of a negative fraction -12/5 and a negative term in parentheses, (-20x).
Determine the sign of the product. The signs      
are the same, so the product is positive.
The image displays the mathematical expression twelve-fifths multiplied by twenty x, written as (12/5)(20x).
Write 20x as a fraction. The mathematical expression 12/5 multiplied by the fraction 20x/1.
Multiply. A fraction with 12 multiplied by 20x in the numerator and 5 multiplied by 1 in the denominator. The '20x' term is highlighted in red.
Rewrite 20 to show the common factor 5
and divide it out.
A mathematical expression showing the cancellation of the number 5 from both the numerator and denominator: (12 * 4 *  (crossed out 5) * x) / ( (crossed out 5) * 1).
Simplify. The image shows the number '48x' in a faded, dark gray font against a white background.

Multiply: 113(−9a).

Solution

−33a

Multiply: 137(−14b).

Solution

−26b

Now that we know how to multiply fractions, we are almost ready to divide. Before we can do that, we need some vocabulary. The reciprocal of a fraction is found by inverting the fraction, placing the numerator in the denominator and the denominator in the numerator. The reciprocal of 23 is 32. Since 4 is written in fraction form as 41, the reciprocal of 4 is 14.

To divide fractions, we multiply the first fraction by the reciprocal of the second.

Fraction Division

If a, b, c, and d are numbers where b≠0,c≠0, and d≠0, then

ab÷cd=ab·dc

To divide fractions, we multiply the first fraction by the reciprocal of the second.

We need to say b≠0, c≠0, and d≠0, to be sure we don’t divide by zero!

Find the quotient: −718÷(−1427).

Solution
A mathematical expression showing the division of two negative fractions: -7/18 divided by (-14/27).
To divide, multiply the first fraction by the      
reciprocal of the second.
A mathematical expression showing the product of two negative fractions: -7/18 multiplied by -27/14.
Determine the sign of the product, and
then multiply.
A fraction with 7 multiplied by 27 in the numerator, and 18 multiplied by 14 in the denominator.
Rewrite showing common factors. Mathematical expression (7*9*3)/(9*2*7*2) with common factors 7 and 9 crossed out, illustrating fraction simplification.
Remove common factors. A fraction with 3 in the numerator and the product of 2 and 2 in the denominator, represented as 3 / (2 * 2).
Simplify. The fraction three-fourths (3/4) is displayed vertically with the number 3 over a horizontal line, and the number 4 below it, set against a plain white background.

Divide: −727÷(−3536).

Solution

415

Divide: −514÷(−1528).

Solution

23

The numerators or denominators of some fractions contain fractions themselves. A fraction in which the numerator or the denominator is a fraction is called a complex fraction.

Complex Fraction

A complex fraction is a fraction in which the numerator or the denominator contains a fraction.

Some examples of complex fractions are:

6733458x256

To simplify a complex fraction, remember that the fraction bar means division. For example, the complex fraction 3458 means 34÷58.

Simplify: x2xy6.

Solution
Step-by-step simplification of a complex rational expression involving division.
x2xy6
Rewrite as division. x2÷xy6
Multiply the first fraction by the reciprocal of the second. x2·6xy
Multiply. x·62·xy
Look for common factors. x·3·22·x·y
Divide common factors and simplify. 3y

Simplify: a8ab6.

Solution

34b

Simplify: p2pq8.

Solution

4q

Add and Subtract Fractions

When we multiplied fractions, we just multiplied the numerators and multiplied the denominators right straight across. To add or subtract fractions, they must have a common denominator.

Fraction Addition and Subtraction

If a, b, and c are numbers where c≠0, then

ac+bc=a+bcandac−bc=a−bc

To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.

The least common denominator (LCD) of two fractions is the smallest number that can be used as a common denominator of the fractions. The LCD of the two fractions is the least common multiple (LCM) of their denominators.

Least Common Denominator

The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.

After we find the least common denominator of two fractions, we convert the fractions to equivalent fractions with the LCD. Putting these steps together allows us to add and subtract fractions because their denominators will be the same!

How to Add or Subtract Fractions

Add: 712+518.

Solution
The expression is 7 by 12 plus 5 by 18. Step 1 is to check if the two numbers have a common denominator. Since they do not, rewrite each fraction with the LCD (least common denominator). For finding the LCD, we write the factors of 12 as 2 times 2 times 2 and the factors of 18 as 2 times 3 times 3. The LCD is 2 times 2 times 3 times 3, which is equal to 36. Step 2 is to add or subtract the fractions. We multiply the numerator and denominator of each fraction by the factor needed to get the denominator to be 36. Do not simplify the equivalent fractions. If you do, you’ll get back to the original fractions and lose the common denominator. We multiply the numerator and denominator of 7 divided by 12, by 3 times. We multiply numerator and denominator of 5 divided by 18 by 2 times. We get the expression 21 by 36 plus 10 by 36. Step 3 is to simplify is possible. Since 31 is prime, its only factors are 1and 31. Since 31 does not go into 36, the answer is simplified.

Add: 712+1115.

Solution

7960

Add: 1315+1720.

Solution

10360

Add or subtract fractions.

  1. Do they have a common denominator?
    • Yes—go to step 2.
    • No—rewrite each fraction with the LCD (least common denominator).
      • Find the LCD.
      • Change each fraction into an equivalent fraction with the LCD as its denominator.
  2. Add or subtract the fractions.
  3. Simplify, if possible.

We now have all four operations for fractions. Table 4 summarizes fraction operations.

Fraction Multiplication Fraction Division
ab·cd=acbd ab÷cd=ab·dc
Multiply the numerators and multiply the denominators Multiply the first fraction by the reciprocal of the second.
Fraction Addition Fraction Subtraction
ac+bc=a+bc ac−bc=a−bc
Add the numerators and place the sum over the common denominator. Subtract the numerators and place the difference over the common denominator.
To multiply or divide fractions, an LCD is NOT needed.
To add or subtract fractions, an LCD is needed.

When starting an exercise, always identify the operation and then recall the methods needed for that operation.

Simplify: ⓐ 5x6−310 ⓑ 5x6·310.

Solution

First ask, “What is the operation?” Identifying the operation will determine whether or not we need a common denominator. Remember, we need a common denominator to add or subtract, but not to multiply or divide.

ⓐ
This table outlines the step-by-step process for subtracting fractions with unlike denominators, from finding the LCD to simplifying the final expression.
What is the operation? The operation is subtraction.
Do the fractions have a common denominator? No. 5x6−310
Find the LCD of 6 and 10 The LCD is 30.
6=2·310=2·5___________LCD=2·3·5LCD=30
Rewrite each fraction as an equivalent fraction with the LCD. 5x·56·5−3·310·3
25x30−930
Subtract the numerators and place the difference over the common denominators. 25x−930
Simplify, if possible. There are no common factors. The fraction is simplified.
ⓑ
Step-by-step guide demonstrating the multiplication and simplification of algebraic fractions, contrasting it with addition regarding LCD.
What is the operation? Multiplication. 5x6·310
To multiply fractions, multiply the numerators and multiply the denominators. 5x·36·10
Rewrite, showing common factors.
Remove common factors.
5x·32·3·2·5
Simplify. x4
Notice, we needed an LCD to add 5x6−310, but not to multiply 5x6·310.

Simplify: ⓐ 3a4−89 ⓑ 3a4·89.

Solution

ⓐ 27a−3236 ⓑ 2a3

Simplify: ⓐ 4k5−16 ⓑ 4k5·16.

Solution

ⓐ 24k−530 ⓑ 2k15

Use the Order of Operations to Simplify Fractions

The fraction bar in a fraction acts as grouping symbol. The order of operations then tells us to simplify the numerator and then the denominator. Then we divide.

Simplify an expression with a fraction bar.

  1. Simplify the expression in the numerator. Simplify the expression in the denominator.
  2. Simplify the fraction.

Where does the negative sign go in a fraction? Usually the negative sign is in front of the fraction, but you will sometimes see a fraction with a negative numerator, or sometimes with a negative denominator. Remember that fractions represent division. When the numerator and denominator have different signs, the quotient is negative.

−13=−13negativepositive=negative
1−3=−13positivenegative=negative

Placement of Negative Sign in a Fraction

For any positive numbers a and b,

−ab=a−b=−ab

Simplify: 4(−3)+6(−2)−3(2)−2.

Solution

The fraction bar acts like a grouping symbol. So completely simplify the numerator and the denominator separately.

This table illustrates the step-by-step evaluation of a complex fractional mathematical expression.
4(−3)+6(−2)−3(2)−2
Multiply. −12+(−12)−6−2
Simplify. −24−8
Divide. 3

Simplify: 8(−2)+4(−3)−5(2)+3.

Solution

4

Simplify: 7(−1)+9(−3)−5(3)−2.

Solution

2

Now we’ll look at complex fractions where the numerator or denominator contains an expression that can be simplified. So we first must completely simplify the numerator and denominator separately using the order of operations. Then we divide the numerator by the denominator as the fraction bar means division.

How to Simplify Complex Fractions

Simplify: (12)24+32.

Solution
The expression is 1 by 2 the whole squared divided by 4 plus 3 squared. Step 1 is to simplify the numerator, which becomes 1 by 4. Step 2 is to simplify the denominator, which becomes 4 plus 9 equals 13. Step 3 is to divide the numerator by the denominator and simplify if possible. Now the expression becomes 1 by 4 divided by 13 by 1, which equals 1 by 4 multiplied by 1 by 13, which equals 1 by 52

Simplify: ( 1 3)223+2.

Solution

190

Simplify: 1+42(14)2.

Solution

272

Simplify complex fractions.

  1. Simplify the numerator.
  2. Simplify the denominator.
  3. Divide the numerator by the denominator. Simplify if possible.

Simplify: 12+2334−16.

Solution

It may help to put parentheses around the numerator and the denominator.

Step-by-step simplification of a complex fraction, demonstrating common denominator usage, fraction arithmetic, and division to arrive at a final integer value.
(12+23)(34−16)
Simplify the numerator (LCD = 6) and simplify the denominator (LCD = 12). (36+46)(912−212)
Simplify. (76)(712)
Divide the numerator by the denominator. 76÷712
Simplify. 76⋅127
Divide out common factors. 7⋅6⋅26⋅7⋅1
Simplify. 2

Simplify: 13+1234−13.

Solution

2

Simplify: 23−1214+13.

Solution

27

Evaluate Variable Expressions with Fractions

We have evaluated expressions before, but now we can evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.

Evaluate 2x2y when x=14 and y=−23.

Solution

Substitute the values into the expression.

The mathematical expression 2x^2y is shown on a white background.
The text reads, 'Substitute 1/4 for x and -2/3 for y.' The fraction 1/4 is colored red and the fraction -2/3 is colored light blue. A mathematical expression showing 2 multiplied by (1/4) squared, then multiplied by (-2/3).
Simplify exponents first. A mathematical expression showing the product of 2, 1/16, and -2/3.
Multiply; divide out the common factors.
Notice we write 16 as 2·2·4 to make it easy to
remove common factors.
A negative fraction displayed with multiplication in the numerator and denominator. Numbers 2, 1, 2 are in the numerator, and 2, 2, 4, 3 are in the denominator. Some '2's are crossed out, showing fraction simplification.
Simplify. The image displays the negative fraction -1/12, represented with a minus sign preceding the fraction bar, 1 as the numerator, and 12 as the denominator, all centered on a white background.

Evaluate 3ab2 when a=−23 and b=−12.

Solution

−12

Evaluate 4c3d when c=−12 and d=−43.

Solution

23

Access this online resource for additional instruction and practice with fractions.

  • Adding Fractions with Unlike Denominators

Key Concepts

  • Equivalent Fractions Property
    If a, b, and c are numbers where b≠0,c≠0, then
    ab=a·cb·canda·cb·c=ab.
  • How to simplify a fraction.
    1. Rewrite the numerator and denominator to show the common factors.
      If needed, factor the numerator and denominator into prime numbers first.
    2. Simplify using the Equivalent Fractions Property by dividing out common factors.
    3. Multiply any remaining factors.
  • Fraction Multiplication
    If a, b, c, and d are numbers where b≠0, and d≠0, then
    ab·cd=acbd.
    To multiply fractions, multiply the numerators and multiply the denominators.
  • Fraction Division
    If a, b, c, and d are numbers where b≠0,c≠0, and d≠0, then
    ab÷cd=ab·dc.
    To divide fractions, we multiply the first fraction by the reciprocal of the second.
  • Fraction Addition and Subtraction
    If a, b, and c are numbers where c≠0, then
    ac+bc=a+bcandac−bc=a−bc.
    To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.
  • How to add or subtract fractions.
    1. Do they have a common denominator?
      • Yes—go to step 2.
      • No—rewrite each fraction with the LCD (least common denominator).
        • Find the LCD.
        • Change each fraction into an equivalent fraction with the LCD as its denominator.
    2. Add or subtract the fractions.
    3. Simplify, if possible.
  • How to simplify an expression with a fraction bar.
    1. Simplify the expression in the numerator. Simplify the expression in the denominator.
    2. Simplify the fraction.
  • Placement of Negative Sign in a Fraction
    For any positive numbers a and b,
    −ab=a−b=−ab.
  • How to simplify complex fractions.
    1. Simplify the numerator.
    2. Simplify the denominator.
    3. Divide the numerator by the denominator. Simplify if possible.

Practice Makes Perfect

Simplify Fractions

In the following exercises, simplify.

−10863

Solution

−127

−10448

120252

Solution

1021

182294

14x221y

Solution

2x23y

24a32b2

−210a2110b2

Solution

−21a211b2

−30x2105y2

Multiply and Divide Fractions

In the following exercises, perform the indicated operation.

−34(−49)

Solution

13

−38·415

(−1415)(920)

Solution

−2150

(−910)(2533)

(−6384)(−4490)

Solution

1130

(−3360)(−4088)

37·21n

Solution

9n

56·30m

34÷x11

Solution

334x

25÷y9

518÷(−1524)

Solution

−49

718÷(−1427)

8u15÷12v25

Solution

10u9v

12r25÷18s35

34÷(−12)

Solution

−116

−15÷(−53)

In the following exercises, simplify.

−8211235

Solution

−109

−9163340

−452

Solution

−25

5310

m3n2

Solution

2m3n

−38−y12

Add and Subtract Fractions

In the following exercises, add or subtract.

712+58

Solution

2924

512+38

712−916

Solution

148

716−512

−1330+2542

Solution

17105

−2330+548

−3956−2235

Solution

−5340

−3349−1835

−23−(−34)

Solution

112

−34−(−45)

x3+14

Solution

4x+312

x5−14


ⓐ 23+16
ⓑ 23÷16

Solution

ⓐ 56 ⓑ 4


ⓐ −25−18
ⓑ −25·18


ⓐ 5n6÷815
ⓑ 5n6−815

Solution

ⓐ 25n16 ⓑ 25n−1630


ⓐ 3a8÷712
ⓑ 3a8−712


ⓐ −4x9−56
ⓑ −4k9·56

Solution

ⓐ −8x−1518 ⓑ −10k27


ⓐ −3y8−43
ⓑ −3y8·43


ⓐ −5a3+(−106)
ⓑ −5a3÷(−106)

Solution

ⓐ −5(a+1)3 ⓑ a


ⓐ 2b5+815
ⓑ 2b5÷815

Use the Order of Operations to Simplify Fractions

In the following exercises, simplify.

5·6−3·44·5−2·3

Solution

97

8·9−7·65·6−9·2

52−323−5

Solution

−8

62−424−6

7·4−2(8−5)9·3−3·5

Solution

116

9·7−3(12−8)8·7−6·6

9(8−2)−3(15−7)6(7−1)−3(17−9)

Solution

52

8(9−2)−4(14−9)7(8−3)−3(16−9)

23+42(23)2

Solution

54

33−32(34)2

(35)2(37)2

Solution

4925

(34)2(58)2

213+15

Solution

154

514+13

78−2312+38

Solution

521

34−3514+25

Mixed Practice

In the following exercises, simplify.

−38÷(−310)

Solution

54

−312÷(−59)

−38+512

Solution

124

−18+712

−715−y4

Solution

−28−15y60

−38−x11

1112a·9a16

Solution

3364

10y13·815y

12+23·512

Solution

79

13+25·34

1−35÷110

Solution

−5

1−56÷112

38−16+34

Solution

2324

25+58−34

12(920−415)

Solution

115

8(1516−56)

58+161924

Solution

1

16+3101430

(59+16)÷(23−12)

Solution

133

(34+16)÷(58−13)

Evaluate Variable Expressions with Fractions

In the following exercises, evaluate.

710−w when
ⓐ w=12 ⓑ w=−12

Solution

ⓐ 15 ⓑ 65

512−w when
ⓐ w=14 ⓑ w=−14

2x2y3 when
x=−23 and y=−12

Solution

−19

8u2v3 when
u=−34 and v=−12

a+ba−b when
a=−3,b=8

Solution

−511

r−sr+s when
r=10,s=−5

Writing Exercises

Why do you need a common denominator to add or subtract fractions? Explain.

Solution

Answers will vary.

How do you find the LCD of 2 fractions?

Explain how you find the reciprocal of a fraction.

Solution

Answers will vary.

Explain how you find the reciprocal of a negative number.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 5 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: simplify fractions, multiply and divide fractions, add and subtract fractions, use the order of operations to simplify fractions, evaluate variable expressions with fractions. The remaining columns are blank.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

complex fraction
A fraction in which the numerator or the denominator is a fraction is called a complex fraction.
denominator
In a fraction, written ab, where b≠0, the denominator b is the number of equal parts the whole has been divided into.
equivalent fractions
Equivalent fractions are fractions that have the same value.
fraction
A fraction is written ab, where b≠0, and a is the numerator and b is the denominator. A fraction represents parts of a whole.
least common denominator
The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.
numerator
In a fraction, written ab, where b≠0, the numerator a indicates how many parts are included.
reciprocal
The reciprocal of a fraction is found by inverting the fraction, placing the numerator in the denominator and the denominator in the numerator.

Decimals

Learning Objectives

By the end of this section, you will be able to:

  • Round decimals
  • Add and subtract decimals
  • Multiply and divide decimals
  • Convert decimals, fractions, and percents
  • Simplify expressions with square roots
  • Identify integers, rational numbers, irrational numbers, and real numbers
  • Locate fractions and decimals on the number line

A more thorough introduction to the topics covered in this section can be found in the Elementary Algebra 2e chapter, Foundations.

Round Decimals

Decimals are another way of writing fractions whose denominators are powers of ten.

0.1=110is “one tenth” 0.01=1100is “one hundredth” 0.001=11000is “one thousandth” 0.0001=110,000is “one ten-thousandth”

Just as in whole numbers, each digit of a decimal corresponds to the place value based on the powers of ten. Figure 1 shows the names of the place values to the left and right of the decimal point.

This table is labeled place value and has 12 columns. The seventh column is blank. Starting from here and going left the columns are labeled: ones, tens, hundreds, thousands, ten thousands, hundred thousands. Starting from the blank column and going right the columns are labeled: tenths, hundredths, thousandths, ten thousandths hundred thousandths. There is a dot under the blank column.

When we work with decimals, it is often necessary to round the number to the nearest required place value. We summarize the steps for rounding a decimal here.

Round decimals.

  1. Locate the given place value and mark it with an arrow.
  2. Underline the digit to the right of the place value.
  3. Is the underlined digit greater than or equal to 5?
    • Yes: add 1 to the digit in the given place value.
    • No: do not change the digit in the given place value
  4. Rewrite the number, deleting all digits to the right of the rounding digit.

Round 18.379 to the nearest ⓐ hundredth ⓑ tenth ⓒ whole number.

Solution

Round 18.379.

ⓐ to the nearest hundredth

Locate the hundredths place with an arrow. An image illustrating the hundredths place in the number 18.379, with an arrow pointing from the text 'hundredths place' to the digit '7'.
Underline the digit to the right of the given
place value.
The number 18.379 is shown, with an arrow pointing from 'hundredths place' to the underlined digit '7', indicating its position in the decimal.
Because 9 is greater than or equal to 5, add 1 to
the 7.
Rounding 18.379: The image illustrates rounding the number 18.379. The digit '7' in the hundredths place has '1' added to it, and the digit '9' in the thousandths place is deleted, resulting in 18.38.
Rewrite the number, deleting all digits to the
right of the rounding digit.
The numbers 18.38 are displayed prominently against a white background.
Notice that the deleted digits were NOT
replaced with zeros.
The text states that 18.379 rounded to the nearest hundredth is 18.38, demonstrating a basic principle of decimal rounding in mathematics.

ⓑ to the nearest tenth

Locate the tenths place with an arrow.      An arrow points from 'tenths place' to the digit '3' in the number 18.379, illustrating the position of the tenths place in a decimal number.
Underline the digit to the right of the
given place value.
An arrow points from the text 'tenths place' to the digit '3' in the number 18.379, indicating that '3' is in the tenths place.
Because 7 is greater than or equal to 5,
add 1 to the 3.
An illustration shows how to transform the number 18.379 by adding 1 to its integer part '18' and deleting its decimal part '.379', resulting in 19.
Rewrite the number, deleting all digits to
the right of the rounding digit.
The number 18.4 is displayed in dark gray text against a plain white background.
Notice that the deleted digits were NOT
replaced with zeros.
The text states, 'So, 18.379 rounded to the nearest tenth is 18.4.'

ⓒ to the nearest whole number

Locate the ones place with an arrow.     An arrow pointing from 'ones place' to the digit 8 in the number 18.379, illustrating the location of the ones place value in a decimal number.
Underline the digit to the right of the
given place value.
An image illustrating the concept of place value, specifically pointing to the 'ones place' in the number 18.379. An arrow from 'ones place' indicates the digit '8'.
Since 3 is not greater than or equal to 5,
do not add 1 to the 8.
Instructions for truncating the number 18.379: delete the decimal part (.379) and do not add 1 to the integer (18).
Rewrite the number, deleting all digits to
the right of the rounding digit.
The number 18 is displayed in a clear, black font against a plain white background, occupying the central upper portion of the image.
Text explaining rounding: 'So, 18.379 rounded to the nearest whole number is 18.'

Round 6.582 to the nearest ⓐ hundredth ⓑ tenth ⓒ whole number.

Solution

ⓐ 6.58 ⓑ 6.6 ⓒ 7

Round 15.2175 to the nearest ⓐ thousandth ⓑ hundredth ⓒ tenth.

Solution

ⓐ 15.218 ⓑ 15.22
ⓒ 15.2

Add and Subtract Decimals

To add or subtract decimals, we line up the decimal points. By lining up the decimal points this way, we can add or subtract the corresponding place values. We then add or subtract the numbers as if they were whole numbers and then place the decimal point in the sum.

Add or subtract decimals.

  1. Determine the sign of the sum or difference.
  2. Write the numbers so the decimal points line up vertically.
  3. Use zeros as placeholders, as needed.
  4. Add or subtract the numbers as if they were whole numbers. Then place the
    decimal point in the answer under the decimal points in the given numbers.
  5. Write the sum or difference with the appropriate sign.

Add or subtract: ⓐ −23.5−41.38 ⓑ 14.65−20.

Solution
ⓐ
Step-by-step calculation demonstrating the subtraction of negative decimal numbers.
−23.5−41.38
The difference will be negative. To subtract, we add the numerals. Write the numbers so the decimal points line up vertically. 23.5+41.38______
Put 0 as a placeholder after the 5 in 23.5.
Remember, 510=50100 so 0.5=0.50.
23.50+41.38______
Add the numbers as if they were whole numbers.
Then place the decimal point in the sum.
23.50+41.38______64.88
Write the result with the correct sign. −23.5−41.38=−64.88
ⓑ
Step-by-step guide on subtracting a decimal from a whole number, illustrating decimal alignment, placeholders, and obtaining a negative result.
14.65−20
The difference will be negative. To subtract, we subtract 14.65 from 20.
Write the numbers so the decimal points line up vertically. 20−14.65______
Remember, 20 is a whole number, so place the decimal point after the 0.
Put in zeros to the right as placeholders. 20.00−14.65______
Subtract and place the decimal point in the answer. 99110101020.00−14.65______________ 5.35
Write the result with the correct sign. 14.65−20=−5.35

Add or subtract: ⓐ −4.8−11.69 ⓑ 9.58−10.

Solution

ⓐ −16.49 ⓑ −0.42

Add or subtract: ⓐ −5.123−18.47 ⓑ 37.42−50.

Solution

ⓐ −23.593 ⓑ −12.58

Multiply and Divide Decimals

When we multiply signed decimals, first we determine the sign of the product and then multiply as if the numbers were both positive. We multiply the numbers temporarily ignoring the decimal point and then count the number of decimal points in the factors and that sum tells us the number of decimal places in the product. Finally, we write the product with the appropriate sign.

Multiply decimals.

  1. Determine the sign of the product.
  2. Write in vertical format, lining up the numbers on the right. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
  3. Place the decimal point. The number of decimal places in the product is the sum of
    the number of decimal places in the factors.
  4. Write the product with the appropriate sign.

Multiply: (−3.9)(4.075).

Solution
(−3.9)(4.075)

The signs are different. The product
will be negative.

The product will be negative.
Write in vertical format, lining up the
numbers on the right.

A vertical multiplication problem shows 4.075 multiplied by 3.9.
Multiply. A long multiplication calculation of 4.075 by 3.9, with partial products leading to a sum of 158925.

Add the number of decimal places in
the factors (1 + 3).
Place the decimal point 4 places from the right.
A step-by-step example of multiplying 4.075 by 3.9, resulting in 15.8925, demonstrating how the decimal places are counted in the product.
The image displays two numbers, (-3.9) and (4.075), indicating their respective number of decimal places: 1 place for -3.9 and 3 places for 4.075.
The signs are the different, so the product is negative. (−3.9)(4.075)=−15.8925

Multiply: −4.5(6.107).

Solution

−27.4815

Multiply: −10.79(8.12).

Solution

−87.6148

Often, especially in the sciences, you will multiply decimals by powers of 10 (10, 100, 1000, etc). If you multiply a few products on paper, you may notice a pattern relating the number of zeros in the power of 10 to number of decimal places we move the decimal point to the right to get the product.

Multiply a decimal by a power of ten.

  1. Move the decimal point to the right the same number of places as the
    number of zeros in the power of 10.
  2. Add zeros at the end of the number as needed.

Multiply: 5.63 by ⓐ 10 ⓑ 100 ⓒ 1000.

Solution

By looking at the number of zeros in the multiple of ten, we see the number of places we need to move the decimal to the right.

ⓐ

The image displays the text '5.63 (10)' in a simple, gray font against a white background.
There is 1 zero in 10, so move the decimal point 1 place to the right.  The number 5.63 with a light blue downward-pointing arrow originating from beneath it.
The number 56.3 is displayed in a digital font against a white background.

ⓑ

The image shows the mathematical expression 5.63(100), representing the multiplication of 5.63 by 100, which equals 563.
There are 2 zeroes in 100, so move the decimal point 2 places to the right. The number 5.63 is shown above a wavy, blue-green arrow pointing to the right, which could symbolize progression, flow, or a mathematical operation like rounding.
The number '563' is displayed in a dark grey font against a plain white background.

ⓒ

A mathematical expression showing the multiplication of 5.63 by 1,000, written as 5.63(1,000).
There are 3 zeroes in 1,000, so move the decimal point 3 place to the right. A light blue wavy arrow points to the right below the number '5.63' on a white background.
A zero must be added to the end. The number 5,630 is displayed in a clear, digital-style font against a white background.

Multiply 2.58 by ⓐ 10 ⓑ 100 ⓒ 1000.

Solution

ⓐ 25.8 ⓑ 258 ⓒ 2,580

Multiply 14.2 by ⓐ 10 ⓑ 100 ⓒ 1000.

Solution

ⓐ 142 ⓑ 1,420 ⓒ 14,200

Just as with multiplication, division of signed decimals is very much like dividing whole numbers. We just have to figure out where the decimal point must be placed and the sign of the quotient. When dividing signed decimals, first determine the sign of the quotient and then divide as if the numbers were both positive. Finally, write the quotient with the appropriate sign.

We review the notation and vocabulary for division:

In the expression a divided by b equals c, a is the dividend, b is the divisor and c is the quotient. This can be written as b right parentheses a overbar, with c on top of the bar. In this case too, a is the dividend, b is the divisor and c is the quotient.

We’ll write the steps to take when dividing decimals for easy reference.

Divide decimals.

  1. Determine the sign of the quotient.
  2. Make the divisor a whole number by “moving” the decimal point all the way to the right. “Move” the decimal point in the dividend the same number of places—adding zeros as needed.
  3. Divide. Place the decimal point in the quotient above the decimal point in the dividend.
  4. Write the quotient with the appropriate sign.

Divide: −25.65÷(−0.06).

Solution

Remember, you can “move” the decimals in the divisor and dividend because of the Equivalent Fractions Property.

A mathematical expression showing the division of two negative decimal numbers: -25.65 ÷ (-0.06).
The signs are the same. The quotient is positive.
Make the divisor a whole number by “moving” the
decimal point all the way to the right.
“Move” the decimal point in the dividend the same
number of places.
A long division problem showing 25.65 divided by 0.06, with light blue arrows indicating the shifting of decimal points to simplify the division by a decimal.
Divide.
Place the decimal point in the quotient above the
decimal point in the dividend.
Long division calculation for 2565.0 divided by 6, showing the step-by-step process resulting in a quotient of 427.5.
Write the quotient with the appropriate sign. A mathematical equation displays the division of -25.65 by -0.06, resulting in a positive value of 427.5.

Divide: −23.492÷(−0.04).

Solution

587.3

Divide: −4.11÷(−0.12).

Solution

34.25

Convert Decimals, Fractions, and Percents

In our work, it is often necessary to change the form of a number. We may have to change fractions to decimals or decimals to percent.

We convert decimals into fractions by identifying the place value of the last (farthest right) digit. In the decimal 0.03. the 3 is in the hundredths place, so 100 is the denominator of the fraction equivalent to 0.03.

0.03=3100

The steps to take to convert a decimal to a fraction are summarized in the procedure box.

Convert a decimal to a proper fraction and a fraction to a decimal.

  1. To convert a decimal to a proper fraction, determine the place value of the final digit.
  2. Write the fraction.
    • numerator—the “numbers” to the right of the decimal point
    • denominator—the place value corresponding to the final digit
  3. To convert a fraction to a decimal, divide the numerator of the fraction by the denominator of the fraction.

Write: ⓐ 0.374 as a fraction ⓑ −58 as a decimal.

Solution

ⓐ

A white background displays the numerical value 0.374 in black text, positioned centrally towards the top of the frame.
Determine the place value of the final digit. An image displaying decimal place values. The number 0.3 is above 'tenths' in red, 7 above 'hundredths' in blue, and 4 above 'thousandths' in orange.
Write the fraction for 0.374:
The numerator is 374.
The denominator is 1,000.
A fraction is displayed with 374 as the numerator and 1000 as the denominator, shown in black text on a white background.
Simplify the fraction. A mathematical expression showing the fraction (2 * 187) divided by (2 * 500), where the multiplication is represented by a middle dot.
Divide out the common factors. The image displays the fraction 187 over 500, written vertically with a horizontal line separating the numerator and the denominator.
The image shows a mathematical equation stating: so, 0.374 = 187/500. This expression converts the decimal 0.374 into its equivalent fraction 187/500.

ⓑ Since a fraction bar means division, we begin by writing the fraction 58 as 85. Now divide.

The division shows that 5 is divided by 8 to yield 0.625. The result concludes that five eights is equal to negative 0.625.

Write: ⓐ 0.234 as a fraction ⓑ −78 as a decimal.

Solution

ⓐ 117500 ⓑ −0.875

Write: ⓐ 0.024 as a fraction ⓑ −38 as a decimal.

Solution

ⓐ 3125 ⓑ −0.375

A percent is a ratio whose denominator is 100. Percent means per hundred. We use the percent symbol, %, to show percent. Since a percent is a ratio, it can easily be expressed as a fraction. Percent means per 100, so the denominator of the fraction is 100. We then change the fraction to a decimal by dividing the numerator by the denominator. After doing this many times, you may see the pattern.

To convert a percent number to a decimal number, we move the decimal point two places to the left.

Figure shows the value 6 percent. An arrow indicates that the decimal is moved two places to the left. Hence the value is equal to 0.06. Similarly, 78 percent is 0.78, 2.7 percent is 0. 027 and 135 percent is 1.35.

To convert a decimal to a percent, remember that percent means per hundred. If we change the decimal to a fraction whose denominator is 100, it is easy to change that fraction to a percent. After many conversions, you may recognize the pattern.

To convert a decimal to a percent, we move the decimal point two places to the right and then add the percent sign.

Figure shows value 0.05. An arrow indicates that the decimal is moved two places to the right. Hence the value becomes 5 percent. Similarly, 0.83 is 83 percent, 1.05 is 105 percent, 0.075 is 7.5 percent and 0.3 is 30 percent.

Convert a percent to a decimal and a decimal to a percent.

  1. To convert a percent to a decimal, move the decimal point two places to the left after removing the percent sign.
  2. To convert a decimal to a percent, move the decimal point two places to the right and then add the percent sign.

Convert each:

ⓐ percent to a decimal: 62%, 135%, and 35.7%.

ⓑ decimal to a percent: 0.51, 1.25, and 0.093.

Solution

ⓐ

Three percentages: 62%, 135%, and 35.7%, each with a light blue, wavy, W-shaped arrow pointing downwards underneath.
Move the decimal point two places to the left. Three decimal numbers are displayed on a white background: 0.62, 1.35, and 0.357.

ⓑ

Three numerical values: 0.51, 1.25, and 0.093, are displayed, each accompanied by a stylized light blue 'W' symbol with an upward arrow, likely representing power output.
Move the decimal point two places to the right. Three distinct percentage values are displayed on a white background: 51%, 125%, and 9.3%.

Convert each:

ⓐ percent to a decimal: 9%, 87%, and 3.9%.

ⓑ decimal to a percent: 0.17, 1.75, and 0.0825.

Solution

ⓐ 0.09, 0.87, 0.039 ⓑ 17%, 175%, 8.25%

Convert each:

ⓐ percent to a decimal: 3%, 91%, and 8.3%.

ⓑ decimal to a percent: 0.41, 2.25, and 0.0925.

Solution

ⓐ 0.03, 0.91, 0.083 ⓑ 41%, 225%, 9.25%

Simplify Expressions with Square Roots

Remember that when a number n is multiplied by itself, we write n2 and read it “n squared.” The result is called the square of a number n. For example, 82 is read “8 squared” and 64 is called the square of 8. Similarly, 121 is the square of 11 because 112 is 121. It will be helpful to learn to recognize the perfect square numbers.

Square of a number

If n2=m, then m is the square of n.

What about the squares of negative numbers? We know that when the signs of two numbers are the same, their product is positive. So the square of any negative number is also positive.

(−3)2=9(−8)2=64(−11)2=121(−15)2=225

Because 102=100, we say 100 is the square of 10. We also say that 10 is a square root of 100. A number whose square is m is called a square root of a number m.

Square Root of a Number

If n2=m, then n is a square root of m.

Notice (−10)2=100 also, so −10 is also a square root of 100. Therefore, both 10 and −10 are square roots of 100. So, every positive number has two square roots—one positive and one negative. The radical sign, m, denotes the positive square root. The positive square root is called the principal square root. When we use the radical sign that always means we want the principal square root.

Square Root Notation

m is read “the square root of m.”

Figure shows the expression square root of m. The square root sign is labeled radical sign and m is labeled radicand.

If m=n2, then m=n, for n≥0.

The square root of m, m, is the positive number whose square is m.

We know that every positive number has two square roots and the radical sign indicates the positive one. We write 100=10. If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, −100=−10. We read −100 as “the opposite of the principal square root of 100.”

Simplify: ⓐ 25 ⓑ 121 ⓒ −144.

Solution
ⓐ
Demonstrates finding the square root of 25, showing both the calculation and its result.
25
Since 52=25 5
ⓑ
Illustrates the calculation of the square root of 121, showing the expression and its numerical value.
121
Since 112=121 11
ⓒ
This table explains and evaluates expressions with a negative sign outside the square root, specifically showing `-\\sqrt{144}` and its resulting value.
−144
The negative is in front of the radical sign. −12

Simplify: ⓐ 36 ⓑ 169 ⓒ −225.

Solution

ⓐ 6 ⓑ 13 ⓒ −15

Simplify: ⓐ 16 ⓑ 196 ⓒ −100.

Solution

ⓐ 4 ⓑ 14 ⓒ −10

Identify Integers, Rational Numbers, Irrational Numbers, and Real Numbers

We have already described numbers as counting numbers, whole numbers, and integers. What is the difference between these types of numbers? Difference could be confused with subtraction. How about asking how we distinguish between these types of numbers?

Counting numbers1,2,3,4,….. Whole numbers0,1,2,3,4,…. Integers….−3,−2,−1,0,1,2,3,….

What type of numbers would we get if we started with all the integers and then included all the fractions? The numbers we would have form the set of rational numbers. A rational number is a number that can be written as a ratio of two integers.

In general, any decimal that ends after a number of digits (such as 7.3 or −1.2684) is a rational number. Simply write the decimal as a mixed number. The decimal for 13 is the number 0.3–. The bar over the 3 indicates that the number 3 repeats infinitely. Continuously has an important meaning in calculus. The number(s) under the bar is called the repeating block and it repeats continuously.

Since all integers can be written as a fraction whose denominator is 1, the integers (and so also the counting and whole numbers. are rational numbers.

Every rational number can be written both as a ratio of integers pq, where p and q are integers and q≠0, and as a decimal that stops or repeats.

Rational Number

A rational number is a number of the form pq, where p and q are integers and q≠0.

Its decimal form stops or repeats.

Are there any decimals that do not stop or repeat? Yes! The number π (the Greek letter pi, pronounced “pie”), which is very important in describing circles, has a decimal form that does not stop or repeat. We use three dots (…) to indicate the decimal does not stop or repeat.

π=3.141592654...

The square root of a number that is not a perfect square is a decimal that does not stop or repeat.

A numbers whose decimal form does not stop or repeat cannot be written as a fraction of integers. We call this an irrational number.

Irrational Number

An irrational number is a number that cannot be written as the ratio of two integers.

Its decimal form does not stop and does not repeat.

Let’s summarize a method we can use to determine whether a number is rational or irrational.

Rational or Irrational

If the decimal form of a number

  • repeats or stops, the number is a rational number.
  • does not repeat and does not stop, the number is an irrational number.

We have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. The irrational numbers are numbers whose decimal form does not stop and does not repeat. When we put together the rational numbers and the irrational numbers, we get the set of real numbers.

Real Number

A real number is a number that is either rational or irrational.

Later in this course we will introduce numbers beyond the real numbers. Figure 2 illustrates how the number sets we’ve used so far fit together.

A chart shows that counting numbers 1, 2, 3 are a part of whole numbers 0, 1, 2, 3. Whole numbers are a part of integers minus 2, minus 1, 0, 1, 2. Integers are a part of rational numbers. Rational numbers along with irrational numbers form the set of real numbers.
This chart shows the number sets that make up the set of real numbers.

Does the term “real numbers” seem strange to you? Are there any numbers that are not “real,” and, if so, what could they be? Can we simplify −25? Is there a number whose square is −25?

()2=−25?

None of the numbers that we have dealt with so far has a square that is −25. Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to −25. The square root of a negative number is not a real number.

Given the numbers −7,145,8,5,5.9,−64, list the ⓐ whole numbers ⓑ integers ⓒ rational numbers ⓓ irrational numbers ⓔ real numbers.

Solution

ⓐ Remember, the whole numbers are 0,1,2,3,…, so 8 is the only whole number given.

ⓑ The integers are the whole numbers and their opposites (which includes 0). So the whole number 8 is an integer, and −7 is the opposite of a whole number so it is an integer, too. Also, notice that 64 is the square of 8 so −64=−8. So the integers are −7,8, and −64.

ⓒ Since all integers are rational, then −7,8, and −64 are rational. Rational numbers also include fractions and decimals that repeat or stop, so 145 and 5.9 are rational. So the list of rational numbers is −7,145,8,5.9,​​ and −64.

ⓓ Remember that 5 is not a perfect square, so 5 is irrational.

ⓔ All the numbers listed are real numbers.

Given the numbers −3,−2,0.3–,95,4,49, list the ⓐ whole numbers ⓑ integers ⓒ rational numbers
ⓓ irrational numbers ⓔ real numbers.

Solution

ⓐ 4,49 ⓑ −3,4,49
ⓒ −3,0.3–,95,4,49 ⓓ −2
ⓔ −3,−2,0.3–,95,4,49

Given numbers −25,−38,−1,6,121,2.041975..., list the ⓐ whole numbers ⓑ integers ⓒ rational numbers ⓓ irrational numbers ⓔ real numbers.

Solution

ⓐ 6,121
ⓑ −25,−1,6,121
ⓒ −25,−38,−1,6,121
ⓓ 2.041975...
ⓔ −25,−38,−1,6,121,2.041975...

Locate Fractions and Decimals on the Number Line

We now want to include fractions and decimals on the number line. Let’s start with fractions and locate 15,−45,3,74,−92,−5 and 83 on the number line.

We’ll start with the whole numbers 3 and −5 because they are the easiest to plot. See Figure 3.

The proper fractions listed are 15 and −45. We know the proper fraction 15 has value less than one and so would be located between 0 and 1. The denominator is 5, so we divide the unit from 0 to 1 into 5 equal parts 15,25,35,45. We plot 15.

Similarly, −45 is between 0 and −1. After dividing the unit into 5 equal parts we plot −45.

Finally, look at the improper fractions 74,92,83. Locating these points may be easier if you change each of them to a mixed number.

74=134−92=−41283=223

Figure 3 shows the number line with all the points plotted.

Figure shows a number line with numbers ranging from minus 6 to 6. Various points on the line are highlighted. From left to right, these are: minus 5, minus 9 by 2, minus 4 by 5, 1 by 5, 4 by 5, 8 by 3 and 3.

Locate and label the following on a number line: 4,34,−14,−3,65,−52, and 73.

Solution

Locate and plot the integers, 4,−3.

Locate the proper fraction 34 first. The fraction 34 is between 0 and 1. Divide the distance between 0 and 1 into four equal parts, then we plot 34. Similarly plot −14.

Now locate the improper fractions 65,−52, and 73. It is easier to plot them if we convert them to mixed numbers and then plot them as described above: 65=115,−52=−212,73=213.

Figure shows a number line with numbers ranging from minus 6 to 6. Various points on the line are highlighted. From left to right, these are: minus 3, minus 5 by 2, minus 1 by 4, 3 by 4, 6 by 5, 7 by 3 and 4.

Locate and label the following on a number line: −1,13,65,−74,92,5,−83.

Solution

Figure shows a number line with numbers ranging from minus 4 to 5. Various points on the line are highlighted. From left to right, these are: minus 8 by 3, minus 7 by 4, minus 1, 1 by 3, 6 by 5, 9 by 2 and 5.

Locate and label the following on a number line: −2,23,75,−74,72,3,−73.

Solution

Figure shows a number line with numbers ranging from minus 4 to 5. Various points on the line are highlighted. From left to right, these are: minus 7 by 3, minus 2, minus 7 by 4, 2 by 3, 7 by 5, 3 and 7 by 2.

Since decimals are forms of fractions, locating decimals on the number line is similar to locating fractions on the number line.

Locate on the number line: ⓐ 0.4 ⓑ −0.74.

Solution

ⓐ The decimal number 0.4 is equivalent to 410, a proper fraction, so 0.4 is located between 0 and 1. On a number line, divide the interval between 0 and 1 into 10 equal parts. Now label the parts 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0. We write 0 as 0.0 and 1 as 1.0, so that the numbers are consistently in tenths. Finally, mark 0.4 on the number line.

Figure shows a number line with numbers ranging from 0.0 to 1. 0.4 is highlighted.

ⓑ The decimal −0.74 is equivalent to −74100, so it is located between 0 and −1. On a number line, mark off and label the hundredths in the interval between 0 and −1.

Figure shows a number line with numbers ranging from minus 1.00 to 0.00. Minus 0.74 is highlighted.

Locate on the number line: ⓐ 0.6 ⓑ −0.25.

Solution

ⓐ

Figure shows a number line with numbers ranging from 0 to 1. 0.6 is highlighted.

ⓑ

Figure shows a number line with numbers ranging from minus 1.00 to 0.00. Minus 0.74 is highlighted, minus 0.25 is highlighted.

Locate on the number line: ⓐ 0.9 ⓑ −0.75.

Solution

ⓐ

Figure shows a number line with numbers ranging from 0 to 1. 0.9 is highlighted.

ⓑ

Figure shows a number line with numbers ranging from minus 1.00 to 0.00. Minus 0.74 is highlighted.

Access this online resource for additional instruction and practice with decimals.

  • Arithmetic Basics: Dividing Decimals

Key Concepts

  • How to round decimals.
    1. Locate the given place value and mark it with an arrow.
    2. Underline the digit to the right of the place value.
    3. Is the underlined digit greater than or equal to 5?
      • Yes: add 1 to the digit in the given place value.
      • No: do not change the digit in the given place value
    4. Rewrite the number, deleting all digits to the right of the rounding digit.
  • How to add or subtract decimals.
    1. Determine the sign of the sum or difference.
    2. Write the numbers so the decimal points line up vertically.
    3. Use zeros as placeholders, as needed.
    4. Add or subtract the numbers as if they were whole numbers. Then place the decimal point in the answer under the decimal points in the given numbers.
    5. Write the sum or difference with the appropriate sign
  • How to multiply decimals.
    1. Determine the sign of the product.
    2. Write in vertical format, lining up the numbers on the right. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
    3. Place the decimal point. The number of decimal places in the product is the sum of the number of decimal places in the factors.
    4. Write the product with the appropriate sign.
  • How to multiply a decimal by a power of ten.
    1. Move the decimal point to the right the same number of places as the number of zeros in the power of 10.
    2. Add zeros at the end of the number as needed.
  • How to divide decimals.
    1. Determine the sign of the quotient.
    2. Make the divisor a whole number by “moving” the decimal point all the way to the right. “Move” the decimal point in the dividend the same number of places—adding zeros as needed.
    3. Divide. Place the decimal point in the quotient above the decimal point in the dividend.
    4. Write the quotient with the appropriate sign.
  • How to convert a decimal to a proper fraction and a fraction to a decimal.
    1. To convert a decimal to a proper fraction, determine the place value of the final digit.
    2. Write the fraction.
      • numerator—the “numbers” to the right of the decimal point
      • denominator—the place value corresponding to the final digit
    3. To convert a fraction to a decimal, divide the numerator of the fraction by the denominator of the fraction.
  • How to convert a percent to a decimal and a decimal to a percent.
    1. To convert a percent to a decimal, move the decimal point two places to the left after removing the percent sign.
    2. To convert a decimal to a percent, move the decimal point two places to the right and then add the percent sign.
  • Square Root Notation
    m is read “the square root of m.”
    If m=n2, then m=n, for n≥0.
    The square root of m, m, is the positive number whose square is m.
  • Rational or Irrational
    If the decimal form of a number
    • repeats or stops, the number is a rational number.
    • does not repeat and does not stop, the number is an irrational number.
  • Real Numbers
    A chart shows that counting numbers 1, 2, 3 are a part of whole numbers 0, 1, 2, 3. Whole numbers are a part of integers minus 2, minus 1, 0, 1, 2. Integers are a part of rational numbers. Rational numbers along with irrational numbers form the set of real numbers.

Practice Makes Perfect

Round Decimals

In the following exercises, round each number to the nearest ⓐ hundredth ⓑ tenth ⓒ whole number.

5.781

Solution

ⓐ 5.78 ⓑ 5.8 ⓒ 6

1.638

0.299

Solution

ⓐ 0.30 ⓑ 0.3 ⓒ 0

0.697

63.479

Solution

ⓐ 63.48 ⓑ 63.5 ⓒ 63

84.281

Add and Subtract Decimals

In the following exercises, add or subtract.

−16.53−24.38

Solution

−40.91

−19.47−32.58

−38.69+31.47

Solution

−7.22

−29.83+19.76

72.5−100

Solution

−27.5

86.2−100

91.75−(−10.462)

Solution

102.212

94.69−(−12.678)

55.01−3.7

Solution

51.31

59.08−4.6

2.51−7.4

Solution

−4.89

3.84−6.1

Multiply and Divide Decimals

In the following exercises, multiply.

(94.69)(−12.678)

Solution

−1200.47982

(−8.5)(1.69)

(−5.18)(−65.23)

Solution

337.8914

(−9.16)(−68.34)

(0.06)(21.75)

Solution

1.305

(0.08)(52.45)

(9.24)(10)

Solution

92.4

(6.531)(10)

(0.025)(100)

Solution

2.5

(0.037)(100)

(55.2)(1000)

Solution

55200

(99.4)(1000)

In the following exercises, divide. Round money monetary answers to the nearest cent.

$117.25÷48

Solution

$2.44

$109.24÷36

1.44÷(−0.3)

Solution

−4.8

−1.15÷(−0.05)

5.2÷2.5

Solution

2.08

14÷0.35

Convert Decimals, Fractions and Percents

In the following exercises, write each decimal as a fraction.

0.04

Solution

125

1.464

0.095

Solution

19200

−0.375

In the following exercises, convert each fraction to a decimal.

1720

Solution

0.85

174

−31025

Solution

−12.4

−1811

In the following exercises, convert each percent to a decimal.

71%

Solution

0.71

150%

39.3%

Solution

0.393

7.8%

In the following exercises, convert each decimal to a percent.

1.56

Solution

156%

3

0.0625

Solution

6.25%

2.254

Simplify Expressions with Square Roots

In the following exercises, simplify.

64

Solution

8

169

144

Solution

12

−4

−100

Solution

−10

−121

Identify Integers, Rational Numbers, Irrational Numbers, and Real Numbers

In the following exercises, list the ⓐ whole numbers, ⓑ integers, ⓒ rational numbers, ⓓ irrational numbers, ⓔ real numbers for each set of numbers.

−8,0,1.95286...,125,36,9

Solution

ⓐ 0,36,9 ⓑ −8,0,36,9 ⓒ −8,0,125,36,9 ⓓ 1.95286..., ⓔ −8,0,1.95286...,125,36,9

−9,−349,−9,0.409—,116,7

−100,−7,−83,−1,0.77,314

Solution

ⓐ none ⓑ −100,−7,−1
ⓒ −100,−7,−83,−1,0.77,314
ⓓ none
ⓔ −100,−7,−83,−1,0.77,314

−6,−52,0,0.714285———,215,14

Locate Fractions and Decimals on the Number Line

In the following exercises, locate the numbers on a number line.

310,72,116,4

Solution

Figure shows a number line with numbers ranging from 0 to 6. Some values are highlighted. From left to right, these are: 3 by 10, 11 by 6, 7 by 2 and 4.

710,52,138,3

34,−34,123,−123,52,−52

Solution

Figure shows a number line with numbers ranging from minus 4 to 4. Some values are highlighted. From left to right, these are: minus 5 by 2, minus 1 and two thirds, minus 3 by 4, 3 by 4, 1 and two thirds, and 5 by 2.

25,−25,134,−134,83,−83

ⓐ 0.8 ⓑ −1.25

Solution

Figure shows a number line with numbers ranging from minus 4 to 4. Two values are highlighted. One is between minus 2 and minus 1. The other is between 0 and 1.

ⓐ −0.9 ⓑ −2.75

ⓐ −1.6 ⓑ 3.25

Solution

Figure shows a number line with numbers ranging from minus 4 to 4. Two values are highlighted. One is between minus 2 and minus 1. The other is between 3 and 4.

ⓐ 3.1 ⓑ −3.65

Writing Exercises

How does knowing about U.S. money help you learn about decimals?

Solution

Answers will vary.

When the Szetos sold their home, the selling price was 500% of what they had paid for the house 30 years ago. Explain what 500% means in this context.

In your own words, explain the difference between a rational number and an irrational number.

Solution

Answers will vary.

Explain how the sets of numbers (counting, whole, integer, rational, irrationals, reals) are related to each other.

Self Check

ⓐ Use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 6 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The statements in the first column are: round decimals, add and subtract decimals, multiply and divide decimals, convert decimals, fractions and percents, simplify expressions with square roots, identify integers, rational numbers, irrational numbers and real numbers, locate fractions and decimals on the number line. The remaining columns are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

irrational number
An irrational number is a number that cannot be written as the ratio of two integers. Its decimal form does not stop and does not repeat.
percent
A percent is a ratio whose denominator is 100.
principal square root
The positive square root is called the principal square root.
rational number
A rational number is a number of the form pq, where p and q are integers and q≠0. Its decimal form stops or repeats.
real number
A real number is a number that is either rational or irrational.
square of a number
If n2=m, then m is the square of n.
square root of a number
If n2=m, then n is a square root of m.

Properties of Real Numbers

Learning Objectives

By the end of this section, you will be able to:

  • Use the commutative and associative properties
  • Use the properties of identity, inverse, and zero
  • Simplify expressions using the Distributive Property

A more thorough introduction to the topics covered in this section can be found in the Elementary Algebra 2e chapter, Foundations.

Use the Commutative and Associative Properties

The order we add two numbers doesn’t affect the result. If we add 8+9 or 9+8, the results are the same—they both equal 17. So, 8+9=9+8. The order in which we add does not matter!

Similarly, when multiplying two numbers, the order does not affect the result. If we multiply 9·8 or 8·9 the results are the same—they both equal 72. So, 9·8=8·9. The order in which we multiply does not matter!

These examples illustrate the Commutative Property.

Commutative Property

of AdditionIfaandbare real numbers, thena+b=b+a. of MultiplicationIfaandbare real numbers, thena·b=b·a.

When adding or multiplying, changing the order gives the same result.

The Commutative Property has to do with order. We subtract 9−8 and 8−9, and see that 9−8≠8−9. Since changing the order of the subtraction does not give the same result, we know that subtraction is not commutative.

Division is not commutative either. Since 12÷3≠3÷12, changing the order of the division did not give the same result. The commutative properties apply only to addition and multiplication!

  Addition and multiplication are commutative.

  Subtraction and division are not commutative.


When adding three numbers, changing the grouping of the numbers gives the same result. For example, (7+8)+2=7+(8+2), since each side of the equation equals 17.

This is true for multiplication, too. For example, (5·13)·3=5·(13·3), since each side of the equation equals 5.

These examples illustrate the Associative Property.

Associative Property

of AdditionIfa,b,andcare real numbers, then(a+b)+c=a+(b+c). of MultiplicationIfa,b,andcare real numbers, then(a·b)·c=a·(b·c).

When adding or multiplying, changing the grouping gives the same result.

The Associative Property has to do with grouping. If we change how the numbers are grouped, the result will be the same. Notice it is the same three numbers in the same order—the only difference is the grouping.

We saw that subtraction and division were not commutative. They are not associative either.

(10−3)−2≠10−(3−2)(24÷4)÷2≠24÷(4÷2) 7−2≠10−16÷2≠24÷2 5≠93≠12

When simplifying an expression, it is always a good idea to plan what the steps will be. In order to combine like terms in the next example, we will use the Commutative Property of addition to write the like terms together.

Simplify: 18p+6q+15p+5q.

Solution
Demonstrates simplifying an algebraic expression by combining like terms, showing steps for reordering and addition.
18p+6q+15p+5q
Use the Commutative Property of addition to reorder so that like terms are together. 18p+15p+6q+5q
Add like terms. 33p+11q

Simplify: 23r+14s+9r+15s.

Solution

32r+29s

Simplify:37m+21n+4m−15n.

Solution

41m+6n

When we have to simplify algebraic expressions, we can often make the work easier by applying the Commutative Property or Associative Property first.

Simplify: (513+34)+14.

Solution
Step-by-step simplification of a fractional expression, demonstrating the utility of changing term grouping for easier calculation.
(513+34)+14
Notice that the last 2 terms have a common denominator, so change the grouping. 513+(34+14)
Add in parentheses first. 513+(44)
Simplify the fraction. 513+1
Add. 1513
Convert to an improper fraction. 1813

Simplify: (715+58)+38.

Solution

1715

Simplify: (29+712)+512.

Solution

129

Use the Properties of Identity, Inverse, and Zero

What happens when we add 0 to any number? Adding 0 doesn’t change the value. For this reason, we call 0 the additive identity. The Identity Property of Addition that states that for any real number a,a+0=a and 0+a=a.

What happens when we multiply any number by one? Multiplying by 1 doesn’t change the value. So we call 1 the multiplicative identity. The Identity Property of Multiplication that states that for any real number a,a·1=a and 1·a=a.

We summarize the Identity Properties here.

Identity Property

of AdditionFor any real numbera:a+0=a0+a=a 0is theadditive identity of MultiplicationFor any real numbera:a·1=a1·a=a 1is themultiplicative identity

What number added to 5 gives the additive identity, 0? We know

Figure shows the expression 5 plus open parentheses minus 5 close parentheses equals 0.

The missing number was the opposite of the number!

We call −a the additive inverse of a. The opposite of a number is its additive inverse. A number and its opposite add to zero, which is the additive identity. This leads to the Inverse Property of Addition that states for any real number a,a+(−a)=0.

What number multiplied by 23 gives the multiplicative identity, 1? In other words, 23 times what results in 1? We know

2 by 3 times 3 by 2 equals 1.

The missing number was the reciprocal of the number!

We call 1a the multiplicative inverse of a. The reciprocal of a number is its multiplicative inverse. This leads to the Inverse Property of Multiplication that states that for any real number a,a≠0,a·1a=1.

We’ll formally state the inverse properties here.

Inverse Property

of AdditionFor any real numbera,a+(−a)=0 −ais theadditive inverseofa A number and itsoppositeadd to zero. of MultiplicationFor any real numbera,a≠0,a·1a=1. 1ais themultiplicative inverseofa. A number and itsreciprocalmultiply to one.

The Identity Property of addition says that when we add 0 to any number, the result is that same number. What happens when we multiply a number by 0? Multiplying by 0 makes the product equal zero.

What about division involving zero? What is 0÷3? Think about a real example: If there are no cookies in the cookie jar and 3 people are to share them, how many cookies does each person get? There are no cookies to share, so each person gets 0 cookies. So, 0÷3=0.

We can check division with the related multiplication fact. So we know 0÷3=0 because 0·3=0.

Now think about dividing by zero. What is the result of dividing 4 by 0? Think about the related multiplication fact:

4 divided by 0 equals question mark means question mark times 0 equals 4.

Is there a number that multiplied by 0 gives 4? Since any real number multiplied by 0 gives 0, there is no real number that can be multiplied by 0 to obtain 4. We conclude that there is no answer to 4÷0 and so we say that division by 0 is undefined.

We summarize the properties of zero here.

Properties of Zero

Multiplication by Zero: For any real number a,

a·0=00·a=0The product of any number and 0 is 0.

Division by Zero: For any real number a, a≠0

0a=0Zero divided by any real number, except itself, is zero. a0is undefinedDivision by zero is undefined.

We will now practice using the properties of identities, inverses, and zero to simplify expressions.

Simplify: −84n+(−73n)+84n.

Solution
This table illustrates the step-by-step simplification of an algebraic expression by re-ordering and combining like terms.
−84n+(−73n)+84n
Notice that the first and third terms are opposites; use the Commutative Property of addition to re-order the terms. −84n+84n+(−73n)
Add left to right. 0+(−73n)
Add. −73n

Simplify: −27a+(−48a)+27a.

Solution

−48a

Simplify: 39x+(−92x)+(−39x).

Solution

−92x

Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals—their product is 1.

Simplify: 715·823·157.

Solution
Demonstrates simplifying a multiplication of fractions using the Commutative Property and reciprocals to reach a solution.
715·823·157
Notice the first and third terms are reciprocals, so use the Commutative Property of multiplication to re-order the factors. 715·157·823
Multiply left to right. 1·823
Multiply. 823

Simplify: 916·549·169.

Solution

549

Simplify: 617·1125·176.

Solution

1125

The next example makes us aware of the distinction between dividing 0 by some number or some number being divided by 0.

Simplify: ⓐ 0n+5, where n≠−5 ⓑ 10−3p0, where 10−3p≠0.

Solution

ⓐ
0n+5 Zero divided by any real number except itself is 0.0

ⓑ
10−3p0 Division by 0 is undefined.undefined

Simplify: ⓐ 0m+7, where m≠−7 ⓑ 18−6c0, where 18−6c≠0.

Solution

ⓐ 0 ⓑ undefined

Simplify: ⓐ 0d−4, where d≠4 ⓑ 15−4q0, where 15−4q≠0.

Solution

ⓐ 0 ⓑ undefined

Simplify Expressions Using the Distributive Property

Suppose that three friends are going to the movies. They each need $9.25—that’s 9 dollars and 1 quarter—to pay for their tickets. How much money do they need all together?

You can think about the dollars separately from the quarters. They need 3 times $9 so $27 and 3 times 1 quarter, so 75 cents. In total, they need $27.75. If you think about doing the math in this way, you are using the Distributive Property.

Distributive Property

Ifa,b,andcare real numbers, thena(b+c)=ab+ac (b+c)a=ba+ca a(b−c)=ab−ac (b−c)a=ba−ca

In algebra, we use the Distributive Property to remove parentheses as we simplify expressions.

Simplify: 3(x+4).

Solution
Steps to simplify the algebraic expression 3(x+4) using the distributive property.
3(x+4)
Distribute. 3·x+3·4
Multiply. 3x+12

Simplify: 4(x+2).

Solution

4x+8

Simplify: 6(x+7).

Solution

6x+42

Some students find it helpful to draw in arrows to remind them how to use the Distributive Property. Then the first step in Example 6 would look like this:

The expression is 3 open parentheses x plus 4 close parentheses. Two arrows originate from 3. One points to x, the other to 4.

Simplify: 8(38x+14).

Solution
The distributive property is applied to the expression 8(3/8x + 1/4), showing 8 multiplied by each term inside the parentheses.
Distribute.     The mathematical expression 8 times 3 over 8x plus 8 times 1 over 4 is shown on a white background.
Multiply. The mathematical expression '3x + 2' is displayed in black text against a white background.

Simplify: 6(56y+12).

Solution

5y+3

Simplify: 12(13n+34).

Solution

4n+9

Using the Distributive Property as shown in the next example will be very useful when we solve money applications in later chapters.

Simplify: 100(0.3+0.25q).

Solution
A mathematical expression displays the distributive property, showing 100 multiplied by (0.3 + 0.25q). Blue arrows indicate that 100 distributes to both 0.3 and 0.25q within the parenthesis.
Distribute.     A mathematical expression showing the sum of two products: 100 multiplied by 0.3, and 100 multiplied by 0.25q.
Multiply. A mathematical expression '30 + 25q' is displayed on a white background. It represents an algebraic equation with constants and a variable 'q'.

Simplify: 100(0.7+0.15p).

Solution

70+15p

Simplify: 100(0.04+0.35d).

Solution

4+35d

When we distribute a negative number, we need to be extra careful to get the signs correct!

Simplify: −11(4−3a).

Solution
Algebraic steps to simplify the expression -11(4-3a) using the distributive property.
−11(4−3a)
Distribute. −11·4−(−11)·3a
Multiply. −44−(−33a)
Simplify. −44+33a

Notice that you could also write the result as 33a−44. Do you know why?

Simplify: −5(2−3a).

Solution

−10+15a

Simplify: −7(8−15y).

Solution

−56+105y

In the next example, we will show how to use the Distributive Property to find the opposite of an expression.

Simplify: −(y+5).

Solution
Steps demonstrating the simplification of the algebraic expression -(y+5) using the distributive property.
−(y+5)
Multiplying by −1 results in the opposite. −1(y+5)
Distribute. −1·y+(−1)·5
Simplify. −y+(−5)
Simplify. −y−5

Simplify: −(z−11).

Solution

−z+11

Simplify: −(x−4).

Solution

−x+4

There will be times when we’ll need to use the Distributive Property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiply using the Distributive Property, which removes the parentheses. The next two examples will illustrate this.

Simplify: 8−2(x+3)

Solution

We follow the order of operations. Multiplication comes before subtraction, so we will distribute the 2 first and then subtract.

Step-by-step simplification of the algebraic expression 8 - 2(x + 3) to its final form -2x + 2.
8−2(x+3)
Distribute. 8−2·x−2·3
Multiply. 8−2x−6
Combine like terms. −2x+2

Simplify: 9−3(x+2).

Solution

3−3x

Simplify: 7x−5(x+4).

Solution

2x−20

Simplify: 4(x−8)−(x+3).

Solution
This table illustrates the step-by-step simplification of an algebraic expression, detailing the operations performed at each stage.
4(x−8)−(x+3)
Distribute. 4x−32−x−3
Combine like terms. 3x−35

Simplify: 6(x−9)−(x+12).

Solution

5x−66

Simplify: 8(x−1)−(x+5).

Solution

7x−13

All the properties of real numbers we have used in this chapter are summarized here.

Commutative Property
When adding or multiplying, changing the order gives the same result

of additionIfa,bare real numbers, thena+b=b+a of multiplicationIfa,bare real numbers, thena·b=b·a
Associative Property
When adding or multiplying, changing the grouping gives the same result.

of additionIfa,b,andcare real numbers, then(a+b)+c=a+(b+c) of multiplicationIfa,b,andcare real numbers, then(a·b)·c=a·(b·c)
Distributive Property

Ifa,b,andcare real numbers, thena(b+c)=ab+ac (b+c)a=ba+ca a(b−c)=ab−ac (b−c)a=ba−ca
Identity Property

of additionFor any real numbera:a+0=a 0is theadditive identity0+a=a of multiplicationFor any real numbera:a·1=a 1is themultiplicative identity1·a=a
Inverse Property

of additionFor any real numbera,a+(−a)=0 −ais theadditive inverseofa A number and itsoppositeadd to zero. of multiplicationFor any real numbera,a≠0a·1a=1 1ais themultiplicative inverseofa A number and itsreciprocalmultiply to one.
Properties of Zero
For any real numbera,a·0=0 0·a=0 For any real numbera,a≠0,0a=0 For any real numbera,a0is undefined

Key Concepts

Commutative Property
When adding or multiplying, changing the order gives the same result

of additionIfa,bare real numbers, thena+b=b+a of multiplicationIfa,bare real numbers, thena·b=b·a
Associative Property
When adding or multiplying, changing the grouping gives the same result.

of additionIfa,b,andcare real numbers, then(a+b)+c=a+(b+c) of multiplicationIfa,b,andcare real numbers, then(a·b)·c=a·(b·c)
Distributive Property

Ifa,b,andcare real numbers, thena(b+c)=ab+ac (b+c)a=ba+ca a(b−c)=ab−ac (b−c)a=ba−ca
Identity Property

of additionFor any real numbera:a+0=a 0is theadditive identity0+a=a of multiplicationFor any real numbera:a·1=a 1is themultiplicative identity1·a=a
Inverse Property

of additionFor any real numbera,a+(−a)=0 −ais theadditive inverseofa A number and itsoppositeadd to zero. of multiplicationFor any real numbera,a≠0a·1a=1 1ais themultiplicative inverseofa A number and itsreciprocalmultiply to one.
Properties of Zero
For any real numbera,a·0=0 0·a=0 For any real numbera,a≠0,0a=0 For any real numbera,a0is undefined

Section Exercises

Practice Makes Perfect

Use the Commutative and Associative Properties

In the following exercises, simplify.

43m+(−12n)+(−16m)+(−9n)

Solution

27m+(−21n)

−22p+17q+(−35p)+(−27q)

38g+112h+78g+512h

Solution

54g+12h

56a+310b+16a+910b

6.8p+9.14q+(−4.37p)+(−0.88q)

Solution

2.43p+8.26q

9.6m+7.22n+(−2.19m)+(−0.65n)

−24·7·38

Solution

−63

−36·11·49

(56+815)+715

Solution

156

(1112+49)+59

17(0.25)(4)

Solution

17

36(0.2)(5)

[2.48(12)](0.5)

Solution

14.88

[9.731(4)](0.75)

12(56p)

Solution

10p

20(35q)

Use the Properties of Identity, Inverse and Zero

In the following exercises, simplify.

19a+44−19a

Solution

44

27c+16−27c

12+78+(−12)

Solution

78

25+512+(−25)

10(0.1d)

Solution

d

100(0.01p)

320·4911·203

Solution

4911

1318·257·1813

0u−4.99, where u≠4.99

Solution

0

0÷(y−16), where x≠16

32−5a0, where 32−5a≠0

Solution

undefined

28−9b0, where 28−9b≠0

(34+910m)÷0, where 34+910m≠0

Solution

undefined

(516n−37)÷0, where 516n−37≠0

Simplify Expressions Using the Distributive Property

In the following exercises, simplify using the Distributive Property.

8(4y+9)

Solution

32y+72

9(3w+7)

6(c−13)

Solution

6c−78

7(y−13)

14(3q+12)

Solution

34q+3

15(4m+20)

9(59y−13)

Solution

5y−3

10(310x−25)

12(14+23r)

Solution

3+8r

12(16+34s)

15·35(4d+10)

Solution

36d+90

18·56(15h+24)

r(s−18)

Solution

rs−18r

u(v−10)

(y+4)p

Solution

yp+4p

(a+7)x

−7(4p+1)

Solution

−28p−7

−9(9a+4)

−3(x−6)

Solution

−3x+18

−4(q−7)

−(3x−7)

Solution

−3x+7

−(5p−4)

16−3(y+8)

Solution

−3y−8

18−4(x+2)

4−11(3c−2)

Solution

−33c+26

9−6(7n−5)

22−(a+3)

Solution

−a+19

8−(r−7)

(5m−3)−(m+7)

Solution

4m−10

(4y−1)−(y−2)

9(8x−3)−(−2)

Solution

72x−25

4(6x−1)−(−8)

5(2n+9)+12(n−3)

Solution

22n+9

9(5u+8)+2(u−6)

14(c−1)−8(c−6)

Solution

6c+34

11(n−7)−5(n−1)

6(7y+8)−(30y−15)

Solution

12y+63

7(3n+9)−(4n−13)

Writing Exercises

In your own words, state the Associative Property of addition.

Solution

Answers will vary.

What is the difference between the additive inverse and the multiplicative inverse of a number?

Simplify 8(x−14) using the Distributive Property and explain each step.

Solution

Answers will vary.

Explain how you can multiply 4($5.97) without paper or calculator by thinking of $5.97 as 6−0.03 and then using the Distributive Property.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: use the commutative and associative properties, use the properties of identity, inverse and zero, simplify expressions using the Distributive Property. The remaining columns are blank.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Chapter Review Exercises

Use the Language of Algebra

Identify Multiples and Factors

Use the divisibility tests to determine whether 180 is divisible by 2, by 3, by 5, by 6, and by 10.

Solution

Divisible by 2,3,5,6,10

Find the prime factorization of 252.

Find the least common multiple of 24 and 40.

Solution

120

In the following exercises, simplify each expression.

24÷3+4(5−2)

7+3[6−4(5−4)]−32

Solution

4

Evaluate an Expression

In the following exercises, evaluate the following expressions.

When x=4, ⓐ x3 ⓑ 5x ⓒ 2x2−5x+3

2x2−4xy−3y2 when x=3, y=1

Solution

3

Simplify Expressions by Combining Like Terms

In the following exercises, simplify the following expressions by combining like terms.

12y+7+2y−5

14x2−9x+11−8x2+8x−6

Solution

6x2−x+5

Translate an English Phrase to an Algebraic Expression

In the following exercises, translate the phrases into algebraic expressions.


ⓐ the sum of 4ab2 and 7a3b2
ⓑ the product of 6y2 and 3y
ⓒ twelve more than 5x
ⓓ 5y less than 8y2


ⓐ eleven times the difference of y and two
ⓑ the difference of eleven times y and two

Solution

ⓐ 11(y−2) ⓑ 11y−2

Dushko has nickels and pennies in his pocket. The number of pennies is four less than five times the number of nickels. Let n represent the number of nickels. Write an expression for the number of pennies.

Integers

Simplify Expressions with Absolute Value

In the following exercise, fill in <,>, or = for each of the following pairs of numbers.


ⓐ −|7|___−|−7|
ⓑ −8___−|−8|
ⓒ |−13|___−13
ⓓ |−12|___−(−12)

Solution

ⓐ = ⓑ = ⓒ > ⓓ =

In the following exercises, simplify.

9−|3(4−8)|

12−3|1−4(4−2)|

Solution

−9

Add and Subtract Integers

In the following exercises, simplify each expression.

−12+(−8)+7


ⓐ 15−7
ⓑ −15−(−7)
ⓒ −15−7
ⓓ 15−(−7)

Solution

ⓐ 8 ⓑ −8 ⓒ −22 ⓓ 22

−11−(−12)+5

ⓐ 23−(−17) ⓑ 23+17

Solution

ⓐ 40 ⓑ 40

−(7−11)−(3−5)

Multiply and Divide Integers

In the following exercise, multiply or divide.

ⓐ −27÷9 ⓑ 120÷(−8) ⓒ 4(−14) ⓓ −1(−17)

Solution

ⓐ −3 ⓑ −15 ⓒ −56 ⓓ 17

Simplify and Evaluate Expressions with Integers

In the following exercises, simplify each expression.

ⓐ (−7)3 ⓑ −73

(7−11)(6−13)

Solution

28

63÷(−9)+(−36)÷(−4)

6−3|4(1−2)−(7−5)|

Solution

−12

(−2)4−24÷(13−5)

For the following exercises, evaluate each expression.

(y+z)2 when
y=−4,z=7

Solution

9

3x2−2xy+4y2 when
x=−2,y=−3

Translate English Phrases to Algebraic Expressions

In the following exercises, translate to an algebraic expression and simplify if possible.

the sum of −4 and −9, increased by 23

Solution

(−4+(−9))+23;10

ⓐ the difference of 17 and −8 ⓑ subtract 17 from −25

Use Integers in Applications

In the following exercise, solve.

Temperature On July 10, the high temperature in Phoenix, Arizona, was 109°, and the high temperature in Juneau, Alaska, was 63°. What was the difference between the temperature in Phoenix and the temperature in Juneau?

Solution

46°

Fractions

Simplify Fractions

In the following exercises, simplify.

204228

−270x3198y2

Solution

−15x311y2

Multiply and Divide Fractions

In the following exercises, perform the indicated operation.

(−1415)(1021)

6x25÷9y20

Solution

8x15y

−49821

Add and Subtract Fractions

In the following exercises, perform the indicated operation.

518+712

Solution

3136

1136−1548

ⓐ 58+34 ⓑ 58÷34

Solution

ⓐ 118 ⓑ 56

ⓐ −3y10−56 ⓑ −3y10·56

Use the Order of Operations to Simplify Fractions

In the following exercises, simplify.

4·3−2·5−6·3+2·3

Solution

−16

4(7−3)−2(4−9)−3(4+2)+7(3−6)

43−42(45)2

Solution

75

Evaluate Variable Expressions with Fractions

In the following exercises, evaluate.

4x2y2 when
x=23 and y=−34

a+ba−b when
a=−4, b=6

Solution

−15

Decimals

Round Decimals

Round 6.738 to the nearest ⓐ hundredth ⓑ tenth ⓒ whole number.

Add and Subtract Decimals

In the following exercises, perform the indicated operation.

−23.67+29.84

Solution

6.17

54.3−100

79.38−(−17.598)

Solution

96.978

Multiply and Divide Decimals

In the following exercises, perform the indicated operation.

(−2.8)(3.97)

(−8.43)(−57.91)

Solution

488.1813

(53.48)(10)

(0.563)(100)

Solution

56.3

$118.35÷2.6

1.84÷(−0.8)

Solution

−2.3

Convert Decimals, Fractions and Percents

In the following exercises, convert each decimal to a fraction.

0.65

−9.6

Solution

−485

In the following exercises, convert each fraction to a decimal.

−58

1411

Solution

1.27¯

In the following exercises, convert each decimal to a percent.

2.43

0.0475

Solution

4.75%

Simplify Expressions with Square Roots

In the following exercises, simplify.

289

−121

Solution

no real number

Identify Integers, Rational Numbers, Irrational Numbers, and Real Numbers

In the following exercise, list the ⓐ whole numbers ⓑ integers ⓒ rational numbers ⓓ irrational numbers ⓔ real numbers for each set of numbers

−8,0,1.95286...,125,36,9

Locate Fractions and Decimals on the Number Line

In the following exercises, locate the numbers on a number line.

34,−34,113,−123,72,−52

Solution

Figure shows a number line with numbers ranging from minus 4 to 4. Some values are highlighted.

ⓐ 3.2 ⓑ −1.35

Properties of Real Numbers

Use the Commutative and Associative Properties

In the following exercises, simplify.

58x+512y+18x+712y

Solution

34x+y

−32·9·58

(1115+38)+58

Solution

11115

Use the Properties of Identity, Inverse and Zero

In the following exercises, simplify.

47+815+(−47)

1315·917·1513

Solution

917

0x−3,x≠3

5x−70,5x−7≠0

Solution

undefined

Simplify Expressions Using the Distributive Property

In the following exercises, simplify using the Distributive Property.

8(a−4)

12(23b+56)

Solution

8b+10

18·56(2x−5)

(x−5)p

Solution

xp−5p

−4(y−3)

12−6(x+3)

Solution

−6x−6

6(3x−4)−(−5)

5(2y+3)−(4y−1)

Solution

6y+16

Practice Test

Find the prime factorization of 756.

Combine like terms: 5n+8+2n−1

Solution

7n+7

Evaluate when x=−2 and y=3: |3x−4y|6

Translate to an algebraic expression and simplify:

ⓐ eleven less than negative eight

ⓑ the difference of −8 and −3, increased by 5

Solution

−8−11;−19

(−8−(−3))+5;0

Dushko has nickels and pennies in his pocket. The number of pennies is seven less than four times the number of nickels. Let n represent the number of nickels. Write an expression for the number of pennies.

Round 28.1458 to the nearest

ⓐ hundredth ⓑ thousandth

Solution

ⓐ 28.15 ⓑ 28.146

Convert

ⓐ 511 to a decimal ⓑ 1.15 to a percent

Locate 35,2.8,and−52 on a number line.

Solution

Figure shows a number line with numbers ranging from minus 4 to 4. Some values are highlighted.

In the following exercises, simplify each expression.

8+3[6−3(5−2)]−42

−(4−9)−(9−5)

Solution

1

56÷(−8)+(−27)÷(−3)

16−2|3(1−4)−(8−5)|

Solution

−8

−5+2(−3)2−9

180204

Solution

1517

−718+512

45÷(−1225)

Solution

−53

9−3·915−9

4(−3+2(3−6))3(11−3(2+3))

Solution

3

513·47·135

−591021

Solution

−76

−4.8+(−6.7)

34.6−100

Solution

−65.4

−12.04·(4.2)

−8÷0.05

Solution

−160

−121

(813+57)+27

Solution

1813

5x+(−8y)−6x+3y

ⓐ 09 ⓑ 110

Solution

ⓐ 0 ⓑ undefined

−3(8x−5)

6(3y−1)−(5y−3)

Solution

13y−3

additive identity
The number 0 is the additive identity because adding 0 to any number does not change its value.
additive inverse
The opposite of a number is its additive inverse.
multiplicative identity
The number 1 is the multiplicative identity because multiplying 1 by any number does not change its value.
multiplicative inverse
The reciprocal of a number is its multiplicative inverse.

Introduction

A white and black drone with its propellers spinning is captured in mid-flight against a clear blue sky, with a range of snow-capped mountains visible in the blurry background.
This drone is flying high in the sky while its pilot remains safely on the ground. (credit: “Unsplash” / Pixabay)

Imagine being a pilot, but not just any pilot—a drone pilot. Drones, or unmanned aerial vehicles, are devices that can be flown remotely. They contain sensors that can relay information to a command center where the pilot is located. Larger drones can also carry cargo. In the near future, several companies hope to use drones to deliver materials and piloting a drone will become an important career. Law enforcement and the military are using drones, rather than sending personnel into dangerous situations. Building and piloting a drone requires the ability to program a set of actions, including taking off, turning, and landing. This, in turn, requires the use of linear equations. In this chapter, you will explore linear equations, develop a strategy for solving them, and relate them to real-world situations.

Use a General Strategy to Solve Linear Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve linear equations using a general strategy
  • Classify equations
  • Solve equations with fraction or decimal coefficients

Before you get started, take this readiness quiz.

Simplify: 32(12x+20).
If you missed this problem, review Example 7 in Properties of Real Numbers.

Solution

18x+30

Simplify: 5−2(n+1).
If you missed this problem, review Example 11 in Properties of Real Numbers.

Solution

3−2n

Find the LCD of 56 and 14.
If you missed this problem, review Example 5 in Fractions.

Solution

12

Solve Linear Equations Using a General Strategy

Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

Solution of an Equation

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation.

Determine Whether a Number is a Solution to an Equation.

  1. Substitute the number for the variable in the equation.
  2. Simplify the expressions on both sides of the equation.
  3. Determine whether the resulting equation is true.
    • If it is true, the number is a solution.
    • If it is not true, the number is not a solution.

Determine whether the values are solutions to the equation: 5y+3=10y−4.

ⓐ y=35 ⓑ y=75

Solution

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

ⓐ
A linear algebraic equation is shown: 5y + 3 = 10y - 4, which involves the variable 'y' and constant terms on both sides of the equality sign, requiring simplification to solve for 'y'.
The image shows the text 'Substitute 3/5 for y.' in a mathematical context, likely instructing the viewer to replace the variable 'y' with the fraction 3/5. The image displays the math problem 5(3/5) + 3 with a question mark over the equals sign, comparing it to 10(3/5) - 4.
Multiply. A mathematical equation asks whether 3 + 3 equals 6 - 4, with a question mark above the equal sign. Calculating both sides reveals that 3 + 3 = 6 and 6 - 4 = 2, so the equality is false.
Simplify. The expression six is not equal to two.

Since y=35 does not result in a true equation, y=35 is not a solution to the equation 5y+3=10y−4.

ⓑ
A mathematical equation is displayed, reading '5y + 3 = 10y - 4' in black text on a white background, representing an algebraic problem.
The text 'Substitute 7/5 for y.' is displayed, instructing to replace the variable 'y' with the fraction seven-fifths. The fraction is highlighted in red. A math problem showing 5 times (7/5) + 3 being compared with 10 times (7/5) - 4, using an equals sign with a question mark.
Multiply. A mathematical equation presented as 7 + 3 ?= 14 - 4, asking to verify if the two sides are equal.
Simplify. The number 10 is shown equal to 10, with a checkmark indicating correctness, against a white background.

Since y=75 results in a true equation, y=75 is a solution to the equation 5y+3=10y−4.

Determine whether the values are solutions to the equation: 9y+2=6y+3.

ⓐ y=43 ⓑ y=13

Solution

ⓐ no ⓑ yes

Determine whether the values are solutions to the equation: 4x−2=2x+1.

ⓐ x=32 ⓑ x=−12

Solution

ⓐ yes ⓑ no

There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.

Linear Equation

A linear equation is an equation in one variable that can be written, where a and b are real numbers and a≠0, as:

ax+b=0

To solve a linear equation it is a good idea to have an overall strategy that can be used to solve any linear equation. In the next example, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

How to Solve a Linear Equation Using a General Strategy

Solve: 7(n−3)−8=−15.

Solution
Step 1 is to simplify each side of the equation, the product of 7 and the quantity n minus 3 minus 8 is equal to negative 15. Use the Distributive Property. The equation first simplifies to 7 n minus 21 minus 8 is equal to negative 15. Then it simplifies to 7 n minus 29 is equal to negative 15. Notice that each side of the equation is now simplified as much as possible. Step 2 is to collect all variable terms on the left side of the equation, 7 n minus 29 is equal to negative 15. Notice there is nothing to do because all n’s are on the left side. Step 3 is to collect all constant terms on the other side of the equation, 7 n minus 29 is equal to negative 15. To get constants only on the right, add 29 to each side. The result is 7 n minus 29 plus 29 is equal to negative 15 plus 29. Simplify. The result is 7 n is equal to 14. Step 4 is to make the coefficient of the equation, 7 n is equal to 14, 1. Divide each side of the equation by 7. The result is 7 n divided by 7 is equal to 14 divided by 7. Simplify. The result is n is equal to 2. Step 5 is to check the solution, n is equal to 2, by substituting into the equation, the product of 7 and the quantity n minus 3 minus 8 is equal to negative 15. Is the product of 7 and the quantity 2 minus 3 minus 8 equal to negative 15? Subtract. Is 7 times negative 1 minus 8 equal to negative 15? Is negative 7 minus 8 equal to negative 15. Negative 15 is equal to negative 15. The solution checks.

Solve: 2(m−4)+3=−1.

Solution

m=2

Solve: 5(a−3)+5=−10.

Solution

a=0

These steps are summarized in the General Strategy for Solving Linear Equations below.

Solve linear equations using a general strategy.

  1. Simplify each side of the equation as much as possible.
    Use the Distributive Property to remove any parentheses.
    Combine like terms.
  2. Collect all the variable terms on one side of the equation.
    Use the Addition or Subtraction Property of Equality.
  3. Collect all the constant terms on the other side of the equation.
    Use the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable term equal to 1.
    Use the Multiplication or Division Property of Equality.
    State the solution to the equation.
  5. Check the solution.
    Substitute the solution into the original equation to make sure the result is a true statement.

Solve: 23(3m−6)=5−m.

Solution
A mathematical equation is shown, displaying (2/3)(3m - 6) = 5 - m, with 'm' representing a variable. The expression on the left side involves a fraction multiplied by a binomial, while the right side shows a constant minus the variable.
Distribute. A mathematical equation displays '2m - 4 = 5 - m' in a bold, dark gray font against a plain white background.
Add m to both sides to get the variables only on the left. An algebraic equation is shown: 2m + m - 4 = 5 - m + m. Variables and constants are displayed in a clean, sans-serif font, with some 'm' terms highlighted in red.
Simplify. A mathematical equation is displayed on a white background. The equation reads '3m - 4 = 5' in black text, solving for the variable 'm'.
Add 4 to both sides to get constants only on the right. An algebraic equation showing 3m - 4 + 4 = 5 + 4, illustrating the addition of 4 to both sides to solve for 'm'.
Simplify. A simple linear equation '3m = 9' is displayed, demonstrating a basic algebraic problem.
Divide both sides by three. A mathematical equation shows both sides being divided by 3 to solve for 'm'. The equation is 3m/3 = 9/3, with the denominator '3' on both sides highlighted in red.
Simplify. The image displays a simple mathematical equation, 'm = 3', rendered in black text on a plain white background. The equation indicates that the variable 'm' is equal to the number '3'.
Check: A mathematical equation is displayed: 2/3 multiplied by the quantity 3m minus 6, which is equal to 5 minus m. The equation is (2/3)(3m - 6) = 5 - m.
Let m=3. A mathematical equation shown as (2/3)(3 * 3 - 6) =? 5 - 3. The numbers '3' (in '3 * 3') and '3' (in '5 - 3') are highlighted in red, indicating a verification or problem-solving context.
A mathematical equation questions if (2/3)(9-6) is equal to 2.
A mathematical problem asking whether two-thirds multiplied by three equals two. The expression is (2/3)(3) ?= 2, with the question mark indicating an inquiry into the equality.
The number two equals two, confirmed with a checkmark, representing a correct and verified mathematical statement or a simple affirmation of equality.

Solve: 13(6u+3)=7−u.

Solution

u=2

Solve: 23(9x−12)=8+2x.

Solution

x=4

We can solve equations by getting all the variable terms to either side of the equal sign. By collecting the variable terms on the side where the coefficient of the variable is larger, we avoid working with some negatives. This will be a good strategy when we solve inequalities later in this chapter. It also helps us prevent errors with negatives.

Solve: 4(x−1)−2=5(2x+3)+6.

Solution
A mathematical equation is displayed: 4(x - 1) - 2 = 5(2x + 3) + 6. This is a linear equation with one variable, x, and involves distribution and simplification to solve for x.
Distribute. A mathematical equation is displayed: 4x - 4 - 2 = 10x + 15 + 6. It is a linear equation with variables on both sides, and it involves subtraction and addition operations.
Combine like terms. The image displays an algebraic equation: 4x - 6 = 10x + 21.
Subtract 4x from each side to get the variables only on
the right since 10>4.
An algebra equation: 4x - 4x - 6 = 10x - 4x + 21. The -4x terms on both sides are highlighted in red, indicating they can be canceled or combined to simplify the equation.
Simplify. A mathematical equation is displayed, showing -6 = 6x + 21, indicating an algebraic problem to solve for x.
Subtract 21 from each side to get the constants on left. A mathematical equation, -6 - 21 = 6x + 21 - 21, showing the step of subtracting 21 from both sides (highlighted in red) to isolate the variable term.
Simplify. An algebraic equation showing -27 equals 6 times x.
Divide both sides by 6. An equation displaying -27/6 = 6x/6, where the common denominator 6 is highlighted in red.
Simplify. A mathematical equation shows a fraction negative nine over two equals x, written as -9/2 = x.
Check: A mathematical equation is displayed against a white background: 4(x - 1) - 2 = 5(2x + 3) + 6.
Let x=−92. A mathematical equation compares two expressions using a question mark over an equals sign: 4(-9/2 - 1) - 2 and 5(2(-9/2) + 3) + 6. The term -9/2 is highlighted in red.
A mathematical equation is shown: 4(-11/2) - 2 =? 5(-9 + 3) + 6, asking to verify if the left side equals the right side.
A mathematical equation on a white background asks to verify if -22 - 2 equals 5 multiplied by -6, plus 6.
The image presents a numerical equation: -24 ?= -30 + 6, questioning the equality between the two sides.
A mathematical equation shows -24 equals -24, followed by a checkmark, indicating the equality is correct.

Solve: 6(p−3)−7=5(4p+3)−12.

Solution

p=−2

Solve: 8(q+1)−5=3(2q−4)−1.

Solution

q=−8

Solve: 10[3−8(2s−5)]=15(40−5s).

Solution
A mathematical equation shows 10 multiplied by the expression [3 minus 8 times (2s minus 5)], which equals 15 multiplied by the expression (40 minus 5s).
Simplify from the innermost parentheses first. An algebraic equation is presented: 10[3 - 16s + 40] = 15(40 - 5s).
Combine like terms in the brackets. A mathematical equation is displayed: 10[43 - 16s] = 15(40 - 5s). The equation features numbers, variables, and arithmetic operations within brackets and parentheses, set against a white background.
Distribute. A mathematical equation is displayed on a white background: 430 - 160s = 600 - 75s. The equation involves numbers and the variable 's' on both sides of the equality sign.
Add 160s to both sides to get the
variables to the right.
An algebraic equation showing a step in solving for 's': 430 - 160s + 160s = 600 - 75s + 160s. The term '+ 160s' is highlighted in red on both sides of the equality.
Simplify. The mathematical equation 430 = 600 + 85s is displayed.
Subtract 600 from both sides to get the
constants to the left.
A mathematical equation, 430 - 600 = 600 + 85s - 600, with the 600 terms highlighted in red, suggesting an operation like cancellation or subtraction.
Simplify. A mathematical equation is displayed, showing '-170 = 85s' against a white background.
Divide both sides by 85. A mathematical equation shows both sides being divided by 85 to solve for 's', with -170/85 on the left and 85s/85 on the right, setting up for -2 = s.
Simplify. A minimalistic image displaying a mathematical equation: -2 = S.
Check: A mathematical equation: 10[3 - 8(2s - 5)] = 15(40 - 5s), presented in a clear, dark font against a white background.
Let s=−2. A mathematical expression featuring nested operations, including multiplication, subtraction, and a power, with some numbers in red indicating negative values, separated by a question mark and followed by another expression.
A mathematical expression reads 10 multiplied by the square of [3 minus 8 times (-4 minus 5)], which is followed by an inequality symbol indicating 'greater than or equal to', then 15 multiplied by (40 plus 10).
A mathematical inequality expression displays '10[3 - 8(-9)]²' on the left side, which is greater than or equal to '15(50)' on the right side. The equation shows a combination of numbers, operators, brackets, and an exponent.
A mathematical equation asks whether 10 multiplied by the square of (3 plus 72) is equal to 750.
A mathematical expression 10(75)^2 with a question mark over the equal sign, asking if it equals 750, is displayed against a white background.
A simple equation 750 = 750 is displayed with a checkmark, confirming its correctness. This image humorously represents agreement, a solved problem, or an undeniable truth.

Solve: 6[4−2(7y−1)]=8(13−8y).

Solution

y=−175

Solve: 12[1−5(4z−1)]=3(24+11z).

Solution

z=0

Classify Equations

Whether or not an equation is true depends on the value of the variable. The equation 7x+8=−13 is true when we replace the variable, x, with the value −3, but not true when we replace x with any other value. An equation like this is called a conditional equation. All the equations we have solved so far are conditional equations.

Conditional Equation

An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

Now let’s consider the equation 7y+14=7(y+2). Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.

Solve:

The image displays the mathematical equation 7y + 14 = 7(y + 2), which illustrates the distributive property or factoring a common term.
Distribute. A mathematical equation displays '7y + 14 = 7y + 14' in a clear, sans-serif font against a white background, representing an algebraic identity.
Subtract 7y to each side to get the y’s to one side. The equation 7y - 7y + 14 = 7y - 7y + 14, which simplifies to 14 = 14, is a mathematical identity.
Simplify—the y’s are eliminated. Text displaying the simple mathematical equality '14 = 14' on a clear white background.
But 14=14 is true.

This means that the equation 7y+14=7(y+2) is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable is called an identity.

Identity

An equation that is true for any value of the variable is called an identity.

The solution of an identity is all real numbers.

What happens when we solve the equation −8z=−8z+9?

Solve:

The image displays the algebraic equation -8z = -8z + 9, which simplifies to 0 = 9, indicating that the equation has no solution.
Add 8z to both sides to leave the constant alone on the right. The equation -8z + 8z = -8z + 8z + 9 is shown. This simplifies to 0 = 9, indicating that this equation has no solution.
Simplify—the z’s are eliminated. The mathematical expression '0 is not equal to 9' is displayed in black text on a white background.
But 0≠9.

Solving the equation −8z=−8z+9 led to the false statement 0=9. The equation −8z=−8z+9 will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

Contradiction

An equation that is false for all values of the variable is called a contradiction.

A contradiction has no solution.

The next few examples will ask us to classify an equation as conditional, an identity, or as a contradiction.

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 6(2n−1)+3=2n−8+5(2n+1).

Solution
A mathematical equation is displayed: 6(2n - 1) + 3 = 2n - 8 + 5(2n + 1). The equation involves linear expressions with the variable 'n' on both sides of the equality.
Distribute. An algebraic equation is shown, displaying '12n - 6 + 3 = 2n - 8 + 10n + 5'.
Combine like terms. A mathematical equation is displayed, showing '12n - 3 = 12n - 3' in a simple, clear font on a white background. The equation represents an identity, as both sides are identical.
Subtract 12n from each side to get the n’s to one side. An algebraic identity: 12n - 12n - 3 = 12n - 12n - 3 simplifies to -3 = -3, true for any value of 'n'.
Simplify. A mathematical equation showing -3 equals -3 is displayed on a white background.
This is a true statement. The equation is an identity.
The solution is all real numbers.

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 4+9(3x−7)=−42x−13+23(3x−2).

Solution

identity; all real numbers

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 8(1−3x)+15(2x+7)=2(x+50)+4(x+3)+1.

Solution

identity; all real numbers

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 8+3(a−4)=0.

Solution
A mathematical equation is displayed: 8 + 3(a - 4) = 0. The equation involves numbers, variables, and basic arithmetic operations set equal to zero.
Distribute. A mathematical equation reads 8 + 3a - 12 = 0 on a white background, in a close-up shot.
Combine like terms. A mathematical equation displays '3q - 4 = 0' on a white background.
Add 4 to both sides. A mathematical equation reads '3a - 4 + 4 = 0 + 4', demonstrating the addition property of equality where 4 is added to both sides of the equation. The numbers 4 and the plus signs are highlighted in red.
Simplify. The mathematical equation '3a = 4' is displayed on a white background.
Divide. A mathematical equation shows '3a over 3 equals 4 over 3', with the number 3 in the denominator of both fractions highlighted in red, indicating a division operation.
Simplify. The mathematical equation displays 'a = 4/3' in a simple, clear font on a white background, representing the value of the variable 'a' as a fraction.
The equation is true when a=43. This is a conditional equation.
The solution is a=43.

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 11(q+3)−5=19.

Solution

conditional equation; q=−911

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 6+14(k−8)=95.

Solution

conditional equation; k=20114

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 5m+3(9+3m)=2(7m−11).

Solution
An algebraic equation is shown: 5m + 3(9 + 3m) = 2(7m - 11).
Distribute. A mathematical equation is displayed, showing '5m + 27 + 9m = 14m - 22' in white text on a black background, representing a linear equation with variable 'm' to be solved.
Combine like terms. A mathematical equation is displayed, showing '14m + 27 = 14m - 22' in black text against a white background.
Subtract 14m from both sides. An algebraic equation: 14m + 27 - 14m = 14m - 22 - 14m, illustrating the step of subtracting 14m from both sides, with the subtracted terms highlighted in red.
Simplify. A mathematical expression '27   ≠   -22' is displayed against a white background, indicating that the number 27 is not equal to -22.
But 27≠−22. The equation is a contradiction.
It has no solution.

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 12c+5(5+3c)=3(9c−4).

Solution

contradiction; no solution

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 4(7d+18)=13(3d−2)−11d.

Solution

contradiction; no solution

We summarize the methods for classifying equations in the table.

Type of equation What happens when you solve it? Solution
Conditional Equation True for one or more values of the variables and false for all other values One or more values
Identity True for any value of the variable All real numbers
Contradiction False for all values of the variable No solution

Solve Equations with Fraction or Decimal Coefficients

We could use the General Strategy to solve the next example. This method would work fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator (LCD) of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called clearing the equation of fractions.

To clear an equation of decimals, we think of all the decimals in their fraction form and then find the LCD of those denominators.

How to Solve Equations with Fraction or Decimal Coefficients

Solve: 112x+56=34.

Solution
Step 1 is to find the least common denominator of all the fractions and decimals in the equation, one-twelfth x plus five-sixth is equal to three-fourths. What is the L C D of one-twelfth, five-sixths, and three-fourths? The L C D is equal to 12. Step 2 is multiply both sides of the equation by the L C D. This clears the fractions and decimals. Multiply both sides of the equation by the L C D, 12. The result is 12 times the quantity one-twelfth x plus five-sixths is equal to 12 times three-fourths. Use the Distributive Property. The result is 12 times one-twelfth x plus 12 times five-sixths is equal to 12 times three-fourths. Simplify. The result is x plus 10 is equal to 9. Notice there are no more fractions. Step 3 is to solve using the General Strategy for Solving Linear Equations. To isolate the variable term, subtract 10. The result is x plus 10 minus 10 is equal to 9 minus 10. Simplify. The result is x is equal to negative 1. Check the solution. Substitute negative into the original equation one-twelfth x plus five-sixths is equal to three-fourth. Is one-twelfth times negative 1 plus five-sixths equal to three-fourths? Is negative one-twelfth plus five-sixths equal to three-fourths? Is negative one-twelfth plus ten-twelfths equal to nine-twelfths? Is nine-twelfths equal to nine-twelfths? Yes. The solution checks.

Solve: 14x+12=58.

Solution

x=12

Solve: 18x+12=14.

Solution

x=−2

Notice in the previous example, once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve. We then used the General Strategy for Solving Linear Equations.

Solve Equations with Fraction or Decimal Coefficients.

  1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
  2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
  3. Solve using the General Strategy for Solving Linear Equations.

Solve: 5=12y+23y−34y.

Solution

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Find the LCD of all fractions in the equation. A mathematical equation is displayed on a white background: 5 = 1/2y + 2/3y - 3/4y.
The LCD is 12.
Multiply both sides of the equation by 12. A mathematical equation shown as 12(5) = 12 * (1/2y + 2/3y - 3/4y). The numbers 12 are highlighted in red, indicating a multiplication on both sides of the equation.
Distribute. An algebraic equation showing 12 multiplied by 5 on the left side, equal to 12 times one-half y, plus 12 times two-thirds y, minus 12 times three-quarters y on the right side.
Simplify—notice, no more fractions. A mathematical equation on a white background reads '60 = 6y + 8y - 9y'.
Combine like terms. A clear image displaying the algebraic equation 60 = 5y. This simple linear equation can be solved by dividing both sides by 5 to find the value of y, which would be 12.
Divide by five. A mathematical equation shows '60 divided by 5 equals 5y divided by 5'. The number 5 in the denominators is highlighted in red.
Simplify. A mathematical equation is displayed on a white background, showing the number '12' equal to the variable 'y', rendered in a simple, sans-serif font.
Check: A mathematical equation is displayed, showing 5 equals one-half y plus two-thirds y minus three-fourths y. The equation involves fractions and a variable 'y'.
Let y=12. A mathematical equation questions whether 5 is equal to the expression: one-half of 12, plus two-thirds of 12, minus three-fourths of 12, with the number 12 highlighted in red.
A mathematical expression 5 =? 6 + 8 - 9 is displayed, asking if 5 is equal to the result of 6 plus 8 minus 9.
The equation 5 = 5 with a checkmark to its right, indicating that the equality is correct. The image displays numerical equality and verification on a white background.

Solve: 7=12x+34x−23x.

Solution

x=12

Solve: −1=12u+14u−23u.

Solution

u=−12

In the next example, we’ll distribute before we clear the fractions.

Solve: 12(y−5)=14(y−1).

Solution
A mathematical equation is shown: one-half times the quantity y minus 5 equals one-fourth times the quantity y minus 1. This can also be written as (1/2)(y-5)=(1/4)(y-1).
Distribute. A mathematical equation is displayed: (1/2) * y - (1/2) * 5 = (1/4) * y - (1/4) * 1. This algebraic expression involves fractions, multiplication, subtraction, and the variable 'y'.
Simplify. A mathematical equation is displayed: 1/2y - 5/2 = 1/4y - 1/4.
Multiply by the LCD, four. An equation showing 4(1/2y - 5/2) = 4(1/4y - 1/4). The number 4 is highlighted in red on both sides of the equation.
Distribute. A mathematical equation shows '4 multiplied by 1/2y minus 4 multiplied by 5/2 equals 4 multiplied by 1/4y minus 4 multiplied by 1/4'.
Simplify. A mathematical equation is displayed, reading '2y - 10 = y - 1'.
Collect the variables to the left. A mathematical equation is displayed, showing '2y - y - 10 = y - y - 1'. The variable 'y' is shown in both black and red, while constants and operators are in black.
Simplify. A mathematical equation is displayed, showing 'y - 10 = -1' in a clear, dark font against a plain white background.
Collect the constants to the right. The equation y - 10 + 10 = -1 + 10 shows a step in solving for y by adding 10 to both sides, effectively canceling out the -10 on the left to isolate y.
Simplify. A simple mathematical equation 'y = 9' is displayed in a clear, dark gray font on a plain white background.
An alternate way to solve this equation is to clear the fractions without distributing first. If you multiply the factors correctly, this method will be easier.
A mathematical equation is displayed, showing 'one half (y minus five) equals one fourth (y minus one)', representing a linear equation to be solved for the variable y.
Multiply by the LCD, 4. A mathematical equation displays the multiplication of 4 by 1/2(y-5) on the left side, and 4 by 1/4(y-1) on the right side, with the number 4 highlighted in red.
Multiply four times the fractions. A mathematical equation is displayed on a white background: 2(y - 5) = 1(y - 1).
Distribute. A mathematical equation is displayed, reading '2y - 10 = y - 1' in a clear, dark gray font against a plain white background.
Collect the variables to the left. A mathematical equation, 2y - y - 10 = y - y - 1, is displayed on a white background. Some 'y' terms are colored red, likely indicating subtraction or grouping of like terms in an algebra problem.
Simplify. A mathematical equation is displayed, reading 'y - 10 = -1'.
Collect the constants to the right. A mathematical equation is displayed on a white background, showing 'y - 10 + 10 = -1 + 10' with the numbers '+ 10' on both sides highlighted in red.
Simplify. The image displays the equation 'y = 9' in the center of a white background. The text is rendered in a clear, dark gray font.
Check: A mathematical equation is displayed: 1/2(y - 5) = 1/4(y - 1).
Let y=9. A mathematical equation asking if 1/2(9-5) equals 1/4(9-1). The numbers 9 and 5 in the first parenthesis are highlighted in red, as are 9 and 1 in the second parenthesis.
Finish the check on your own.

Solve: 15(n+3)=14(n+2).

Solution

n=2

Solve: 12(m−3)=14(m−7).

Solution

m=−1

When you multiply both sides of an equation by the LCD of the fractions, make sure you multiply each term by the LCD—even if it does not contain a fraction.

Solve: 4q+32+6=3q+54

Solution
A mathematical equation is shown: (4q + 3) / 2 + 6 = (3q + 5) / 4. It is an algebraic expression involving the variable 'q' and fractions.
Multiply both sides by the LCD, 4. A mathematical equation shows '4 multiplied by the quantity (4q+3)/2 plus 6', which is equal to '4 multiplied by the quantity (3q+5)/4'.
Distribute. An algebraic equation: 4 multiplied by the fraction (4q + 3) / 2, plus 4 multiplied by 6, equals 4 multiplied by the fraction (3q + 5) / 4.
Simplify. A mathematical equation is displayed on a white background: 2(4q + 3) + 24 = 3q + 5.
An image displays the algebraic equation 8q + 6 + 24 = 3q + 5 in black text on a white background. The equation involves a variable 'q' and various integer constants on both sides of the equality.
A mathematical equation is displayed: 8q + 30 = 3q + 5.
Collect the variables to the left. A mathematical equation is displayed, showing '8q - 3q + 30 = 3q - 3q + 5'. Some 'q' terms are in black, while others are in red.
Simplify. A white background displays the mathematical equation 5q + 30 = 5, rendered in clear, dark gray text.
Collect the constants to the right. An algebra equation 5q + 30 - 30 = 5 - 30, demonstrating the step of subtracting 30 from both sides to isolate the variable, with the subtracted 30s highlighted in red.
Simplify. A mathematical equation is displayed on a white background, showing '5q = -25' in black text.
Divide both sides by five. Solving for q: Both sides of the equation 5q = -25 are divided by 5, resulting in 5q/5 = -25/5 to find the value of q.
Simplify. The image displays the mathematical equation 'q = -5' in black text against a plain white background, presenting a simple assignment of the value negative five to the variable q.
Check: A mathematical equation is displayed: (4q + 3) / 2 + 6 = (3q + 5) / 4.
Let q=−5. A mathematical equation asking to compare two expressions: (4(-5)+3)/2+6 on the left, and (3(-5)+5)/4 on the right, separated by a question mark and an equals sign.
Finish the check on your own.

Solve: 3r+56+1=4r+33.

Solution

r=1

Solve: 2s+32+1=3s+24.

Solution

s=−8

Some equations have decimals in them. This kind of equation may occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, 0.7=710 and 0.29=29100. So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.

The next example uses an equation that is typical of the ones we will see in the money applications in a later section. Notice that we will clear all decimals by multiplying by the LCD of their fraction form.

Solve: 0.25x+0.05(x+3)=2.85.

Solution

Look at the decimals and think of the equivalent fractions:

0.25=25100,0.05=5100,2.85=285100.

Notice, the LCD is 100. By multiplying by the LCD we will clear the decimals from the equation.

A mathematical equation is displayed, reading 0.25x + 0.05(x + 3) = 2.85.
Distribute first. A mathematical equation is displayed with the expression 0.25x + 0.05x + 0.15 = 2.85.
Combine like terms. A mathematical equation is displayed, reading '0.30x + 0.15 = 2.85' against a white background.
To clear decimals, multiply by 100. A mathematical equation is displayed, showing 100 multiplied by the quantity (0.30x + 0.15) on the left side, which equals 100 multiplied by 2.85 on the right side.
Distribute. A mathematical equation is displayed, showing 30x + 15 = 285.
Subtract 15 from both sides. An algebraic equation showing the step of subtracting 15 from both sides: 30x + 15 - 15 = 285 - 15.
Simplify. A mathematical equation shows '30x = 270' against a white background.
Divide by 30. The equation 30x/30 = 270/30 illustrates a step in solving for 'x' where both sides are divided by 30 to isolate the variable, leading to x = 9.
Simplify. The mathematical equation 'x=9' is displayed on a plain white background, presenting a simple statement of equality.
Check it yourself by substituting x=9 into the original equation.

Solve: 0.25n+0.05(n+5)=2.95.

Solution

n=9

Solve: 0.10d+0.05(d−5)=2.15.

Solution

d=16

Key Concepts

  • How to determine whether a number is a solution to an equation
    1. Substitute the number in for the variable in the equation.
    2. Simplify the expressions on both sides of the equation.
    3. Determine whether the resulting equation is true.
      If it is true, the number is a solution.
      If it is not true, the number is not a solution.
  • How to Solve Linear Equations Using a General Strategy
    1. Simplify each side of the equation as much as possible.
      Use the Distributive Property to remove any parentheses.
      Combine like terms.
    2. Collect all the variable terms on one side of the equation.
      Use the Addition or Subtraction Property of Equality.
    3. Collect all the constant terms on the other side of the equation.
      Use the Addition or Subtraction Property of Equality.
    4. Make the coefficient of the variable term equal to 1.
      Use the Multiplication or Division Property of Equality.
      State the solution to the equation.
    5. Check the solution.
      Substitute the solution into the original equation to make sure the result is a true statement.
  • How to Solve Equations with Fraction or Decimal Coefficients
    1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
    2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
    3. Solve using the General Strategy for Solving Linear Equations.

Practice Makes Perfect

Solve Equations Using the General Strategy

In the following exercises, determine whether the given values are solutions to the equation.

6y+10=12y

ⓐ y=53 ⓑ y=−12

Solution

ⓐ yes ⓑ no

4x+9=8x

ⓐ x=−78 ⓑ x=94

8u−1=6u

ⓐ u=−12 ⓑ u=12

Solution

ⓐ no ⓑ yes

9v−2=3v

ⓐ v=−13 ⓑ v=13

In the following exercises, solve each linear equation.

15(y−9)=−60

Solution

y=5

−16(3n+4)=32

−(w−12)=30

Solution

w=−18

−(t−19)=28

51+5(4−q)=56

Solution

q=3

−6+6(5−k)=15

3(10−2x)+54=0

Solution

x=14

−2(11−7x)+54=4

23(9c−3)=22

Solution

c=4

35(10x−5)=27

15(15c+10)=c+7

Solution

c=52

14(20d+12)=d+7

3(4n−1)−2=8n+3

Solution

n=2

9(2m−3)−8=4m+7

12+2(5−3y)=−9(y−1)−2

Solution

y=−5

−15+4(2−5y)=−7(y−4)+4

5+6(3s−5)=−3+2(8s−1)

Solution

s=10

−12+8(x−5)=−4+3(5x−2)

4(p−4)−(p+7)=5(p−3)

Solution

p=−4

3(a−2)−(a+6)=4(a−1)

4[5−8(4c−3)]=12(1−13c)−8

Solution

c=−4

5[9−2(6d−1)]=11(4−10d)−139

3[−9+8(4h−3)]=2(5−12h)−19

Solution

h=34

3[−14+2(15k−6)]=8(3−5k)−24

5[2(m+4)+8(m−7)]=2[3(5+m)−(21−3m)]

Solution

m=6

10[5(n+1)+4(n−1)]=11[7(5+n)−(25−3n)]

Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

23z+19=3(5z−9)+8z+46

Solution

identity; all real numbers

15y+32=2(10y−7)−5y+46

18(5j−1)+29=47

Solution

conditional equation;j=25

24(3d−4)+100=52

22(3m−4)=8(2m+9)

Solution

conditional equation; m=165

30(2n−1)=5(10n+8)

7v+42=11(3v+8)−2(13v−1)

Solution

contradiction; no solution

18u−51=9(4u+5)−6(3u−10)

45(3y−2)=9(15y−6)

Solution

contradiction; no solution

60(2x−1)=15(8x+5)

9(14d+9)+4d=13(10d+6)+3

Solution

identity; all real numbers

11(8c+5)−8c=2(40c+25)+5

Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation with fraction coefficients.

14x−12=−34

Solution

x=−1

34x−12=14

56y−23=−32

Solution

y=−1

56y−13=−76

12a+38=34

Solution

a=34

58b+12=−34

2=13x−12x+23x

Solution

x=4

2=35x−13x+25x

13w+54=w−14

Solution

w=94

12a−14=16a+112

13b+15=25b−35

Solution

b=12

13x+25=15x−25

14(p−7)=13(p+5)

Solution

p=−41

15(q+3)=12(q−3)

12(x+4)=34

Solution

x=−52

13(x+5)=56

4n+84=n3

Solution

n=−3

3p+63=p2

3x+42+1=5x+108

Solution

x=−2

10y−23+3=10y+19

7u−14−1=4u+85

Solution

u=3

3v−62+5=11v−45

In the following exercises, solve each equation with decimal coefficients.

0.4x+0.6=0.5x−1.2

Solution

x=18

0.7x+0.4=0.6x+2.4

0.9x−1.25=0.75x+1.75

Solution

x=20

1.2x−0.91=0.8x+2.29

0.05n+0.10(n+8)=2.15

Solution

n=9

0.05n+0.10(n+7)=3.55

0.10d+0.25(d+5)=4.05

Solution

d=8

0.10d+0.25(d+7)=5.25

Everyday Math

Fencing Micah has 74 feet of fencing to make a dog run in his yard. He wants the length to be 2.5 feet more than the width. Find the length, L, by solving the equation 2L+2(L−2.5)=74.

Solution

L=19.75 feet

Stamps Paula bought $22.82 worth of 49-cent stamps and 21-cent stamps. The number of 21-cent stamps was eight less than the number of
49-cent stamps. Solve the equation
0.49s+0.21​(s−8)​​=22.82 for s, to find the number of 49-cent stamps Paula bought.

Writing Exercises

Using your own words, list the steps in the general strategy for solving linear equations.

Solution

Answers will vary.

Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.

What is the first step you take when solving the equation 3−7(y−4)=38? Why is this your first step?

Solution

Answers will vary.

If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?

If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

Solution

Answers will vary.

For the equation 0.35x+2.1=3.85, how do you clear the decimal?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve linear equations using a general strategy. In row 3, the I can was classify equations. In row 4, the I can was solve equations with fraction or decimal coefficients.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

conditional equation
An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.
contradiction
An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.
identity
An equation that is true for any value of the variable is called an Identity. The solution of an identity is all real numbers.
linear equation
A linear equation is an equation in one variable that can be written, where a and b are real numbers and a≠0, as ax+b=0.
solution of an equation
A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

Use a Problem Solving Strategy

Learning Objectives

By the end of this section, you will be able to:

  • Use a problem solving strategy for word problems
  • Solve number word problems
  • Solve percent applications
  • Solve simple interest applications

Before you get started, take this readiness quiz.

Translate “six less than twice x” into an algebraic expression.
If you missed this problem, review Example 8 in Use the Language of Algebra.

Solution

2x−6

Convert 4.5% to a decimal.
If you missed this problem, review Example 7 in Decimals.

Solution

0.045

Convert 0.6 to a percent.
If you missed this problem, review Example 7 in Decimals.

Solution

60%

Have you ever had any negative experiences in the past with word problems? When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. Realize that your negative experiences with word problems are in your past. To move forward you need to calm your fears and change your negative feelings.

Start with a fresh slate and begin to think positive thoughts. Repeating some of the following statements may be helpful to turn your thoughts positive. Thinking positive thoughts is a first step towards success.

  I think I can! I think I can!

  While word problems were hard in the past, I think I can try them now.

  I am better prepared now—I think I will begin to understand word problems.

  I am able to solve equations because I practiced many problems and I got help when I needed it—I can try that
  with word problems.

  It may take time, but I can begin to solve word problems.

You are now well prepared and you are ready to succeed. If you take control and believe you can be successful, you will be able to master word problems.

Use a Problem Solving Strategy for Word Problems

Now that we can solve equations, we are ready to apply our new skills to word problems. We will develop a strategy we can use to solve any word problem successfully.

Normal yearly snowfall at the local ski resort is 12 inches more than twice the amount it received last season. The normal yearly snowfall is 62 inches. What was the snowfall last season at the ski resort?

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. What was the snowfall last season?
Step 3. Name what we are looking for and
choose a variable to represent it.
Let s= the snowfall last season.
Step 4. Translate.
Restate the problem in one sentence with all the important information.
The text reads: 'The normal snowfall was twelve more than twice the amount last year.' Light blue brackets segment the text into phrases.
Translate into an equation. A mathematical equation is displayed on a white background, which reads '62 = 2s + 12'.
Step 5. Solve the equation. A mathematical equation is displayed horizontally, reading '62 = 2s + 12'. The numbers and operators are rendered in a dark gray font against a clean white background.
Subtract 12 from each side. A mathematical equation is shown with the step of subtracting 12 from both sides highlighted in red: 62 - 12 = 2s + 12 - 12.
Simplify. A mathematical equation is displayed, showing '50 = 2s' in dark gray text against a plain white background.
Divide each side by two. A mathematical equation shows '50 divided by 2 equals 2s divided by 2'. The number 2 in the denominator of both fractions is colored red. It illustrates a step in solving for 's'.
Simplify. The image shows a mathematical equation on a white background, displaying '25 = s' in a simple, clear font, indicating that the value of 's' is 25.
Step 6. Check: First, is our answer reasonable?
Yes, having 25 inches of snow seems OK.
The problem says the normal snowfall is twelve
inches more than twice the number of last season.
Twice 25 is 50 and 12 more than that is 62.
Step 7. Answer the question. The snowfall last season was 25 inches.

Guillermo bought textbooks and notebooks at the bookstore. The number of textbooks was three more than twice the number of notebooks. He bought seven textbooks. How many notebooks did he buy?

Solution

He bought two notebooks.

Gerry worked Sudoku puzzles and crossword puzzles this week. The number of Sudoku puzzles he completed is eight more than twice the number of crossword puzzles. He completed 22 Sudoku puzzles. How many crossword puzzles did he do?

Solution

He did seven crosswords puzzles.

We summarize an effective strategy for problem solving.

Use a Problem Solving Strategy for word problems.

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what you are looking for.
  3. Name what you are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
  5. Solve the equation using proper algebra techniques.
  6. Check the answer in the problem to make sure it makes sense.
  7. Answer the question with a complete sentence.

Solve Number Word Problems

We will now apply the problem solving strategy to “number word problems.” Number word problems give some clues about one or more numbers and we use these clues to write an equation. Number word problems provide good practice for using the Problem Solving Strategy.

The sum of seven times a number and eight is thirty-six. Find the number.

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. the number
Step 3. Name what you are looking for and
choose a variable to represent it.
Let n = the number.
Step 4. Translate:
Restate the problem as one sentence.
Translate into an equation.
The image shows a mathematical word problem: 'The sum of seven times a number and 8 is 36.' Light blue brackets underneath highlight 'The sum of seven times a number and 8', 'is', and '36' respectively.
A mathematical equation is displayed, showing '7n + 8 = 36' centered on a white background.
Step 5. Solve the equation.
Subtract eight from each side and simplify.
Divide each side by seven and simplify.
A mathematical equation is displayed on a white background, reading '7n + 8 = 36' in black text.
A mathematical equation shows '7n = 28' on a white background, where 'n' is an unknown variable in a multiplication problem.
The text 'n = 4' is displayed in black on a white background, positioned in the top right corner of the image.
Step 6. Check.
Is the sum of seven times four plus eight equal to 36?
7·4+8=?3628+8=?3636=36✓
Step 7. Answer the question. The number is 4.

Did you notice that we left out some of the steps as we solved this equation? If you’re not yet ready to leave out these steps, write down as many as you need.

The sum of four times a number and two is fourteen. Find the number.

Solution

3

The sum of three times a number and seven is twenty-five. Find the number.

Solution

6

Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far, we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.

The sum of two numbers is negative fifteen. One number is nine less than the other. Find the numbers.

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. two numbers
Step 3. Name what you are looking for by
choosing a variable to represent the first
number.
“One number is nine less than the other.”

Let n=1st number.

n−9=2nd number
Step 4. Translate.
Write as one sentence.
Translate into an equation.

The sum of two numbers is negative fifteen.
A visual illustrating the translation of the word problem '1st number + 2nd number is negative fifteen' into the algebraic equation 'n + n - 9 = -15'.
Step 5. Solve the equation.
Combine like terms.
Add nine to each side and simplify.
Simplify.
A mathematical equation is displayed on a white background, reading 'n + n - 9 = -15'.
A mathematical equation is displayed against a white background, reading '2n - 9 = -15'.
The mathematical equation '2n = -6' is displayed on a white background, representing a simple algebraic problem.
The image shows the text 'n = -3 1st number' in black on a white background. The text is centrally aligned and occupies the upper right portion of the image.

The image displays the mathematical expression 'n - 9' followed by the text '2nd number' in a simple, clear font on a white background.
The image features the mathematical expression '3 - 9' prominently displayed on a clean white background. The number '3' is rendered in a distinct reddish-orange hue, while the minus sign and the number '9' are both in black.
The number -12 is displayed prominently on a stark white background.
Step 6. Check.
Is −12 nine less than −3?
−3−9=?−12−12=−12✓
Is their sum −15?
−3+(−12)=?−15−15=−15✓
Step 7. Answer the question. The numbers are −3 and −12.

The sum of two numbers is negative twenty-three. One number is seven less than the other. Find the numbers.

Solution

−15,−8

The sum of two numbers is negative eighteen. One number is forty more than the other. Find the numbers.

Solution

−29,11

Some number problems involve consecutive integers. Consecutive integers are integers that immediately follow each other. Examples of consecutive integers are:

1,2,3,4−10,−9,−8,−7150,151,152,153

Notice that each number is one more than the number preceding it. Therefore, if we define the first integer as n, the next consecutive integer is n+1. The one after that is one more than n+1, so it is n+1+1, which is n+2.

n1stintegern+12ndconsecutive integern+23rdconsecutive integeretc.

We will use this notation to represent consecutive integers in the next example.

Find three consecutive integers whose sum is −54.

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. three consecutive integers
Step 3. Name each of the three numbers Let n=1st integer.
n+1=2nd consecutive integer
n+2=3rd consecutive integer
Step 4. Translate.
Restate as one sentence.
Translate into an equation.

The sum of the three integers is −54.
The image displays a mathematical equation: n + n + 1 + n + 2 = -54. The equation involves the variable 'n' and constants, summing to a negative number. It appears to be a problem involving consecutive integers.
Step 5. Solve the equation.
Combine like terms.
Subtract three from each side.
Divide each side by three.
A mathematical equation showing the sum of three consecutive integers, represented by 'n + n + 1 + n + 2', equaling -54.
A mathematical equation is displayed against a white background, reading 3n + 3 = -54.
A mathematical equation '3n = -57' is displayed on a white background.
The image shows the expression 'n = -19 1st integer' in black text on a white background, indicating that 'n' is assigned the value -19, identified as the first integer.

A white background displays the mathematical expression 'n + 1' followed by the text '2nd integer' in gray font. The text is horizontally aligned and centered in the frame.
The image displays a simple arithmetic problem: -19 + 1, against a plain white background.
The image displays a white background with the numbers '-.18' in a dark grey color, slightly out of focus.

The text 'n + 2 3rd integer' is displayed on a white background, representing mathematical notation or a sequence.
A mathematical expression shows '-19 + 2' in grey text on a white background.
The number -17 is displayed.
Step 6. Check.
−19+(−18)+(−17)=−54−54=−54✓
Step 7. Answer the question. The three consecutive integers are
−17,−18, and −19.

Find three consecutive integers whose sum is −96.

Solution

−33,−32,−31

Find three consecutive integers whose sum is −36.

Solution

−13,−12,−11

Now that we have worked with consecutive integers, we will expand our work to include consecutive even integers and consecutive odd integers. Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

24, 26, 28
−12,−10,−8

Notice each integer is two more than the number preceding it. If we call the first one n, then the next one is n+2. The one after that would be n+2+2 or n+4.

n1steven integern+22ndconsecutive even integern+43rdconsecutive even integeretc.

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers 63, 65, and 67.

63, 65, 67
n,n+2,n+4
n1stodd integern+22ndconsecutive odd integern+43rdconsecutive odd integeretc.

Does it seem strange to have to add two (an even number) to get the next odd number? Do we get an odd number or an even number when we add 2 to 3? to 11? to 47?

Whether the problem asks for consecutive even numbers or odd numbers, you do not have to do anything different. The pattern is still the same—to get to the next odd or the next even integer, add two.

Find three consecutive even integers whose sum is 120.

Solution
A 7-step method for solving word problems, illustrated by finding three consecutive even integers that sum to 120.
Step 1. Read the problem.
Step 2. Identify what you are looking for. three consecutive even integers
Step 3. Name. Letn=1steven integer.n+2=2ndconsecutive even integern+4=3rdconsecutive even integer
Step 4. Translate.
Restate as one sentence.
Translate into an equation.
The sum of the three even integers is120.n+n+2+n+4=120
Step 5. Solve the equation.
Combine like terms.
Subtract 6 from each side.
Divide each side by 3.
n+n+2+n+4=1203n+6=1203n=114n=381stintegern+22ndinteger38+240n+43rdinteger38+442
Step 6. Check.
38+40+42=?120120=120✓
Step 7. Answer the question. The three consecutive integers are 38, 40, and 42.

Find three consecutive even integers whose sum is 102.

Solution

32, 34, 36

Find three consecutive even integers whose sum is −24.

Solution

−10,−8,−6

When a number problem is in a real life context, we still use the same strategies that we used for the previous examples.

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. How much does the husband earn?
Step 3. Name.
Choose a variable to represent the amount the husband earns.
The wife earns $16,000 less than twice that.
Let h = the amount the husband earns.
Step 4. Translate.
Restate the problem in one sentence with
all the important information.
Translate into an equation.
2h − 16,000 = the amount the wife
Together the husband and wife earn $110,000.
Step 5. Solve the equation.
Combine like terms.
Add 16,000 to both sides and simplify.
Divide each side by three.
h+2h−16,000=110,000h+2h−16,000=110,0003h−16,000=110,0003h=126,000h=42,000
$42,000 amount husband earns
2h−16,000amount wife earns2(42,000)−16,00084,000−16,00068,000
Step 6. Check:
If the wife earns $68,000 and the husband earns $42,000, is that $110,000? Yes!
Step 7. Answer the question. The husband earns $42,000 a year.

According to the National Automobile Dealers Association, the average cost of a car in 2014 was $28,400. This was $1,600 less than six times the cost in 1975. What was the average cost of a car in 1975?

Solution

The average cost was $5,000.

US Census data shows that the median price of new home in the U.S. in November 2014 was $280,900. This was $10,700 more than 14 times the price in November 1964. What was the median price of a new home in November 1964?

Solution

The median price was $19,300.

Solve Percent Applications

There are several methods to solve percent equations. In algebra, it is easiest if we just translate English sentences into algebraic equations and then solve the equations. Be sure to change the given percent to a decimal before you use it in the equation.

Translate and solve:

ⓐ What number is 45% of 84?ⓑ 8.5% of what amount is $4.76? ⓒ 168 is what percent of 112?

Solution
ⓐ
A math problem displayed, asking to calculate 45% of 84. Light blue brackets are used below parts of the text to visually separate the terms in the question.
Translate into algebra. Let n = the number. A mathematical equation is displayed on a white background, showing n equals 0.45 multiplied by 84.
Multiply. The image shows the text 'n = 37.8' in black font on a white background, likely representing a numerical value for a variable 'n'.
37.8 is 45% of 84.


ⓑ
A word problem asks: '8.5% of what amount is $4.76?' Each part of the question is underlined with a light blue bracket.
Translate. Let n = the amount. A mathematical equation is displayed, showing '0.085 multiplied by n equals 4.76' on a white background.
Multiply. A mathematical equation displays '0.085n = 4.76' on a white background.
Divide both sides by 0.085 and simplify. The text 'n = 56' is displayed in black against a plain white background, positioned towards the top right corner of the image. The text is clear and readable.
8.5% of $56 is $4.76


ⓒ
We are asked to find percent, so we must
have our result in percent form.
A mathematical question written in black text on a white background asks: '168 is what percent of 112?' Light blue brackets underline each part of the phrase to highlight individual components.
Translate into algebra. Let p = the percent. A mathematical equation displaying 168 equals p multiplied by 112.
Multiply. A mathematical equation is displayed on a white background, reading '168 = 112 p'.
Divide both sides by 112 and simplify. A simple mathematical equation is displayed on a white background, showing '1.5 = p' in black text.
Convert to percent. A mathematical equation shows '150% = p' displayed in black text against a plain white background.
168 is 150% of 112.

Translate and solve: ⓐ What number is 45% of 80? ⓑ 7.5% of what amount is $1.95? ⓒ 110 is what percent of 88?

Solution

ⓐ 36 ⓑ $26 ⓒ 125%

Translate and solve: ⓐ What number is 55% of 60? ⓑ 8.5% of what amount is $3.06? ⓒ 126 is what percent of 72?

Solution

ⓐ 33 ⓑ $36 ⓒ 175%

Now that we have a problem solving strategy to refer to, and have practiced solving basic percent equations, we are ready to solve percent applications. Be sure to ask yourself if your final answer makes sense—since many of the applications we will solve involve everyday situations, you can rely on your own experience.

The label on Audrey’s yogurt said that one serving provided 12 grams of protein, which is 24% of the recommended daily amount. What is the total recommended daily amount of protein?

Solution
What are you asked to find? What total amount of protein is recommended?
Choose a variable to represent it. Let a= total amount of protein.
Write a sentence that gives the
information to find it.
The text reads '12 g is 24% of the total amount.' Underlines and brackets visually segment the phrase into its mathematical components.
Translate into an equation. A mathematical equation is displayed on a white background, showing '12 = 0.24 * q' where 'q' is a variable to be solved.
Solve. A simple mathematical equation is displayed with the number 50 equal to the variable 'a' on a white background.
Check: Does this make sense?
Yes, 24% is about 14 of the total and
12 is about 14 of 50.
Write a complete sentence to answer the question. The amount of protein that is recommended is 50 g.

One serving of wheat square cereal has 7 grams of fiber, which is 28% of the recommended daily amount. What is the total recommended daily amount of fiber?

Solution

25 grams

One serving of rice cereal has 190 mg of sodium, which is 8% of the recommended daily amount. What is the total recommended daily amount of sodium?

Solution

2,375 mg

Remember to put the answer in the form requested. In the next example we are looking for the percent.

Veronica is planning to make muffins from a mix. The package says each muffin will be 240 calories and 60 calories will be from fat. What percent of the total calories is from fat?

Solution
What are you asked to find? What percent of the total calories is fat?
Choose a variable to represent it. Let p= percent of fat.
Write a sentence that gives the
information to find it.
A mathematical question asks, 'What percent of 240 is 60?' The words are underlined with light blue brackets, visually breaking down the problem into segments.
Translate the sentence into an equation. A mathematical equation is displayed, showing 'p * 240 = 60' with the variable 'p' multiplied by 240, equaling 60. The equation is rendered in black text against a white background.
Multiply. A partial mathematical equation is displayed on a white background, showing '240 p = 60' in black text.
Divide both sides by 240. The image displays the equation 'p = 0.25' in black text against a white background, representing a numerical probability or a parameter value.
Put in percent form. A clean white background with a mathematical expression 'p = 25%' displayed in the top right corner. The 'p' is a lowercase letter, followed by an equals sign, then the number '25' and a percent symbol.
Check: does this make sense?
Yes, 25% is one-fourth; 60 is one-fourth
of 240. So, 25% makes sense.
Write a complete sentence to answer the question. Of the total calories in each muffin, 25% is fat.

Mitzi received some gourmet brownies as a gift. The wrapper said each brownie was 480 calories, and had 240 calories of fat. What percent of the total calories in each brownie comes from fat? Round the answer to the nearest whole percent.

Solution

50%

The mix Ricardo plans to use to make brownies says that each brownie will be 190 calories, and 76 calories are from fat. What percent of the total calories are from fat? Round the answer to the nearest whole percent.

Solution

40%

It is often important in many fields—business, sciences, pop culture—to talk about how much an amount has increased or decreased over a certain period of time. This increase or decrease is generally expressed as a percent and called the percent change.

To find the percent change, first we find the amount of change, by finding the difference of the new amount and the original amount. Then we find what percent the amount of change is of the original amount.

Find percent change.

  1. Find the amount of change.
    change=new amount−original amount
  2. Find what percent the amount of change is of the original amount.
    change is what percent of the original amount?

Recently, the California governor proposed raising community college fees from $36 a unit to $46 a unit. Find the percent change. (Round to the nearest tenth of a percent.)

Solution
Find the amount of change. 46−36=10
Find the percent. Change is what percent of the original amount?
Let p= the percent. A math problem asks: 10 is what percent of 36? The words are bracketed underneath to show the structure of the question.
Translate to an equation. The image displays a mathematical equation: '10 = p · 36', where 'p' is a variable, and the dot signifies multiplication. The equation is set against a plain white background.
Simplify. The image displays a mathematical equation: 10 = 36p. This equation involves the numbers 10 and 36, and the variable 'p', indicating a linear relationship between them.
Divide both sides by 36. A mathematical expression shows '0.278 ×≈× p' against a plain white background, indicating that 0.278 is approximately equal to p.
Change to percent form; round to the
nearest tenth
The image displays a mathematical expression '27.8% ≈ p' in black font on a plain white background, indicating that 27.8 percent is approximately equal to the variable p.
Write a complete sentence to answer
the question.
The new fees are approximately a 27.8% increase
over the old fees.
Remember to round the division to the nearest thousandth in order to round the percent to the nearest tenth.

Find the percent change. (Round to the nearest tenth of a percent.) In 2011, the IRS increased the deductible mileage cost to 55.5 cents from 51 cents.

Solution

8.8%

Find the percent change. (Round to the nearest tenth of a percent.) In 1995, the standard bus fare in Chicago was $1.50. In 2008, the standard bus fare was 2.25.

Solution

50%

Applications of discount and mark-up are very common in retail settings.

When you buy an item on sale, the original price has been discounted by some dollar amount. The discount rate, usually given as a percent, is used to determine the amount of the discount. To determine the amount of discount, we multiply the discount rate by the original price.

The price a retailer pays for an item is called the original cost. The retailer then adds a mark-up to the original cost to get the list price, the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost.

Discount

amount of discount=discount rate·original pricesale price=original amount–discount price

The sale price should always be less than the original price.

Mark-up

amount of mark-up=mark-up rate·original pricelist price=original cost+mark-up

The list price should always be more than the original cost.

Liam’s art gallery bought a painting at an original cost of $750. Liam marked the price up 40%. Find ⓐ the amount of mark-up and ⓑ the list price of the painting.

Solution
ⓐ

Identify what you are asked to find, and
choose a variable to represent it.
What is the amount of mark-up?
Let m= the amount of mark-up.
Write a sentence that gives the
information to find it.
The mark-up is 40% of the $750 original cost
Translate into an equation. A mathematical equation is displayed, showing 'm = 0.40 x 750' in black text against a white background.
Solve the equation. The text 'm = 300' is displayed in black characters against a plain white background, centered in the frame. The numbers and equal sign are clear and legible.
Write a complete sentence. The mark-up on the painting was $300.


ⓑ
Identify what you are asked to find, and
choose a variable to represent it.
What is the list price?
Let p= the list price.
Write a sentence that gives the
information to find it.
A graphic displays the formula 'The list price is original cost plus the mark-up,' with each component of the phrase underlined and connected by blue brackets. It illustrates how retail prices are determined.
Translate into an equation. A mathematical equation is displayed on a white background, which reads 'p = 750 + 300'. The equation shows the variable 'p' is equal to the sum of 750 and 300.
Solve the equation. The image displays the equation 'p = 1,050' in a clear, dark gray font against a plain white background, indicating a numerical value assigned to the variable 'p'.
Check. Is the list price more than the original cost?
Is $1,050 more than $750? Yes.
Write a complete sentence. The list price of the painting was $1,050.

Find ⓐ the amount of mark-up and ⓑ the list price: Jim’s music store bought a guitar at original cost $1,200. Jim marked the price up 50%.

Solution

ⓐ $600 ⓑ $1,800

Find ⓐ the amount of mark-up and ⓑ the list price: The Auto Resale Store bought Pablo’s Toyota for $8,500. They marked the price up 35%.

Solution

ⓐ $2,975 ⓑ $11,475

Solve Simple Interest Applications

Interest is a part of our daily lives. From the interest earned on our savings to the interest we pay on a car loan or credit card debt, we all have some experience with interest in our lives.

The amount of money you initially deposit into a bank is called the principal, P, and the bank pays you interest, I. When you take out a loan, you pay interest on the amount you borrow, also called the principal.

In either case, the interest is computed as a certain percent of the principal, called the rate of interest, r. The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable t, (for time) represents the number of years the money is saved or borrowed.

Interest is calculated as simple interest or compound interest. Here we will use simple interest.

Simple Interest

If an amount of money, P, called the principal, is invested or borrowed for a period of t years at an annual interest rate r, the amount of interest, I, earned or paid is

I=interestI=PrtwhereP=principalr=ratet=time

Interest earned or paid according to this formula is called simple interest.

The formula we use to calculate interest is I=Prt. To use the formula we substitute in the values for variables that are given, and then solve for the unknown variable. It may be helpful to organize the information in a chart.

Areli invested a principal of $950 in her bank account that earned simple interest at an interest rate of 3%. How much interest did she earn in five years?

Solution

I=?P=$950r=3%t=5years

This table outlines the step-by-step process for calculating simple interest, including the formula, substitution, simplification, and final answer.
Identify what you are asked to find, and choose a variable to represent it. What is the simple interest?
LetI=interest.
Write the formula. I=Prt
Substitute in the given information. I=(950)(0.03)(5)
Simplify. I=142.5
Check.
Is $142.50 a reasonable amount of interest on $950?
Yes.
Write a complete sentence. The interest is $142.50.

Nathaly deposited $12,500 in her bank account where it will earn 4% simple interest. How much interest will Nathaly earn in five years?

Solution

He will earn $2,500.

Susana invested a principal of $36,000 in her bank account that earned simple interest at an interest rate of 6.5%. How much interest did she earn in three years?

Solution

She earned $7,020.

There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we do not know the rate.

Hang borrowed $7,500 from her parents to pay her tuition. In five years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of simple interest?

Solution

I=$1500P=$7500r=?t=5years

Step-by-step guide and mathematical solution for calculating the simple interest rate.
Identify what you are asked to find, and choose a variable to represent it. What is the rate of simple interest?
Letr=rate of interest.
Write the formula.
Substitute in the given information.
Multiply.
Divide.
Change to percent form.
I=Prt 1,500=(7,500)r(5) 1,500=37,500r 0.04=r 4%=r
Check.
I=Prt1,500=?(7,500)(0.04)(5)1,500=1,500✓
Write a complete sentence. The rate of interest was 4%.

Jim lent his sister $5,000 to help her buy a house. In three years, she paid him the $5,000, plus $900 interest. What was the rate of simple interest?

Solution

The rate of simple interest was 6%.

Loren lent his brother $3,000 to help him buy a car. In four years, his brother paid him back the $3,000 plus $660 in interest. What was the rate of simple interest?

Solution

The rate of simple interest was 5.5%.

In the next example, we are asked to find the principal—the amount borrowed.

Sean’s new car loan statement said he would pay $4,866.25 in interest from a simple interest rate of 8.5% over five years. How much did he borrow to buy his new car?

Solution

I=4,866.25P=?r=8.5%t=5years

This table outlines the steps and mathematical work involved in calculating the principal amount using the simple interest formula.
Identify what you are asked to find, and choose a variable to represent it. What is the amount borrowed (the principal)?
LetP=principal borrowed.
Write the formula.
Substitute in the given information.
Multiply.
Divide.
I=Prt4,866.25=P(0.085)(5)4,866.25=0.425P11,450=P
Check.
I=Prt4,866.25=?(11,450)(0.085)(5)4,866.25=4,866.25✓
Write a complete sentence. The principal was $11,450.

Eduardo noticed that his new car loan papers stated that with a 7.5% simple interest rate, he would pay $6,596.25 in interest over five years. How much did he borrow to pay for his car?

Solution

He paid $17,590.

In five years, Gloria’s bank account earned $2,400 interest at 5% simple interest. How much had she deposited in the account?

Solution

She deposited $9,600.

Access this online resource for additional instruction and practice with using a problem solving strategy.

  • Begining Arithmetic Problems

Key Concepts

  • How To Use a Problem Solving Strategy for Word Problems
    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what you are looking for.
    3. Name what you are looking for. Choose a variable to represent that quantity.
    4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
    5. Solve the equation using proper algebra techniques.



    6. Check the answer in the problem to make sure it makes sense.
    7. Answer the question with a complete sentence.
  • How To Find Percent Change
    1. Find the amount of change
      change=new amount−original amount
    2. Find what percent the amount of change is of the original amount.
      change is what percent of the original amount?
  • Discount
    amount of discount=discount rate·original pricesale price=original amount−discount
  • Mark-up
    amount of mark-up=mark-up rate·original costlist price=original cost+mark up
  • Simple Interest
    If an amount of money, P, called the principal, is invested or borrowed for a period of t years at an annual interest rate r, the amount of interest, I, earned or paid is:
    I=interestI=PrtwhereP=principalr=ratet=time

Practice Makes Perfect

Use a Problem Solving Strategy for Word Problems

List five positive thoughts you can say to yourself that will help you approach word problems with a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often.

Solution

Answers will vary.

List five negative thoughts that you have said to yourself in the past that will hinder your progress on word problems. You may want to write each one on a small piece of paper and rip it up to symbolically destroy the negative thoughts.

In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question.

There are 16 girls in a school club. The number of girls is four more than twice the number of boys. Find the number of boys.

Solution

six boys

There are 18 Cub Scouts in Troop 645. The number of scouts is three more than five times the number of adult leaders. Find the number of adult leaders.

Huong is organizing paperback and hardback books for her club’s used book sale. The number of paperbacks is 12 less than three times the number of hardbacks. Huong had 162 paperbacks. How many hardback books were there?

Solution

58 hardback books

Jeff is lining up children’s and adult bicycles at the bike shop where he works. The number of children’s bicycles is nine less than three times the number of adult bicycles. There are 42 adult bicycles. How many children’s bicycles are there?

Solve Number Word Problems

In the following exercises, solve each number word problem.

The difference of a number and 12 is three. Find the number.

Solution

15

The difference of a number and eight is four. Find the number.

The sum of three times a number and eight is 23. Find the number.

Solution

5

The sum of twice a number and six is 14. Find the number.

The difference of twice a number and seven is 17. Find the number.

Solution

12

The difference of four times a number and seven is 21. Find the number.

Three times the sum of a number and nine is 12. Find the number.

Solution

−5

Six times the sum of a number and eight is 30. Find the number.

One number is six more than the other. Their sum is 42. Find the numbers.

Solution

18, 24

One number is five more than the other. Their sum is 33. Find the numbers.

The sum of two numbers is 20. One number is four less than the other. Find the numbers.

Solution

8, 12

The sum of two numbers is 27. One number is seven less than the other. Find the numbers.

One number is 14 less than another. If their sum is increased by seven, the result is 85. Find the numbers.

Solution

32, 46

One number is 11 less than another. If their sum is increased by eight, the result is 71. Find the numbers.

The sum of two numbers is 14. One number is two less than three times the other. Find the numbers.

Solution

4, 10

The sum of two numbers is zero. One number is nine less than twice the other. Find the numbers.

The sum of two consecutive integers is 77. Find the integers.

Solution

38, 39

The sum of two consecutive integers is 89. Find the integers.

The sum of three consecutive integers is 78. Find the integers.

Solution

25, 26, 27

The sum of three consecutive integers is 60. Find the integers.

Find three consecutive integers whose sum is −36.

Solution

−11,−12,−13

Find three consecutive integers whose sum is −3.

Find three consecutive even integers whose sum is 258.

Solution

84, 86, 88

Find three consecutive even integers whose sum is 222.

Find three consecutive odd integers whose sum is −213.

Solution

−69,−71,−73

Find three consecutive odd integers whose sum is −267.

Philip pays $1,620 in rent every month. This amount is $120 more than twice what his brother Paul pays for rent. How much does Paul pay for rent?

Solution

$750

Marc just bought an SUV for $54,000. This is $7,400 less than twice what his wife paid for her car last year. How much did his wife pay for her car?

Laurie has $46,000 invested in stocks and bonds. The amount invested in stocks is $8,000 less than three times the amount invested in bonds. How much does Laurie have invested in bonds?

Solution

$13,500

Erica earned a total of $50,450 last year from her two jobs. The amount she earned from her job at the store was $1,250 more than three times the amount she earned from her job at the college. How much did she earn from her job at the college?

Solve Percent Applications

In the following exercises, translate and solve.

ⓐ What number is 45% of 120? ⓑ 81 is 75% of what number? ⓐ What percent of 260 is 78?

Solution

ⓐ 54 ⓑ 108 ⓐ 30%

ⓐ What number is 65% of 100? ⓑ 93 is 75% of what number? ⓐ What percent of 215 is 86?

ⓐ 250% of 65 is what number? ⓑ 8.2% of what amount is $2.87? ⓐ 30 is what percent of 20?

Solution

ⓐ 162.5 ⓑ $35 ⓐ 150%

ⓐ 150% of 90 is what number? ⓑ 6.4% of what amount is $2.88? ⓐ 50 is what percent of 40?

In the following exercises, solve.

Geneva treated her parents to dinner at their favorite restaurant. The bill was $74.25. Geneva wants to leave 16% of the total bill as a tip. How much should the tip be?

Solution

$11.88

When Hiro and his co-workers had lunch at a restaurant near their work, the bill was $90.50. They want to leave 18% of the total bill as a tip. How much should the tip be?

One serving of oatmeal has 8 grams of fiber, which is 33% of the recommended daily amount. What is the total recommended daily amount of fiber?

Solution

24.2 g

One serving of trail mix has 67 grams of carbohydrates, which is 22% of the recommended daily amount. What is the total recommended daily amount of carbohydrates?

A bacon cheeseburger at a popular fast food restaurant contains 2070 milligrams (mg) of sodium, which is 86% of the recommended daily amount. What is the total recommended daily amount of sodium?

Solution

2407 mg

A grilled chicken salad at a popular fast food restaurant contains 650 milligrams (mg) of sodium, which is 27% of the recommended daily amount. What is the total recommended daily amount of sodium?

The nutrition fact sheet at a fast food restaurant says the fish sandwich has 380 calories, and 171 calories are from fat. What percent of the total calories is from fat?

Solution

45%

The nutrition fact sheet at a fast food restaurant says a small portion of chicken nuggets has 190 calories, and 114 calories are from fat. What percent of the total calories is from fat?

Emma gets paid $3,000 per month. She pays $750 a month for rent. What percent of her monthly pay goes to rent?

Solution

25%

Dimple gets paid $3,200 per month. She pays $960 a month for rent. What percent of her monthly pay goes to rent?

In the following exercises, solve.

Tamanika received a raise in her hourly pay, from $15.50 to $17.36. Find the percent change.

Solution

12%

Ayodele received a raise in her hourly pay, from $24.50 to $25.48. Find the percent change.

Annual student fees at the University of California rose from about $4,000 in 2000 to about $12,000 in 2010. Find the percent change.

Solution

200%

The price of a share of one stock rose from $12.50 to $50. Find the percent change.

A grocery store reduced the price of a loaf of bread from $2.80 to $2.73. Find the percent change.

Solution

−2.5%

The price of a share of one stock fell from $8.75 to $8.54. Find the percent change.

Hernando’s salary was $49,500 last year. This year his salary was cut to $44,055. Find the percent change.

Solution

−11%

In ten years, the population of Detroit fell from 950,000 to about 712,500. Find the percent change.

In the following exercises, find ⓐ the amount of discount and ⓑ the sale price.

Janelle bought a beach chair on sale at 60% off. The original price was $44.95.

Solution

ⓐ $26.97 ⓑ $17.98

Errol bought a skateboard helmet on sale at 40% off. The original price was $49.95.

In the following exercises, find ⓐ the amount of discount and ⓑ the discount rate (Round to the nearest tenth of a percent if needed.)

Larry and Donna bought a sofa at the sale price of $1,344. The original price of the sofa was $1,920.

Solution

ⓐ $576 ⓑ 30%

Hiroshi bought a lawnmower at the sale price of $240. The original price of the lawnmower is $300.

In the following exercises, find ⓐ the amount of the mark-up and ⓑ the list price.

Daria bought a bracelet at original cost $16 to sell in her handicraft store. She marked the price up 45%. What was the list price of the bracelet?

Solution

ⓐ $7.20 ⓑ $23.20

Regina bought a handmade quilt at original cost $120 to sell in her quilt store. She marked the price up 55%. What was the list price of the quilt?

Tom paid $0.60 a pound for tomatoes to sell at his produce store. He added a 33% mark-up. What price did he charge his customers for the tomatoes?

Solution

ⓐ $0.20 ⓑ $0.80

Flora paid her supplier $0.74 a stem for roses to sell at her flower shop. She added an 85% mark-up. What price did she charge her customers for the roses?

Solve Simple Interest Applications

In the following exercises, solve.

Casey deposited $1,450 in a bank account that earned simple interest at an interest rate of 4%. How much interest was earned in two years?

Solution

$116

Terrence deposited $5,720 in a bank account that earned simple interest at an interest rate of 6%. How much interest was earned in four years?

Robin deposited $31,000 in a bank account that earned simple interest at an interest rate of 5.2%. How much interest was earned in three years?

Solution

$4836

Carleen deposited $16,400 in a bank account that earned simple interest at an interest rate of 3.9% How much interest was earned in eight years?

Hilaria borrowed $8,000 from her grandfather to pay for college. Five years later, she paid him back the $8,000, plus $1,200 interest. What was the rate of simple interest?

Solution

3%

Kenneth lent his niece $1,200 to buy a computer. Two years later, she paid him back the $1,200, plus $96 interest. What was the rate of simple interest?

Lebron lent his daughter $20,000 to help her buy a condominium. When she sold the condominium four years later, she paid him the $20,000, plus $3,000 interest. What was the rate of simple interest?

Solution

3.75%

Pablo borrowed $50,000 to start a business. Three years later, he repaid the $50,000, plus $9,375 interest. What was the rate of simple interest?

In 10 years, a bank account that paid 5.25% simple interest earned $18,375 interest. What was the principal of the account?

Solution

$35,000

In 25 years, a bond that paid 4.75% simple interest earned $2,375 interest. What was the principal of the bond?

Joshua’s computer loan statement said he would pay $1,244.34 in simple interest for a three-year loan at 12.4%. How much did Joshua borrow to buy the computer?

Solution

$3345

Margaret’s car loan statement said she would pay $7,683.20 in simple interest for a five-year loan at 9.8%. How much did Margaret borrow to buy the car?

Everyday Math

Tipping At the campus coffee cart, a medium coffee costs $1.65. MaryAnne brings $2.00 with her when she buys a cup of coffee and leaves the change as a tip. What percent tip does she leave?

Solution

21.2%

Tipping Four friends went out to lunch and the bill came to $53.75 They decided to add enough tip to make a total of $64, so that they could easily split the bill evenly among themselves. What percent tip did they leave?

Writing Exercises

What has been your past experience solving word problems? Where do you see yourself moving forward?

Solution

Answers will vary.

Without solving the problem “44 is 80% of what number” think about what the solution might be. Should it be a number that is greater than 44 or less than 44? Explain your reasoning.

After returning from vacation, Alex said he should have packed 50% fewer shorts and 200% more shirts. Explain what Alex meant.

Solution

Answers will vary.

Because of road construction in one city, commuters were advised to plan that their Monday morning commute would take 150% of their usual commuting time. Explain what this means.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objective of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was use a problem-solving strategy for word problems. In row 3, the I can was solve number problems. In row 4, the I can was solve percent applications. In row 5, the I can was solve simple interest applications.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Solve a Formula for a Specific Variable

Learning Objectives

By the end of this section, you will be able to:

  • Solve a formula for a specific variable
  • Use formulas to solve geometry applications

Before you get started, take this readiness quiz.

Evaluate 2(x+3) when x=5.
If you missed this problem, review Example 6 in Use the Language of Algebra.

Solution

16

The length of a rectangle is three less than the width. Let w represent the width. Write an expression for the length of the rectangle.
If you missed this problem, review Example 10 in Use the Language of Algebra.

Solution

w−3

Evaluate 12bh when b=14 and h=9.
If you missed this problem, review Example 10 in Fractions.

Solution

63

Solve a Formula for a Specific Variable

We have all probably worked with some geometric formulas in our study of mathematics. Formulas are used in so many fields, it is important to recognize formulas and be able to manipulate them easily.

It is often helpful to solve a formula for a specific variable. If you need to put a formula in a spreadsheet, it is not unusual to have to solve it for a specific variable first. We isolate that variable on one side of the equals sign with a coefficient of one and all other variables and constants are on the other side of the equal sign.

Geometric formulas often need to be solved for another variable, too. The formula V=13πr2h is used to find the volume of a right circular cone when given the radius of the base and height. In the next example, we will solve this formula for the height.

Solve the formula V=13πr2h for h.

Solution
Write the formula. The image displays the mathematical formula for the volume of a cone, which is V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height of the cone.
Remove the fraction on the right. A mathematical equation displays 3 multiplied by V (volume) on the left side, equaling 3 multiplied by (1/3)πr²h (volume of a cone formula) on the right side, highlighting the multiplication by 3.
Simplify. Mathematical equation showing a cone's volume formula: 3V = pi times r squared times h. V represents volume, r is the radius, and h is the height.
Divide both sides by πr2. A mathematical equation shows h equals 3V divided by pi r squared, which is the formula for the height of a cone or pyramid given its volume and radius.

We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula h=3Vπr2.

Use the formula A=12bh to solve for b.

Solution

b=2Ah

Use the formula A=12bh to solve for h.

Solution

h=2Ab

In the sciences, we often need to change temperature from Fahrenheit to Celsius or vice versa. If you travel in a foreign country, you may want to change the Celsius temperature to the more familiar Fahrenheit temperature.

Solve the formula C=59(F−32) for F.

Solution
Write the formula. A mathematical formula for converting temperature from Fahrenheit (F) to Celsius (C) is displayed: C = 5/9(F - 32).
Remove the fraction on the right. An equation (9/5)C = (9/5) * (5/9)(F - 32), showing a step in temperature conversion. Red numbers highlight fractions being multiplied on both sides.
Simplify. A mathematical equation showing the conversion formula from Fahrenheit to Celsius, specifically (9/5)C = (F - 32). The equation is presented in a clear, typeset format against a white background.
Add 32 to both sides. A mathematical formula for converting Celsius to Fahrenheit, which reads as '9/5 C + 32 = F'.

We can now use the formula F=95C+32 to find the Fahrenheit temperature when we know the Celsius temperature.

Solve the formula F=95C+32 for C.

Solution

C=59(F−32)

Solve the formula A=12h(b+B) for b.

Solution

b=2A−Bhh

The next example uses the formula for the surface area of a right cylinder.

Solve the formula S=2πr2+2πrh for h.

Solution
Write the formula. A mathematical formula is displayed: S = 2πr² + 2πrh, representing the surface area of a cylinder with radius 'r' and height 'h'.
Isolate the h term by subtracting 2πr2 from each side. A mathematical equation is displayed, showing S minus two pi r squared equals two pi r squared minus two pi r squared plus two pi r h, written in black and red text on a white background.
Simplify. Mathematical formula showing the relationship between total surface area (S), radius (r), height (h), and pi (π) for a cylinder: S - 2πr² = 2πrh.
Solve for h by dividing both sides by 2πr. Mathematical equation: (S - 2πr^2) / (2πr) = (2πrh) / (2πr), an intermediate step in calculating the height of a cylinder from its surface area.
Simplify. A mathematical equation showing h = (S - 2πr^2) / (2πr), likely representing the height (h) of a cylinder derived from its total surface area (S) and radius (r).

Solve the formula A=P+Prt for t.

Solution

t=A−PPr

Solve the formula A=P+Prt for r.

Solution

r=A−PPt

Sometimes we might be given an equation that is solved for y and need to solve it for x, or vice versa. In the following example, we’re given an equation with both x and y on the same side and we’ll solve it for y.

Solve the formula 8x+7y=15 for y.

Solution
We will isolate y on one side of the equation. A mathematical equation is displayed with a white background, reading '8x + 7y = 15'. The numbers and variables are in a dark gray font.
Subtract 8x from both sides to isolate the term with y. An equation is shown: 8x - 8x + 7y = 15 - 8x, demonstrating the step of subtracting 8x from both sides to isolate the 'y' term for simplification.
Simplify. A mathematical equation is displayed, showing 7y = 15 - 8x. This linear equation relates variables 'y' and 'x' with constant terms.
Divide both sides by 7 to make the coefficient of y one. An equation showing 7y divided by 7 on the left, and (15 - 8x) divided by 7 on the right. The number 7 in the denominator is red on both sides.
Simplify. A mathematical equation shows y equals the fraction (15 minus 8x) all over 7. It is written as y = (15 - 8x) / 7.

Solve the formula 4x+7y=9 for y.

Solution

y=9−4x7

Solve the formula 5x+8y=1 for y.

Solution

y=1−5x8

Use Formulas to Solve Geometry Applications

In this objective we will use some common geometry formulas. We will adapt our problem solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

Solve geometry applications.

  1. Read the problem and make sure all the words and ideas are understood.
  2. Identify what you are looking for.
  3. Name what we are looking for by choosing a variable to represent it. Draw the figure and label it with the given information.
  4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

When we solve geometry applications, we often have to use some of the properties of the figures. We will review those properties as needed.

The next example involves the area of a triangle. The area of a triangle is one-half the base times the height. We can write this as A=12bh, where b = length of the base and h = height.

The figure is a triangle with its height shown. Its base is b and its height is h. The formula for the area of the triangle is A is equal to one-half times b times h.

The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. height of a triangle
Step 3. Name.
Choose a variable to represent it. Let h= the height.
Draw the figure and label it with the given information. Area = 126 sq. in.
A white equilateral triangle is depicted with a dashed line representing its height, labeled 'h', extending from the top vertex perpendicularly to the base. The base of the triangle is labeled '18 in'.
Step 4. Translate.
Write the appropriate formula. A=12bh
Substitute in the given information. 126=12·18·h
Step 5. Solve the equation. 126=9h
Divide both sides by 9. 14=h
Step 6. Check.

A=12bh126=?12·18·14126=126✓
Step 7. Answer the question. The height of the triangle is 14 inches.

The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?

Solution

The window’s height is 12 meters.

A triangular tent door has an area of 15 square feet. The height is five feet. What is the length of the base?

Solution

The length of the base is 6 feet.

In the next example, we will work with a right triangle. To solve for the measure of each angle, we need to use two triangle properties. In any triangle, the sum of the measures of the angles is 180°. We can write this as a formula: m∠A+m∠B+m∠C=180. Also, since the triangle is a right triangle, we remember that a right triangle has one 90° angle.

Here, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.

The measure of one angle of a right triangle is 40 degrees more than the measure of the smallest angle. Find the measures of all three angles.

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. the measures of all three angles
Step 3. Name. Choose a variable to represent it. Leta=1stangle.a+40=2ndangle90=3rdangle (the right angle)
Draw the figure and label it with the given information. A right-angled triangle ABC is shown, with the right angle at C. Angle B is labeled 'a' and angle A is labeled 'a + 40'.
Step 4. Translate.
Write the appropriate formula. A mathematical equation displays the sum of the measures of angles A, B, and C as 180 degrees, represented as m∠A + m∠B + m∠C = 180, on a white background.
Substitute into the formula. A mathematical equation is displayed, showing 'a + (a + 40) + 90 = 180' in bold, dark gray text on a white background.
Step 5. Solve the equation. A math problem demonstrating the steps to find three angles. It begins with 2a + 130 = 180, solves for 'a' (25 for the first angle), then calculates the second angle (a+40=65), and lists the third angle as 90.
Step 6. Check.

25+65+90=?180180=180✓
Step 7. Answer the question. The three angles measure 25°,65°, and 90°.

The measure of one angle of a right triangle is 50 more than the measure of the smallest angle. Find the measures of all three angles.

Solution

The measures of the angles are 20°, 70°, and 90°.

The measure of one angle of a right triangle is 30 more than the measure of the smallest angle. Find the measures of all three angles.

Solution

The measures of the angles are 30°, 60°, and 90°.

The next example uses another important geometry formula. The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. Writing the formula in every exercise and saying it aloud as you write it may help you memorize the Pythagorean Theorem.

The Pythagorean Theorem

In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.

The figure is a right triangle with sides a and b, and a hypotenuse c. a squared plus b squared is equal to c squared. In a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.

We will use the Pythagorean Theorem in the next example.

Use the Pythagorean Theorem to find the length of the other leg in

This figure is a right triangle with one leg that is 12 units and a hypotenuse that is 13 units.
Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. the length of the leg of the triangle
Step 3. Name.
Choose a variable to represent it. Let a = the leg of the triangle.
Label side a. A right-angled triangle with one leg labeled 'a', the other leg labeled '12', and the hypotenuse labeled '13'. A square symbol indicates the right angle.
Step 4. Translate.
Write the appropriate formula.
Substitute.
a2+b2=c2a2+122=132
Step 5. Solve the equation.
Isolate the variable term.
Use the definition of square root.
Simplify.
a2+144=169a2=25a=25a=5
Step 6. Check.
A clear step-by-step verification of the Pythagorean theorem using the 5-12-13 triangle, demonstrating that 5 squared plus 12 squared equals 13 squared, resulting in 169 = 169.
Step 7. Answer the question. The length of the leg is 5.

Use the Pythagorean Theorem to find the length of the leg in the figure.

The figure is a right triangle with legs that are b units and 15 units, and a hypotenuse that is 17 units.
Solution

The length of the leg is 8.

Use the Pythagorean Theorem to find the length of the leg in the figure.

The figure is a right triangle with legs that are b units and 9 units, and a hypotenuse that is 15 units.
Solution

The length of the leg is 12.

The next example is about the perimeter of a rectangle. Since the perimeter is just the distance around the rectangle, we find the sum of the lengths of its four sides—the sum of two lengths and two widths. We can write is as P=2L+2W where L is the length and W is the width. To solve the example, we will need to define the length in terms of the width.

The length of a rectangle is six centimeters more than twice the width. The perimeter is 96 centimeters. Find the length and width.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. the length and the width
Step 3. Name. Choose a variable to represent the width.
The length is six more than twice the width.
Let w= width.
2w+6= length

A geometric illustration features a rectangle. The width of the rectangle is labeled as 'W', and its length is labeled with the expression '2W + 6', indicating the relationship between its dimensions.

P=96 cm
Step 4. Translate.
Write the appropriate formula. The formula P = 2L + 2W is displayed, representing the perimeter of a rectangle, where P is perimeter, L is length, and W is width. This fundamental equation is a cornerstone of geometry.
Substitute in the given information. A mathematical equation is displayed, showing 96 = 2(2W + 6) + 2W, which is an algebraic expression likely intended for solving for the variable W.
Step 5. Solve the equation. This image illustrates an algebraic solution for finding the width (W=14) and subsequent length (34 cm) of an object, starting from a perimeter equation and showing each calculation step.
Step 6. Check.

A rectangle is shown with a height of 14 cm and a width of 34 cm, indicating its dimensions.

P=2L+2W96=?2·34+2·1496=96✓
Step 7. Answer the question. The length is 34 cm and the width is 14 cm.

The length of a rectangle is seven more than twice the width. The perimeter is 110 inches. Find the length and width.

Solution

The length is 39 inches and the width is 16 inches.

The width of a rectangle is eight yards less than twice the length. The perimeter is 86 yards. Find the length and width.

Solution

The length is 17 yards and the width is 26 yards.

The next example is about the perimeter of a triangle. Since the perimeter is just the distance around the triangle, we find the sum of the lengths of its three sides. We can write this as P=a+b+c, where a, b, and c are the lengths of the sides.

One side of a triangle is three inches more than the first side. The third side is two inches more than twice the first. The perimeter is 29 inches. Find the length of the three sides of the triangle.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. the lengths of the three sides of a triangle
Step 3. Name. Choose a variable to
represent the length of the first side.
Letx=length of1stside.x+3=length of2ndside2x+2=length of3rdside
A triangle with side lengths x, x+3, and 2x+2, with the text 'Perimeter is 29 in.' displayed next to it.
Step 4. Translate.
Write the appropriate formula.
Substitute in the given information.
A mathematical formula is shown, P = a + b + c, typically representing the perimeter of a shape or a sum of three variables.
A mathematical equation is displayed, showing 29 equal to the sum of x, (x + 3), and (2x + 2).
Step 5. Solve the equation. Algebraic problem solution: solving 29 = 4x + 5 for x, then calculating side lengths x+3 and 2x+2 by substituting the value of x.
Step 6. Check.
A scalene triangle with side lengths labeled as 6, 9, and 14.
29=?6+9+14
29=29✓
Step 7. Answer the question. The lengths of the sides of the triangle
are 6, 9, and 14 inches.

One side of a triangle is seven inches more than the first side. The third side is four inches less than three times the first. The perimeter is 28 inches. Find the length of the three sides of the triangle.

Solution

The lengths of the sides of the triangle are 5, 11 and 12 inches.

One side of a triangle is three feet less than the first side. The third side is five feet less than twice the first. The perimeter is 20 feet. Find the length of the three sides of the triangle.

Solution

The lengths of the sides of the triangle are 4, 7 and 9 feet.

The perimeter of a rectangular soccer field is 360 feet. The length is 40 feet more than the width. Find the length and width.

The figure is an illustration of rectangular soccer field.
Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. the length and width of the soccer field
Step 3. Name. Choose a variable to represent it.
The length is 40 feet more than the width.
Draw the figure and label it with the
given information.
Let w = width.
w+40= length
A soccer field diagram illustrates a perimeter problem. The field has a perimeter of 360 feet, with one side labeled 'W' and the other 'W + 40', setting up an algebraic equation.
Step 4. Translate.
Write the appropriate formula and
substitute.
The mathematical formula P = 2L + 2W, which represents the perimeter of a rectangle where P is the perimeter, L is the length, and W is the width, is displayed on a white background.
A mathematical equation is displayed, stating 360 equals 2 multiplied by the quantity of w plus 40, plus 2 multiplied by w. The equation is '360 = 2(w + 40) + 2w'.
Step 5. Solve the equation. Calculation showing the steps to determine the width (70) and length (110) of a field, derived from the equation 360 = 2w + 80 + 2w.
Step 6. Check.

P=2L+2W360=?2(110)+2(70)360=360✓
Step 7. Answer the question. The length of the soccer field is 110 feet
and the width is 70 feet.

The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.

Solution

The length of the swimming pool is 70 feet and the width is 30 feet.

The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.

Solution

The length of the garden is 90 yards and the width is 60 yards.

Applications of these geometric properties can be found in many everyday situations as shown in the next example.

Kelvin is building a gazebo and wants to brace each corner by placing a 10” piece of wood diagonally as shown.

The figure is an illustration of a gazebo whose corner forms a right triangle with a 10 inch piece of wood that is placed diagonally to brace it.

How far from the corner should he fasten the wood if wants the distances from the corner to be equal? Approximate to the nearest tenth of an inch.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. the distance from the corner that the
bracket should be attached
Step 3. Name. Choose a variable to represent it.
Draw the figure and label it with the given
information.
Let x= the distance from the corner.
A right-angled triangle with two equal sides labeled 'x' and a hypotenuse of 10. This diagram is used to illustrate the Pythagorean theorem or an isosceles right triangle.
Step 4. Translate.
Write the appropriate formula and substitute.

a2+b2=c2
x2+x2=102
Step 5. Solve the equation.
Isolate the variable.
Use the definition of square root.
Simplify. Approximate to the nearest tenth.
2x2=100x2=50x=50x≈7.1
Step 6. Check.
a2+b2=c2(7.1)2+(7.1)2≈102Yes.
Step 7. Answer the question. Kelvin should fasten each piece of wood
approximately 7.1” from the corner.

John puts the base of a 13-foot ladder five feet from the wall of his house as shown in the figure. How far up the wall does the ladder reach?

The figure is an illustration that shows a ladder placed against the wall of a house. The ladder forms a right triangle with the side of the house. The ladder is 13 feet long and the base of the ladder is 5 feet from the wall of the house.
Solution

The ladder reaches 12 feet.

Randy wants to attach a 17-foot string of lights to the top of the 15 foot mast of his sailboat, as shown in the figure. How far from the base of the mast should he attach the end of the light string?

The figure is an illustration of a sailboat that has a 15 foot mast. A string of lights that are 17 feet long are placed diagonally from the top of the mast.
Solution

He should attach the lights 8 feet from the base of the mast.

Access this online resource for additional instruction and practice with solving for a variable in literal equations.

  • Solving Literal Equations

Key Concepts

  • How To Solve Geometry Applications
    1. Read the problem and make sure all the words and ideas are understood.
    2. Identify what you are looking for.
    3. Name what you are looking for by choosing a variable to represent it. Draw the figure and label it with the given information.
    4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.
  • The Pythagorean Theorem
    • In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.
      The figure is a right triangle with sides a and b, and a hypotenuse c with the formula, a squared plus b squared is equal to c squared.

Practice Makes Perfect

Solve a Formula for a Specific Variable

In the following exercises, solve the given formula for the specified variable.

Solve the formula C=πd for d.

Solution

d=Cπ

Solve the formula C=πd for π.

Solve the formula V=LWH for L.

Solution

L=VWH

Solve the formula V=LWH for H.

Solve the formula A=12bh for b.

Solution

b=2Ah

Solve the formula A=12bh for h.

Solve the formula
A=12d1d2 for d1.

Solution

d1=2Ad2

Solve the formula
A=12d1d2 for d2.

Solve the formula
A=12h(b1+b2) for b1.

Solution

b1=2Ah−b2

Solve the formula
A=12h(b1+b2) for b2.

Solve the formula
h=54t+12at2 for a.

Solution

a=2h−108tt2

Solve the formula
h=48t+12at2 for a.

Solve 180=a+b+c for a.

Solution

a=180−b−c

Solve 180=a+b+c for c.

Solve the formula
A=12pl+B for p.

Solution

p=2A−2Bl

Solve the formula
A=12pl+B for l.

Solve the formula
P=2L+2W for L.

Solution

L=P−2W2

Solve the formula
P=2L+2W for W.

In the following exercises, solve for the formula for y.

Solve the formula
8x+y=15 for y.

Solution

y=15−8x

Solve the formula
9x+y=13 for y.

Solve the formula
−4x+y=−6 for y.

Solution

y=−6+4x

Solve the formula
−5x+y=−1 for y.

Solve the formula
x−y=−4 for y.

Solution

y=4+x

Solve the formula
x−y=−3 for y.

Solve the formula
4x+3y=7 for y.

Solution

y=7−4x3

Solve the formula
3x+2y=11 for y.

Solve the formula
2x+3y=12 for y.

Solution

y=12−2x3

Solve the formula
5x+2y=10 for y.

Solve the formula
3x−2y=18 for y.

Solution

y=18−3x−2

Solve the formula
4x−3y=12 for y.

Use Formulas to Solve Geometry Applications

In the following exercises, solve using a geometry formula.

A triangular flag has area 0.75 square feet and height 1.5 foot. What is its base?

Solution

1 foot

A triangular window has area 24 square feet and height six feet. What is its base?

What is the base of a triangle with area 207 square inches and height 18 inches?

Solution

23 inches

What is the height of a triangle with area 893 square inches and base 38 inches?

The two smaller angles of a right triangle have equal measures. Find the measures of all three angles.

Solution

45°,45°,90°

The measure of the smallest angle of a right triangle is 20° less than the measure of the next larger angle. Find the measures of all three angles.

The angles in a triangle are such that one angle is twice the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

Solution

30°,60°,90°

The angles in a triangle are such that one angle is 20 more than the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

The figure is a right triangle with sides 9 units and 12 units.
Solution

15

The figure is a right triangle with sides 16 units and 12 units.
The figure is a right triangle with sides 15 units and 20 units.
Solution

25

The figure is a right triangle with sides 5 units and 12 units.

In the following exercises, use the Pythagorean Theorem to find the length of the leg. Round to the nearest tenth if necessary.

The figure is a right triangle with sides 6 units and 10 units.
Solution

8

The figure is a right triangle with a side that is 9 units and a hypotenuse that is 13 units.
The figure is a right triangle with a side that is 5 units and a hypotenuse that is 13 units.
Solution

12

The figure is a right triangle with a side that is 16 units and a hypotenuse that is 20 units.
The figure is a right triangle with a side that is 8 units and a hypotenuse that is 13 units.
Solution

10.2

The figure is a right triangle with sides that are both 6 units.
The figure is a right triangle with sides that are 5 units and 11 units.
Solution

9.8

The figure is a right triangle with sides that are 5 units and 7 units.

In the following exercises, solve using a geometry formula.

The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.

Solution

18 meters, 11 meters

The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width.

The width of the rectangle is 0.7 meters less than the length. The perimeter of a rectangle is 52.6 meters. Find the dimensions of the rectangle.

Solution

13.5 m, 12.8 m

The length of the rectangle is 1.1 meters less than the width. The perimeter of a rectangle is 49.4 meters. Find the dimensions of the rectangle.

The perimeter of a rectangle of 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.

Solution

25 ft, 50 ft

The length of the rectangle is three times the width. The perimeter of a rectangle is 72 feet. Find the length and width of the rectangle.

The length of the rectangle is three meters less than twice the width. The perimeter of a rectangle is 36 meters. Find the dimensions of the rectangle.

Solution

l = 11 m, w = 7 m

The length of a rectangle is five inches more than twice the width. The perimeter is 34 inches. Find the length and width.

The perimeter of a triangle is 39 feet. One side of the triangle is one foot longer than the second side. The third side is two feet longer than the second side. Find the length of each side.

Solution

12 ft, 13 ft, 14 ft

The perimeter of a triangle is 35 feet. One side of the triangle is five feet longer than the second side. The third side is three feet longer than the second side. Find the length of each side.

One side of a triangle is twice the smallest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.

Solution

3 ft, 6 ft, 8 ft

One side of a triangle is three times the smallest side. The third side is three feet more than the shortest side. The perimeter is 13 feet. Find the lengths of all three sides.

The perimeter of a rectangular field is 560 yards. The length is 40 yards more than the width. Find the length and width of the field.

Solution

120 yd, 160 yd

The perimeter of a rectangular atrium is 160 feet. The length is 16 feet more than the width. Find the length and width of the atrium.

A rectangular parking lot has perimeter 250 feet. The length is five feet more than twice the width. Find the length and width of the parking lot.

Solution

40 ft, 85 ft

A rectangular rug has perimeter 240 inches. The length is 12 inches more than twice the width. Find the length and width of the rug.

In the following exercises, solve. Approximate answers to the nearest tenth, if necessary.

A 13-foot string of lights will be attached to the top of a 12-foot pole for a holiday display as shown. How far from the base of the pole should the end of the string of lights be anchored?

The figure is an illustration that shows a 13 foot string of lights attached diagonally to the top of a 12 foot pole.
Solution

5 feet

am wants to put a banner across her garage door diagonally, as shown, to congratulate her son for his college graduation. The garage door is 12 feet high and 16 feet wide. Approximately how long should the banner be to fit the garage door?

The figure is an illustration of a banner positioned diagonally across a garage door that is 12 feet high and 16 feet wide.

Chi is planning to put a diagonal path of paving stones through her flower garden as shown. The flower garden is a square with side 10 feet. What will the length of the path be?

The figure is an illustration of a diagonal path of stones through a square garden with 10 foot sides.
Solution

14.1 feet

Brian borrowed a 20-foot extension ladder to use when he paints his house. If he sets the base of the ladder six feet from the house as shown, how far up will the top of the ladder reach?

The figure is an illustration of a house that has a ladder against it. The ladder is 20 feet. Its base is positioned 6 feet from the house.

Everyday Math

Converting temperature While on a tour in Greece, Tatyana saw that the temperature was 40° Celsius. Solve for F in the formula C=59(F−32) to find the Fahrenheit temperature.

Solution

104°F

Converting temperature Yon was visiting the United States and he saw that the temperature in Seattle one day was 50° Fahrenheit. Solve for C in the formula F=95C+32 to find the Celsius temperature.

Christa wants to put a fence around her triangular flowerbed. The sides of the flowerbed are six feet, eight feet and 10 feet. How many feet of fencing will she need to enclose her flowerbed?

Solution

240 ft

Jose just removed the children’s play set from his back yard to make room for a rectangular garden. He wants to put a fence around the garden to keep the dog out. He has a 50-foot roll of fence in his garage that he plans to use. To fit in the backyard, the width of the garden must be 10 feet. How long can he make the other side?

Writing Exercises

If you need to put tile on your kitchen floor, do you need to know the perimeter or the area of the kitchen? Explain your reasoning.

Solution

Answers will vary.

If you need to put a fence around your backyard, do you need to know the perimeter or the area of the backyard? Explain your reasoning.

Look at the two figures below.

A figure of a rectangle with a width that is 2 units and a length that is 8 units and a square with sides that are 4 units.

ⓐ Which figure looks like it has the larger area? Which looks like it has the larger perimeter?
ⓑ Now calculate the area and perimeter of each figure. Which has the larger area? Which has the larger perimeter?
ⓒ Were the results of part (b) the same as your answers in part (a)? Is that surprising to you?

Solution

ⓐ Answers will vary. ⓑ The areas are the same. The 2×8 rectangle has a larger perimeter than the 4×4 square.
ⓒ Answers will vary.

Write a geometry word problem that relates to your life experience, then solve it and explain all your steps.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve a formula for a specific variable. In row 3, the I can was use formulas to solve geometry applications.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Solve Mixture and Uniform Motion Applications

Learning Objectives

By the end of this section, you will be able to:

  • Solve coin word problems
  • Solve ticket and stamp word problems
  • Solve mixture word problems
  • Solve uniform motion applications

Before you get started, take this readiness quiz.

Simplify: 0.25x+0.10(x+4).
If you missed this problem, review Example 8 in Properties of Real Numbers.

Solution

0.35x+0.4

The number of adult tickets is three more than twice the number of children tickets. Let c represent the number of children tickets. Write an expression for the number of adult tickets.
If you missed this problem, review Example 11 in Use the Language of Algebra.

Solution

2c+3

Convert 4.2% to a decimal.
If you missed this problem, review Example 7 in Decimals.

Solution

0.042

Solve Coin Word Problems

Using algebra to find the number of nickels and pennies in a piggy bank may seem silly. You may wonder why we just don’t open the bank and count them. But this type of problem introduces us to some techniques that will be useful as we move forward in our study of mathematics.

A photo of coins; separate stacks of nickels, pennies, quarters, and dimes

If we have a pile of dimes, how would we determine its value? If we count the number of dimes, we’ll know how many we have—the number of dimes. But this does not tell us the value of all the dimes. Say we counted 23 dimes, how much are they worth? Each dime is worth $0.10—that is the value of one dime. To find the total value of the pile of 23 dimes, multiply 23 by $0.10 to get $2.30.

The number of dimes times the value of each dime equals the total value of the dimes.

number·value=totalvalue23·$0.10=$2.30

This method leads to the following model.

Total Value of Coins

For the same type of coin, the total value of a number of coins is found by using the model

number·value=totalvalue
  • number is the number of coins
  • value is the value of each coin
  • total value is the total value of all the coins

If we had several types of coins, we could continue this process for each type of coin, and then we would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.

Jesse has $3.02 worth of pennies and nickels in his piggy bank. The number of nickels is three more than eight times the number of pennies. How many nickels and how many pennies does Jesse have?

Solution
Step 1. Read the problem.
Determine the types of coins involved.
Create a table.
Write in the value of each type of coin.

pennies and nickels

Pennies are worth $0.01.
Nickels are worth $0.05.
Step 2. Identify what we are looking for. the number of pennies and nickels
Step 3. Name. Represent the number of each type of coin using variables.
The number of nickels is defined in terms of the
number of pennies, so start with pennies.
The number of nickels is three more than eight times
the number of pennies.


Let p= number of pennies.

8p+3= number of nickels
In the chart, multiply the number and the value to
get the total value of each type of coin.
A table showing the setup for a coin problem involving pennies and nickels, with 'Type', 'Number', 'Value ($)', and 'Total Value ($)' columns. The total value of all coins is $3.02.
Step 4. Translate. Write the equation by adding the total value of all the types of coins.
A mathematical equation is displayed on a white background: 0.01p + 0.05(8p + 3) = 3.02. It's a linear equation with one variable 'p' and decimal coefficients.
Step 5. Solve the equation. A step-by-step solution to an algebraic equation is displayed. The equation 0.01p + 0.40p + 0.15 = 3.02 is solved, combining like terms and isolating 'p', resulting in p = 7 pennies.
How many nickels? Steps to solve 8p+3 by substituting p=7, showing 8(7)+3 = 59, which is then contextualized as 59 nickels.
Step 6. Check the answer in the problem and make sure it makes sense.
Jesse has 7 pennies and 59 nickels.
Is the total value $3.02?
7(0.01)+59(0.05)=?3.023.02=3.02✓

Jesse has $6.55 worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse have?

Solution

Jess has 41 nickels and 18 quarters.

Elane has $7.00 total in dimes and nickels in her coin jar. The number of dimes that Elane has is seven less than three times the number of nickels. How many of each coin does Elane have?

Solution

Elane has 22 nickels and 59 dimes.

The steps for solving a coin word problem are summarized below.

Solve coin word problems.

  1. Read the problem. Make sure all the words and ideas are understood.
    • Determine the types of coins involved.
    • Create a table to organize the information.
      • Label the columns “type,” “number,” “value,” and “total value.”
      • List the types of coins.
      • Write in the value of each type of coin.
      • Write in the total value of all the coins. This chart has two columns and four rows. The first row is a header and it labels the first column “Type” and the second column “Number times Value in dollars is equal to Total Value in dollars.” The second header column is subdivided into three columns for the “number,” “value,” and “total value.” The total value column has an additional row. The chart is empty.
  2. Identify what you are looking for.
  3. Name what you are looking for. Choose a variable to represent that quantity.
    • Use variable expressions to represent the number of each type of coin and write them in the table.
    • Multiply the number times the value to get the total value of each type of coin.
  4. Translate into an equation.
    • It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.
    • Write the equation by adding the total values of all the types of coins.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Solve Ticket and Stamp Word Problems

Problems involving tickets or stamps are very much like coin problems. Each type of ticket and stamp has a value, just like each type of coin does. So to solve these problems, we will follow the same steps we used to solve coin problems.

Danny paid $15.75 for stamps. The number of 49-cent stamps was five less than three times the number of 35-cent stamps. How many 49-cent stamps and how many 35-cent stamps did Danny buy?

Solution
Step 1. Determine the types of stamps involved. 49-cent stamps and 35-cent stamps
Step 2. Identify we are looking for. the number of 49-cent stamps and the number of 35-cent stamps
Step 3. Write variable expressions to represent the number of each type of stamp. Let x = number of 35-cent stamps.
“The number of 49-cent stamps was five less
than three times the number of 35-cent
stamps.”

3x−5= number of 49-cent stamps
A table displays a stamp problem, detailing the number, value, and total value of 49-cent and 35-cent stamps. The quantities are expressed using 'x', and the combined total value is $15.75.
Step 4. Write the equation from the total values. A mathematical equation is displayed: 0.49(3x - 5) + 0.35x = 15.75. The equation involves multiplication of decimals with a binomial, addition of a decimal with a variable, and an equality to a decimal number.
Step 5. Solve the equation. A step-by-step solution to a linear equation involving decimals. The problem starts with 1.47x - 2.45 + 0.35x = 15.75, simplifies to 1.82x - 2.45 = 15.75, and then to 1.82x = 18.2.
How many 49-cent stamps? Calculation of stamps: x=10 for 35-cent stamps. 49-cent stamps are derived from 3x-5, resulting in 25 after substituting x=10.
Step 6. Check.
10(0.35)+25(0.49)=?15.753.50+12.25=?15.7515.75=15.75✓
 
Step 7. Answer the question with a complete sentence. Danny bought ten 35-cent stamps and twenty-five 49-cent stamps.

Eric paid $19.88 for stamps. The number of 49-cent stamps was eight more than twice the number of 35-cent stamps. How many 49-cent stamps and how many 35-cent stamps did Eric buy?

Solution

Eric bought thirty-two 49-cent stamps and twelve 35-cent stamps.

Kailee paid $14.74 for stamps. The number of 49-cent stamps was four less than three times the number of 20-cent stamps. How many 49-cent stamps and how many 20-cent stamps did Kailee buy?

Solution

Kailee bought twenty-six 49-cent stamps and ten 20-cent stamps.

In most of our examples so far, we have been told that one quantity is four more than twice the other, or something similar. In our next example, we have to relate the quantities in a different way.

Suppose Aniket sold a total of 100 tickets. Each ticket was either an adult ticket or a child ticket. If he sold 20 child tickets, how many adult tickets did he sell?

  Did you say “80”? How did you figure that out? Did you subtract 20 from 100?

If he sold 45 child tickets, how many adult tickets did he sell?

  Did you say “55”? How did you find it? By subtracting 45 from 100?

Now, suppose Aniket sold x child tickets. Then how many adult tickets did he sell? To find out, we would follow the same logic we used above. In each case, we subtracted the number of child tickets from 100 to get the number of adult tickets. We now do the same with x.

We have summarized this in the table.

This table has two columns and four rows. The first row is a header row and it labels each column, “Child tickets” and “Adult tickets.” In row two, the number of child tickets was 20 and the number of adult tickets was 80. In row two, the number of child tickets was 45 and the number of adult tickets was 55. In row three, the number of child tickets was x and the number of adult tickets was 100 minus x.

We will apply this technique in the next example.

A whale-watching ship had 40 paying passengers on board. The total revenue collected from tickets was $1,196. Full-fare passengers paid $32 each and reduced-fare passengers paid $26 each. How many full-fare passengers and how many reduced-fare passengers were on the ship?

Solution
Step 1. Determine the types of tickets involved. full-fare tickets and reduced-fare tickets
Step 2. Identify what we are looking for. the number of full-fare tickets and reduced-fare tickets
Step 3. Name. Represent the number of each type of ticket using variables. Let f = the number of full-fare tickets.
40−f= the number of reduced-fare tickets
We know the total number of tickets sold was 40. This means the number of reduced-fare tickets is 40 less the number of full-fare tickets.
Multiply the number times the value to get the total value of each type of ticket.
A table illustrating the calculation of total value for full-fare and reduced-fare items, with 'f' representing the number of full-fare items, leading to a total value of $1,196.
Step 4. Translate. Write the equation by adding the total values of each type of ticket. A mathematical equation is displayed: 32f + 26(40 - f) = 1,196.
Step 5. Solve the equation. A step-by-step algebraic solution showing the calculation for 'f', which represents the number of full-fare tickets, resulting in f = 26. The equation starts with 32f + 1,040 - 26f = 1,196.
How many reduced-fare? A mathematical step-by-step calculation showing '40 - f', then substituting 'f' with '26' to get '40 - 26', and finally resulting in '14 reduced-fare tickets'.
Step 6. Check the answer.
There were 26 full-fare tickets at $32 each and 14 reduced-fare tickets at $26 each. Is the total value $116?
26·32=83214·26=364——1,196✓
 
Step 7. Answer the question. They sold 26 full-fare and 14 reduced-fare tickets.

During her shift at the museum ticket booth, Leah sold 115 tickets for a total of $1,163. Adult tickets cost $12 and student tickets cost $5. How many adult tickets and how many student tickets did Leah sell?

Solution

84 adult tickets, 31 student tickets

Galen sold 810 tickets for his church’s carnival for a total revenue of $2,820. Children’s tickets cost $3 each and adult tickets cost $5 each. How many children’s tickets and how many adult tickets did he sell?

Solution

615 children’s tickets and 195 adult tickets

Solve Mixture Word Problems

Now we’ll solve some more general applications of the mixture model. In mixture problems, we are often mixing two quantities, such as raisins and nuts, to create a mixture, such as trail mix. In our tables we will have a row for each item to be mixed as well as one for the final mixture.

Henning is mixing raisins and nuts to make 25 pounds of trail mix. Raisins cost $4.50 a pound and nuts cost $8 a pound. If Henning wants his cost for the trail mix to be $6.60 a pound, how many pounds of raisins and how many pounds of nuts should he use?

Solution
Step 1. Determine what is being mixed. The 25 pounds of trail mix will come from mixing raisins and nuts.
Step 2. Identify what we are looking for. the number of pounds of raisins and nuts
Step 3. Represent the number of each type of ticket using variables.

As before, we fill in a chart to organize our information.
We enter the price per pound for each item.
We multiply the number times the value to get the total value.
Let x= number of pounds of raisins.
25−x= number of pounds of nuts
A table detailing the quantities, unit prices, and total values for raisins, nuts, and trail mix, setting up an algebra problem often seen in mixture calculations.
Notice that the last column in the table gives
the information for the total amount of the
mixture.
Step 4. Translate into an equation. The value of the raisins plus the value of the nuts will be
the value of the trail mix.
Step 5. Solve the equation. A mathematical equation is displayed: 4.5x + 8(25 - x) = 2.5(6.60).
  A step-by-step solution to an algebraic equation resulting in x = 10 pounds of raisins. The problem starts with 4.5x + 200 - 8x = 165, simplifies to -3.5x = -35, and concludes with x = 10.
Find the number of pounds of nuts. A mathematical solution showing '25 - x' then '25 - 10' (with 10 in red), leading to '15 pounds of nuts', illustrating variable substitution in a subtraction problem.
Step 6. Check.
4.5(10)+8(15)=?25(6.60)45+120=?165165=165✓
 
Step 7. Answer the question. Henning mixed ten pounds of raisins with 15 pounds of nuts.

Orlando is mixing nuts and cereal squares to make a party mix. Nuts sell for $7 a pound and cereal squares sell for $4 a pound. Orlando wants to make 30 pounds of party mix at a cost of $6.50 a pound, how many pounds of nuts and how many pounds of cereal squares should he use?

Solution

Orlando mixed five pounds of cereal squares and 25 pounds of nuts.

Becca wants to mix fruit juice and soda to make a punch. She can buy fruit juice for $3 a gallon and soda for $4 a gallon. If she wants to make 28 gallons of punch at a cost of $3.25 a gallon, how many gallons of fruit juice and how many gallons of soda should she buy?

Solution

Becca mixed 21 gallons of fruit punch and seven gallons of soda.

Solve Uniform Motion Applications

When you are driving down the interstate using your cruise control, the speed of your car stays the same—it is uniform. We call a problem in which the speed of an object is constant a uniform motion application. We will use the distance, rate, and time formula, D=rt, to compare two scenarios, such as two vehicles travelling at different rates or in opposite directions.

Our problem solving strategies will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the coin, ticket, and stamp applications.

The steps are listed here for easy reference:

Solve a uniform motion application.

  1. Read the problem. Make sure all the words and ideas are understood.
    • Draw a diagram to illustrate what is happening.
    • Create a table to organize the information.
      • Label the columns rate, time, distance.
      • List the two scenarios.
      • Write in the information you know.
    This chart has two columns and four rows. The first row is a header and it labels the second column “Rate times Times is equal to Distance.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The Distance column has an additional row. The chart is empty.
  2. Identify what you are looking for.
  3. Name what you are looking for. Choose a variable to represent that quantity.
    • Complete the chart.
    • Use variable expressions to represent that quantity in each row.
    • Multiply the rate times the time to get the distance.
  4. Translate into an equation.
    • Restate the problem in one sentence with all the important information.
    • Then, translate the sentence into an equation.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne two hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

  • Draw a diagram to illustrate what it happening. Shown below is a sketch of what is happening in the example.

    The figure shows the uniform motion of Dennis’ and Wayne’s ride along the bike path from Riverside Park. The path for Dennis is represented by an arrow labeled “7 miles per hour” and “1.5 hours”. The path for Wayne is represented by a second arrow of the same length and in the same direction labeled “2 hours”. A bracket represents the distance between Riverside Park and the beach.
  • Create a table to organize the information.
    • Label the columns “Rate,” “Time,” and “Distance.”
    • List the two scenarios.
    • Write in the information you know.
      This chart has two columns and three rows. The first row is a header and it labels the second column “Rate in miles per hours times Time in hours is equal to Distance in miles.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The first column is a header and labels the second row “Dennis” and the third row “Wayne.” In row 2, the time is 1.5 hours. In row 3, the time is 2 hours.

Step 2. Identify what you are looking for.

  You are asked to find the speed of both bikers.

  Notice that the distance formula uses the word “rate,” but it is more common to use “speed”

  when we talk about vehicles in everyday English.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

  • Complete the chart
  • Use variable expressions to represent that quantity in each row.
    We are looking for the speed of the bikers. Let’s let r represent Wayne’s speed. Since Dennis’ speed is 7 mph faster, we represent that as r+7
    r+7=Dennis’ speedr=Wayne’s speed
    Fill in the speeds into the chart.
    This chart has two columns and three rows. The first row is a header and it labels the second column “Rate in miles per hours times Time in hours is equal to Distance in miles.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The first column is a header and labels the second row “Dennis” and the third row “Wayne.” In row 2, the rate is the expression r plus 7 and the time is 1.5 hours. In row 3, the rate is r and the time is 2 hours.
  • Multiply the rate times the time to get the distance.
    This chart has two columns and three rows. The first row is a header and it labels the second column “Rate in miles per hours times Time in hours is equal to Distance in miles.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The first column is a header and labels the second row “Dennis” and the third row “Wayne.” In row 2, the rate is the expression r plus 7, the time is 1.5 hours, and the distance is 1.5 times the quantity r plus 7. In row 3, the rate is r, the time is 2 hours, and the distance is 2 r.

Step 4. Translate into an equation.

  • Restate the problem in one sentence with all the important information.
  • Then, translate the sentence into an equation.
    The equation to model this situation will come from the relation between the distances. Look at the diagram we drew above. How is the distance travelled by Dennis related to the distance travelled by Wayne?
    Since both bikers leave from Riverside and travel to the beach, they travel the same distance. So we write:

    The figure shows that the distance travelled by Dennis equals the distance travelled by Wayne, and when translated into an equation, the result is 1.5 times the quantity r plus 7 is equal to 2 r.

Step 5. Solve the equation using algebra techniques.

Now solve this equation. A step-by-step solution to the algebraic equation 1.5(r + 7) = 2r, demonstrating the distributive property, combining like terms, and isolating the variable 'r' to find the solution r = 21.
So Wayne’s speed is 21 mph.
Find Dennis’ speed. Substituting r with 21 in the expression r+7 leads to 21+7, which equals 28, demonstrating variable evaluation.
Dennis’ speed 28 mph.

Step 6. Check the answer in the problem and make sure it makes sense.

Dennis28mph(1.5hours)=42milesWayne21mph(2hours)=42miles✓

Step 7. Answer the question with a complete sentence.

Wayne rode at 21 mph and Dennis rode at 28 mph.

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.

Solution

The speed of the local train is 48 mph and the speed of the express train is 60 mph.

Jeromy can drive from his house in Cleveland to his college in Chicago in 4.5 hours. It takes his mother six hours to make the same drive. Jeromy drives 20 miles per hour faster than his mother. Find Jeromy’s speed and his mother’s speed.

Solution

Jeromy drove at a speed of 80 mph and his mother drove 60 mph.

In Example 5, we had two bikers traveling the same distance. In the next example, two people drive toward each other until they meet.

Carina is driving from her home in Anaheim to Berkeley on the same day her brother is driving from Berkeley to Anaheim, so they decide to meet for lunch along the way in Buttonwillow. The distance from Anaheim to Berkeley is 395 miles. It takes Carina three hours to get to Buttonwillow, while her brother drives four hours to get there. Carina’s average speed is 15 miles per hour faster than her brother’s average speed. Find Carina’s and her brother’s average speeds.

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

  • Draw a diagram to illustrate what it happening. Below shows a sketch of what is happening in the example.
    The figure shows the uniform motion of Carina and her brother using arrows. The arrow for Carina is labeled “3 hours.” The arrow for Carina’s brother is pointed in the opposite direction and is labeled “15 miles per hour” and “4 hours.” Where the arrows meet is labeled “lunch.” The path of Carina and her Brother is represented by a bracket and labeled “410 miles.”
  • Create a table to organize the information.
    • Label the columns rate, time, distance.
    • List the two scenarios.
    • Write in the information you know.
      This chart has two columns and four rows. The first row is a header and it labels the second column “Rate in miles per hours times Time in hours is equal to Distance in miles.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The first column is a header and labels the second row “Carina” and the third row “Brother.” In row 2, the the time is 3 hours. In row 3, the time is 4 hours. In row 4, the distance is 410 miles.

Step 2. Identify what we are looking for.

  We are asked to find the average speeds of Carina and her brother.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

  • Complete the chart.
  • Use variable expressions to represent that quantity in each row.
    We are looking for their average speeds. Let’s let r represent the average speed of Carina's brother. Since Carina’s speed is 15 mph faster, we represent that as r+15.
    Fill in the speeds into the chart.
  • Multiply the rate times the time to get the distance. This chart has two columns and four rows. The first row is a header and it labels the second column “Rate in miles per hours times Time in hours is equal to Distance in miles.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The first column is a header and labels the second row “Carina” and the third row “Brother.” In row 2, the rate is r, the time is 3 hours, and the distance is 3 r. In row 3, the rate is the expression r plus 15, the time is 4 hours, and the distance is 4 times the quantity r plus 15. In row 4, the distance is 410 miles.

Step 4. Translate into an equation.

  • Restate the problem in one sentence with all the important information.
  • Then, translate the sentence into an equation.
    Again, we need to identify a relationship between the distances in order to write an equation. Look at the diagram we created above and notice the relationship between the distance Carina traveled and the distance her brother traveled.
    The distance Carina traveled plus the distance her brother travel must add up to 410 miles. So we write:

    The figure shows that the distance travelled by Carina plus the distance travelled by her brother is equal to 410, and when translated into an equation, the result is 3 r plus 4 time the quantity r plus 15 is equal to 410.

Step 5. Solve the equation using algebra techniques.

Now solve this equation. A step-by-step solution of the algebraic equation 3(r + 15) + 4r = 395, showing the process to simplify and solve for 'r', resulting in r = 50.
So Carina’s brother's speed was 50 mph.
Carina’s speed is r+15. Calculation of r + 15, where r is replaced by 50, showing 50 + 15 = 65. The number 50 is highlighted in red.
Carina’s speed was 65 mph.

Step 6. Check the answer in the problem and make sure it makes sense.
Carina drove65mph(3hours)=195milesHer brother drove50mph(4hours)=200miles—————395miles✓

Step 7. Answer the question with a complete sentence.

Carina drove 65 mph and her brother 50 mph.

Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove one and a half hours while his parents drove one hour to get to the restaurant. Christopher’s average speed was ten miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?

Solution

Christopher’s speed was 50 mph and his parents’ speed was 40 mph.

Ashley goes to college in Minneapolis, 234 miles from her home in Sioux Falls. She wants her parents to bring her more winter clothes, so they decide to meet at a restaurant on the road between Minneapolis and Sioux Falls. Ashley and her parents both drove two hours to the restaurant. Ashley’s average speed was seven miles per hour faster than her parents’ average speed. Find Ashley’s and her parents’ average speed.

Solution

Ashley’s parents drove 55 mph and Ashley drove 62 mph.

As you read the next example, think about the relationship of the distances traveled. Which of the previous two examples is more similar to this situation?

Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at 70 mph and the truck traveling east has an average speed of 60 mph. How long will they travel before they are 325 miles apart?

Solution

Step 1. Read the problem. Make all the words and ideas are understood.

  • Draw a diagram to illustrate what it happening.
    The figure shows the uniform motion of two truck drivers using arrows. The arrow for the truck driver travelling west is labeled “70 miles per hour.” The arrow for truck driver travelling east is pointed in the opposite direction and is labeled “60 miles per hour.” Where the arrows meet is labeled “Rest stop.” The path of truck drivers is represented by a bracket and labeled “325 miles.”
  • Create a table to organize the information.
    • Label the columns rate, time, distance.
    • List the two scenarios.
    • Write in the information you know.

      The chart has two columns and four rows. The first row is a header and it labels the second column “Rate in miles per hour times Time in hours is equal to Distance in miles.” The first column is a header column and it labels the second row “West” and the third row “East” The second header column is subdivided into three columns for the “Rate,” “Time,” and “Distance.” The fourth row only gives the total distance the truck drivers travelled. In row 2, the truck driver travelling west has a rate 70 miles per hour. In row 3, truck driver travelling east has a rate 60 miles per hour. In row 4, the total distance travelled by the truck drivers is 325.

Step 2. Identify what we are looking for.

We are asked to find the amount of time the trucks will travel until they are 325 miles apart.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

  • Complete the chart.
  • Use variable expressions to represent that quantity in each row.
    We are looking for the time travelled. Both trucks will travel the same amount of time.
    Let’s call the time t. Since their speeds are different, they will travel different distances.

  • Multiply the rate times the time to get the distance. The chart has two columns and four rows. The first row is a header and it labels the second column “Rate in miles per hour times Time in hours is equal to Distance in miles.” The first column is a header column and it labels the second row “West” and the third row “East” The second header column is subdivided into three columns for the “Rate,” “Time,” and “Distance.” The fourth row only gives the total distance the truck drivers travelled. In row 2, the truck driver travelling west has a rate 70 miles per hour, a time t hours, and a distance of 70 t. In row 3, truck driver travelling east has a rate 60 miles per hour, a time t, and a distance of 60 t. In row 4, the total distance travelled by the truck drivers is 325.

Step 4. Translate into an equation.

  • Restate the problem in one sentence with all the important information.
  • Then, translate the sentence into an equation.
    We need to find a relation between the distances in order to write an equation. Looking at the diagram, what is the relationship between the distances each of the trucks will travel?
    The distance travelled by the truck going west plus the distance travelled by the truck going east must add up to 325 miles. So we write:

    The figure shows that the distance travelled by the westbound truck plus the distance travelled by the eastbound truck is equal to 325, and when translated into an equation, the result is 70 t plus 60 t is equal to 325.

Step 5. Solve the equation using algebra techniques.

Now solve this equation70t+60t=325130t=325t=2.5

So it will take the trucks 2.5 hours to be 325 miles apart.

Step 6. Check the answer in the problem and make sure it makes sense.

Truck going West70mph(2.5hours)=175milesTruck going East60mph(2.5hours)=150miles—————325miles✓

Step 7. Answer the question with a complete sentence.
It will take the trucks 2.5 hours to be 325 miles apart.

Pierre and Monique leave their home in Portland at the same time. Pierre drives north on the turnpike at a speed of 75 miles per hour while Monique drives south at a speed of 68 miles per hour. How long will it take them to be 429 miles apart?

Solution

Pierre and Monique will be 429 miles apart in 3 hours.

Thanh and Nhat leave their office in Sacramento at the same time. Thanh drives north on I-5 at a speed of 72 miles per hour. Nhat drives south on I-5 at a speed of 76 miles per hour. How long will it take them to be 330 miles apart?

Solution

Thanh and Nhat will be 330 miles apart in 2.2 hours.

It is important to make sure that the units match when we use the distance rate and time formula. For instance, if the rate is in miles per hour, then the time must be in hours.

When Naoko walks to school, it takes her 30 minutes. If she rides her bike, it takes her 15 minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What is her speed walking and her speed riding her bike?

Solution

First, we draw a diagram that represents the situation to help us see what is happening.

The figure shows the uniform motion when Naoko walks to school and when she rides her bike to school. Her walk to school is represented by an arrow labeled “30 minutes.” Her ride to school is represented by a second arrow of the same length and in the same direction labeled “3 miles per hour faster” and “15 minutes.” A bracket represents the distance between her house and the school.

We are asked to find her speed walking and riding her bike. Let’s call her walking speed r. Since her biking speed is three miles per hour faster, we will call that speed r+3. We write the speeds in the chart.

The speed is in miles per hour, so we need to express the times in hours, too, in order for the units to be the same. Remember, 1 hour is 60 minutes. So:

30 minutes is3060or12hour15 minutes is1560or14hour

We write the times in the chart.

Next, we multiply rate times time to fill in the distance column.

This chart has two columns and three rows. The first row is a header and it labels the second column “Rate in miles per hours times Time in hours is equal to Distance in miles.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The first column is a header and labels the second row “Walk” and the third row “Bike.” In row 2, the rate is r, the time is one-half hour, and the distance is one-half r. In row 3, the rate is the expression r plus 3, the time is one-fourth hour, and the distance is one-fourth times the quantity r plus 3.

The equation will come from the fact that the distance from Naoko’s home to her school is the same whether she is walking or riding her bike.
So we say:

An equation stating 'distance walked = distance covered by bike' with each phrase underlined by a light blue bracket.
Translate to an equation. A mathematical equation on a white background showing one-half r equals one-fourth multiplied by the sum of r and 3, written as (1/2)r = (1/4)(r + 3).
Solve this equation. A mathematical equation shows one-half r equals one-fourth multiplied by the sum of r and three. The equation is 1/2r = 1/4(r + 3).
Clear the fractions by multiplying by the LCD of all the fractions in the equation. An algebraic equation is shown, with '8 times 1/2 times r' on the left side and '8 times 1/4 times (r plus 3)' on the right side, separated by an equals sign.
Simplify. Step-by-step algebraic solution for a word problem, calculating walking speed 'r' as 3 mph and biking speed 'r+3' as 6 mph from the equation 4r = 2(r+3).
6
Let’s check if this works.
Walk 3 mph (0.5 hour)=1.5milesBike 6 mph (0.25 hour)=1.5miles

Yes, either way Naoko travels 1.5 miles to school.

Naoko’s walking speed is 3 mph and her speed riding her bike is 6 mph.

Suzy takes 50 minutes to hike uphill from the parking lot to the lookout tower. It takes her 30 minutes to hike back down to the parking lot. Her speed going downhill is 1.2 miles per hour faster than her speed going uphill. Find Suzy’s uphill and downhill speeds.

Solution

Suzy’s speed uphill is 1.8 mph and downhill is three mph.

Llewyn takes 45 minutes to drive his boat upstream from the dock to his favorite fishing spot. It takes him 30 minutes to drive the boat back downstream to the dock. The boat’s speed going downstream is four miles per hour faster than its speed going upstream. Find the boat’s upstream and downstream speeds.

Solution

The boat’s speed upstream is eight mph and downstream is12 mph.

In the distance, rate and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.

Cruz is training to compete in a triathlon. He left his house at 6:00 and ran until 7:30. Then he rode his bike until 9:45. He covered a total distance of 51 miles. His speed when biking was 1.6 times his speed when running. Find Cruz’s biking and running speeds.

Solution

A diagram will help us model this trip.

The figure shows the uniform motion of Cruz’s training for a triathlon. Cruz’s path is represented by an arrow labeled “run” which starts at 6 a m and extends to 7:30 a m and a second arrow labeled “rode” and “1.6 times faster” which starts at 7:30 a m and extends to 9:45 a m. A bracket represents the distance Cruz covers and is labeled “51 miles.”

Next, we create a table to organize the information. We know the total distance is 51 miles. We are looking for the rate of speed for each part of the trip. The rate while biking is 1.6 times the rate of running. If we let r = the rate running, then the rate biking is 1.6r.

The times here are given as clock times. Cruz started from home at 6:00 a.m. and started biking at 7:30 a.m. So he spent 1.5 hours running. Then he biked from 7:30 a.m until 9:45 a.m. So he spent 2.25 hours biking.

Now, we multiply the rates by the times.

The chart has two columns and four rows. The first row is a header and it labels the second column “Rate in miles per hour times Time in hours is equal to Distance in miles.” The first column is a header column and it labels the second row “run” and the third row “bike” The second header column is subdivided into three columns for the “Rate,” “Time,” and “Distance.” The fourth row only gives the total distance covered. In row 2, the rate is r, the time is 1.5 hours, and the distance is 1.5 r. In row 3, the rate is 1.6, the time is 2.25 hours, and the distance is 2.25 times 1.6 r. In row 4, the total distance covered is 51 miles.

By looking at the diagram, we can see that the sum of the distance running and the distance biking is 255 miles.

An equation displays 'distance running + distance biking = 51', indicating the sum of distances covered by running and biking equals 51 units. Underlines beneath 'distance running' and 'distance biking' highlight each term.
Translate to an equation. A mathematical equation is displayed on a white background: 1.5r + 2.25(1.6r) = 51. The variables and numbers are clearly visible in black text.
Solve this equation. A step-by-step solution to a word problem, calculating running and biking speeds. It shows the equation 1.5r + 2.25(1.6r) = 51, solves for r=10 mph running, then finds the biking speed as 1.6(10) = 16 mph.
Check.
Run10mph(1.5hours)=15miBike16mph(2.25hours)=36mi——51mi

Hamilton loves to travel to Las Vegas, 255 miles from his home in Orange County. On his last trip, he left his house at 2:00 p.m. The first part of his trip was on congested city freeways. At 4:00 pm, the traffic cleared and he was able to drive through the desert at a speed 1.75 times as fast as when he drove in the congested area. He arrived in Las Vegas at 6:30 p.m. How fast was he driving during each part of his trip?

Solution

Hamilton drove 40 mph in the city and 70 mph in the desert.

Phuong left home on his bicycle at 10:00. He rode on the flat street until 11:15, then rode uphill until 11:45. He rode a total of 31 miles. His speed riding uphill was 0.6 times his speed on the flat street. Find his speed biking uphill and on the flat street.

Solution

Phuong rode uphill at a speed of 12 mph and on the flat street at 20 mph.

Key Concepts

  • Total Value of Coins
    For the same type of coin, the total value of a number of coins is found by using the model
    number·value=totalvalue
    • number is the number of coins
    • value is the value of each coin
    • total value is the total value of all the coins




  • How to solve coin word problems.
    1. Read the problem. Make sure all the words and ideas are understood.
      Determine the types of coins involved.
      Create a table to organize the information.
        Label the columns “type,” “number,” “value,” “total value.”
        List the types of coins.
        Write in the value of each type of coin.
        Write in the total value of all the coins.
      This chart has two columns and four rows. The first row is a header and it labels the first column “Type” and the second column “Number times Value in dollars is equal to Total Value in dollars.” The second header column is subdivided into three columns for the “number,” “value,” and “total value.” The total value column has an additional row. The chart is empty.
    2. Identify what you are looking for.
    3. Name what you are looking for. Choose a variable to represent that quantity.
      Use variable expressions to represent the number of each type of coin and write them in the table.
      Multiply the number times the value to get the total value of each type of coin.
    4. Translate into an equation.
      It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.
      Write the equation by adding the total values of all the types of coins.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.


    7. Answer the question with a complete sentence.
  • How To Solve a Uniform Motion Application
    1. Read the problem. Make sure all the words and ideas are understood.
      Draw a diagram to illustrate what it happening.
      Create a table to organize the information.
        Label the columns rate, time, distance.
        List the two scenarios.
        Write in the information you know.
      This chart has two columns and four rows. The first row is a header and it labels the second column “Rate times Times is equal to Distance.” The second header column is subdivided into three columns for “Rate,” “Time,” and “Distance.” The Distance column has an additional row. The chart is empty.
    2. Identify what you are looking for.
    3. Name what you are looking for. Choose a variable to represent that quantity.
      Complete the chart.
      Use variable expressions to represent that quantity in each row.
      Multiply the rate times the time to get the distance.
    4. Translate into an equation.
      Restate the problem in one sentence with all the important information.
      Then, translate the sentence into an equation.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.

Practice Makes Perfect

Solve Coin Word Problems

In the following exercises, solve each coin word problem.

Michaela has $2.05 in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?

Solution

nine nickels, 16 dimes

Liliana has $2.10 in nickels and quarters in her backpack. She has 12 more nickels than quarters. How many coins of each type does she have?

In a cash drawer there is $125 in $5 and $10 bills. The number of $10 bills is twice the number of $5 bills. How many of each type of bill is in the drawer?

Solution

ten $10 bills, five $5 bills

Sumanta has $175 in $5 and $10 bills in his drawer. The number of $5 bills is three times the number of $10 bills. How many of each are in the drawer?

Chi has $11.30 in dimes and quarters. The number of dimes is three more than three times the number of quarters. How many of each are there?

Solution

63 dimes, 20 quarters

Alison has $9.70 in dimes and quarters. The number of quarters is eight more than four times the number of dimes. How many of each coin does she have?

Mukul has $3.75 in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and nine more nickels than quarters. How many of each coin are in his pocket?

Solution

16 nickels, 12 dimes, seven quarters

Vina has $4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin are in her purse?

Solve Ticket and Stamp Word Problems

In the following exercises, solve each ticket or stamp word problem.

The first day of a water polo tournament the total value of tickets sold was $17,610. One-day passes sold for $20 and tournament passes sold for $30. The number of tournament passes sold was 37 more than the number of day passes sold. How many day passes and how many tournament passes were sold?

Solution

330 day passes, 367 tournament passes

At the movie theater, the total value of tickets sold was $2,612.50. Adult tickets sold for $10 each and senior/child tickets sold for $7.50 each. The number of senior/child tickets sold was 25 less than twice the number of adult tickets sold. How many senior/child tickets and how many adult tickets were sold?

Julie went to the post office and bought both $0.41 stamps and $0.26 postcards. She spent $51.40. The number of stamps was 20 more than twice the number of postcards. How many of each did she buy?

Solution

40 postcards, 100 stamps

Jason went to the post office and bought both $0.41 stamps and $0.26 postcards and spent $10.28 The number of stamps was four more than twice the number of postcards. How many of each did he buy?

Hilda has $210 worth of $10 and $12 stock shares. The number of $10 shares is five more than twice the number of $12 shares. How many of each type of share does she have?

Solution

15 $10 shares, five $12 shares

Mario invested $475 in $45 and $25 stock shares. The number of $25 shares was five less than three times the number of $45 shares. How many of each type of share did he buy?

The ice rink sold 95 tickets for the afternoon skating session, for a total of $828. General admission tickets cost $10 each and youth tickets cost $8 each. How many general admission tickets and how many youth tickets were sold?

Solution

34 general, 61 youth

For the 7:30 show time, 140 movie tickets were sold. Receipts from the $13 adult tickets and the $10 senior tickets totaled $1,664. How many adult tickets and how many senior tickets were sold?

The box office sold 360 tickets to a concert at the college. The total receipts were $4,170. General admission tickets cost $15 and student tickets cost $10. How many of each kind of ticket was sold?

Solution

114 general, 246 student

Last Saturday, the museum box office sold 281 tickets for a total of $3,954. Adult tickets cost $15 and student tickets cost $12. How many of each kind of ticket was sold?

Solve Mixture Word Problems

In the following exercises, solve each mixture word problem.

Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. Macadamia nuts cost $9 per pound and almonds cost $5.25 per pound. How many pounds of macadamia nuts and how many pounds of almonds should Macario use for the mixture to cost $6.50 per pound to make?

Solution

Four pounds of macadamia nuts, eight pounds almonds

Carmen wants to tile the floor of his house. He will need 1,000 square feet of tile. He will do most of the floor with a tile that costs $1.50 per square foot, but also wants to use an accent tile that costs $9.00 per square foot. How many square feet of each tile should he plan to use if he wants the overall cost to be $3 per square foot?

Riley is planning to plant a lawn in his yard. He will need nine pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs $3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound?

Solution

3.6 lbs Bermuda seed, 5.4 lbs Fescue seed

Vartan was paid $25,000 for a cell phone app that he wrote and wants to invest it to save for his son’s education. He wants to put some of the money into a bond that pays 4% annual interest and the rest into stocks that pay 9% annual interest. If he wants to earn 7.4% annual interest on the total amount, how much money should he invest in each account?

Vern sold his 1964 Ford Mustang for $55,000 and wants to invest the money to earn him 5.8% interest per year. He will put some of the money into Fund A that earns 3% per year and the rest in Fund B that earns 10% per year. How much should he invest into each fund if he wants to earn 5.8% interest per year on the total amount?

Solution

$33,000 in Fund A, $22,000 in Fund B

Dominic pays 7% interest on his $15,000 college loan and 12% interest on his $11,000 car loan. What average interest rate does he pay on the total $26,000 he owes? (Round your answer to the nearest tenth of a percent.)

Liam borrowed a total of $35,000 to pay for college. He pays his parents 3% interest on the $8,000 he borrowed from them and pays the bank 6.8% on the rest. What average interest rate does he pay on the total $35,000? (Round your answer to the nearest tenth of a percent.)

Solution

5.9%

Solve Uniform Motion Applications

In the following exercises, solve.

Lilah is moving from Portland to Seattle. It takes her three hours to go by train. Mason leaves the train station in Portland and drives to the train station in Seattle with all Lilah’s boxes in his car. It takes him 2.4 hours to get to Seattle, driving at 15 miles per hour faster than the speed of the train. Find Mason’s speed and the speed of the train.

Kathy and Cheryl are walking in a fundraiser. Kathy completes the course in 4.8 hours and Cheryl completes the course in eight hours. Kathy walks two miles per hour faster than Cheryl. Find Kathy’s speed and Cheryl’s speed.

Solution

Kathy 5 mph, Cheryl 3 mph

Two busses go from Sacramento to San Diego. The express bus makes the trip in 6.8 hours and the local bus takes 10.2 hours for the trip. The speed of the express bus is 25 mph faster than the speed of the local bus. Find the speed of both busses.

A commercial jet and a private airplane fly from Denver to Phoenix. It takes the commercial jet 1.6 hours for the flight, and it takes the private airplane 2.6 hours. The speed of the commercial jet is 210 miles per hour faster than the speed of the private airplane. Find the speed of both airplanes to the nearest 10 mph.

Solution

commercial 550 mph, private plane 340 mph

Saul drove his truck three hours from Dallas towards Kansas City and stopped at a truck stop to get dinner. At the truck stop he met Erwin, who had driven four hours from Kansas City towards Dallas. The distance between Dallas and Kansas City is 542 miles, and Erwin’s speed was eight miles per hour slower than Saul’s speed. Find the speed of the two truckers.

Charlie and Violet met for lunch at a restaurant between Memphis and New Orleans. Charlie had left Memphis and drove 4.8 hours towards New Orleans. Violet had left New Orleans and drove two hours towards Memphis, at a speed 10 miles per hour faster than Charlie’s speed. The distance between Memphis and New Orleans is 394 miles. Find the speed of the two drivers.

Solution

Violet 65 mph, Charlie 55 mph

Sisters Helen and Anne live 332 miles apart. For Thanksgiving, they met at their other sister’s house partway between their homes. Helen drove 3.2 hours and Anne drove 2.8 hours. Helen’s average speed was four miles per hour faster than Anne’s. Find Helen’s average speed and Anne’s average speed.

Ethan and Leo start riding their bikes at the opposite ends of a 65-mile bike path. After Ethan has ridden 1.5 hours and Leo has ridden two hours, they meet on the path. Ethan’s speed is six miles per hour faster than Leo’s speed. Find the speed of the two bikers.

Solution

Ethan 22 mph, Leo 16 mph

Elvira and Aletheia live 3.1 miles apart on the same street. They are in a study group that meets at a coffee shop between their houses. It took Elvira half an hour and Aletheia two-thirds of an hour to walk to the coffee shop. Aletheia’s speed is 0.6 miles per hour slower than Elvira’s speed. Find both women’s walking speeds.

DaMarcus and Fabian live 23 miles apart and play soccer at a park between their homes. DaMarcus rode his bike for three-quarters of an hour and Fabian rode his bike for half an hour to get to the park. Fabian’s speed was six miles per hour faster than DaMarcus’ speed. Find the speed of both soccer players.

Solution

DaMarcus 16 mph, Fabian 22 mph

Cindy and Richard leave their dorm in Charleston at the same time. Cindy rides her bicycle north at a speed of 18 miles per hour. Richard rides his bicycle south at a speed of 14 miles per hour. How long will it take them to be 96 miles apart?

Matt and Chris leave their uncle’s house in Phoenix at the same time. Matt drives west on I-60 at a speed of 76 miles per hour. Chris drives east on I-60 at a speed of 82 miles per hour. How many hours will it take them to be 632 miles apart?

Solution

four hours

Two busses leave Billings at the same time. The Seattle bus heads west on I-90 at a speed of 73 miles per hour while the Chicago bus heads east at a speed of 79 miles an hour. How many hours will it take them to be 532 miles apart?

Two boats leave the same dock in Cairo at the same time. One heads north on the Mississippi River while the other heads south. The northbound boat travels four miles per hour. The southbound boat goes eight miles per hour. How long will it take them to be 54 miles apart?

Solution

4.5 hours

Lorena walks the path around the park in 30 minutes. If she jogs, it takes her 20 minutes. Her jogging speed is 1.5 miles per hour faster than her walking speed. Find Lorena’s walking speed and jogging speed.

Julian rides his bike uphill for 45 minutes, then turns around and rides back downhill. It takes him 15 minutes to get back to where he started. His uphill speed is 3.2 miles per hour slower than his downhill speed. Find Julian’s uphill and downhill speed.

Solution

uphill 1.6 mph, downhill 4.8 mph

Cassius drives his boat upstream for 45 minutes. It takes him 30 minutes to return downstream. His speed going upstream is three miles per hour slower than his speed going downstream. Find his upstream and downstream speeds.

It takes Darline 20 minutes to drive to work in light traffic. To come home, when there is heavy traffic, it takes her 36 minutes. Her speed in light traffic is 24 miles per hour faster than her speed in heavy traffic. Find her speed in light traffic and in heavy traffic.

Solution

light traffic 54 mph, heavy traffic 30 mph

At 1:30, Marlon left his house to go to the beach, a distance of 7.6 miles. He rode his skateboard until 2:15, and then walked the rest of the way. He arrived at the beach at 3:00. Marlon’s speed on his skateboard is 2.5 times his walking speed. Find his speed when skateboarding and when walking.

Aaron left at 9:15 to drive to his mountain cabin 108 miles away. He drove on the freeway until 10:45 and then drove on a mountain road. He arrived at 11:05. His speed on the freeway was three times his speed on the mountain road. Find Aaron’s speed on the freeway and on the mountain road.

Solution

freeway 67 mph, mountain road 22.3 mph

Marisol left Los Angeles at 2:30 to drive to Santa Barbara, a distance of 95 miles. The traffic was heavy until 3:20. She drove the rest of the way in very light traffic and arrived at 4:20. Her speed in heavy traffic was 40 miles per hour slower than her speed in light traffic. Find her speed in heavy traffic and in light traffic.

Lizette is training for a marathon. At 7:00 she left her house and ran until 8:15 then she walked until 11:15. She covered a total distance of 19 miles. Her running speed was five miles per hour faster than her walking speed. Find her running and walking speeds.

Solution

running eight mph, walking three mph

Everyday Math

John left his house in Irvine at 8:35 a.m. to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50 a.m.. At 5:30 p.m. he left the meeting and drove home. He arrived home at 7:18 p.m.

ⓐ What was his average speed on the drive from Irvine to Los Angeles?

ⓑ What was his average speed on the drive from Los Angeles to Irvine?

ⓒ What was the total time he spent driving to and from this meeting?

Sarah wants to arrive at her friend’s wedding at 3:00. The distance from Sarah’s house to the wedding is 95 miles. Based on usual traffic patterns, Sarah predicts she can drive the first 15 miles at 60 miles per hour, the next 10 miles at 30 miles per hour, and the remainder of the drive at 70 miles per hour.

ⓐ How long will it take Sarah to drive the first 15 miles?

ⓑ How long will it take Sarah to drive the next 10 miles?

ⓒ How long will it take Sarah to drive the rest of the trip?

ⓓ What time should Sarah leave her house?

Solution

ⓐ 15 minutes ⓑ 20 minutes ⓒ one hour (d) 1:25

Writing Exercises

Suppose you have six quarters, nine dimes, and four pennies. Explain how you find the total value of all the coins.

Do you find it helpful to use a table when solving coin problems? Why or why not?

Solution

Answers will vary.

In the table used to solve coin problems, one column is labeled “number” and another column is labeled “value.” What is the difference between the “number” and the “value”?

When solving a uniform motion problem, how does drawing a diagram of the situation help you?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve coin word problems. In row 3, the I can was use formulas to solve ticket and stamp word problems. In row 4, the I can was solve mixture word problems. In row 5, the I can was solve uniform motion applications.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Solve Linear Inequalities

Learning Objectives

By the end of this section, you will be able to:

  • Graph inequalities on the number line
  • Solve linear inequalities
  • Translate words to an inequality and solve
  • Solve applications with linear inequalities

Before you get started, take this readiness quiz.

Translate from algebra to English: 15>x.
If you missed this problem, review Example 3 in Use the Language of Algebra.

Solution

15 is greater than x.

Translate to an algebraic expression: 15 is less than x.
If you missed this problem, review Example 8 in Use the Language of Algebra.

Solution

15<x

Graph Inequalities on the Number Line

What number would make the inequality x>3 true? Are you thinking, “x could be four”? That’s correct, but x could be 6, too, or 37, or even 3.001. Any number greater than three is a solution to the inequality x>3.

We show all the solutions to the inequality x>3 on the number line by shading in all the numbers to the right of three, to show that all numbers greater than three are solutions. Because the number three itself is not a solution, we put an open parenthesis at three.

We can also represent inequalities using interval notation. There is no upper end to the solution to this inequality. In interval notation, we express x>3 as (3,∞). The symbol ∞ is read as “infinity.” It is not an actual number.

Figure 1 shows both the number line and the interval notation.

The figure shows the inquality, x is greater than 3, graphed on a number line from negative 5 to 5. There is shading that starts at 3 and extends to numbers to its right. The solution for the inequality is written in interval notation. It is the interval from 3 to infinity, not including 3.
The inequality x>3 is graphed on this number line and written in interval notation.

We use the left parenthesis symbol, (, to show that the endpoint of the inequality is not included. The left bracket symbol, [, shows that the endpoint is included.

The inequality x≤1 means all numbers less than or equal to one. Here we need to show that one is a solution, too. We do that by putting a bracket at x=1. We then shade in all the numbers to the left of one, to show that all numbers less than one are solutions. See Figure 2.

There is no lower end to those numbers. We write x≤1 in interval notation as (−∞,1]. The symbol −∞ is read as “negative infinity.” Figure 2 shows both the number line and interval notation.

The figure shows the inquality, x is less than or equal to l, graphed on a number line from negative 5 to 5. There is shading that starts at 1 and extends to numbers to its left. The solution for the inequality is written in interval notation. It is the interval from negative infinity to one, including 1.
The inequality x≤1 is graphed on this number line and written in interval notation.

Inequalities, Number Lines, and Interval Notation

The figure shows that the solution of the inequality x is greater than a is indicated on a number line with a left parenthesis at a and shading to the right, and that the solution in interval notation is the interval from a to infinity enclosed in parentheses. It shows the solution of the inequality x is greater than or equal to a is indicated on a number line with an left bracket at a and shading to the right, and that the solution in interval notation is the interval a to infinity within a left bracket and right parenthesis. It shows that the solution of the inequality x is less than a is indicated on a number line with a right parenthesis at a and shading to the left, and that the solution in interval notation is the the interval negative infinity to a within parentheses. It shows that the solution of the inequality x is less than or equal to a is indicated on anumber line with a right bracket at a and shading to the left, and that the solution in interval notation is negative infinity to a within a left parenthesis and right bracket.

The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals.

Graph each inequality on the number line and write in interval notation.

ⓐ x≥−3 ⓑ x<2.5 ⓒ x≤−35

Solution
ⓐ
The mathematical inequality 'x is greater than or equal to -3' is displayed on a white background.
Shade to the right of −3, and put a bracket at −3.   A number line graph depicting the inequality x >= -3. The number line shows integer markings from -4 to -1. A light blue shaded region starts with a square bracket at -3 and extends to the right, indicating values greater than or equal to -3.
Write in interval notation. A mathematical interval notation is displayed, showing a closed interval starting at -3 and extending to positive infinity, written as [-3, ∞).
ⓑ
A white background displays the mathematical inequality x < 2.5 in black font, indicating that the variable x is less than two point five.
Shade to the left of 2.5 and put a parenthesis at 2.5. A number line graph representing the inequality x < 2.5 or the interval (-∞, 2.5). The shaded arrow extends left from an open parenthesis at 2.5.
Write in interval notation. The mathematical interval notation '(-∞, 2.5)' is shown in black text against a white background.
ⓒ
The image displays the mathematical inequality x <= -3/5, indicating that the variable x is less than or equal to negative three-fifths.
Shade to the left of −35, and put a bracket at −35. A number line graph representing the inequality x is less than or equal to -3/5, with a closed bracket at -3/5 and a thick blue arrow pointing to the left, towards negative infinity.
Write in interval notation. A mathematical interval notation is displayed, showing a set of real numbers from negative infinity up to and including -3/5, represented as (-∞, -3/5].

Graph each inequality on the number line and write in interval notation: ⓐ x>2 ⓑ x≤−1.5 ⓒ x≥34.

Solution

ⓐ
The graph of the inequality x is greater than 2 is indicated on a number line with a left parenthesis at 2 and shading to the right. The solution in interval notation is the interval from 2 to infinity enclosed within parentheses.

ⓑ
The graph of the inequality x is less than or equal to negative 1.5 is indicated on a number line with a right bracket at negative 1.5 and shading to the left. The solution in interval notation is the interval from negative infinity to negative 1.5 enclosed within a left parenthesis and right bracket.

ⓒ
The graph of the inequality x is greater than or rqual to three-fourths is indicated on a number line with a left bracket at three-fourths and shading to the right. The solution in interval notation is the interval from three-fourths to infinity enclosed within a left bracket and left parentheses.

Graph each inequality on the number line and write in interval notation: ⓐ x≤−4 ⓑ x≥0.5 ⓒ x<−23.

Solution

ⓐ
The graph of the inequality x is less than or equal to negative 4 is indicated on a number line with a right bracket at negative 4 and shading to the left. The solution in interval notation is the interval from negative infinity to negative 4 enclosed within an left parenthesis and right bracket.

ⓑ
The graph of the inequality x is greater than or equal to 0.5 is indicated on a number line with a left bracket at 0.5 and shading to the right. The solution in interval notation is the interval from 0.5 to infinity enclosed within a left bracket and right parenthesis.

ⓒ
The graph of the inequality x is less than negative two-thirds is indicated on a number line with a right parenthesis at negative two-thirds and shading to the left. The solution in interval notation is the interval from negative infinity to negative two-thirds enclosed within parentheses.

What numbers are greater than two but less than five? Are you thinking say, 2.5,3,323,4,4.99? We can represent all the numbers between two and five with the inequality 2<x<5. We can show 2<x<5 on the number line by shading all the numbers between two and five. Again, we use the parentheses to show the numbers two and five are not included. See Figure 3.

The graph of the inequality 2 is less than x which is less than 5 shows open circles a 2 and 5 and shading in between.

Graph each inequality on the number line and write in interval notation.

ⓐ −3<x<4 ⓑ −6≤x<−1 ⓒ 0≤x≤2.5

Solution
ⓐ
A mathematical inequality is displayed on a white background, stating '-3 < x < 4' in black text, indicating that x is a value greater than -3 and less than 4.
Shade between −3 and 4.
Put a parentheses at −3 and 4.
A number line shows an open interval from -3 to 4, meaning numbers greater than -3 and less than 4 are included. The number line is marked from -4 to 5, with integer labels.
Write in interval notation. The mathematical coordinate notation (-3, 4).
ⓑ
A mathematical inequality is displayed on a white background, reading '-6  <=  X < -1'.
Shade between −6 and −1.
Put a bracket at −6, and
a parenthesis at −1.
A number line showing the interval [-6, -1), with -6 included and -1 excluded, highlighted in teal.
Write in interval notation. Opening bracket negative six negative one close parenthesis.
ⓒ
A mathematical inequality states that 0 is less than or equal to x, and x is less than or equal to 2.5.
Shade between 0 and 2.5.
Put a bracket at 0 and at 2.5.
A number line showing the closed interval from 0.0 to 2.5, represented by a shaded teal segment with square brackets at both ends.
Write in interval notation. A cropped image displays a white background with a vertical segment of black text, which appears to be part of a mathematical notation or number sequence: '[0, 2.5]'

Graph each inequality on the number line and write in interval notation:

ⓐ −2<x<1 ⓑ −5≤x<−4 ⓒ 1≤x≤4.25

Solution

ⓐ
Negative 2 is less x which is less than 1. There are open circles at negative 2 and 1 and shading between negative 2 and 1 on the number line. Put parentheses at negative 2 and 1. Write in interval notation.

ⓑ
Negative 5 is less than or equal to x which is less than negative 4. There is a closed circle at negative 6 and an open circle at negative 4 and shading between negative 5 and negative 4 on the number line. Put a bracket at negative 5 and a parenthesis at negative 4. Write in interval notation.

ⓒ
1 is less than or equal to x which is less than 4.25. There is closed circle at 1 and a closed circle at 4.25 and shading between 1 and 4.25 on the number line. Put brackets at 1 and 4.25. Write in interval notation.

Graph each inequality on the number line and write in interval notation:

ⓐ −6<x<2 ⓑ −3≤x<−1 ⓒ 2.5≤x≤6

Solution

ⓐ
Negative 6 is less than x which is less than 2. There is an open circle at negative 6 and an open circle at 2 and shading between negative 6 and 2 on the number line. Put parentheses at negative 6 and 2. Write in interval notation.

ⓑ
Negative 3 is less than or equal to x which is less than negative 1. There is a closed circle at negative 3 and an open circle at negative 1 and shading between negative 3 and negative 1 on the number line. Put a bracket at negative 3 and a parenthesis at negative 1. Write in interval notation.

ⓒ
2.5 is less than or equal to x which is less thanor equal to 6. There is a closed circle at 2.5 and a closed circle at 6 and shading between 2.5 and 6 on the number line. Put brackets at 2.5 and 6. Write in interval notation.

Solve Linear Inequalities

A linear inequality is much like a linear equation—but the equal sign is replaced with an inequality sign. A linear inequality is an inequality in one variable that can be written in one of the forms, ax+b<c,ax+b≤c,ax+b>c, or ax+b≥c.

Linear Inequality

A linear inequality is an inequality in one variable that can be written in one of the following forms where a, b, and c are real numbers and a≠0:

ax+b<c,ax+b≤c,ax+b>c,ax+b≥c.

When we solved linear equations, we were able to use the properties of equality to add, subtract, multiply, or divide both sides and still keep the equality. Similar properties hold true for inequalities.

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality. For example:

Negative 4 is less than 2. Negative 4 minus 5 is less than 2 minus 5. Negative 9 is less than negative 3, which is true. Negative 4 is less than 2. Negative 4 plus 7 is less than 2 plus 7. 3 is less than 9, which is true.

Notice that the inequality sign stayed the same.

This leads us to the Addition and Subtraction Properties of Inequality.

Addition and Subtraction Property of Inequality

For any numbers a, b, and c, if a<b,then

a+c<b+ca−c<b−c

For any numbers a, b, and c, if a>b,then

a+c>b+ca−c>b−c

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality.

What happens to an inequality when we divide or multiply both sides by a constant?

Let’s first multiply and divide both sides by a positive number.

10 is less than 15. 10 times 5 is less than 15 times 5. 50 is less than 75 is true. 10 is less than 15. 10 divided by 5 is less than 15 divided by 5. 2 is less than 3 is true.

The inequality signs stayed the same.

Does the inequality stay the same when we divide or multiply by a negative number?

10 is less than 15 10 times negative 5 is blank 15 times negative 5? Negative 50 is blank negative 75. Negative 50 is greater than negative 75. 10 is less than 15. 10 divided by negative 5 is blank 15 divided by negative 5. Negative 2 is blank negative 3. Negative 2 is blank negative 3.

Notice that when we filled in the inequality signs, the inequality signs reversed their direction.

When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.

This gives us the Multiplication and Division Property of Inequality.

Multiplication and Division Property of Inequality

For any numbers a, b, and c,

multiply or divide by a positiveifa<bandc>0,thenac<bcandac<bc.ifa>bandc>0,thenac>bcandac>bc.multiply or divide by a negativeifa<bandc<0,thenac>bcandac>bc.ifa>bandc<0,thenac<bcandac<bc.

When we divide or multiply an inequality by a:

  • positive number, the inequality stays the same.
  • negative number, the inequality reverses.

Sometimes when solving an inequality, as in the next example, the variable ends upon the right. We can rewrite the inequality in reverse to get the variable to the left.

x>ahas the same meaning asa<x

Think about it as “If Xander is taller than Andy, then Andy is shorter than Xander.”

Solve each inequality. Graph the solution on the number line, and write the solution in interval notation.

ⓐ x−38≤34 ⓑ 9y<​​54 ⓒ −15<35z

Solution
ⓐ
A mathematical inequality expression is presented, showing 'x minus three-eighths is less than or equal to three-fourths'.
Add 38 to both sides of the inequality. An algebraic inequality is displayed, showing 'x - 3/8 + 3/8 <= 3/4 + 3/8' on a white background.
Simplify. A mathematical inequality is shown, displaying 'x' is less than or equal to '9/8'.
Graph the solution on the number line.       A number line shows the inequality x <= 1 9/8. The dark line extends from negative infinity to the point labeled '1 9/8', indicated by a closed bracket. The point '1 9/8' is positioned between 0 and 2 on the line.
Write the solution in interval notation. A mathematical interval notation showing all real numbers x such that x is less than or equal to 9/8, expressed as (-∞, 9/8].
ⓑ
A mathematical inequality is displayed on a white background, reading '9y < 54' in black text.
Divide both sides of the inequality by 9; since  
9 is positive, the inequality stays the same.
A mathematical inequality shows '9y over 9 is less than 54 over 9' set against a white background.
Simplify. A mathematical inequality, 'y < 6', is displayed in the center of a white background.
Graph the solution on the number line. A number line shows an interval less than 6. An open parenthesis at 6 indicates that 6 is not included, and a thick blue line with a left arrow shows that the interval extends infinitely to the left.
Write the solution in interval notation. A mathematical interval notation is displayed as '(-infinity, 6)', indicating all real numbers less than 6.
ⓒ
An image showing the mathematical inequality -15 < 3/5 z.
Multiply both sides of the inequality by 53.
Since 53 is positive, the inequality stays the same.
An algebraic inequality is displayed: (5/3)(-15) < (5/3)(3/5z). It represents a mathematical expression involving fractions, multiplication, and an unknown variable 'z'.
Simplify. A mathematical inequality is shown, stating '-25 < Z' in gray text against a plain white background, indicating that -25 is less than Z.
Rewrite with the variable on the left. The mathematical inequality 'z > -25' is displayed on a white background.
Graph the solution on the number line. A number line graph showing an inequality where a blue arrow starts with an open parenthesis-like mark at -25 and extends to the right, indicating values greater than -25.
Write the solution in interval notation. The mathematical interval notation '(-25, ')' indicating all real numbers greater than -25, extending to positive infinity.

Solve each inequality, graph the solution on the number line, and write the solution in interval notation:

ⓐ p−34≥16 ⓑ 9c>72 ⓒ 24≤38m

Solution

ⓐ
p is less than eleven-twelfths. The solution on the number line has a right bracket at eleven-twelfths with shading to the right. The solution in interval notation is, eleven-twelfths to infinity within a bracket and parenthesis.

ⓑ

c is less than 8. The solution on the number line has a left bracket at 8 with shading to the right. The solution in interval notation is, 8 to infinity within parentheses.

ⓒ
m is greater than or equal to 8. The solution on the number line has a right bracket at 64 with shading to the right. The solution in interval notation is, 64 to infinity within a bracket and parentheses.

Solve each inequality, graph the solution on the number line, and write the solution in interval notation:

ⓐ r−13≤712 ⓑ 12d≤​60 ⓒ −24<43n

Solution

ⓐ
r is less than or equal to eleven-twelfths. The solution on the number line has a left bracket at eleven-twelfths with shading to the left. The solution in interval notation is negative infinity to eleven-twelfths within a parenthesis and a bracket.

ⓑ

c is less than or equal to 5. The solution on the number line has a right bracket at 5 with shading to the left. The solution in interval notation is negative infinity to 5 within a parentheses and a bracket.

ⓒ
n is greater than negative 18. The solution on the number line has a left parenthesis at negative 18 with shading to the right. The solution in interval notation is negative 18 to infinity within parentheses.

Be careful when you multiply or divide by a negative number—remember to reverse the inequality sign.

Solve each inequality, graph the solution on the number line, and write the solution in interval notation.

ⓐ −13m≥65 ⓑ n−2≥8

Solution
ⓐ
A mathematical inequality is shown on a white background: -13m >= 65.
Divide both sides of the inequality by −13.
Since −13 is a negative, the inequality reverses.
A mathematical inequality showing '-13m divided by -13 is less than or equal to 65 divided by -13'.
Simplify. A mathematical inequality displays 'm is less than or equal to -5' on a white background.
Graph the solution on the number line. A number line shows a dark blue arrow pointing left from a vertical bracket at -5, indicating all numbers less than -5. The number line is marked with integers -7, -6, -5, and -4.
Write the solution in interval notation. A mathematical notation for an interval, showing '(-∞, -5]'. This represents all real numbers from negative infinity up to and including -5.
ⓑ
A mathematical inequality displays 'n divided by negative two is greater than or equal to eight' on a white background.
Multiply both sides of the inequality by −2.
Since −2 is a negative, the inequality reverses.
An algebraic inequality is displayed: -2 multiplied by the fraction n over -2, is less than or equal to -2 multiplied by 8. The inequality sign is highlighted in red.
Simplify. The mathematical inequality n <= -16 is displayed centered on a white background.
Graph the solution on the number line. A number line graph showing the interval x <= -16. A closed bracket is at -16, and a dark blue line extends to the left with an arrow, indicating all numbers less than or equal to -16.
Write the solution in interval notation. A mathematical interval notation is displayed, showing '(-∞, -16]' in grey text against a white background.

Solve each inequality, graph the solution on the number line, and write the solution in interval notation:

ⓐ −8q<32 ⓑ k−12≤15.

Solution

ⓐ

q is greater than or equal to negative 4. The solution on the number line has a left parenthesis at negative 4 with shading to the right. The solution in interval notation is negative 4 to infinity within parentheses.

ⓑ
k is greater than or equal to negative 180. The solution on the number line has a left bracket at negative 180 with shading to the right. The solution in interval notation is negative 180 to infinity within a bracket and a parenthesis.

Solve each inequality, graph the solution on the number line, and write the solution in interval notation:

ⓐ −7r≤​−70 ⓑ u−4≥−16.

Solution

ⓐ

r is greater than or equal to 10. The solution on the number line has a left bracket at 10 with shading to the right. The solution in interval notation is 10 to infinity within a bracket and parenthesis.

ⓑ
u is less than or equal to 64. The solution on the number line has a right bracket at 64 with shading to the left. The solution in interval notation is negative infinity to 64 within parenthesis and a bracket.

Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but make sure to pay close attention when we multiply or divide to isolate the variable.

Solve the inequality 6y≤11y+17, graph the solution on the number line, and write the solution in interval notation.

Solution
A mathematical inequality is shown, displaying the expression '6y <= 11y + 17' on a white background. This inequality involves a variable 'y' and integers.
Subtract 11y from both sides to collect
the variables on the left.
Algebraic equation showing six y minus eleven y less than or equal to eleven y minus eleven y plus seventeen.
Simplify. A mathematical inequality is shown in black text on a white background, reading '-5y ', followed by the less than or equal to symbol, and then '17'.
Divide both sides of the inequality by −5,
and reverse the inequality.
A mathematical inequality is shown, with a fraction on each side of a greater than or equal to sign. The left side is '-5y over -5' and the right side is '17 over -5'.
Simplify. The image displays the mathematical inequality y >= 17/-5.
Graph the solution on the number line. A number line shows an interval starting at -17/5 (approximately -3.4) with a left square bracket, indicating it's included, and extending to the right with a blue arrow, representing all numbers greater than or equal to -17/5.
Write the solution in interval notation. Closed interval from negative seventeen over five to positive infinity, represented by bracketed left endpoint and parenthesis on the right.

Solve the inequality, graph the solution on the number line, and write the solution in interval notation: 3q≥7q−23.

Solution
q is less than or equal to 23 divided by 4. The solution on the number line has a right bracket at 23 divided by 4 with shading to the left. The solution in interval notation is negative infinity to 23 divided by 4 within a parenthesis and a bracket.

Solve the inequality, graph the solution on the number line, and write the solution in interval notation: 6x<10x+19.

Solution
x is greater than negative 19 divided by 4. The solution on the number line has a left parenthesis at negative 19 divided by 4 with shading to the right. The solution in interval notation is negative 19 divided by 4 to infinity within parentheses.

When solving inequalities, it is usually easiest to collect the variables on the side where the coefficient of the variable is largest. This eliminates negative coefficients and so we don’t have to multiply or divide by a negative—which means we don’t have to remember to reverse the inequality sign.

Solve the inequality 8p+3(p−12)>7p−28, graph the solution on the number line, and write the solution in interval notation.

Solution
8p+3(p−12)>7p−28
Simplify each side as much as possible.
Distribute. 8p+3p−36>7p−28
Combine like terms. 11p−36>7p−28
Subtract 7p from both sides to collect the
variables on the left, since 11>7.
11p−36−7p>7p−28−7p
Simplify. 4p−36>−28
Add 36 to both sides to collect the
constants on the right.
4p−36+36>−28+36
Simplify. 4p>8
Divide both sides of the inequality by
4; the inequality stays the same.
4p4>​84
Simplify. p>2
Graph the solution on the number line. A number line shows an open circle or bracket at 2, with a shaded line and arrow extending to the right, indicating all numbers greater than 2.
Write the solution in interval notation. (2,∞)

Solve the inequality 9y+2(y+6)>5y−24, graph the solution on the number line, and write the solution in interval notation.

Solution

y is greater than negative 6. The solution on the number line has a left parenthesis at negative 6 with shading to the right. The solution in interval notation is negative 6 to infinity within parentheses.

Solve the inequality 6u+8(u−1)>10u+32, graph the solution on the number line, and write the solution in interval notation.

Solution

u is greater than negative 10. The solution on the number line has a left parenthesis at 10 with shading to the right. The solution in interval notation is 10 to infinity within parentheses.

Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.

Solve the inequality 8x−2(5−x)<4(x+9)+6x, graph the solution on the number line, and write the solution in interval notation.

Solution
Simplify each side as much as possible. 8x−2(5−x)<4(x+9)+6x
Distribute. 8x−10+2x<4x+36+6x
Combine like terms. 10x−10<10x+36
Subtract 10x from both sides to collect
the variables on the left.
10x−10−10x<10x+36−10x
Simplify. −10<36
The x’s are gone, and we have a true
statement.
The inequality is an identity.
The solution is all real numbers.
Graph the solution on the number line. A number line displaying integers -1, 0, 1, and 2, with arrows indicating infinite extension in both directions.
Write the solution in interval notation. (−∞,∞)

Solve the inequality 4b−3(3−b)>5(b−6)+2b, graph the solution on the number line, and write the solution in interval notation.

Solution

The inequality is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Solve the inequality 9h−7(2−h)<8(h+11)+8h, graph the solution on the number line, and write the solution in interval notation.

Solution

The inequality is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

We can clear fractions in inequalities much as we did in equations. Again, be careful with the signs when multiplying or dividing by a negative.

Solve the inequality 13a−18a>524a​+34, graph the solution on the number line, and write the solution in interval notation.

Solution
A mathematical inequality is displayed showing the expression (1/3)a - (1/8)a > (5/24)a + (3/4).
Multiply both sides by the LCD, 24,
to clear the fractions.
An image showing the mathematical inequality 24(1/3 a - 1/8 a) > 24(5/24 a + 3/4), with the number 24 highlighted in red on both sides.
Simplify. A mathematical inequality is displayed: 8a - 3a > 5a + 18.
Combine like terms. A mathematical inequality '5a > 5a + 18' is displayed, illustrating a logically impossible statement that has no solution for 'a'.
Subtract 5a from both sides to collect the
variables on the left.
A mathematical expression reads 5a - 5a > 5a - 5a + 18. This inequality simplifies to 0 > 18, which is a false statement, indicating no solution or an impossible condition.
Simplify. The image displays a mathematical inequality '0 > 18' in black text on a white background, which is a false statement as zero is not greater than eighteen.
The statement is false. The inequality is a contradiction.
There is no solution.
Graph the solution on the number line. A number line displays integers from -1 to 2, with tick marks and corresponding labels for each integer.
Write the solution in interval notation. There is no solution.

Solve the inequality 14x−112x>16x+78, graph the solution on the number line, and write the solution in interval notation.

Solution

The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.

Solve the inequality 25z−13z<115z​+35, graph the solution on the number line, and write the solution in interval notation.

Solution

The inequality is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Translate to an Inequality and Solve

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like “more than” and “less than.” But others are not as obvious. Table 16 shows some common phrases that indicate inequalities.

> ≥ < ≤
is greater than

is more than

is larger than

exceeds
is greater than or equal to

is at least

is no less than

is the minimum
is less than

is smaller than

has fewer than

is lower than
is less than or equal to

is at most

is no more than

is the maximum

Translate and solve. Then graph the solution on the number line, and write the solution in interval notation.

Twenty-seven less thanxis at least 48.
Solution
The image shows the phrase 'Twenty-seven less than x is at least 48.' written in black text on a white background. A light blue bracket highlights 'is at least' below the text.
Translate. A mathematical inequality expression is presented on a white background, which reads 'x - 27 >= 48'.
Solve—add 27 to both sides. A mathematical inequality showing the step of adding 27 to both sides to isolate the variable x: x - 27 + 27 >= 48 + 27.
Simplify. A mathematical inequality is shown against a white background, displaying the expression 'X   Y' where Y is 75.
Graph on the number line. A number line displays values 73 to 77. A black arrow extends left from 75, and a blue arrow extends right from 75. The number 75 is marked by a dashed vertical line, indicating a central point or threshold.
Write in interval notation. The mathematical notation shows the interval [75, ∞), representing all real numbers greater than or equal to 75. The square bracket indicates that 75 is included in the set, while the infinity symbol (∞) with a parenthesis indicates an unbounded upper limit.

Translate and solve. Then graph the solution on the number line, and write the solution in interval notation.

Nineteen less than p is no less than 47.

Solution

p minus 19 is greater than or equal to 47. Its solution is p is greater than or equal to 66. The solution on the number line has a left bracket at 66 with shading to the right. The solution in interval notation is 66 to infinity within a bracket and a parenthesis.

Translate and solve. Then graph the solution on the number line, and write the solution in interval notation.

Four more than a is at most 15.

Solution

a plus 4 is less than or equal to 15. Its solution is a is less than or equal to 11. The solution on the number line has a right bracket at 11with shading to the left. The solution in interval notation is negative infinity to 11 within a parenthesis and bracket.

Solve Applications with Linear Inequalities

Many real-life situations require us to solve inequalities. The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations.

We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality.

Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down.

Dawn won a mini-grant of $4,000 to buy tablet computers for her classroom. The tablets she would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets Dawn can buy?

Solution
A seven-step process for solving an inequality problem, illustrated by determining the maximum number of tablets Dawn can purchase.
Step 1. Read the problem.
Step 2. Identify what you are looking for. the maximum number of tablets Dawn can buy
Step 3. Name what you are looking for.
Choose a variable to represent that quantity. Letn=the number of tablets.
Step 4. Translate. Write a sentence that gives the information to find it. $254.12 times the number of tablets is no more than $4,000.
Translate into an inequality. 254.12n≤4000
Step 5. Solve the inequality.
But n must be a whole number of tablets, so round to 15.
n≤15.74n≤15
Step 6. Check the answer in the problem and make sure it makes sense.
Rounding down the price to $250, 15 tablets would cost $3,750, while 16 tablets would be $4,000. So a maximum of 15 tablets at $254.12 seems reasonable.
Step 7. Answer the question with a complete sentence. Dawn can buy a maximum of 15 tablets.

Angie has $20 to spend on juice boxes for her son’s preschool picnic. Each pack of juice boxes costs $2.63. What is the maximum number of packs she can buy?

Solution

Angie can buy 7 packs of juice.

Daniel wants to surprise his girlfriend with a birthday party at her favorite restaurant. It will cost $42.75 per person for dinner, including tip and tax. His budget for the party is $500. What is the maximum number of people Daniel can have at the party?

Solution

Daniel can have 11 people at the party.

Taleisha’s phone plan costs her $28.80 a month plus $0.20 per text message. How many text messages can she send/receive and keep her monthly phone bill no more than $50?

Solution
A step-by-step guide to solving a word problem using inequalities, demonstrated with an example of Taleisha's text message bill calculation.
Step 1. Read the problem.
Step 2. Identify what you are looking for. the number of text messages Taleisha can make
Step 3. Name what you are looking for.
Choose a variable to represent that quantity. Lett=the number of text messages.
Step 4. Translate Write a sentence that gives the information to find it. $28.80 plus $0.20 times the number of text messages is less than or equal to $50.
Translate into an inequality. 28.80+0.20t≤50
Step 5. Solve the inequality. 0.2t≤21.2t≤106text messages
Step 6. Check the answer in the problem and make sure it makes sense.
Yes,28.80+0.20(106)=50.
Step 7. Write a sentence that answers the question. Taleisha can send/receive no more than 106 text messages to keep her bill no more than $50.

Sergio and Lizeth have a very tight vacation budget. They plan to rent a car from a company that charges $75 a week plus $0.25 a mile. How many miles can they travel during the week and still keep within their $200 budget?

Solution

Sergio and Lizeth can travel no more than 500 miles.

Rameen’s heating bill is $5.42 per month plus $1.08 per therm. How many therms can Rameen use if he wants his heating bill to be a maximum of $87.50.

Solution

Rameen can use no more than 76 therms.

Profit is the money that remains when the costs have been subtracted from the revenue. In the next example, we will find the number of jobs a small businesswoman needs to do every month in order to make a certain amount of profit.

Felicity has a calligraphy business. She charges $2.50 per wedding invitation. Her monthly expenses are $650. How many invitations must she write to earn a profit of at least $2,800 per month?

Solution
A step-by-step guide demonstrating how to solve a word problem using inequalities, exemplified by calculating minimum invitations needed.
Step 1. Read the problem.
Step 2. Identify what you are looking for. the number of invitations Felicity needs to write
Step 3. Name what you are looking for.
Choose a variable to represent it.
Letj=the number of invitations.
Step 4. Translate. Write a sentence that gives the information to find it. $2.50 times the number of invitations minus $650 is at least $2,800.
Translate into an inequality. 2.50j−650≥2,800
Step 5. Solve the inequality. 2.5j≥3,450j≥1,380invitations
Step 6. Check the answer in the problem and make sure it makes sense.
If Felicity wrote 1400 invitations, her profit would be
2.50(1400) − 650, or $2,850. This is more than $2800.
Step 7. Write a sentence that answers the question. Felicity must write at least 1,380 invitations.

Caleb has a pet sitting business. He charges $32 per hour. His monthly expenses are $2,272. How many hours must he work in order to earn a profit of at least $800 per month?

Solution

Caleb must work at least 96 hours.

Elliot has a landscape maintenance business. His monthly expenses are $1,100. If he charges $60 per job, how many jobs must he do to earn a profit of at least $4,000 a month?

Solution

Elliot must work at least 85 jobs.

There are many situations in which several quantities contribute to the total expense. We must make sure to account for all the individual expenses when we solve problems like this.

Malik is planning a six-day summer vacation trip. He has $840 in savings, and he earns $45 per hour for tutoring. The trip will cost him $525 for airfare, $780 for food and sightseeing, and $95 per night for the hotel. How many hours must he tutor to have enough money to pay for the trip?

Solution
A 7-step process for solving problems, illustrated with an example involving an inequality calculation.
Step 1. Read the problem.
Step 2. Identify what you are looking for. the number of hours Malik must tutor
Step 3. Name what you are looking for.
Choose a variable to represent that quantity. Leth=the number of hours.
Step 4. Translate. Write a sentence that gives the information to find it. The expenses must be less than or equal to the income. The cost of airfare plus the cost of food and sightseeing and the hotel bill must be less than the savings plus the amount earned tutoring.
Translate into an inequality. 525+780+95(6)≤840+45h
Step 5. Solve the inequality. 1,875≤840+45h1,035≤45h23≤hh≥23
Step 6. Check the answer in the problem and make sure it makes sense.
We substitute 23 into the inequality.
1,875≤840+45h1,875≤840+45(23)1,875≤1875
Step 7. Write a sentence that answers the question. Malik must tutor at least 23 hours.

Brenda’s best friend is having a destination wedding and the event will require 3 nights in a hotel. Brenda has $500 in savings and can earn $15 an hour babysitting. She expects to pay $350 airfare, $375 for food and entertainment and $60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip?

Solution

Brenda must babysit at least 27 hours.

Josue wants to go on a 10-night road trip with friends next spring. It will cost him $180 for gas, $450 for food, and $49 per night to share a motel room. He has $520 in savings and can earn $30 per driveway shoveling snow. How many driveways must he shovel to have enough money to pay for the trip?

Solution

Josue must shovel at least 20 driveways.

Key Concepts

  • Inequalities, Number Lines, and Interval Notation
    x>ax≥ax<ax≤a
    The figure shows that the solution of the inequality x is greater than a is indicated on a number line with a left parenthesis at a and shading to the right, and that the solution in interval notation is the interval from a to infinity enclosed in parentheses. It shows the solution of the inequality x is greater than or equal to a is indicated on a number line with an left bracket at a and shading to the right, and that the solution in interval notation is the interval a to infinity within a left bracket and right parenthesis. It shows that the solution of the inequality x is less than a is indicated on a number line with a right parenthesis at a and shading to the left, and that the solution in interval notation is the the interval negative infinity to a within parentheses. It shows that the solution of the inequality x is less than or equal to a is indicated on anumber line with a right bracket at a and shading to the left, and that the solution in interval notation is negative infinity to a within a left parenthesis and right bracket.
  • Linear Inequality
    • A linear inequality is an inequality in one variable that can be written in one of the following forms where a, b, and c are real numbers and a≠0:
      ax+b<c,ax+b≤c,ax+b>c,ax+b≥c.
  • Addition and Subtraction Property of Inequality
    • For any numbers a, b, and c, if a<b,then
      a+c<b+ca−c<b−ca+c>b+ca−c>b−c
    • We can add or subtract the same quantity from both sides of an inequality and still keep the inequality.
  • Multiplication and Division Property of Inequality
    • For any numbers a, b, and c,
      multiply or divide by apositiveifa<bandc>0,thenac<bcandac<bc.ifa>bandc>0,thenac>bcandac>bc.multiply or divide by anegativeifa<bandc<0,thenac>bcandac>bc.ifa>bandc<0,thenac<bcandac<bc.
  • Phrases that indicate inequalities
    > ≥ < ≤
    is greater than

    is more than

    is larger than

    exceeds
    is greater than or equal to

    is at least

    is no less than

    is the minimum
    is less than

    is smaller than

    has fewer than

    is lower than
    is less than or equal to

    is at most

    is no more than

    is the maximum

Practice Makes Perfect

Graph Inequalities on the Number Line

In the following exercises, graph each inequality on the number line and write in interval notation.


ⓐ x<−2
ⓑ x≥−3.5
ⓒ x≤23


ⓐ x>3
ⓑ x≤−0.5
ⓒ x≥13

Solution


ⓐ
The solution for x is greater than 3 on a number line has a left bracket 3 with shading to the right. The solution in interval notation is 3 to infinity within parentheses.
ⓑ
The solution for x is less than or equal to negative 0.5 on a number line has a right bracket at negative 0.5 with shading to the left. The solution in interval notation is negative infinity to negative 0.5 within a parenthesis and a bracket. ⓒ
The solution for x is greater than or equal to one-third on a number line has a left bracket at one-third with shading to the right. The solution in interval notation is one-third to infinity within a bracket and a parenthesis.


ⓐ x≥−4
ⓑ x<2.5
ⓒ x>−32


ⓐ x≤5
ⓑ x≥−1.5
ⓒ x<−73

Solution


ⓐ
The solution for x is less than or equal to 5 on a number line has a right bracket with shading to the left. The solution in interval notation is negative infinity to 5 within a parenthesis and a bracket.
ⓑ
The solution for x is greater than or equal to negative 1.5 on a number line has a left bracket with shading to the right. The solution in interval notation is negative 1.5 to infinity within a bracket and a parenthesis.
ⓒ
The solution for x is less than negative seven-thirds on a number line has a right parenthesis with shading to the left. The solution in interval notation is negative infinity to negative seven-thirds within parentheses.


ⓐ −5<x<2
ⓑ −3≤x<1
ⓒ 0≤x≤1.5


ⓐ −2<x<0
ⓑ −5≤x<−3
ⓒ 0≤x≤3.5

Solution


ⓐ
Negative 2 is less than x which is less than 0. There is an open circle at negative 2 and an open circle at 0 and shading between negative 2 and 0 on the number line. The interval notation is negative 2 and 0 within parentheses.
ⓑ
Negative 5 is less than or equal to x which is less than negative 3. There is a closed circle at negative 5 and an open circle at negative 3 and shading between negative 5 and negative 3 on the number line. The interval notation is negative 5 and negative 3 within a bracket and a parenthesis.
ⓒ
0 is less than or equal to x which is less than or equal to 3.5. There is a closed circle at 0 and a closed circle at 3.5 and shading between 0 and 3.5 on the number line. The interval notation is 0 and 3.5 within brackets.


ⓐ −1<x<3
ⓑ −3<x≤−2
ⓒ −1.25≤x≤0


ⓐ −4<x<2
ⓑ −5<x≤−2
ⓒ −3.75≤x≤0

Solution


ⓐ
Negative 4 is less than x which is less than 2. There is a open circle at negative 4 and an open circle at 2 and shading between negative 4 and 2 on the number line. The interval notation is negative 4 and 2 within parentheses.
ⓑ
Negative 5 is less than x which is less than or equal to 2. There is a open circle at negative 5 and a closed circle at negative 2 and shading between negative 5 and negative 2 on the number line. The interval notation is negative 5 and negative 2 within a parenthesis and a bracket.
ⓒ
Negative 3.75 is less than or equal to x which is less than or equal to 0. There is a closed circle at negative 3.75 and a closed circle at 0 and shading between negative 3.75 and 0 on the number line. The interval notation is negative 3.75 and 0 within brackets.

Solve Linear Inequalities

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.


ⓐ a+34≥710
ⓑ 8x>72
ⓒ 20>25h


ⓐ b+78≥16
ⓑ 6y<48
ⓒ 40<58k

Solution


ⓐ
The solution is b is greater than or equal to negative seventeen twenty-fourths. The solution on a number line has a left bracket negative seventeen twenty-fourths with shading to the right. The solution in interval notation is negative seventeen twenty-fourths to infinity within a bracket and a parenthesis.
ⓑ
The solution is y is less than 8. The solution on a number line has a right parenthesis at 8 with shading to the left. The solution in interval notation is negative infinity to 8 within parentheses.
ⓒ
The solution is k is greater than 64. The solution on a number line has a left parenthesis at 64 with shading to the right. The solution in interval notation is 64 to infinity within parentheses.


ⓐ f−1320<−512
ⓑ 9t≥−27
ⓒ 76j≥42


ⓐ g−1112<−518
ⓑ 7s<−28
ⓒ 94g≤36

Solution


ⓐ
The solution is g is less than twenty-three thirty-sixths. The solution on a number line has a right parenthesis at twenty-three thirty-sixths with shading to the left. The solution in interval notation is negative infinity to twenty-three thirty-sixths within parentheses.
ⓑ
The solution is s is less than negative 4. The solution on a number line has a right parenthesis at negative 4 with shading to the left. The solution in interval notation is negative infinity to negative 4 within parentheses.
ⓒ
The solution is g is less than or equal to 16. The solution on a number line has a right bracket at 16 with shading to the left. The solution in interval notation is negative infinity to 16 within parenthesis and a bracket.


ⓐ −5u≥65
ⓑ a−3≤9


ⓐ −8v≤96
ⓑ b−10≥30

Solution


ⓐ
The solution is v is greater than or equal to negative 12. The solution on a number line has a left bracket with shading to the right. The solution in interval notation is negative 12 to infinity within a bracket and a parenthesis.
ⓑ
The solution is b is less than or equal to negative 300. The solution on a number line has a right bracket at negative 300 with shading to the left. The solution in interval notation is negative infinity to negative 300 within a parenthesis and a bracket.


ⓐ −9c<126
ⓑ −25<p−5


ⓐ −7d>105
ⓑ −18>q−6

Solution


ⓐ
The solution is d is less than negative 15. The solution on a number line has a right parentheses with shading to the left. The solution in interval notation is negative infinity to negative 15 within parentheses.
ⓑ
The solution is q is greater than 108. The solution on a number line has a left parentheses at 108 with shading to the right. The solution in interval notation is 108 to infinity within parentheses.

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

4v≥9v−40

5u≤8u−21

Solution


The solution is u is greater than or equal to 7. The solution on a number line has a left bracket at 7 with shading to the right. The solution in interval notation is 7 to infinity within a a bracket and a parenthesis.

13q<7q−29

9p>14p−18

Solution


The solution is p is less than eighteen fifths. The solution on a number line has a right parenthesis at eighteen fifths with shading to the left. The solution in interval notation negative infinity to eighteen fifths within parentheses.

12x+3(x+7)>10x−24

9y+5(y+3)<4y−35

Solution


The solution is y is less than negative 5. The solution on a number line has a right parenthesis at negative 5 with shading to the left. The solution in interval notation is negative infinity to negative 5 within parentheses.

6h−4(h−1)≤7h−11

4k−(k−2)≥7k−26

Solution


The solution is k is less than or equal to 7. The solution on a number line has a right bracket at 7with shading to the left. The solution in interval notation is negative infinity to 7 within a parenthesis and a bracket.

8m−2(14−m)≥​7(m−4)+3m

6n−12(3−n)≤9(n−4)+9n

Solution


The inequality is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

34b−13b<512b−12

9u+5(2u−5)≥12(u−1)+7u

Solution


The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.

23g−12(g−14)≤16(g+42)

45h−23(h−9)≥115(2h+90)

Solution


The inequality is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

56a−14a>712a+23

12v+3(4v−1)≤19(v−2)+5v

Solution


The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

15k≤−40

35k≥−77

Solution


The solution is k is greater than or equal to negative eleven fifthss. The solution on a number line has a left bracket at negative eleven fifths with shading to the right. The solution in interval notation is negative eleven fifths to negative infinity within a bracket and a parenthesis.

23p−2(6−5p)>3(11p−4)

18q−4(10−3q)<5(6q−8)

Solution


The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.

−94x≥−512

−218y≤−1528

Solution


The solution is y is greater than or equal to ten twenty-ninths. The solution on a number line has a left bracket at ten twenty-ninths with shading to the right. The solution in interval notation is ten twenty-ninths to infinity within a bracket and a parenthesis.

c+34<−99

d+29>−61

Solution


The solution is g is greater than negative 90. The solution on a number line has a left parenthesis at negative 90 with shading to the right. The solution in interval notation is negative 90 to infinity within parentheses.

m18≥−4

n13≤−6

Solution


The solution is n is less than or equal to negative 78. The solution on a number line has a right bracket at negative 78 with shading to the left. The solution in interval notation is negative infinity to negative 78 within a parenthesis and a bracket.

Translate to an Inequality and Solve

In the following exercises, translate and solve. Then graph the solution on the number line and write the solution in interval notation.

Three more than h is no less than 25.

Six more than k exceeds 25.

Solution


The inequality is k plus 6 is greater than 25. Its solution is k is greater than 19. The solution on a number line has a left parenthesis at 19 with shading to the right. The solution in interval notation is 19 to infinity within parentheses.

Ten less than w is at least 39.

Twelve less than x is no less than 21.

Solution


The inequality is x minus 12 is greater than or equal to 21. Its solution is x is greater than or equal to 33. The solution on a number line has a left bracket at 33 with shading to the right. The solution in interval notation is 33 to infinity within a bracket and a parenthesis.

Negative five times r is no more than 95.

Negative two times s is lower than 56.

Solution


The inequality is negative 2 s is less than 56. Its solution is s is greater than negative 28. The solution on a number line has a left parenthesis at negative 28 with shading to the right. The solution in interval notation is negative 28 to infinity within parentheses.

Nineteen less than b is at most −22.

Fifteen less than a is at least −7.

Solution


The inequality is a minus 15 is greater than or equal to negative 7. Its solution is a is greater than or equal to 8. The solution on a number line has a left bracket at 8 with shading to the right. The solution in interval notation is 8 to infinity within a bracket and a parenthesis.

Solve Applications with Linear Inequalities

In the following exercises, solve.

Alan is loading a pallet with boxes that each weighs 45 pounds. The pallet can safely support no more than 900 pounds. How many boxes can he safely load onto the pallet?

The elevator in Yehire’s apartment building has a sign that says the maximum weight is 2100 pounds. If the average weight of one person is 150 pounds, how many people can safely ride the elevator?

Solution

A maximum of 14 people can safely ride in the elevator.

Andre is looking at apartments with three of his friends. They want the monthly rent to be no more than $2,360. If the roommates split the rent evenly among the four of them, what is the maximum rent each will pay?

Arleen got a $20 gift card for the coffee shop. Her favorite iced drink costs $3.79. What is the maximum number of drinks she can buy with the gift card?

Solution

five drinks

Teegan likes to play golf. He has budgeted $60 next month for the driving range. It costs him $10.55 for a bucket of balls each time he goes. What is the maximum number of times he can go to the driving range next month?

Ryan charges his neighbors $17.50 to wash their car. How many cars must he wash next summer if his goal is to earn at least $1,500?

Solution

86 cars

Keshad gets paid $2,400 per month plus 6% of his sales. His brother earns $3,300 per month. For what amount of total sales will Keshad’s monthly pay be higher than his brother’s monthly pay?

Kimuyen needs to earn $4,150 per month in order to pay all her expenses. Her job pays her $3,475 per month plus 4% of her total sales. What is the minimum Kimuyen’s total sales must be in order for her to pay all her expenses?

Solution

$16,875

Andre has been offered an entry-level job. The company offered him $48,000 per year plus 3.5% of his total sales. Andre knows that the average pay for this job is $62,000. What would Andre’s total sales need to be for his pay to be at least as high as the average pay for this job?

Nataly is considering two job offers. The first job would pay her $83,000 per year. The second would pay her $66,500 plus 15% of her total sales. What would her total sales need to be for her salary on the second offer be higher than the first?

Solution

$110,000

Jake’s water bill is $24.80 per month plus $2.20 per ccf (hundred cubic feet) of water. What is the maximum number of ccf Jake can use if he wants his bill to be no more than $60?

Kiyoshi’s phone plan costs $17.50 per month plus $0.15 per text message. What is the maximum number of text messages Kiyoshi can use so the phone bill is no more than $56.60?

Solution

260 messages

Marlon’s TV plan costs $49.99 per month plus $5.49 per first-run movie. How many first-run movies can he watch if he wants to keep his monthly bill to be a maximum of $100?

Kellen wants to rent a banquet room in a restaurant for her cousin’s baby shower. The restaurant charges $350 for the banquet room plus $32.50 per person for lunch. How many people can Kellen have at the shower if she wants the maximum cost to be $1,500?

Solution

35 people

Moshde runs a hairstyling business from her house. She charges $45 for a haircut and style. Her monthly expenses are $960. She wants to be able to put at least $1,200 per month into her savings account order to open her own salon. How many “cut & styles” must she do to save at least $1,200 per month?

Noe installs and configures software on home computers. He charges $125 per job. His monthly expenses are $1,600. How many jobs must he work in order to make a profit of at least $2,400?

Solution

32 jobs

Katherine is a personal chef. She charges $115 per four-person meal. Her monthly expenses are $3,150. How many four-person meals must she sell in order to make a profit of at least $1,900?

Melissa makes necklaces and sells them online. She charges $88 per necklace. Her monthly expenses are $3,745. How many necklaces must she sell if she wants to make a profit of at least $1,650?

Solution

62 necklaces

Five student government officers want to go to the state convention. It will cost them $110 for registration, $375 for transportation and food, and $42 per person for the hotel. There is $450 budgeted for the convention in the student government savings account. They can earn the rest of the money they need by having a car wash. If they charge $5 per car, how many cars must they wash in order to have enough money to pay for the trip?

Cesar is planning a four-day trip to visit his friend at a college in another state. It will cost him $198 for airfare, $56 for local transportation, and $45 per day for food. He has $189 in savings and can earn $35 for each lawn he mows. How many lawns must he mow to have enough money to pay for the trip?

Solution

seven lawns

Alonzo works as a car detailer. He charges $175 per car. He is planning to move out of his parents’ house and rent his first apartment. He will need to pay $120 for application fees, $950 for security deposit, and first and last months’ rent at $1,140 per month. He has $1,810 in savings. How many cars must he detail to have enough money to rent the apartment?

Eun-Kyung works as a tutor and earns $60 per hour. She has $792 in savings. She is planning an anniversary party for her parents. She would like to invite 40 guests. The party will cost her $1,520 for food and drinks and $150 for the photographer. She will also have a favor for each of the guests, and each favor will cost $7.50. How many hours must she tutor to have enough money for the party?

Solution

20 hours

Everyday Math

Maximum load on a stage In 2014, a high school stage collapsed in Fullerton, California, when 250 students got on stage for the finale of a musical production. Two dozen students were injured. The stage could support a maximum of 12,750 pounds. If the average weight of a student is assumed to be 140 pounds, what is the maximum number of students who could safely be on the stage?

Maximum weight on a boat In 2004, a water taxi sank in Baltimore harbor and five people drowned. The water taxi had a maximum capacity of 3,500 pounds (25 people with average weight 140 pounds). The average weight of the 25 people on the water taxi when it sank was 168 pounds per person. What should the maximum number of people of this weight have been?

Solution

20 people

Wedding budget Adele and Walter found the perfect venue for their wedding reception. The cost is $9850 for up to 100 guests, plus $38 for each additional guest. How many guests can attend if Adele and Walter want the total cost to be no more than $12,500?

Shower budget Penny is planning a baby shower for her daughter-in-law. The restaurant charges $950 for up to 25 guests, plus $31.95 for each additional guest. How many guests can attend if Penny wants the total cost to be no more than $1,500?

Solution

42 guests

Writing Exercises

Explain why it is necessary to reverse the inequality when solving −5x>10.

Explain why it is necessary to reverse the inequality when solving n−3<12.

Solution

Answers will vary.

Find your last month’s phone bill and the hourly salary you are paid at your job. Calculate the number of hours of work it would take you to earn at least enough money to pay your phone bill by writing an appropriate inequality and then solving it. Do you feel this is an appropriate number of hours? Is this the appropriate phone plan for you?

Find out how many units you have left, after this term, to achieve your college goal and estimate the number of units you can take each term in college. Calculate the number of terms it will take you to achieve your college goal by writing an appropriate inequality and then solving it. Is this an acceptable number of terms until you meet your goal? What are some ways you could accelerate this process?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was graph inequalities on the number line. In row 3, the I can was solve linear inequalities. In row 4, the I can was translate words to an inequality and solve. In row 5, the I can was solve applications with linear inequalities.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Solve Compound Inequalities

Learning Objectives

By the end of this section, you will be able to:

  • Solve compound inequalities with “and”
  • Solve compound inequalities with “or”
  • Solve applications with compound inequalities

Before you get started, take this readiness quiz.

Simplify: 25(x+10).
If you missed this problem, review Example 7 in Properties of Real Numbers.

Solution

25x+4

Simplify: −(x−4).
If you missed this problem, review Example 10 in Properties of Real Numbers.

Solution

−x+4

Solve Compound Inequalities with “and”

Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. A compound inequality is made up of two inequalities connected by the word “and” or the word “or.” For example, the following are compound inequalities.

x+3>−4and4x−5≤3
2(y+1)<0ory−5≥−2

Compound Inequality

A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”

To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions.

To solve a compound inequality with the word “and,” we look for all numbers that make both inequalities true. To solve a compound inequality with the word “or,” we look for all numbers that make either inequality true.

Let’s start with the compound inequalities with “and.” Our solution will be the numbers that are solutions to both inequalities known as the intersection of the two inequalities. Consider the intersection of two streets—the part where the streets overlap—belongs to both streets.

The figure is an illustration of two streets with their intersection shaded

To find the solution of the compound inequality, we look at the graphs of each inequality and then find the numbers that belong to both graphs—where the graphs overlap.

For the compound inequality x>−3 and x≤2, we graph each inequality. We then look for where the graphs “overlap”. The numbers that are shaded on both graphs, will be shaded on the graph of the solution of the compound inequality. See Figure 1.

The figure shows the graph of x is greater than negative 3 with a left parenthesis at negative 3 and shading to its right, the graph of x is less than or equal to 2 with a bracket at 2 and shading to its left, and the graph of x is greater than negative 3 and x is less than or equal to 2 with a left parenthesis at negative 3 and a right parenthesis at 2 and shading between negative 3 and 2. Negative 3 and 2 are marked by lines on each number line.

We can see that the numbers between −3 and 2 are shaded on both of the first two graphs. They will then be shaded on the solution graph.

The number −3 is not shaded on the first graph and so since it is not shaded on both graphs, it is not included on the solution graph.

The number two is shaded on both the first and second graphs. Therefore, it is be shaded on the solution graph.

This is how we will show our solution in the next examples.

Solve 6x−3<9 and 2x+9≥3. Graph the solution and write the solution in interval notation.

Solution
6x−3<9 and 2x+9≥3
Step 1. Solve each
inequality.
6x−3<9 2x+9≥3
6x<12 2x≥−6
x<2 and x≥−3
Step 2. Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs. This image displays three number lines. The first shows x < 2, the second shows x >= -3, and the third shows their intersection, which is -3 <= x < 2, illustrating how to combine two inequalities.
Step 3. Write the solution in interval notation. [−3,2)
All the numbers that make both inequalities true are the solution to the compound inequality.

Solve the compound inequality. Graph the solution and write the solution in interval notation: 4x−7<9 and 5x+8≥3.

Solution

The solution is negative 1 is less than or equal to x which is less than 4. On a number line it is shown with a closed circle at negative 1 and an open circle at 4 with shading in between the closed and open circles. Its interval notation is negative 1 to 4 within a bracket and a parenthesis.

Solve the compound inequality. Graph the solution and write the solution in interval notation: 3x−4<5 and 4x+9≥1.

Solution

The solution is negative 2 is less than or equal to x which is less than 3. On a number line it is shown with a closed circle at negative 2 and an open circle at 3 with shading in between the closed and open circles. Its interval notation is negative 2 to 3 within a bracket and a parenthesis.

Solve a compound inequality with “and.”

  1. Solve each inequality.
  2. Graph each solution. Then graph the numbers that make both inequalities true.
    This graph shows the solution to the compound inequality.
  3. Write the solution in interval notation.

Solve 3(2x+5)≤18 and 2(x−7)<−6. Graph the solution and write the solution in interval notation.

Solution
3(2x+5)≤18 and 2(x−7)<−6
Solve each
inequality.
6x+15≤18 2x−14<−6
6x≤3 2x<8
x≤12 and x<4
Graph each
solution.
Graph of two inequalities on number lines. The first shows x less than or equal to 1/2. The second shows x less than 4.
Graph the numbers
that make both
inequalities true.
A number line shows the inequality 'x < 1/2 and x < 4'. A dashed red line marks 1/2 with an open parenthesis facing left, and a thick dark arrow extends left from there, illustrating the solution x < 1/2.
Write the solution
in interval notation.
(−∞,12]

Solve the compound inequality. Graph the solution and write the solution in interval notation: 2(3x+1)≤20 and 4(x−1)<2.

Solution

The solution is x is less than three-halves. On a number line it is shown with an open circle at three-halves with shading to its left. Its interval notation is negative infinity to three-halves within a parentheses.

Solve the compound inequality. Graph the solution and write the solution in interval notation: 5(3x−1)≤10 and 4(x+3)<8.

Solution

The solution is x is less than negative 1. On a number line it is shown with an open circle at 1 with shading to its left. Its interval notation is negative infinity to negative 1 within parentheses.

Solve 13x−4≥−2 and −2(x−3)≥4. Graph the solution and write the solution in interval notation.

Solution
13x−4≥−2 and −2(x−3)≥4
Solve each inequality. 13x−4≥−2 −2x+6≥4
13x≥2 −2x≥−2
x≥6 and x≤1
Graph each solution. Two number lines illustrate inequalities. The top line shows x '>= ' 6, with a light blue arrow starting at 6 and pointing right. The bottom line shows x '<=' 1, with a dark blue arrow starting at 1 and pointing left.
Graph the numbers that
make both inequalities
true.
A number line from -1 to 9 illustrates the contradictory inequalities 'x greater than or equal to 6 and x less than or equal to 1'
There are no numbers that make both inequalities true.

This is a contradiction so there is no solution.

Solve the compound inequality. Graph the solution and write the solution in interval notation: 14x−3≥−1 and −3(x−2)≥2.

Solution

The inequality is a contradiction. So, there is no solution. As a result, there is no graph of the number line or interval notation.

Solve the compound inequality. Graph the solution and write the solution in interval notation: 15x−5≥−3 and −4(x−1)≥−2.

Solution

The inequality is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

Sometimes we have a compound inequality that can be written more concisely. For example, a<x and x<b can be written simply as a<x<b and then we call it a double inequality. The two forms are equivalent.

Double Inequality

A double inequality is a compound inequality such as a<x<b. It is equivalent to a<x and x<b.

Other forms:a<x<bis equivalent toa<xandx<ba≤x≤bis equivalent toa≤xandx≤ba>x>bis equivalent toa>xandx>ba≥x≥bis equivalent toa≥xandx≥b

To solve a double inequality we perform the same operation on all three “parts” of the double inequality with the goal of isolating the variable in the center.

Solve −4≤3x−7<8. Graph the solution and write the solution in interval notation.

Solution
A mathematical inequality is displayed: -4 is less than or equal to 3x minus 7, which is less than 8.
Add 7 to all three parts. An algebraic inequality '-4 <= 3x - 7 < 8' is shown with '+7' being added to all three parts, resulting in '-4 + 7 <= 3x - 7 + 7 < 8 + 7'.
Simplify. A mathematical inequality is shown, reading '3 ×<= 3x ×< 15' against a white background.
Divide each part by three. A mathematical inequality shown, with 3/3 less than or equal to 3x/3, which is less than 15/3. The denominators are highlighted in red, indicating a division step in solving the inequality.
Simplify. An indistinct mathematical inequality 1 <= x < 5.
Graph the solution. A number line shows an interval starting at 1 (inclusive) and ending at 5 (exclusive). The interval is represented by a solid teal line between a square bracket at 1 and a parenthesis at 5, on a number line ranging from -2 to 8.
Write the solution in interval notation. The range 1 inclusive to 5 exclusive is displayed.

When written as a double inequality, 1≤x<5, it is easy to see that the solutions are the numbers caught between one and five, including one, but not five. We can then graph the solution immediately as we did above.

Another way to graph the solution of 1≤x<5 is to graph both the solution of x≥1 and the solution of x<5. We would then find the numbers that make both inequalities true as we did in previous examples.

Solve the compound inequality. Graph the solution and write the solution in interval notation: −5≤4x−1<7.

Solution

The solution is negative 1 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 1 and an open circle at 2 with shading between the closed and open circles. Its interval notation is negative 1 to 2 within a bracket and a parenthesis.

Solve the compound inequality. Graph the solution and write the solution in interval notation: −3<2x−5≤1.

Solution

The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 with shading between the closed and open circles. Its interval notation is negative 1 to 3 within a parenthesis and a bracket.

Solve Compound Inequalities with “or”

To solve a compound inequality with “or”, we start out just as we did with the compound inequalities with “and”—we solve the two inequalities. Then we find all the numbers that make either inequality true.

Just as the United States is the union of all of the 50 states, the solution will be the union of all the numbers that make either inequality true. To find the solution of the compound inequality, we look at the graphs of each inequality, find the numbers that belong to either graph and put all those numbers together.

To write the solution in interval notation, we will often use the union symbol, ∪ to show the union of the solutions shown in the graphs.

Solve a compound inequality with “or.”

  1. Solve each inequality.
  2. Graph each solution. Then graph the numbers that make either inequality true.
  3. Write the solution in interval notation.

Solve 5−3x≤−1 or 8+2x≤5. Graph the solution and write the solution in interval notation.

Solution
5−3x≤−1 or 8+2x≤5
Solve each inequality. 5−3x≤−1 8+2x≤5
−3x≤−6 2x≤−3
x≥2 or x≤−32
Graph each solution. Two number lines display inequalities. The top line shows x >= 2, shaded light blue to the right. The bottom line shows x <= -3/2, shaded dark blue to the left.
Graph numbers that
make either inequality
true.
A number line illustrating the solution set for the inequality x '>= 2 or x <= -3/2, with a dark blue ray extending left from -3/2 (inclusive) and a light blue ray extending right from 2 (inclusive).
(−∞,−32]∪[2,∞)

Solve the compound inequality. Graph the solution and write the solution in interval notation: 1−2x≤−3 or 7+3x≤4.

Solution

The solution is x is greater than or equal to 2 or x is less than or equal to 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 2 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 2 and infinity within a bracket and a parenthesis.

Solve the compound inequality. Graph the solution and write the solution in interval notation: 2−5x≤−3 or 5+2x≤3.

Solution

The solution is x is greater than or equal to 1 or x is less than or equal to negative 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 1 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 1 and infinity within a bracket and a parenthesis.

Solve 23x−4≤3 or 14(x+8)≥−1. Graph the solution and write the solution in interval notation.

Solution
23x−4≤3 or 14(x+8)≥−1
Solve each
inequality.
3(23x−4)≤3(3) 4·14(x+8)≥4·(−1)
2x−12≤9 x+8≥−4
2x≤21 x≥−12
x≤212
x≤212 or x≥−12
Graph each
solution.
Two number lines display inequalities. The top line graphs x <= 21/2 (10.5), shaded left from a closed bracket at 10.5. The bottom line graphs x >= -12, shaded right from a closed bracket at -12.
Graph numbers
that make either
inequality true.
A number line graph representing the solution to the inequality 'x ×<= 21/2 or x ×>= -12'. The entire number line is highlighted in blue, indicating that all real numbers are solutions.
The solution covers all real numbers.
(−∞,∞)

Solve the compound inequality. Graph the solution and write the solution in interval notation: 35x−7≤−1 or 13(x+6)≥−2.

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Solve the compound inequality. Graph the solution and write the solution in interval notation: 34x−3≤3 or 25(x+10)≥0.

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Solve Applications with Compound Inequalities

Situations in the real world also involve compound inequalities. We will use the same problem solving strategy that we used to solve linear equation and inequality applications.

Recall the problem solving strategies are to first read the problem and make sure all the words are understood. Then, identify what we are looking for and assign a variable to represent it. Next, restate the problem in one sentence to make it easy to translate into a compound inequality. Last, we will solve the compound inequality.

Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the summer, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $57.06 and $171.02. How many hcf can the owner use if he wants his usage to stay in the normal range?

Solution
Identify what we are looking for. The number of hcf he can use and stay in the “normal usage” billing range.
Name what we are looking for. Let x= the number of hcf he can use.
Translate to an inequality. Bill is $24.72 plus $1.54 times the number of hcf he uses or 24.72+1.54x.
An image displays the text 'His bill will be between or equal to $57.06 and $171.02' with a light blue brace underneath it. Below the brace, the mathematical inequality '57.06 ', followed by the less than or equal to symbol, ' 24.74 + 1.54x ', then the less than or equal to symbol again, and finally ' 171.02' is shown.
Solve the inequality. The image illustrates the step-by-step solution of a compound inequality, beginning with 57.06  A  24.74 + 1.54x  A  171.02, subtracting 24.72 from all parts, and then dividing by 1.54 to isolate x, resulting in 21  A  x  A  95.
Answer the question. The property owner can use 21–95 hcf and still fall within the “normal usage” billing range.

Due to the drought in California, many communities now have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the summer, a property owner will pay $24.72 plus $1.32 per hcf for Conservation Usage. The bill for Conservation Usage would be between or equal to $31.32 and $52.12. How many hcf can the owner use if she wants her usage to stay in the conservation range?

Solution

The homeowner can use 5–20 hcf and still fall within the “conservation usage” billing range.

Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the winter, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $49.36 and $86.32. How many hcf will he be allowed to use if he wants his usage to stay in the normal range?

Solution

The homeowner can use 16–40 hcf and still fall within the “normal usage” billing range.

Access this online resource for additional instruction and practice with solving compound inequalities.

  • Compound inequalities

Key Concepts

  • How to solve a compound inequality with “and”
    1. Solve each inequality.
    2. Graph each solution. Then graph the numbers that make both inequalities true. This graph shows the solution to the compound inequality.
    3. Write the solution in interval notation.
  • Double Inequality
    • A double inequality is a compound inequality such as a<x<b. It is equivalent to a<x and x<b.
      Other forms:a<x<bis equivalent toa<xandx<ba≤x≤bis equivalent toa≤xandx≤ba>x>bis equivalent toa>xandx>ba≥x≥bis equivalent toa≥xandx≥b
  • How to solve a compound inequality with “or”
    1. Solve each inequality.
    2. Graph each solution. Then graph the numbers that make either inequality true.
    3. Write the solution in interval notation.

Practice Makes Perfect

Solve Compound Inequalities with “and”

In the following exercises, solve each inequality, graph the solution, and write the solution in interval notation.

x<3 and x≥1

x≤4 and x>−2

Solution

The solution is negative 2 is less than x which is less than or equal to 4. Its graph has an open circle at 1negative 2 and a closed circle at 4 with shading between the open and closed circles. Its interval notation is negative 2 to 4 within a parenthesis and a bracket.

x≥−4 and x≤−1

x>−6 and x<−3

Solution

The solution is negative 6 is less than x which is less than negative 3. Its graph has an open circle at negative 6 and an open circle at negative 3 with shading between open circles. Its interval notation is negative 6 to negative 3 within parentheses.

5x−2<8 and 6x+9≥3

4x−1<7 and 2x+8≥4

Solution

The solution is negative 2 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 2 and an open circle at 2 with shading between the closed and open circles. Its interval notation is negative 2 to 2 within a bracket and a parenthesis.

4x+6≤2 and
2x+1≥−5

4x−2≤4 and
7x−1>−8

Solution

The solution is negative 1 is less than x which is less than or equal to three-halves. Its graph has an open circle at negative 1 and a closed circle at three-halves with shading between the open and closed circles. Its interval notation is negative 1 to three-halves within a parenthesis and a bracket.

2x−11<5 and
3x−8>−5

7x−8<6 and
5x+7>−3

Solution

The solution is negative 2 is less than x which is less than 2. Its graph has an open circle at negative 2 and an open circle at 2 with shading between the open circles. Its interval notation is negative 2 to 2 within parentheses.

4(2x−1)≤12 and
2(x+1)<4

5(3x−2)≤5 and
3(x+3)<3

Solution

The solution is x is less than negative 2. Its graph has an open circle at negative 2 and is shaded to the left. Its interval notation is negative infinity to negative 2 within parentheses.

3(2x−3)>3 and
4(x+5)≥4

−3(x+4)<0 and
−1(3x−1)≤7

Solution

The solution is x is greater than or equal to negative 2. Its graph has a closed circle at negative 2 and is shaded to the right. Its interval notation is negative 2 to infinity within a bracket and a parenthesis.

12(3x−4)≤1 and
13(x+6)≤4

34(x−8)≤3 and
15(x−5)≤3

Solution

The solution is x is less than or equal to 12. Its graph has a closed circle at 12 and is shaded to the left. Its interval notation is negative infinity to 12 within a parenthesis and a bracket.

5x−2≤3x+4 and
3x−4≥2x+1

34x−5≥−2 and
−3(x+1)≥6

Solution

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

23x−6≥−4 and
−4(x+2)≥0

12(x−6)+2<−5 and
4−23x<6

Solution

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

−5≤4x−1<7

−3<2x−5≤1

Solution

The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 and is shaded between the open and closed circles. Its interval notation is 1 to 3 within a parenthesis and a bracket.

5<4x+1<9

−1<3x+2<8

Solution

The solution is negative 1 is less than x which is less than 2. Its graph has an open circle at negative 1 an open circle at 2 and is shaded between. Its interval notation is negative 1 to 2 within parentheses.

−8<5x+2≤−3

−6≤4x−2<−2

Solution

The solution is negative 1 is less than or equal to x which is less than or 0. Its graph has a closed circle at negative 1 and an open circle at 0 and is shaded between the closed and open circles. Its interval notation is negative 1 to 0 within a bracket and a parenthesis.

Solve Compound Inequalities with “or”

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

x≤−2 or x>3

x≤−4 or x>−3

Solution

The solution is x is less than or equal to negative 4 or x is greater than negative 3. The graph of the solutions on a number line has a closed circle at negative 4 and shading to the left and an open circle at negative 3 with shading to the right. The interval notation is the union of negative infinity to negative 4 within a parenthesis and a bracket and negative 3 and infinity within parentheses.

x<2 or x≥5

x<0 or x≥4

Solution

The solution is x is less than 0 or x is greater than or equal to 2. The graph of the solutions on a number line has an open circle at 0 and shading to the left and a closed circle at 4 with shading to the right. The interval notation is the union of negative infinity to 0 within parentheses and 4 to infinity within a bracket and parenthesis.

2+3x≤4 or
5−2x≤−1

4−3x≤−2 or
2x−1≤−5

Solution

The solution is x is less than or equal to negative 2 or x is greater than or equal to 2. The graph of the solutions on a number line has a closed circle at negative 2 and shading to the left and a closed circle at 2 with shading to the right. The interval notation is the union of negative infinity to negative 2 within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

2(3x−1)<4 or
3x−5>1

3(2x−3)<−5 or
4x−1>3

Solution

The solution is x is less than two-thirds or x is greater than 1. The graph of the solutions on a number line has an open circle at two-thirds and shading to the left and an open circle at 1 with shading to the right. The interval notation is the union of negative infinity to two-thirds within parentheses and 1 and infinity within parentheses.

34x−2>4 or 4(2−x)>0

23x−3>5 or 3(5−x)>6

Solution

The solution is x is less than 3 or x is greater than 12. The graph of the solutions on a number line has an open circle at 3 and shading to the left and an open circle at 12 with shading to the right. The interval notation is the union of negative infinity to 3 within parentheses and 12 and infinity within parentheses.

3x−2>4 or 5x−3≤7

2(x+3)≥0 or
3(x+4)≤6

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

12x−3≤4 or
13(x−6)≥−2

34x+2≤−1 or
12(x+8)≥−3

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Mixed practice

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

3x+7≤1 and
2x+3≥−5

6(2x−1)>6 and
5(x+2)≥0

Solution

The solution is x is less than 1. Its graph has an open circle at negative 1 is shaded to the right. Its interval notation is 1 to infinity within parentheses.

4−7x≥−3 or
5(x−3)+8>3

12x−5≤3 or
14(x−8)≥−3

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

−5≤2x−1<7

15(x−5)+6<4 and
3−23x<5

Solution

The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.

4x−2>6 or
3x−1≤−2

6x−3≤1 and
5x−1>−6

Solution

The solution is negative 1 is less than x which is less than or equal to two-thirds. Its graph has an open circle at negative 1 and a closed circle at two-thirds and is shaded between the open and closed circles. Its interval notation is negative 1 to two-thirds within a parenthesis and a bracket.

−2(3x−4)≤2 and
−4(x−1)<2

−5≤3x−2≤4

Solution

The solution is negative 1 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 1 and a closed circle at 2 and is shaded between the closed circles. Its interval notation is negative 1 to 4 within brackets.

Solve Applications with Compound Inequalities

In the following exercises, solve.

Penelope is playing a number game with her sister June. Penelope is thinking of a number and wants June to guess it. Five more than three times her number is between 2 and 32. Write a compound inequality that shows the range of numbers that Penelope might be thinking of.

Gregory is thinking of a number and he wants his sister Lauren to guess the number. His first clue is that six less than twice his number is between four and forty-two. Write a compound inequality that shows the range of numbers that Gregory might be thinking of.

Solution

5≤n≤24

Andrew is creating a rectangular dog run in his back yard. The length of the dog run is 18 feet. The perimeter of the dog run must be at least 42 feet and no more than 72 feet. Use a compound inequality to find the range of values for the width of the dog run.

Elouise is creating a rectangular garden in her back yard. The length of the garden is 12 feet. The perimeter of the garden must be at least 36 feet and no more than 48 feet. Use a compound inequality to find the range of values for the width of the garden.

Solution

6≤w≤12

Everyday Math

Blood Pressure A person’s blood pressure is measured with two numbers. The systolic blood pressure measures the pressure of the blood on the arteries as the heart beats. The diastolic blood pressure measures the pressure while the heart is resting.

ⓐ Let x be your systolic blood pressure. Research and then write the compound inequality that shows you what a normal systolic blood pressure should be for someone your age.

ⓑ Let y be your diastolic blood pressure. Research and then write the compound inequality that shows you what a normal diastolic blood pressure should be for someone your age.

Body Mass Index (BMI) is a measure of body fat is determined using your height and weight.

ⓐ Let x be your BMI. Research and then write the compound inequality to show the BMI range for you to be considered normal weight.

ⓑ Research a BMI calculator and determine your BMI. Is it a solution to the inequality in part (a)?

Solution

ⓐ answers vary ⓑ answers vary

Writing Exercises

In your own words, explain the difference between the properties of equality and the properties of inequality.

Explain the steps for solving the compound inequality 2−7x≥−5 or 4(x−3)+7>3.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve compound inequalities with “and.” In row 3, the I can was solve compound inequalities with “or.” In row 4, the I can was solve applications with compound inequalities.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

compound inequality
A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”

Solve Absolute Value Inequalities

Learning Objectives

By the end of this section, you will be able to:

  • Solve absolute value equations
  • Solve absolute value inequalities with “less than”
  • Solve absolute value inequalities with “greater than”
  • Solve applications with absolute value

Before you get started, take this readiness quiz.

Evaluate: −|7|.
If you missed this problem, review Example 1 in Integers.

Solution

−7

Fill in <,>, or = for each of the following pairs of numbers.
ⓐ |−8|___−|−8| ⓑ 12___−|−12| ⓒ |−6|___−6 ⓓ −(−15)___−|−15|
If you missed this problem, review Example 1 in Integers.

Solution

a. >; b. >; c. >; d. >

Simplify: 14−2|8−3(4−1)|.
If you missed this problem, review Example 2 in Integers.

Solution

12

Solve Absolute Value Equations

As we prepare to solve absolute value equations, we review our definition of absolute value.

Absolute Value

The absolute value of a number is its distance from zero on the number line.

The absolute value of a number n is written as |n| and |n|≥0 for all numbers.

Absolute values are always greater than or equal to zero.

We learned that both a number and its opposite are the same distance from zero on the number line. Since they have the same distance from zero, they have the same absolute value. For example:

−5 is 5 units away from 0, so |−5|=5.

5 is 5 units away from 0, so |5|=5.

Figure 1 illustrates this idea.

The figure is a number line with tick marks at negative 5, 0, and 5. The distance between negative 5 and 0 is given as 5 units, so the absolute value of negative 5 is 5. The distance between 5 and 0 is 5 units, so the absolute value of 5 is 5.
The numbers 5 and −5 are both five units away from zero.

For the equation |x|=5, we are looking for all numbers that make this a true statement. We are looking for the numbers whose distance from zero is 5. We just saw that both 5 and −5 are five units from zero on the number line. They are the solutions to the equation.

If|x|=5thenx=−5orx=5

The solution can be simplified to a single statement by writing x=±5. This is read, “x is equal to positive or negative 5”.

We can generalize this to the following property for absolute value equations.

Absolute Value Equations

For any algebraic expression, u, and any positive real number, a,

if|u|=athenu=−aoru=a

Remember that an absolute value cannot be a negative number.

Solve: ⓐ |x|=8 ⓑ |y|=−6 ⓒ |z|=0

Solution
ⓐ
Illustrates how to write equivalent equations for the absolute value equation |x|=8, showing its two solutions.
|x|=8
Write the equivalent equations. x=−8orx=8
x=±8
ⓑ
This table illustrates the absolute value equation |y| = -6, demonstrating it has no solution as absolute values are always non-negative.
|y|=−6
No solution
Since an absolute value is always positive, there are no solutions to this equation.
ⓒ
Steps for solving the absolute value equation |z|=0, illustrating the transformation to equivalent equations and the unique solution z=0.
|z|=0
Write the equivalent equations. z=−0orz=0
Since −0 = 0, z=0
Both equations tell us that z = 0 and so there is only one solution.

Solve: ⓐ |x|=2 ⓑ |y|=−4 ⓒ |z|=0

Solution

ⓐ ±2 ⓑ no solution ⓒ 0

Solve: ⓐ |x|=11 ⓑ |y|=−5 ⓒ |z|=0

Solution

ⓐ ±11 ⓑ no solution ⓒ 0

To solve an absolute value equation, we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations.

How to Solve Absolute Value Equations

Solve |5x−4|−3=8.

Solution
Step 1 is to isolate the absolute value expression. The difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Add 3 to both sides. The result is the absolute value of the quantity 5 x minus 4 is equal to 11. Step 2 is to write the equivalent equations, 5 x minus 4 is equal to negative 11 and 5 x minus 4 is equal to 11. Step 3 is to solve each equation. Add 4 to each side. 5 x is equal to negative 7 or 5 x is equal to 15. Divide each side by 5. The result is x is equal to negative seven-fifths or x is equal to 3. Step 4 is to check each solution. Substitute 3 and negative seven-fifths into the original equation, the difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Substitute 3 for x. Is the difference between the absolute value of the quantity 5 times 3 minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity 15 minus 4 and 3 equal to 8? Is the difference between the absolute value of the 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to 3 checks. Substitute negative seven-fifths for x. Is the difference between the absolute value of the quantity 5 times negative seven-fifths minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity negative 7 minus 4 and 3 equal to 8? Is the difference between the absolute value of the negative 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to negative seven-fifths checks.

Solve: |3x−5|−1=6.

Solution

x=4,x=−23

Solve: |4x−3|−5=2.

Solution

x=−1,x=52

The steps for solving an absolute value equation are summarized here.

Solve absolute value equations.

  1. Isolate the absolute value expression.
  2. Write the equivalent equations.
  3. Solve each equation.
  4. Check each solution.

Solve 2|x−7|+5=9.

Solution
2|x−7 |+5=9
Isolate the absolute value expression.       2|x−7 |=4
|x−7 |=2
Write the equivalent equations. x−7=−2 or x−7=2
Solve each equation. x=5 or x=9
Check:
Two solutions for an absolute value equation: checking x=5 and x=9 in 2|x-7|+5=9, both result in a true statement, 9=9, confirming they are valid solutions.

Solve: 3|x−4|−4=8.

Solution

x=8,x=0

Solve: 2|x−5|+3=9.

Solution

x=8,x=2

Remember, an absolute value is always positive!

Solve: |23x−4|+11=3.

Solution
Steps for solving an absolute value equation resulting in no solution.
|23x−4|+11=3
Isolate the absolute value term. |23x−4|=−8
An absolute value cannot be negative. No solution

Solve: |34x−5|+9=4.

Solution

No solution

Solve: |56x+3|+8=6.

Solution

No solution

Some of our absolute value equations could be of the form |u|=|v| where u and v are algebraic expressions. For example, |x−3|=|2x+1|.

How would we solve them? If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The property for absolute value equations says that for any algebraic expression, u, and a positive real number, a, if |u|=a, then u=−a or u=a.

This tells us that

if |u|=|v| then u=−v oru=v

This leads us to the following property for equations with two absolute values.

Equations with Two Absolute Values

For any algebraic expressions, u and v,

if|u|=|v|thenu=−voru=v

When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed.

Solve: |5x−1|=|2x+3|.

Solution
This table illustrates the step-by-step process of solving the absolute value equation |5x - 1| = |2x + 3|, from setting up equivalent equations to finding the solutions.
|5x−1|=|2x+3|
Write the equivalent equations.
Solve each equation.
5x−1=−(2x+3)or5x−1=2x+35x−1=−2x−3or3x−1=37x−1=−33x=47x=−2x=43x=−27orx=43
Check.
We leave the check to you.

Solve: |7x−3|=|3x+7|.

Solution

x=−25, x=52

Solve: |6x−5|=|3x+4|.

Solution

x=3, x=19

Solve Absolute Value Inequalities with “Less Than”

Let’s look now at what happens when we have an absolute value inequality. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work.

Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation |x|=5, we saw that both 5 and −5 are five units from zero on the number line. They are the solutions to the equation.

|x|=5x=−5orx=5

What about the inequality |x|≤5? Where are the numbers whose distance is less than or equal to 5? We know −5 and 5 are both five units from zero. All the numbers between −5 and 5 are less than five units from zero. See Figure 2.

The figure is a number line with negative 5, 0, and 5 displayed. There is a left bracket at negative 5 and a right bracket at 5. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x which is less than or equal to 5.

In a more general way, we can see that if |u|≤a, then −a≤u≤a. See Figure 3.

The figure is a number line with negative a 0, and a displayed. There is a left bracket at negative a and a right bracket at a. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is less than or equal to a, then negative a is less than or equal to u which is less than or equal to a.

This result is summarized here.

Absolute Value Inequalities with < or ≤

For any algebraic expression, u, and any positive real number, a,

if|u|<a,then−a<u<aif|u|≤a,then−a≤u≤a

After solving an inequality, it is often helpful to check some points to see if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete check, it often helps verify the solution.

Solve |x|<7. Graph the solution and write the solution in interval notation.

Solution
A mathematical inequality is shown in the center of a white background, which states the absolute value of x is less than 7, written as |x| < 7.
Write the equivalent inequality. A mathematical inequality is displayed with the expression -7 < |X| < 7. The characters are rendered in a standard mathematical font on a white background.
Graph the solution. A number line illustrates the open interval (-7, 7), where all real numbers between -7 and 7 are included, but the endpoints -7 and 7 are excluded. The interval is highlighted in teal.
Write the solution using interval notation. The image displays the mathematical ordered pair or interval notation '(-7, 7)' centered on a plain white background.

Check:

To verify, check a value in each section of the number line showing the solution. Choose numbers such as −8, 1, and 9.

The figure is a number line with a left parenthesis at negative 7, a right parenthesis at 7 and shading between the parentheses. The values negative 8, 1, and 9 are marked with points. The absolute value of negative 8 is less than 7 is false. It does not satisfy the absolute value of x is less than 7. The absolute value of 1 is less than 7 is true. It does satisfy the absolute value of x is less than 7. The absolute value of 9 is less than 7 is false. It does not satisfy the absolute value of x is less than 7.

Graph the solution and write the solution in interval notation: |x|<9.

Solution

The solution is negative 9 is less than x which is less than 9. The number line shows open circles at negative 9 and 9 with shading in between the circles. The interval notation is negative 9 to 9 within parentheses.

Graph the solution and write the solution in interval notation: |x|<1.

Solution

The solution is negative 1 is less than x which is less than 1. The number line shows open circles at negative 1 and 1 with shading in between the circles. The interval notation is negative 1 to 1 within parentheses.

Solve |5x−6|≤4. Graph the solution and write the solution in interval notation.

Solution
Step 1. Isolate the absolute value expression.
It is isolated.
|5x−6|≤4
Step 2. Write the equivalent compound inequality. −4≤5x−6≤4
Step 3. Solve the compound inequality. 2≤5x≤10
25≤x≤2
Step 4. Graph the solution. A number line displaying the closed interval from 2/5 to 2.0, inclusive, with additional markers at 0 and 1.0.
Step 5. Write the solution using interval notation. [25,2]
Check:
The check is left to you.

Solve |2x−1|≤5. Graph the solution and write the solution in interval notation:

Solution

The solution is negative 2 is less than or equal to x which is less than or equal to 3. The number line shows closed circles at negative 2 and 3 with shading between the circles. The interval notation is negative 2 to 3 within brackets.

Solve |4x−5|≤3. Graph the solution and write the solution in interval notation:

Solution

The solution is one-half is less than or equal to x which is less than or equal to 2. The number line shows closed circles at one-half and 2 with shading between the circles. The interval notation is one-half to 2 within brackets.

Solve absolute value inequalities with < or ≤.

  1. Isolate the absolute value expression.
  2. Write the equivalent compound inequality.
    |u|<ais equivalent to−a<u<a|u|≤ais equivalent to−a≤u≤a
  3. Solve the compound inequality.
  4. Graph the solution
  5. Write the solution using interval notation.

Solve Absolute Value Inequalities with “Greater Than”

What happens for absolute value inequalities that have “greater than”? Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line.

We started with the inequality |x|≤5. We saw that the numbers whose distance is less than or equal to five from zero on the number line were −5 and 5 and all the numbers between −5 and 5. See Figure 4.

The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its right and a right bracket at 5 with shading to its left. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x is less than or equal to 5.

Now we want to look at the inequality |x|≥5. Where are the numbers whose distance from zero is greater than or equal to five?

Again both −5 and 5 are five units from zero and so are included in the solution. Numbers whose distance from zero is greater than five units would be less than −5 and greater than 5 on the number line. See Figure 5.

The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its left and a left bracket at 5 with shading to its right. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is greater than or equal to 5, then x is less than or equal to negative 5 or x is greater than or equal to 5.

In a more general way, we can see that if |u|≥a, then u≤−a or u≥a. See Figure 6.

The figure is a number line with negative a, 0, and a displayed. There is a right bracket at negative a that has shading to its left and a left bracket at a with shading to its right. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is greater than or equal to a, then u is less than or equal to negative a or u is greater than or equal to a.

This result is summarized here.

Absolute Value Inequalities with > or ≥

For any algebraic expression, u, and any positive real number, a,

if|u|>a,thenu<−aoru>aif|u|≥a,thenu≤−aoru≥a

Solve |x|>4. Graph the solution and write the solution in interval notation.

Solution
|x|>4
Write the equivalent inequality. x<−4orx>4
Graph the solution. A number line illustrates two inequalities: x < -4, shown in dark teal from -10 to -4 (exclusive), and x > 4, shown in light blue from 4 (exclusive) to 10. Arrows indicate extension to negative and positive infinity.
Write the solution using interval notation. (−∞,−4)∪(4,∞)
Check:

To verify, check a value in each section of the number line showing the solution. Choose numbers such as −6, 0, and 7.

The figure is a number line with a right parenthesis at negative 4 with shading to its left and a left parenthesis at 4 shading to its right. The values negative 6, 0, and 7 are marked with points. The absolute value of negative 6 is greater than negative 4 is true. It does not satisfy the absolute value of x is greater than 4. The absolute value of 0 is greater than 4 is false. It does not satisfy the absolute value of x is greater than 4. The absolute value of 7 is less than 4 is true. It does satisfy the absolute value of x is greater than 4.

Solve |x|>2. Graph the solution and write the solution in interval notation.

Solution

The solution is x is less than negative 2 or x is greater than 2. The number line shows an open circle at negative 2 with shading to its left and an open circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 2 to infinity within parentheses.

Solve |x|>1. Graph the solution and write the solution in interval notation.

Solution

The solution is x is less than negative 1 or x is greater than 1. The number line shows an open circle at negative 1 with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to negative 1 within parentheses and 1 to infinity within parentheses.

Solve |2x−3|≥5. Graph the solution and write the solution in interval notation.

Solution
|2x−3|≥5
Step 1. Isolate the absolute value expression. It is isolated.
Step 2. Write the equivalent compound inequality. 2x−3≤−5or2x−3≥5
Step 3. Solve the compound inequality. 2x≤−2or2x≥8
x≤−1orx≥4
Step 4. Graph the solution. A number line illustrates two separate intervals: numbers less than -3 (dark blue arrow extending left) and numbers greater than 5 (light blue arrow extending right), with labels at 0, 5, and 10.
Step 5. Write the solution using interval notation. (−∞,−1]∪[4,∞)
Check:
The check is left to you.

Solve |4x−3|≥5. Graph the solution and write the solution in interval notation.

Solution

The solution is x is less than or equal to negative one-half or x is greater than or equal 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis

Solve |3x−4|≥2. Graph the solution and write the solution in interval notation.

Solution

The solution is x is less than or equal to two-thirds or x is greater than or equal 2. The number line shows a closed circle at two-thirds with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to two-thirds within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

Solve absolute value inequalities with > or ≥.

  1. Isolate the absolute value expression.
  2. Write the equivalent compound inequality.
    |u|>ais equivalent tou<−aoru>a|u|≥ais equivalent tou≤−aoru≥a
  3. Solve the compound inequality.
  4. Graph the solution
  5. Write the solution using interval notation.

Solve Applications with Absolute Value

Absolute value inequalities are often used in the manufacturing process. An item must be made with near perfect specifications. Usually there is a certain tolerance of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected.

|actual-ideal|≤tolerance

The ideal diameter of a rod needed for a machine is 60 mm. The actual diameter can vary from the ideal diameter by 0.075 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Solution
Demonstrates solving an absolute value inequality problem, from setup to finding the final measurement range.
Let x = the actual measurement.
Use an absolute value inequality to express this situation. |actual-ideal|≤tolerance
|x−60|≤0.075
Rewrite as a compound inequality. −0.075≤x−60≤0.075
Solve the inequality. 59.925≤x≤60.075
Answer the question. The diameter of the rod can be between 59.925 mm and 60.075 mm.

The ideal diameter of a rod needed for a machine is 80 mm. The actual diameter can vary from the ideal diameter by 0.009 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Solution

The diameter of the rod can be between 79.991 and 80.009 mm.

The ideal diameter of a rod needed for a machine is 75 mm. The actual diameter can vary from the ideal diameter by 0.05 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Solution

The diameter of the rod can be between 74.95 and 75.05 mm.

Access this online resource for additional instruction and practice with solving linear absolute value equations and inequalities.

  • Solving Linear Absolute Value Equations and Inequalities

Key Concepts

  • Absolute Value
    The absolute value of a number is its distance from 0 on the number line.
    The absolute value of a number n is written as |n| and |n|≥0 for all numbers.
    Absolute values are always greater than or equal to zero.
  • Absolute Value Equations
    For any algebraic expression, u, and any positive real number, a,
    if|u|=athenu=−aoru=a
    Remember that an absolute value cannot be a negative number.
  • How to Solve Absolute Value Equations
    1. Isolate the absolute value expression.
    2. Write the equivalent equations.
    3. Solve each equation.
    4. Check each solution.
  • Equations with Two Absolute Values
    For any algebraic expressions, u and v,
    if|u|=|v|thenu=−voru=v
  • Absolute Value Inequalities with < or ≤
    For any algebraic expression, u, and any positive real number, a,
    if|u|<a,then−a<u<aif|u|≤a,then−a≤u≤a
  • How To Solve Absolute Value Inequalities with < or ≤
    1. Isolate the absolute value expression.
    2. Write the equivalent compound inequality.
      |u|<ais equivalent to−a<u<a|u|≤ais equivalent to−a≤u≤a
    3. Solve the compound inequality.
    4. Graph the solution
    5. Write the solution using interval notation
  • Absolute Value Inequalities with > or ≥
    For any algebraic expression, u, and any positive real number, a,
    if|u|>a,thenu<−aoru>aif|u|≥a,thenu≤−aoru≥a
  • How To Solve Absolute Value Inequalities with > or ≥
    1. Isolate the absolute value expression.
    2. Write the equivalent compound inequality.
      |u|>ais equivalent tou<−aoru>a|u|≥ais equivalent tou≤−aoru≥a
    3. Solve the compound inequality.
    4. Graph the solution
    5. Write the solution using interval notation

Section Exercises

Practice Makes Perfect

Solve Absolute Value Equations

In the following exercises, solve.

ⓐ |x|=6 ⓑ |y|=−3 ⓒ |z|=0

ⓐ |x|=4 ⓑ |y|=−5 ⓒ |z|=0

Solution

ⓐ x=4,x=−4 ⓑ no solution ⓒ z=0

ⓐ |x|=7 ⓑ |y|=−11 ⓒ |z|=0

ⓐ |x|=3 ⓑ |y|=−1 ⓒ |z|=0

Solution

ⓐ x=3,x=−3 ⓑ no solution ⓒ z=0

|2x−3|−4=1

|4x−1|−3=0

Solution

x=1,x=−12

|3x−4|+5=7

|4x+7|+2=5

Solution

x=−1,x=−52

4|x−1|+2=10

3|x−4|+2=11

Solution

x=7,x=1

3|4x−5|−4=11

3|x+2|−5=4

Solution

x=1,x=−5

−2|x−3|+8=−4

−3|x−4|+4=−5

Solution

x=7,x=1

|34x−3|+7=2

|35x−2|+5=2

Solution

no solution

|12x+5|+4=1

|14x+3|+3=1

Solution

no solution

|3x−2|=|2x−3|

|4x+3|=|2x+1|

Solution

x=−1,x=−23

|6x−5|=|2x+3|

|6−x|=|3−2x|

Solution

x=−3,x=3

Solve Absolute Value Inequalities with “less than”

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

|x|<5

|x|<1

Solution

The solution is negative 1 is less than x which is less than 1. The number line shows an open circle at negative 1, an open circle at 1, and shading between the circles. The interval notation is negative 1 to 1 within parentheses.

|x|≤8

|x|≤3

Solution

The solution is negative 3 is less than or equal to x which is less than or equal to 3. The number line shows a closed circle at negative 3, a closed circle at 3, and shading between the circles. The interval notation is negative 3 to 3 within brackets.

|3x−3|≤6

|2x−5|≤3

Solution

The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading between the circles. The interval notation is 1 to 4 within brackets.

|2x+3|+5<4

|3x−7|+3<1

Solution

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

|4x−3|<1

|6x−5|<7

Solution

The solution is negative one-third is less than x which is less than 2. The number line shows an open circle at negative one-half, an open circle at 2, and shading between the circles. The interval notation is negative one-third to 2 within parentheses.

|x−4|≤−1

|5x+1|≤−2

Solution

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

Solve Absolute Value Inequalities with “greater than”

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

|x|>3

|x|>6

Solution

The solution is x is less than negative 6 or x is greater than 6. The number line shows an open circle at negative 6 with shading to its left and an open circle at 6 with shading to its right. The interval notation is the union of negative infinity to negative 6 within parentheses and 6 to infinity within parentheses

|x|≥2

|x|≥5

Solution

The solution is x is less then or equal to negative 5 or x is greater than or equal to 5. The number line shows an open circle at negative 5 with shading to its left and an open circle at 5 with shading to its right. The interval notation is the union of negative infinity to negative 5 within parentheses and 5 to infinity within parentheses.

|3x−8|>−1

|x−5|>−2

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

|3x−2|>4

|2x−1|>5

Solution

The solution is x is less than negative 2 or x is greater than 3. The number line shows an open circle at negative 2 with shading to its left and an open circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 3 to infinity within parentheses.

|x+3|≥5

|x−7|≥1

Solution

The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to 6 within parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.

3|x|+4≥1

5|x|+6≥1

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

In the following exercises, solve. For each inequality, also graph the solution and write the solution in interval notation.

2|x+6|+4=8

|6x−5|=|2x+3|

Solution

x=2,x=14

|3x−4|≥2

|2x−5|+2=3

Solution

x=3,x=2

|4x−3|<5

|3x+1|−3=7

Solution

x=3,x=−113

|7x+2|+8<4

5|2x−1|−3=7

Solution

x=32,x=−12

|8−x|=|4−3x|

|x−7|>−3

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Solve Applications with Absolute Value

In the following exercises, solve.

A chicken farm ideally produces 200,000 eggs per day. But this total can vary by as much as 25,000 eggs. What is the maximum and minimum expected production at the farm?

An organic juice bottler ideally produces 215,000 bottle per day. But this total can vary by as much as 7,500 bottles. What is the maximum and minimum expected production at the bottling company?

Solution

The minimum to maximum expected production is 207,500 to 2,225,000 bottles

In order to insure compliance with the law, Miguel routinely overshoots the weight of his tortillas by 0.5 gram. He just received a report that told him that he could be losing as much as $100,000 per year using this practice. He now plans to buy new equipment that guarantees the thickness of the tortilla within 0.005 inches. If the ideal thickness of the tortilla is 0.04 inches, what thickness of tortillas will be guaranteed?

At Lilly’s Bakery, the ideal weight of a loaf of bread is 24 ounces. By law, the actual weight can vary from the ideal by 1.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined?

Solution

The acceptable weight is 22.5 to 25.5 ounces.

Writing Exercises

Write a graphical description of the absolute value of a number.

In your own words, explain how to solve the absolute value inequality, |3x−2|≥4.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve absolute value equations. In row 3, the I can was solve absolute value inequalities with “less than.” In row 4, the I can was solve absolute value inequalities with “greater than.” In row 5, the I can was solve applications with absolute value.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Chapter Review Exercises

Use a General Strategy to Solve Linear Equations

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, determine whether each number is a solution to the equation.

10x−1=5x,x=15

−12n+5=8n,n=−54

Solution

no

In the following exercises, solve each linear equation.

6(x+6)=24

−(s+4)=18

Solution

s=−22

23−3(y−7)=8

13(6m+21)=m−7

Solution

m=−14

4(3.5y+0.25)=365

0.25(q−8)=0.1(q+7)

Solution

q=18

8(r−2)=6(r+10)

5+7(2−5x)=2(9x+1)−(13x−57)

Solution

x=−1

(9n+5)−(3n−7)=20−(4n−2)

2[−16+5(8k−6)]=8(3−4k)−32

Solution

k=34

Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

17y−3(4−2y)=11(y−1)+12y−1

9u+32=15(u−4)−3(2u+21)

Solution

contradiction; no solution

−8(7m+4)=−6(8m+9)

Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation.

25n−110=710

Solution

n=2

34a−13=12a+56

12(k+3)=13(k+16)

Solution

k=23

5y−13+4=−8y+46

0.8x−0.3=0.7x+0.2

Solution

x=5

0.10d+0.05(d−4)=2.05

Use a Problem-Solving Strategy

Use a Problem Solving Strategy for Word Problems

In the following exercises, solve using the problem solving strategy for word problems.

Three-fourths of the people at a concert are children. If there are 87 children, what is the total number of people at the concert?

Solution

There are 116 people.

There are nine saxophone players in the band. The number of saxophone players is one less than twice the number of tuba players. Find the number of tuba players.

Solve Number Word Problems

In the following exercises, solve each number word problem.

The sum of a number and three is forty-one. Find the number.

Solution

38

One number is nine less than another. Their sum is negative twenty-seven. Find the numbers.

One number is two more than four times another. Their sum is negative thirteen. Find the numbers.

Solution

−3,−10

The sum of two consecutive integers is −135. Find the numbers.

Find three consecutive even integers whose sum is 234.

Solution

76, 78, 80

Find three consecutive odd integers whose sum is 51.

Koji has $5,502 in his savings account. This is $30 less than six times the amount in his checking account. How much money does Koji have in his checking account?

Solution

$922

Solve Percent Applications

In the following exercises, translate and solve.

What number is 67% of 250?

12.5% of what number is 20?

Solution

160

What percent of 125 is 150?

In the following exercises, solve.

The bill for Dino’s lunch was $19.45. He wanted to leave 20% of the total bill as a tip. How much should the tip be?

Solution

$3.89

Dolores bought a crib on sale for $350. The sale price was 40% of the original price. What was the original price of the crib?

Jaden earns $2,680 per month. He pays $938 a month for rent. What percent of his monthly pay goes to rent?

Solution

35%

Angel received a raise in his annual salary from $55,400 to $56,785. Find the percent change.

Rowena’s monthly gasoline bill dropped from $83.75 last month to $56.95 this month. Find the percent change.

Solution

32%

Emmett bought a pair of shoes on sale at 40% off from an original price of $138. Find ⓐ the amount of discount and ⓑ the sale price.

Lacey bought a pair of boots on sale for $95. The original price of the boots was $200. Find ⓐ the amount of discount and ⓑ the discount rate. (Round to the nearest tenth of a percent, if needed.)

Solution

ⓐ $105 ⓑ 52.5%

Nga and Lauren bought a chest at a flea market for $50. They re-finished it and then added a 350% mark-up. Find ⓐ the amount of the mark-up and ⓑ the list price.

Solve Simple Interest Applications

In the following exercises, solve.

Winston deposited $3,294 in a bank account with interest rate 2.6% How much interest was earned in five years?

Solution

$428.22

Moira borrowed $4,500 from her grandfather to pay for her first year of college. Three years later, she repaid the $4,500 plus $243 interest. What was the rate of interest?

Jaime’s refrigerator loan statement said he would pay $1,026 in interest for a four-year loan at 13.5%. How much did Jaime borrow to buy the refrigerator?

Solution

$1,900

Solve a formula for a Specific Variable

Solve a Formula for a Specific Variable

In the following exercises, solve the formula for the specified variable.

Solve the formula
V=LWH for L.

Solve the formula
A=12d1d2 for d2.

Solution

d2=2Ad1

Solve the formula
h=48t+12at2 for a.

Solve the formula
4x−3y=12 for y.

Solution

y=4x3−4

Use Formulas to Solve Geometry Applications

In the following exercises, solve using a geometry formula.

What is the height of a triangle with area 67.5 square meters and base 9 meters?

The measure of the smallest angle in a right triangle is 45° less than the measure of the next larger angle. Find the measures of all three angles.

Solution

22.5°,67.5°,90°

The perimeter of a triangle is 97 feet. One side of the triangle is eleven feet more than the smallest side. The third side is six feet more than twice the smallest side. Find the lengths of all sides.

Find the length of the hypotenuse.

The figure is a right triangle with a base of 10 units and a height of 24 units.
Solution

26

Find the length of the missing side. Round to the nearest tenth, if necessary.

The figure is a right triangle with a height of 15 units and a hypotenuse of 17 units.

Sergio needs to attach a wire to hold the antenna to the roof of his house, as shown in the figure. The antenna is eight feet tall and Sergio has 10 feet of wire. How far from the base of the antenna can he attach the wire? Approximate to the nearest tenth, if necessary.

The figure is a right triangle with a height of 8 feet and a hypotenuse of 10 feet.
Solution

6 feet

Seong is building shelving in his garage. The shelves are 36 inches wide and 15 inches tall. He wants to put a diagonal brace across the back to stabilize the shelves, as shown. How long should the brace be?

The figure illustrates rectangular shelving whose width of 36 inch and height of 15 inches forms a right triangle with a diagonal brace.

The length of a rectangle is 12 cm more than the width. The perimeter is 74 cm. Find the length and the width.

Solution

24.5 cm, 12.5 cm

The width of a rectangle is three more than twice the length. The perimeter is 96 inches. Find the length and the width.

The perimeter of a triangle is 35 feet. One side of the triangle is five feet longer than the second side. The third side is three feet longer than the second side. Find the length of each side.

Solution

9 ft, 14 ft, 12 ft

Solve Mixture and Uniform Motion Applications

Solve Coin Word Problems

In the following exercises, solve.

Paulette has $140 in $5 and $10 bills. The number of $10 bills is one less than twice the number of $5 bills. How many of each does she have?

Lenny has $3.69 in pennies, dimes, and quarters. The number of pennies is three more than the number of dimes. The number of quarters is twice the number of dimes. How many of each coin does he have?

Solution

nine pennies, six dimes, 12 quarters

Solve Ticket and Stamp Word Problems

In the following exercises, solve each ticket or stamp word problem.

Tickets for a basketball game cost $2 for students and $5 for adults. The number of students was three less than 10 times the number of adults. The total amount of money from ticket sales was $619. How many of each ticket were sold?

125 tickets were sold for the jazz band concert for a total of $1,022. Student tickets cost $6 each and general admission tickets cost $10 each. How many of each kind of ticket were sold?

Solution

57 students, 68 general admission

Yumi spent $34.15 buying stamps. The number of $0.56 stamps she bought was 10 less than four times the number of $0.41 stamps. How many of each did she buy?

Solve Mixture Word Problems

In the following exercises, solve.

Marquese is making 10 pounds of trail mix from raisins and nuts. Raisins cost $3.45 per pound and nuts cost $7.95 per pound. How many pounds of raisins and how many pounds of nuts should Marquese use for the trail mix to cost him $6.96 per pound?

Solution

2.2 lbs of raisins, 7.8 lbs of nuts

Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tile. She will use basic tiles that cost $8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be $10 per square foot?

Enrique borrowed $23,500 to buy a car. He pays his uncle 2% interest on the $4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total $23,500? (Round your answer to the nearest tenth of a percent.)

Solution

9.7%

Solve Uniform Motion Applications

In the following exercises, solve.

When Gabe drives from Sacramento to Redding it takes him 2.2 hours. It takes Elsa two hours to drive the same distance. Elsa’s speed is seven miles per hour faster than Gabe’s speed. Find Gabe’s speed and Elsa’s speed.

Louellen and Tracy met at a restaurant on the road between Chicago and Nashville. Louellen had left Chicago and drove 3.2 hours towards Nashville. Tracy had left Nashville and drove 4 hours towards Chicago, at a speed one mile per hour faster than Louellen’s speed. The distance between Chicago and Nashville is 472 miles. Find Louellen’s speed and Tracy’s speed.

Solution

Louellen 65 mph, Tracy 66 mph

Two busses leave Amarillo at the same time. The Albuquerque bus heads west on the I-40 at a speed of 72 miles per hour, and the Oklahoma City bus heads east on the I-40 at a speed of 78 miles per hour. How many hours will it take them to be 375 miles apart?

Kyle rowed his boat upstream for 50 minutes. It took him 30 minutes to row back downstream. His speed going upstream is two miles per hour slower than his speed going downstream. Find Kyle’s upstream and downstream speeds.

Solution

upstream 3 mph, downstream 5 mph

At 6:30, Devon left her house and rode her bike on the flat road until 7:30. Then she started riding uphill and rode until 8:00. She rode a total of 15 miles. Her speed on the flat road was three miles per hour faster than her speed going uphill. Find Devon’s speed on the flat road and riding uphill.

Anthony drove from New York City to Baltimore, which is a distance of 192 miles. He left at 3:45 and had heavy traffic until 5:30. Traffic was light for the rest of the drive, and he arrived at 7:30. His speed in light traffic was four miles per hour more than twice his speed in heavy traffic. Find Anthony’s driving speed in heavy traffic and light traffic.

Solution

heavy traffic 32 mph, light traffic 68 mph

Solve Linear Inequalities

Graph Inequalities on the Number Line

In the following exercises, graph the inequality on the number line and write in interval notation.

x<−1

x≥−2.5

Solution

The solution is x is greater than or equal to negative 2.5. The number line shows a left bracket at negative 2.5 with shading to its right. The interval notation is negative 2.5 to infinity within a bracket and a parenthesis.

x≤54

x>2

Solution

The solution is x is greater than 2. The number line shows a left parenthesis at 2 with shading to its right. The interval notation is 2 to infinity within parentheses.

−2<x<0

−5≤x<−3

Solution

The solution is negative 5 is less than or equal to x which is less than negative 3. The number line shows a closed circle at negative 5, an open circle at negative 3, and shading between the circles. The interval notation is negative 5 to negative 3 within a bracket and a parenthesis.

0≤x≤3.5

Solve Linear Inequalities

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

n−12≤23

Solution

The solution is n is less than or equal to 35. The number line shows a a right bracket at 35 with shading to its left. The interval notation is negative infinity to 35 within a parenthesis and a bracket.

a+23≥712

9x>54

Solution

The solution is x is greater than 6. The number line shows a left parenthesis at 6 with shading to its right. The interval notation is 6 to infinity within parentheses.

q−2≥−24

6p>15p−30

Solution

The solution is p is less than ten-thirds. The number line shows a right parenthesis at ten-thirds with shading to its left. The interval notation is negative infinity to ten-thirds within parentheses.

9h−7(h−1)≤4h−23

5n−15(4−n)≤10(n−6)+10n

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

38a−112a>512a+34

Translate Words to an Inequality and Solve

In the following exercises, translate and solve. Then write the solution in interval notation and graph on the number line.

Five more than z is at most 19.

Solution

The inequality is z plus 5 is less than or equal to 19. Its solution is z is less than or equal to 14. The number line shows a right bracket at 14 with shading to its left. The interval notation is negative infinity to 14 within a parenthesis and a bracket.

Three less than c is at least 360.

Nine times n exceeds 42.

Solution

The inequality is 9 n is greater than 42. Its solution is n is greater than fourteen-thirds. The number line shows a left parentheses at fourteen-thirds with shading to its right. The interval notation is fourteen-thirds to infinity within parentheses.

Negative two times a is no more than eight.

Solve Applications with Linear Inequalities

In the following exercises, solve.

Julianne has a weekly food budget of $231 for her family. If she plans to budget the same amount for each of the seven days of the week, what is the maximum amount she can spend on food each day?

Solution

$33 per day

Rogelio paints watercolors. He got a $100 gift card to the art supply store and wants to use it to buy 12″ × 16″ canvases. Each canvas costs $10.99. What is the maximum number of canvases he can buy with his gift card?

Briana has been offered a sales job in another city. The offer was for $42,500 plus 8% of her total sales. In order to make it worth the move, Briana needs to have an annual salary of at least $66,500. What would her total sales need to be for her to move?

Solution

at least $300,000

Renee’s car costs her $195 per month plus $0.09 per mile. How many miles can Renee drive so that her monthly car expenses are no more than $250?

Costa is an accountant. During tax season, he charges $125 to do a simple tax return. His expenses for buying software, renting an office, and advertising are $6,000. How many tax returns must he do if he wants to make a profit of at least $8,000?

Solution

at least 112 jobs

Jenna is planning a five-day resort vacation with three of her friends. It will cost her $279 for airfare, $300 for food and entertainment, and $65 per day for her share of the hotel. She has $550 saved towards her vacation and can earn $25 per hour as an assistant in her uncle’s photography studio. How many hours must she work in order to have enough money for her vacation?

Solve Compound Inequalities

Solve Compound Inequalities with “and”

In each of the following exercises, solve each inequality, graph the solution, and write the solution in interval notation.

x≤5 and x>−3

Solution

The solution is negative 3 is less than x which is less than or equal to 5. The number line shows an open circle at negative 3 and a closed circle at 5. The interval notation is negative 3 to 5 within a parenthesis and a bracket.

4x−2≤4 and
7x−1>−8

5(3x−2)≤5 and
4(x+2)<3

Solution

The solution is negative x is less than negative five-fourths. The number line shows an open circle at negative five-fourths with shading to its left. The interval notation is negative infinity to negative five-fourths within parentheses.

34(x−8)≤3 and
15(x−5)≤3

34x−5≥−2 and
−3(x+1)≥6

Solution

The solution is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation

−5≤4x−1<7

Solve Compound Inequalities with “or”

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

5−2x≤−1 or
6+3x≤4

Solution

The solution is x is less than negative two-thirds or x is greater than or equal to 3. The number line shows a closed circle at negative two-thirds with shading to its left and a closed circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative two-thirds within a parenthesis and a bracket and 3 to infinity within a bracket and a parenthesis.

3(2x−3)<−5 or
4x−1>3

34x−2>4 or 4(2−x)>0

Solution

The solution is x is less than 2 or x is greater than 8. The number line shows an open circle at 2 with shading to its left and an open circle at 8 with shading to its right. The interval notation is the union of negative infinity to 8 within parentheses and 8 to infinity within parentheses.

2(x+3)≥0 or
3(x+4)≤6

12x−3≤4 or
13(x−6)≥−2

Solution

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Solve Applications with Compound Inequalities

In the following exercises, solve.

Liam is playing a number game with his sister Audry. Liam is thinking of a number and wants Audry to guess it. Five more than three times her number is between 2 and 32. Write a compound inequality that shows the range of numbers that Liam might be thinking of.

Elouise is creating a rectangular garden in her back yard. The length of the garden is 12 feet. The perimeter of the garden must be at least 36 feet and no more than 48 feet. Use a compound inequality to find the range of values for the width of the garden.

Solution

6≤w≤12

Solve Absolute Value Inequalities

Solve Absolute Value Equations

In the following exercises, solve.

|x|=8

|y|=−14

Solution

no solution

|z|=0

|3x−4|+5=7

Solution

x=2,x=23

4|x−1|+2=10

−2|x−3|+8=−4

Solution

x=9,x=−3

|12x+5|+4=1

|6x−5|=|2x+3|

Solution

x=2,x=14

Solve Absolute Value Inequalities with “less than”

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

|x|≤8

|2x−5|≤3

Solution

The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading in between the circles. The interval notation is 1 to 4 within brackets.

|6x−5|<7

|5x+1|≤−2

Solution

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

Solve Absolute Value Inequalities with “greater than”

In the following exercises, solve. Graph the solution and write the solution in interval notation.

|x|>6

|x|≥2

Solution

The solution is x is less than negative 2 or x is greater than 6. The number line shows a closed circle at negative 2 with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

|x−5|>−2

|x−7|≥1

Solution

The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to negative 6 within a parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.

3|x|+4≥1

Solve Applications with Absolute Value

In the following exercises, solve.

A craft beer brewer needs 215,000 bottle per day. But this total can vary by as much as 5,000 bottles. What is the maximum and minimum expected usage at the bottling company?

Solution

The minimum to maximum expected usage is 210,000 to 220,000 bottles

At Fancy Grocery, the ideal weight of a loaf of bread is 16 ounces. By law, the actual weight can vary from the ideal by 1.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined?

Practice Test

In the following exercises, solve each equation.

−5(2x+1)=45

Solution

x=−5

14(12m+28)=6+2(3m+1)

8(3a+5)−7(4a−3)=20−3a

Solution

a=41

0.1d+0.25(d+8)=4.1

14n−3(4n+5)=−9+2(n−8)

Solution

contradiction; no solution

3(3u+2)+4[6−8(u−1)]=3(u−2)

34x−23=12x+56

Solution

x=6

|3x−4|=8

|2x−1|=|4x+3|

Solution

x=−2,x=−13

Solve the formula
x+2y=5 for y.

In the following exercises, graph the inequality on the number line and write in interval notation.

x≥−3.5

Solution

The inequality is x is greater than or equal to negative 3.5. The number line shows a left bracket at negative 3.5 and shading to the right. The interval notation is negative 3.5 to infinity within a bracket and a parenthesis.

x<114

−2≤x<5

Solution

The inequality is negative two is less than or equal to x which is less than 5. The number line shows a closed circle at negative 2 and an open circle at 5 with shading between the circles. The interval notation is negative 2 to 5 within a bracket and a parenthesis.

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

8k≥5k−120

3c−10(c−2)<5c+16

Solution

The solution is c is greater than one-third. The number line shows a left parenthesis at one-third with shading to its right. The interval notation is one-third to infinity within parentheses.

34x−5≥−2 and
−3(x+1)≥6

3(2x−3)<−5 or
4x−1>3

Solution

The solution is x is less than two-thirds or x is greater than 1. The number line shows an open circle at two-thirds with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to two-thirds within parentheses and 1 to infinity within parentheses.

12x−3≤4 or
13(x−6)≥−2

|4x−3|≥5

Solution

The solution is x is less than or equal to negative one-half or x is greater than or equal to 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and bracket and 2 to infinity within a bracket and a parenthesis.

In the following exercises, translate to an equation or inequality and solve.

Four less than twice x is 16.

Find the length of the missing side.

The figure is a right triangle with a base of 6 units and a height of 9 units.
Solution

10.8

One number is four more than twice another. Their sum is −47. Find the numbers.

The sum of two consecutive odd integers is −112. Find the numbers.

Solution

−57,−55

Marcus bought a television on sale for $626.50 The original price of the television was $895. Find ⓐ the amount of discount and ⓑ the discount rate.

Bonita has $2.95 in dimes and quarters in her pocket. If she has five more dimes than quarters, how many of each coin does she have?

Solution

12 dimes, seven quarters

Kim is making eight gallons of punch from fruit juice and soda. The fruit juice costs $6.04 per gallon and the soda costs $4.28 per gallon. How much fruit juice and how much soda should she use so that the punch costs $5.71 per gallon?

The measure of one angle of a triangle is twice the measure of the smallest angle. The measure of the third angle is three times the measure of the smallest angle. Find the measures of all three angles.

Solution

30°,60°,90°

The length of a rectangle is five feet more than four times the width. The perimeter is 60 feet. Find the dimensions of the rectangle.

Two planes leave Dallas at the same time. One heads east at a speed of 428 miles per hour. The other plane heads west at a speed of 382 miles per hour. How many hours will it take them to be 2,025 miles apart?

Solution

2.5 hours

Leon drove from his house in Cincinnati to his sister’s house in Cleveland, a distance of 252 miles. It took him 412 hours. For the first half hour, he had heavy traffic, and the rest of the time his speed was five miles per hour less than twice his speed in heavy traffic. What was his speed in heavy traffic?

Sara has a budget of $1,000 for costumes for the 18 members of her musical theater group. If all the costumes are the same price, what is the maximum she can spend for each costume?

Solution

At most $55.56 per costume.

Introduction

A black virtual reality headset is displayed on a clear glass head model, showcasing its design and strap details in an indoor setting with a blurred background.
This odd-looking headgear provides the user with a virtual world. (credit: fill/Pixabay)

Imagine visiting a faraway city or even outer space from the comfort of your living room. It could be possible using virtual reality. This technology creates realistic images that make you feel as if you are truly immersed in the scene and even enable you to interact with them. It is being developed for fun applications, such as video games, but also for architects to plan buildings, car companies to design prototypes, the military to train, and medical students to learn.

Developing virtual reality devices requires modeling the environment using graphs and mathematical relationships. In this chapter, you will graph different relationships and learn ways to describe and analyze graphs.

Graph Linear Equations in Two Variables

Learning Objectives

By the end of this section, you will be able to:

  • Plot points in a rectangular coordinate system
  • Graph a linear equation by plotting points
  • Graph vertical and horizontal lines
  • Find the x- and y-intercepts
  • Graph a line using the intercepts

Before you get started, take this readiness quiz.

Evaluate 5x−4 when x=−1.
If you missed this problem, review Example 6 in Use the Language of Algebra.

Solution

−9

Evaluate 3x−2y when x=4,y=−3.
If you missed this problem, review Example 10 in Integers.

Solution

18

Solve for y: 8−3y=20.
If you missed this problem, review Example 2 in Use a General Strategy to Solve Linear Equations.

Solution

y=−4

Plot Points on a Rectangular Coordinate System

Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the xy-plane or the “coordinate plane.”

The rectangular coordinate system is formed by two intersecting number lines, one horizontal and one vertical. The horizontal number line is called the x-axis. The vertical number line is called the y-axis. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See Figure 1.

This figure shows a square grid. A horizontal number line in the middle is labeled x. A vertical number line in the middle is labeled y. The number lines intersect at zero and together divide the square grid into 4 equally sized smaller squares. The square in the top right is labeled I. The square in the top left is labeled II. The square in the bottom left is labeled III. The square in the bottom right is labeled IV.

In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the x-coordinate of the point, and the second number is the y-coordinate of the point. The phrase “ordered pair” means that the order is important.

Ordered Pair

An ordered pair, (x,y) gives the coordinates of a point in a rectangular coordinate system. The first number is the x-coordinate. The second number is the y-coordinate.

This figure shows the expression (x, y). The variable x is labeled x-coordinate. The variable y is labeled y-coordinate.

What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is (0,0). The point (0,0) has a special name. It is called the origin.

The Origin

The point (0,0) is called the origin. It is the point where the x-axis and y-axis intersect.

We use the coordinates to locate a point on the xy-plane. Let’s plot the point (1,3) as an example. First, locate 1 on the x-axis and lightly sketch a vertical line through x=1. Then, locate 3 on the y-axis and sketch a horizontal line through y=3. Now, find the point where these two lines meet—that is the point with coordinates (1,3). See Figure 2.

This figure shows a point plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point (1, 3) is labeled. A dashed vertical line goes through the point and intersects the x-axis at xplus1. A dashed horizontal line goes through the point and intersects the y-axis at yplus3.

Notice that the vertical line through x=1 and the horizontal line through y=3 are not part of the graph. We just used them to help us locate the point (1,3).

When one of the coordinate is zero, the point lies on one of the axes. In Figure 3 the point (0,4) is on the y-axis and the point (−2,0) is on the x-axis.

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point (negative 2, 0) is labeled and lies on the x-axis. The point (0, 4) is labeled and lies on the y-axis.

Points on the Axes

Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates (a,0).

Points with an x-coordinate equal to 0 are on the y-axis, and have coordinates (0,b).

Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:

ⓐ (−5,4) ⓑ (−3,−4) ⓒ (2,−3) ⓓ (0,−1) ⓔ (3,52).

Solution

The first number of the coordinate pair is the x-coordinate, and the second number is the y-coordinate. To plot each point, sketch a vertical line through the x-coordinate and a horizontal line through the y-coordinate. Their intersection is the point.
ⓐ Since x=−5, the point is to the left of the y-axis. Also, since y=4, the point is above the x-axis. The point (−5,4) is in Quadrant II.
ⓑ Since x=−3, the point is to the left of the y-axis. Also, since y=−4, the point is below the x-axis. The point (−3,−4) is in Quadrant III.
ⓒ Since x=2, the point is to the right of the y-axis. Since y=−3, the point is below the x-axis. The point (2,−3) is in Quadrant IV.
ⓓ Since x=0, the point whose coordinates are (0,−1) is on the y-axis.
ⓔ Since x=3, the point is to the right of the y-axis. Since y=52, the point is above the x-axis. (It may be helpful to write 52 as a mixed number or decimal.) The point (3,52) is in Quadrant I.
This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The following points are labeled: (3, 5 divided by 2), negative 5, 4), (negative 3, negative 4), (0, negative 1), and (2, negative 3).

Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
ⓐ (−2,1) ⓑ (−3,−1) ⓒ (4,−4) ⓓ (−4,4) ⓔ (−4,32)

Solution
This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point labeled a is 2 units to the left of the origin and 1 unit above the origin and is located in quadrant II. The point labeled b is 3 units to the left of the origin and 1 unit below the origin and is located in quadrant III. The point labeled c is 4 units to the right of the origin and 4 units below the origin and is located in quadrant IV. The point labeled d is 4 units to the left of the origin and 4 units above the origin and is located in quadrant II. The point labeled e is 4 units to the left of the origin and 1 and a half units above the origin and is located in quadrant II.

Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
ⓐ (−4,1) ⓑ (−2,3) ⓒ (2,−5) ⓓ (−2,5) ⓔ (−3,52)

Solution
This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point labeled a is 4 units to the left of the origin and 1 unit above the origin and is located in quadrant II. The point labeled b is 2 units to the left of the origin and 3 units above the origin and is located in quadrant II. The point labeled c is 2 units to the right of the origin and 5 units below the origin and is located in quadrant IV. The point labeled d is 2 units to the left of the origin and 5 units above the origin and is located in quadrant II. The point labeled e is 3 units to the left of the origin and 2 and a half units above the origin and is located in quadrant II.

The signs of the x-coordinate and y-coordinate affect the location of the points. You may have noticed some patterns as you graphed the points in the previous example. We can summarize sign patterns of the quadrants in this way:

Quadrants

Quadrant IQuadrant IIQuadrant IIIQuadrant IV (x,y)(x,y)(x,y)(x,y) (+,+)(−,+)(−,−)(+,−)
This figure shows the x y-coordinate plane with the four quadrants labeled. In the top right of the plane is quadrant I labeled (plus, plus). In the top left of the plane is quadrant II labeled (minus, plus). In the bottom left of the plane is quadrant III labeled (minus, minus). In the bottom right of the plane is quadrant IV labeled (plus, minus).

Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation you got exactly one solution. But equations can have more than one variable. Equations with two variables may be of the form Ax+By=C. An equation of this form is called a linear equation in two variables.

Linear Equation

An equation of the form Ax+By=C, where A and B are not both zero, is called a linear equation in two variables.

Here is an example of a linear equation in two variables, x and y.

This figure shows the equation A x plus B y plus C. Below this is the equation x plus 4 y plus 8. Below this are the equations A plus 1, B plus 4, C plus 8. B and 4 are the same color in all the equations. C and 8 are the same color in all the equations.

The equation y=−3x+5 is also a linear equation. But it does not appear to be in the form Ax+By=C. We can use the Addition Property of Equality and rewrite it in Ax+By=C form.

y=−3x+5 Add to both sides.y+3x=−3x+5+3x Simplify.y+3x=5 Use the Commutative Property to put it in Ax+By=Cform.3x+y=5

By rewriting y=−3x+5 as 3x+y=5, we can easily see that it is a linear equation in two variables because it is of the form Ax+By=C. When an equation is in the form Ax+By=C, we say it is in standard form of a linear equation.

Standard Form of Linear Equation

A linear equation is in standard form when it is written Ax+By=C.

Most people prefer to have A, B, and C be integers and A≥0 when writing a linear equation in standard form, although it is not strictly necessary.

Linear equations have infinitely many solutions. For every number that is substituted for x there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair (x,y). When we substitute these values of x and y into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.

Solution of a Linear Equation in Two Variables

An ordered pair (x,y) is a solution of the linear equationAx+By=C, if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.

Linear equations have infinitely many solutions. We can plot these solutions in the rectangular coordinate system. The points will line up perfectly in a straight line. We connect the points with a straight line to get the graph of the equation. We put arrows on the ends of each side of the line to indicate that the line continues in both directions.

A graph is a visual representation of all the solutions of the equation. It is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to that equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation. Points not on the line are not solutions!

Graph of a Linear Equation

The graph of a linear equation Ax+By=C is a straight line.

  • Every point on the line is a solution of the equation.
  • Every solution of this equation is a point on this line.

The graph of y=2x−3 is shown.

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line has arrows on both ends and goes through the points (negative 3, negative 9), (negative 2, negative 7), (negative 1, negative 5), (0, negative 3), (1, negative 1), (2, 1), (3, 3), (4, 5), (5, 7), and (6, 9). The line is labeled y plus 2 x minus 3.

For each ordered pair, decide:

ⓐ Is the ordered pair a solution to the equation?

ⓑ Is the point on the line?

A: (0,–3) B: (3,3) C: (2,−3) D: (−1,−5)

Solution

Substitute the x- and y-values into the equation to check if the ordered pair is a solution to the equation.

ⓐ
Example A shows the ordered pair (0, negative 3). Under this is the equation y plus 2 x minus 3. Under this is the equation negative 3 equals 2 times 0 minus 3. The negative 3 and 0 are colored the same as the negative 3 and 0 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the equation negative 3 plus negative 3. Below this is the statement (0, negative 3) is a solution. Example B shows the ordered pair (3, 3). Under this is the equation y plus 2 x minus 3. Under this is the equation 3 equals 2 times 3 minus 3. The 3 and 3 are colored the same as the 3 and 3 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the equation 3 plus 3. Below this is the statement (3, 3) is a solution. Example C shows the ordered pair (2, negative 3). Under this is the equation y plus 2 x minus 3. Under this is the equation negative 3 equals 2 times 2 minus 3. The negative 3 and 2 are colored the same as the negative 3 and 2 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the inequality negative 3 is not equal to 1. Below this is the statement (2, negative 3) is not a solution. Example D shows the ordered pair (negative 1, negative 5). Under this is the equation y plus 2 x minus 3. Under this is the equation negative 5 equals 2 times negative 1 minus 3. The negative 1 and negative 5 are colored the same as the negative 1 and negative 5 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the equation negative 5 plus negative 5. Below this is the statement (negative 1, negative 5) is a solution.

ⓑ Plot the points (0,−3),(3,3),(2,−3), and (−1,−5).
This figure shows the graph of the linear equation y plus 2 x minus 3 and some points graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line has arrows on both ends and goes through the points (negative 1, negative 5), (0, negative 3), and (3, 3). The point (2, negative 3) is also plotted but not on the line.
The points (0,3),(3,−3), and (−1,−5) are on the line y=2x−3, and the point (2,−3) is not on the line.
The points that are solutions to y=2x−3 are on the line, but the point that is not a solution is not on the line.

Use graph of y=3x−1. For each ordered pair, decide:

ⓐ Is the ordered pair a solution to the equation?
ⓑ Is the point on the line?

A (0,−1) B (2,5)

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line has arrows on both ends and goes through the points (negative 3, negative 10), (negative 2, negative 7), (negative 1, negative 4), (0, negative 1), (1, 2), (2, 5), and (3, 8). The line is labeled y plus 3 x minus 1.
Solution

ⓐ yes, yes ⓑ yes, yes

Use graph of y=3x−1. For each ordered pair, decide:

ⓐ Is the ordered pair a solution to the equation?
ⓑ Is the point on the line?

A(3,−1) B(−1,−4)

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line has arrows on both ends and goes through the points (negative 3, negative 10), (negative 2, negative 7), (negative 1, negative 4), (0, negative 1), (1, 2), (2, 5), and (3, 8). The line is labeled y plus 3 x minus 1.
Solution

ⓐ no, no ⓑ yes, yes

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The first method we will use is called plotting points, or the Point-Plotting Method. We find three points whose coordinates are solutions to the equation and then plot them in a rectangular coordinate system. By connecting these points in a line, we have the graph of the linear equation.

How to Graph a Linear Equation by Plotting Points

Graph the equation y=2x+1 by plotting points.

Solution
Step 1 is to Find three points whose coordinates are solutions to the equation. You can choose any values for x or y. In this case since y is isolated on the left side of the equations, it is easier to choose values for x. Choosing x plus 0. We substitute this into the equation y plus 2 x plus 1 to get y plus 2 times 0 plus 1. This simplifies to y plus 0 plus 1. So y plus 1. Choosing x plus 1. We substitute this into the equation y plus 2 x plus 1 to get y plus 2 times 1 plus 1. This simplifies to y plus 2 plus 1. So y plus 3. Choosing x plus negative 2. We substitute this into the equation y plus 2 x plus 1 to get y plus 2 times negative 2 plus 1. This simplifies to y plus negative 4 plus 1. The y plus negative 3. Next we want to organize the solutions in a table. For this problem we will put the three solutions we just found in a table. The table has 5 rows and 3 columns. The first row is a title row with the equation y plus 2 x plus 1. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, 1, and (0, 1). The fourth row has the numbers 1, 3, and (1, 3). The fifth row has the numbers negative 2, negative 3, and (negative 2, negative 3). Step 2 is to plot the points in a rectangular coordinate system. Plot: (0, 1), (1, 3), (negative 2, negative 3). The figure then shows a graph of some points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The points (0, 1), (1, 3), and (negative 2, negative 3) are plotted. Check that the points line up. If they do not, carefully check your work! Do the point line up? Yes, the points in this example line up. Step 3 is to draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. This line is the graph of y plus 2 x plus 1. The figure shows the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The points (negative 2, negative 3), (0, 1), and (1, 3) are plotted. The straight line goes through the three points and has arrows on both ends.

Graph the equation by plotting points: y=2x−3.

Solution


This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 2, negative 7), (negative 1, negative 5), (0, negative 3), (1, negative 1), (2, 1), (3, 3), (4, 5), and (5, 7).

Graph the equation by plotting points: y=−2x+4.

Solution


This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 2, 8), (negative 1, 6), (0, 4), (1, 2), (2, 0), (3, negative 2), (4, negative 4), (5, negative 6) and (6, negative 8).

The steps to take when graphing a linear equation by plotting points are summarized here.

Graph a linear equation by plotting points.

  1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
  2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
  3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.

It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you only plot two points and one of them is incorrect, you can still draw a line but it will not represent the solutions to the equation. It will be the wrong line.

If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. Look at the difference between these illustrations.

The figure shows two images. In the first image there are three points with a straight line going through all three. In the second image there are three points that do not all lie on a straight line.

When an equation includes a fraction as the coefficient of x, we can still substitute any numbers for x. But the arithmetic is easier if we make “good” choices for the values of x. This way we will avoid fractional answers, which are hard to graph precisely.

Graph the equation: y=12x+3.

Solution
Find three points that are solutions to the equation. Since this equation has the fraction 12 as a coefficient of x, we will choose values of x carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of two a good choice for values of x? By choosing multiples of 2 the multiplication by 12 simplifies to a whole number
The first set of equations starts with x plus 0. Under this is the equation y plus 1 half x plus 3. Under this is the equation y plus 1 half times 0 plus 3. Below this is the equation y plus 0 plus 3. Below this is the equation y plus 3. The second set of equations starts with x plus 2. Under this is the equation y plus 1 half x plus 3. Under this is the equation y plus 1 half times 2 plus 3. Below this is the equation y plus 1 plus 3. Below this is the equation y plus 4. The third set of equations starts with x plus 4. Under this is the equation y plus 1 half x plus 3. Under this is the equation y plus 1 half times 4 plus 3. Below this is the equation y plus 2 plus 3. Below this is the equation y plus 5.
The points are shown in Table 1.
y=12x+3
x y (x,y)
0 3 (0,3)
2 4 (2,4)
4 5 (4,5)

Plot the points, check that they line up, and draw the line.
The figure shows the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (0, 3), (2, 4), and (4, 5) are plotted. The straight line goes through the three points and has arrows on both ends. The line is labeled y plus 1 divided by 2 times x plus 3.

Graph the equation: y=13x−1.

Solution


This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 12, negative 5), (negative 9, negative 4), (negative 6, negative 3), (negative 3, negative 2), (0, negative 1), (3, 0), (6, 1), (9, 2), and (12, 3).

Graph the equation: y=14x+2.

Solution


This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 12, negative 1), (negative 8, 0), (negative 4, 1), (0, 2), (4, 3), (8, 4), and (12, 5).

Graph Vertical and Horizontal Lines

Some linear equations have only one variable. They may have just x and no y, or just y without an x. This changes how we make a table of values to get the points to plot.

Let’s consider the equation x=−3. This equation has only one variable, x. The equation says that x is always equal to−3, so its value does not depend on y. No matter what is the value of y, the value of x is always −3.

So to make a table of values, write −3 in for all the x-values. Then choose any values for y. Since x does not depend on y, you can choose any numbers you like. But to fit the points on our coordinate graph, we’ll use 1, 2, and 3 for the y-coordinates. See Table 2.

x=−3
x y (x,y)
−3 1 (−3,1)
−3 2 (−3,2)
−3 3 (−3,3)

Plot the points from the table and connect them with a straight line. Notice that we have graphed a vertical line.

The figure shows the graph of a straight vertical line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (negative 3, 1), (negative 3, 2), and (negative 3, 3) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled x plus negative 3.

What if the equation has y but no x? Let’s graph the equation y=4. This time the y-value is a constant, so in this equation, y does not depend on x. Fill in 4 for all the y’s in Table 3 and then choose any values for x. We’ll use 0, 2, and 4 for the x-coordinates.

y=4
x y (x,y)
0 4 (0,4)
2 4 (2,4)
4 4 (4,4)

In this figure, we have graphed a horizontal line passing through the y-axis at 4.

The figure shows the graph of a straight horizontal line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (0, 4), (2, 4), and (4, 4) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled y plus 4.

Vertical and Horizontal Lines

A vertical line is the graph of an equation of the form x=a.

The line passes through the x-axis at (a,0).

A horizontal line is the graph of an equation of the form y=b.

The line passes through the y-axis at (0,b).

Graph: ⓐ x=2 ⓑ y=−1.

Solution
ⓐ The equation has only one variable, x, and x is always equal to 2. We create a table where x is always 2 and then put in any values for y. The graph is a vertical line passing through the x-axis at 2.
x=2
x y (x,y)
2 1 (2,1)
2 2 (2,2)
2 3 (2,3)

The figure shows the graph of a straight vertical line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (2, 1), (2, 2), and (2, 3) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled x plus 2.
ⓑ Similarly, the equation y=−1 has only one variable, y. The value of y is constant. All the ordered pairs in the next table have the same y-coordinate. The graph is a horizontal line passing through the y-axis at −1.
y=−1
x y (x,y)
0 −1 (0,−1)
3 −1 (3,−1)
−3 −1 (−3,−1)

The figure shows the graph of a straight horizontal line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (negative 3, negative 1), (0, negative 1), and (3, negative 1) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled y plus negative 1.

Graph the equations: ⓐ x=5 ⓑ y=−4.

Solution

ⓐ
The figure shows the graph of a straight vertical line on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (5, negative 3), (5, negative 2), (5, negative 1), (5, 0), (5, 1), (5, 2), and (5, 3).

ⓑ
The figure shows the graph of a straight horizontal line on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 3, negative 4), (negative 2, negative 4), (negative 1, negative 4), (0, negative 4), (1, negative 4), (2, negative 4), and (3, negative 4).

Graph the equations: ⓐ x=−2 ⓑ y=3.

Solution

ⓐ
The figure shows the graph of a straight vertical line on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 2, negative 3), (negative 2, negative 2), (negative 2, negative 1), (negative 2, 0), (negative 2, 1), (negative 2, 2), and (negative 2, 3).

ⓑ
The figure shows the graph of a straight horizontal line on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 3, 3), (negative 2, 3), (negative 1, 3), (0, 3), (1, 3), (2, 3), and (3, 3).

What is the difference between the equations y=4x and y=4?

The equation y=4x has both x and y. The value of y depends on the value of x, so the y -coordinate changes according to the value of x. The equation y=4 has only one variable. The value of y is constant, it does not depend on the value of x, so the y-coordinate is always 4.

This figure has two tables. The first table has 5 rows and 3 columns. The first row is a title row with the equation y plus 4 x. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, 0, and (0, 0). The fourth row has the numbers 1, 4, and (1, 4). The fifth row has the numbers 2, 8, and (2, 8). The second table has 5 rows and 3 columns. The first row is a title row with the equation y plus 4. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, 4, and (0, 4). The fourth row has the numbers 1, 4, and (1, 4). The fifth row has the numbers 2, 4, and (2, 4). The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 7 to 7. The horizontal line goes through the points (0, 4), (1, 4), and (2,4) and is labeled y plus 4. The slanted line goes through the points (0, 0), (1, 4), and (2, 8) and is labeled y plus 4 x.

Notice, in the graph, the equation y=4x gives a slanted line, while y=4 gives a horizontal line.

Graph y=−3x and y=−3 in the same rectangular coordinate system.

Solution

We notice that the first equation has the variable x, while the second does not. We make a table of points for each equation and then graph the lines. The two graphs are shown.

This figure has two tables. The first table has 5 rows and 3 columns. The first row is a title row with the equation y plus negative 3 x. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, 0, and (0, 0). The fourth row has the numbers 1, negative 3, and (1, negative 3). The fifth row has the numbers 2, negative 6, and (2, neg ative 6). The second table has 5 rows and 3 columns. The first row is a title row with the equation y plus negative 3. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, negative 3, and (0, negative 3). The fourth row has the numbers 1, negative 3, and (1, negative 3). The fifth row has the numbers 2, negative 3, and (2, negative 3).


The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 7 to 7. The horizontal line goes through the points (0, negative 3), (1, negative 3), and (2, negative 3) and is labeled y plus negative 3. The slanted line goes through the points (0, 0), (1, negative 3), and (2, negative 6) and is labeled y plus negative 3 x.

Graph the equations in the same rectangular coordinate system: y=−4x and y=−4.

Solution


The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 12 to 12. The horizontal line goes through the points (0, negative 4), (1, negative 4), and (2, negative 4). The slanted line goes through the points (0, 0), (1, negative 4), and (2, negative 8).

Graph the equations in the same rectangular coordinate system: y=3 and y=3x.

Solution


The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 12 to 12. The horizontal line goes through the points (0, 3), (1, 3), and (2, 3). The slanted line goes through the points (0, 0), (1, 3), and (2, 6).

Find x- and y-intercepts

Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.

At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x-axis and the y-axis. These points are called the intercepts of a line.

Intercepts of a Line

The points where a line crosses the x-axis and the y-axis are called the intercepts of the line.

Let’s look at the graphs of the lines.

The figure shows four graphs of different equations. In example a the graph of 2 x plus y plus 6 is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The points (0, 6) and (3, 0) are plotted and labeled. A straight line goes through both points and has arrows on both ends. In example b the graph of 3 x minus 4 y plus 12 is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The points (0, negative 3) and (4, 0) are plotted and labeled. A straight line goes through both points and has arrows on both ends. In example c the graph of x minus y plus 5 is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The points (0, negative 5) and (5, 0) are plotted and labeled. A straight line goes through both points and has arrows on both ends. In example d the graph of y plus negative 2 x is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The point (0, 0) is plotted and labeled. A straight line goes through this point and the points (negative 1, 2) and (1, negative 2) and has arrows on both ends.

First, notice where each of these lines crosses the x-axis. See Table 6.

Now, let’s look at the points where these lines cross the y-axis.

Figure The line crosses
the x-axis at:
Ordered pair
for this point
The line crosses
the y-axis at:
Ordered pair
for this point
Figure (a) 3 (3,0) 6 (0,6)
Figure (b) 4 (4,0) −3 (0,−3)
Figure (c) 5 (5,0) −5 (0,5)
Figure (d) 0 (0,0) 0 (0,0)
General Figure a (a,0) b (0,b)

Do you see a pattern?

For each line, the y-coordinate of the point where the line crosses the x-axis is zero. The point where the line crosses the x-axis has the form (a,0) and is called the x-intercept of the line. The x-intercept occurs when y is zero.

In each line, the x-coordinate of the point where the line crosses the y-axis is zero. The point where the line crosses the y-axis has the form (0,b) and is called the y-intercept of the line. The y-intercept occurs when x is zero.

x-intercept and y-intercept of a Line

The x-intercept is the point (a,0) where the line crosses the x-axis.

The y-intercept is the point (0,b) where the line crosses the y-axis.

The table has 3 rows and 2 columns. The first row is a header row with the headers x and y. The second row contains a and 0. The third row contains 0 and b.

Find the x- and y-intercepts on each graph shown.

The figure has three graphs. Figure a shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 8, 6), (negative 4, 4), (0, 2), (4, 0), (8, negative 2). Figure b shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (0, negative 6), (2, 0), and (4, 6). Figure c shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 5, 0), (negative 3, negative 3), (0, negative 5), (1, negative 6), and (2, negative 7).
Solution

ⓐ The graph crosses the x-axis at the point (4,0). The x-intercept is (4,0).
The graph crosses the y-axis at the point (0,2). The y-intercept is (0,2).

ⓑ The graph crosses the x-axis at the point (2,0). The x-intercept is (2,0).
The graph crosses the y-axis at the point (0,−6). The y-intercept is (0,−6).

ⓒ The graph crosses the x-axis at the point (−5,0). The x-intercept is (−5,0).
The graph crosses the y-axis at the point (0,−5). The y-intercept is (0,−5).

Find the x- and y-intercepts on the graph.

This figure a shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 10 to 10. The line goes through the points (negative 6, negative 8), (negative 4, negative 6), (negative 2, negative 4), (0, negative 2), (2, 0), (4, 2), (6, 4), (8, 6).
Solution

x-intercept: (2,0),
y-intercept: (0,−2)

Find the x- and y-intercepts on the graph.

This figure a shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 10 to 10. The line goes through the points (negative 6, 6), (negative 3, 4), (0, 2), (3, 0), (6, negative 2), and (9, negative 4).
Solution

x-intercept: (3,0),
y-intercept: (0,2)

Recognizing that the x-intercept occurs when y is zero and that the y-intercept occurs when x is zero, gives us a method to find the intercepts of a line from its equation. To find the x-intercept, let y=0 and solve for x. To find the y-intercept, let x=0 and solve for y.

Find the x- and y-intercepts from the Equation of a Line

Use the equation of the line. To find:

  • the x-intercept of the line, let y=0 and solve for x.
  • the y-intercept of the line, let x=0 and solve for y.

Find the intercepts of 2x+y=8.

Solution
We will let y=0 to find the x-intercept, and let x=0 to find the y-intercept. We will fill in a table, which reminds us of what we need to find.
The figure has a table with 4 rows and 2 columns. The first row is a title row with the equation 2 x plus y plus 8. The second row is a header row with the headers x and y. The third row is labeled x-intercept and has the first column blank and a 0 in the second column. The fourth row is labeled y-intercept and has a 0 in the first column and the second column blank.
To find the x-intercept, let y=0.
A mathematical equation is displayed on a white background, reading '2x + y = 8' in black text.
Let y=0. A mathematical equation shows '2x + 0 = 8' with the number zero highlighted in red. This illustrates an algebraic problem where '0' is irrelevant to the value of '2x'.
Simplify. The image displays a simple algebraic equation: 2x = 8. It presents a basic problem in mathematics where the goal is to solve for the variable 'x'.
The image displays the equation 'x = 4' in black text against a white background.
The x-intercept is: (4,0)
To find the y-intercept, let x=0.
A mathematical equation is displayed, reading '2x + y = 8' in a bold, black font against a white background.
Let x=0. A mathematical equation is displayed: 2 multiplied by 0, plus y, equals 8. The '0' is highlighted with a red outline, indicating a substitution or point of focus in the calculation.
Simplify. A mathematical equation is displayed, showing '0 + y = 8' in a clear, black font against a white background.
The image displays a mathematical equation 'y = 8' in black text against a white background, representing a horizontal line in a coordinate plane where the y-value is constant at 8.
The y-intercept is: (0,8)

The intercepts are the points (4,0) and (0,8) as shown in the table.
2x+y=8
x y
4 0
0 8

Find the intercepts: 3x+y=12.

Solution

x-intercept: (4,0),
y-intercept: (0,12)

Find the intercepts: x+4y=8.

Solution

x-intercept: (8,0),
y-intercept: (0,2)

Graph a Line Using the Intercepts

To graph a linear equation by plotting points, you need to find three points whose coordinates are solutions to the equation. You can use the x- and y- intercepts as two of your three points. Find the intercepts, and then find a third point to ensure accuracy. Make sure the points line up—then draw the line. This method is often the quickest way to graph a line.

How to Graph a Line Using the Intercepts

Graph –x+2y=6 using the intercepts.

Solution
Step 1 is to find the x and y-intercepts of the line. To find the x-intercept let y plus 0 and solve for x. The equation negative x plus 2 y plus 6 becomes negative x plus 2 times 0 plus 6. This simplifies to negative x plus 6. This is equivalent to x plus negative 6. The x-intercept is (negative 6, 0). To find the y-intercept let x plus 0 and solve for y. The equation negative x plus 2 y plus 6 becomes negative 0 plus 2 y plus 6. This simplifies to negative 2 y plus 6. This is equivalent to y plus 3. The y-intercept is (0, 3). Step 2 is to find another solution to the equation. We’ll use x plus 2. The equation negative x plus 2 y plus 6 becomes negative 2 plus 2 y plus 6. This simplifies to 2 y plus 8. This is equivalent to y plus 4. The third point is (2, 4). Step 3 is to plot the three points. The figure shows a table with 4 rows and 3 columns. The first row is a header row with the headers x, y, and (x, y). The second row contains negative 6, 0, and (negative 6, 0). The third row contains 0, 3, and (0, 3). The fourth row contains 2, 4, and (2, 4). The figure also has a graph of the three points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The three points (negative 6, 0), (0, 3), and (2, 4) are plotted and labeled. Step 4 is to draw the line. The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The straight line goes through the points (negative 6, 0), (0, 3), and (2, 4).

Graph using the intercepts: x–2y=4.

Solution
The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The straight line goes through the points (negative 4, negative 4), (negative 2, negative 3), (0, negative 2), (2, negative 1), (4, 0), (6, 1), and (8, 2).

Graph using the intercepts: –x+3y=6.

Solution


The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The straight line goes through the points (negative 9, negative 1), (negative 6, 0), (negative 3, 1), (0, 2), (3, 3), (6, 4), and (9, 5).

The steps to graph a linear equation using the intercepts are summarized here.

Graph a linear equation using the intercepts.

  1. Find the x- and y-intercepts of the line.
    • Let y=0 and solve for x.
    • Let x=0 and solve for y.
  2. Find a third solution to the equation.
  3. Plot the three points and check that they line up.
  4. Draw the line.

Graph 4x−3y=12 using the intercepts.

Solution
Find the intercepts and a third point.
To find the x-intercept let y plus 0 and solve for x. The equation 4 x minus 3 y plus 12 becomes 4 x minus 3 times 0 plus 12. This simplifies to negative 4 x plus 12. This is equivalent to x plus 3. To find the y-intercept let x plus 0 and solve for y. The equation 4 x minus 3 y plus 12 becomes 4 times 0 minus 3 y plus 12. This simplifies to negative 3 y plus 12. This is equivalent to y plus negative 4. To find the third point let y plus 4 and solve for x. The equation 4 x minus 3 y plus 12 becomes 4 x minus 3 times 4 plus 12. This simplifies to negative 4 x plus 24. This is equivalent to x plus 6.

We list the points in the table and show the graph.
4x−3y=12
x y (x,y)
3 0 (3,0)
0 −4 (0,−4)
6 4 (6,4)

The figure shows a graph of the equation 4 x minus 3 y plus 12 on the x y-coordinate plane. The x and y-axes run from negative 7 to 7. The straight line goes through the points (0, negative 4), (3, 0), and (6, 4).

Graph using the intercepts: 5x−2y=10.

Solution


The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The straight line goes through the points (0, negative 5), (2, 0), and (4, 5).

Graph using the intercepts: 3x−4y=12.

Solution


The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The straight line goes through the points (negative 4, negative 6), (0, negative 3), (4, 0), and (8, 3).

When the line passes through the origin, the x-intercept and the y-intercept are the same point.

Graph y=5x using the intercepts.

Solution
To find the x-intercept let y plus 0 and solve for x. The equation y plus 5 x becomes 0 plus 5 x. This simplifies to 0 plus x. The x-intercept is (0, 0). To find the y-intercept let x plus 0 and solve for y. The equation y plus 5 x becomes y plus 5 times 0. This simplifies to y plus 0. The y-intercept is also (0, 0).


This line has only one intercept. It is the point (0,0).
To ensure accuracy, we need to plot three points. Since the x- and y-intercepts are the same point, we need two more points to graph the line.

To find a second point let x plus 1 and solve for y. The equation y plus 5 x becomes y plus 5 times 1. This simplifies to y plus 5. To find a third point let x plus negative 1 and solve for y. The equation y plus 5 x becomes y plus 5 times negative 1. This simplifies to y plus negative 5

The resulting three points are summarized in the table.

y=5x
x y (x,y)
0 0 (0,0)
1 5 (1,5)
−1 −5 (−1,−5)

Plot the three points, check that they line up, and draw the line.

The figure shows a graph of the equation y plus 5 x on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The straight line goes through the points (negative 1, negative 5), (0, 0), and (1, 5).

Graph using the intercepts: y=4x.

Solution


The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The straight line goes through the points (negative 1, negative 4), (0, 0), and (1, 4).

Graph the intercepts: y=−x.

Solution


The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The straight line goes through the points (negative 1, 1), (0, 0), and (1, negative 1).

Key Concepts

  • Points on the Axes
    • Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates (a,0).
    • Points with an x-coordinate equal to 0 are on the y-axis, and have coordinates (0,b).
  • Quadrant
    Quadrant IQuadrant IIQuadrant IIIQuadrant IV (x,y)(x,y)(x,y)(x,y) (+,+)(−,+)(−,−)(+,−)
    This figure shows the x y-coordinate plane with the four quadrants labeled. In the top right of the plane is quadrant I labeled (plus, plus). In the top left of the plane is quadrant II labeled (minus, plus). In the bottom left of the plane is quadrant III labeled (minus, minus). In the bottom right of the plane is quadrant IV labeled (plus, minus).
  • Graph of a Linear Equation: The graph of a linear equation Ax+By=C is a straight line.
    Every point on the line is a solution of the equation.
    Every solution of this equation is a point on this line.
  • How to graph a linear equation by plotting points.
    1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
    2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
    3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.
  • x-intercept and y-intercept of a Line
    • The x-intercept is the point (a,0) where the line crosses the x-axis.
    • The y-intercept is the point (0,b) where the line crosses the y-axis. The table has 3 rows and 2 columns. The first row is a header row with the headers x and y. The second row contains a and 0. The x-intercept occurs when y is zero. The third row contains 0 and b. The y-intercept occurs when x is zero.
  • Find the x- and y-intercepts from the Equation of a Line
    • Use the equation of the line. To find:
      the x-intercept of the line, let y=0 and solve for x.
      the y-intercept of the line, let x=0 and solve for y.
  • How to graph a linear equation using the intercepts.
    1. Find the x- and y-intercepts of the line.
      Let y=0 and solve for x.
      Let x=0 and solve for y.
    2. Find a third solution to the equation.
    3. Plot the three points and check that they line up.
    4. Draw the line

Practice Makes Perfect

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.

ⓐ (−4,2) ⓑ (−1,−2) ⓒ (3,−5) ⓓ (−3,0) ⓔ (53,2)

Solution

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point labeled a is 4 units to the left of the origin and 2 units above the origin and is located in quadrant II. The point labeled b is 1 unit to the left of the origin and 2 units below the origin and is located in quadrant III. The point labeled c is 3 units to the right of the origin and 5 units below the origin and is located in quadrant IV. The point labeled d is 3 units to the left of the origin and 5 units above the origin and is located in quadrant II. The point labeled e is 1 and a half units to the right of the origin and 2 units above the origin and is located in quadrant I.

ⓐ (−2,−3) ⓑ (3,−3) ⓒ (−4,1) ⓓ (4,−1) ⓔ (32,1)

ⓐ (3,−1) ⓑ (−3,1) ⓒ (−2,0) ⓓ (−4,−3) ⓔ (1,145)

Solution

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point labeled a is 3 units to the right of the origin and 1 unit below the origin and is located in quadrant IV. The point labeled b is 3 units to the left of the origin and 1 unit above the origin and is located in quadrant II. The point labeled c is 2 units to the left of the origin and 2 units above the origin and is located in quadrant II. The point labeled d is 4 units to the left of the origin and 3 units below the origin and is located in quadrant III. The point labeled e is 1 unit to the right of the origin and 3 and 4 fifths units above the origin and is located in quadrant I.

ⓐ (−1,1) ⓑ (−2,−1) ⓒ (2,0) ⓓ (1,−4) ⓔ (3,72)

In the following exercises, for each ordered pair, decide

ⓐ is the ordered pair a solution to the equation? ⓑ is the point on the line?

y=x+2;
A: (0,2); B: (1,2); C: (−1,1); D: (−3,−1).

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 3, negative 1), (negative 2, 0), (negative 1, 1), (0, 2), (1, 3), (2, 4), and (3, 5).
Solution

ⓐ A: yes, B: no, C: yes, D: yes ⓑ A: yes, B: no, C: yes, D: yes

y=x−4;
A: (0,−4); B: (3,−1); C: (2,2); D: (1,−5).

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 3, negative 7), (negative 2, negative 6), (negative 1, negative 5), (0, negative 4), (1, negative 3), (2, negative 2), and (3, negative 1).

y=12x−3;
A: (0,−3); B: (2,−2); C: (−2,−4); D: (4,1)

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 4, negative 5), (negative 2, negative 4), (0, negative 3), (2, negative 2), (4, negative 1), and (6, 0).
Solution

ⓐ A: yes, B: yes, C: yes, D: no ⓑ A: yes, B: yes, C: yes, D: no

y=13x+2;
A: (0,2); B: (3,3); C: (−3,2); D: (−6,0).

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 6, 0), (negative 3, 1), (0, 2), (3, 3), (6, 4), and (9, 5).

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

y=x+2

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 3, negative 1), (negative 2, 0), (negative 1, 1), (0, 2), (1, 3), (2, 4), and (3, 5).

y=x−3

y=3x−1

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 3, negative 10), (negative 2, negative 7), (negative 1, negative 4), (0, negative 1), (1, 2), (2, 5), (3, 8), and (4, 11).

y=−2x+2

y=−x−3

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 3, 0), (negative 2, negative 1), (negative 1, negative 2), (0, negative 3), (1, negative 4), (2, negative 5), (3, negative 6), and (4, negative 7).

y=−x−2

y=2x

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 3, negative 6), (negative 2, negative 4), (negative 1, negative 2), (0, 0), (1, 2), (2, 4), and (3, 6).

y=−2x

y=12x+2

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 6, negative 2), (negative 4, 0), (negative 2, 1), (0, 2), (2, 3), (4, 4), and (6, 5).

y=13x−1

y=43x−5

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 3, negative 9), (0, negative 5), (3, negative 1), (6, 3), and (9, 7).

y=32x−3

y=−25x+1

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 10, 5), (negative 5, 3), (0, 1), (5, negative 1), and (10, negative 3).

y=−45x−1

y=−32x+2

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 4, 8), (negative 2, 5), (0, 2), (2, negative 1), (4, negative 4), and (6, negative 7).

y=−53x+4

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

ⓐ x=4 ⓑ y=3

Solution

ⓐ
This figure shows a vertical straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (4, negative 1), (4, 0), and (4, 1).

ⓑ
This figure shows a horizontal straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 1, 3), (0, 3), and (1, 3).

ⓐ x=3 ⓑ y=1

ⓐ x=−2 ⓑ y=−5

Solution

ⓐ
This figure shows a vertical straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 2, negative 1), (negative 2, 0), and (negative 2, 1).

ⓑ
This figure shows a horizontal straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 1, negative 5), (0, negative 5), and (1, negative 5).

ⓐ x=−5 ⓑ y=−2

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

y=2x and y=2

Solution

The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 12 to 12. The horizontal line goes through the points (0, 2), (1, 2), and (2, 2). The slanted line goes through the points (0, 0), (1, 2), and (2, 4).

y=5x and y=5

y=−12x and y=−12

Solution

The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 12 to 12. The horizontal line goes through the points (0, negative 1 divided 2), (1, negative 1 divided 2), and (2, negative 1 divided 2). The slanted line goes through the points (0, 0), (1, negative 1 divided 2), and (2, negative 1).

y=−13x and y=−13

Find x- and y-Intercepts

In the following exercises, find the x- and y-intercepts on each graph.


The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 6, 9), (negative 3, 6), (0, 3), (3, 0), and (6, negative 3).

Solution

(3,0),(0,3)


The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 6, 4), (negative 4, 2), (negative 2, 0), (0, negative 2), (2, negative 4), and (4, negative 6).


The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 1, negative 6), (0, negative 5), (2, negative 3), (5, 0), and (7, 2).

Solution

(5,0),(0,−5)


The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 2, negative 4), (negative 1, negative 2), (0, 0), (1, 2), and (2, 4).

In the following exercises, find the intercepts for each equation.

x−y=5

Solution

(5,0),(0,−5)

x−y=−4

3x+y=6

Solution

(2,0),(0,6)

x−2y=8

4x−y=8

Solution

(2,0),(0,−8)

5x−y=5

2x+5y=10

Solution

(5,0),(0,2)

3x−2y=12

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

−x+4y=8

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 8, 0), (0, 2), (4, 3), and (8, 4).

x+2y=4

x+y=−3

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 3, 0), (0, negative 3), and (3, negative 6).

x−y=−4

4x+y=4

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (0, 4), (1, 0), and (2, negative 4).

3x+y=3

3x−y=−6

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 2, 0), (negative 1, 3), and (0, 6).

2x−y=−8

2x+4y=12

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (0, 3), (2, 2), and (6, 0).

3x−2y=6

2x−5y=−20

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 10, 0), (0, 4), and (10, 8).

3x−4y=−12

y=−2x

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 1, 2), (0, 0), and (1, negative 2).

y=5x

y=x

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 1, negative 1), (0, 0), and (1, 1).

y=−x

Mixed Practice

In the following exercises, graph each equation.

y=32x

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 2, negative 3), (0, 0), and (2, 3).

y=−23x

y=−12x+3

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 2, 4), (0, 3), (2, 2), (4, 1), and (6, 0).

y=14x−2

4x+y=2

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 1, 6), (0, 2), (1, negative 2), and (2, negative 4).

5x+2y=10

y=−1

Solution

The figure shows a horizontal straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 2, negative 1), (0, negative 1), and (1, negative 1).

x=3

Writing Exercises

Explain how you would choose three x-values to make a table to graph the line y=15x−2.

Solution

Answers will vary.

What is the difference between the equations of a vertical and a horizontal line?

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation 4x+y=−4? Why?

Solution

Answers will vary.

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation y=23x−2? Why?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 6 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “plot points on a rectangular coordinate system”, “graph a linear equation by plotting points”, “graph vertical and horizontal lines”, “find x and y intercepts”, and “graph a line using intercepts”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ If most of your checks were:

Confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

With some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

No, I don’t get it. This is a warning sign and you must address it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

horizontal line
A horizontal line is the graph of an equation of the form y=b. The line passes through the y-axis at (0,b).
intercepts of a line
The points where a line crosses the x-axis and the y-axis are called the intercepts of the line.
linear equation
An equation of the form Ax+By=C, where A and B are not both zero, is called a linear equation in two variables.
ordered pair
An ordered pair, (x,y) gives the coordinates of a point in a rectangular coordinate system. The first number is the x-coordinate. The second number is the y-coordinate.
origin
The point (0,0) is called the origin. It is the point where the x-axis and y-axis intersect.
solution of a linear equation in two variables
An ordered pair (x,y) is a solution of the linear equationAx+By=C, if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.
standard form of a linear equation
A linear equation is in standard form when it is written Ax+By=C.
vertical line
A vertical line is the graph of an equation of the form x=a. The line passes through the x-axis at (a,0).

Slope of a Line

Learning Objectives

By the end of this section, you will be able to:

  • Find the slope of a line
  • Graph a line given a point and the slope
  • Graph a line using its slope and intercept
  • Choose the most convenient method to graph a line
  • Graph and interpret applications of slope–intercept
  • Use slopes to identify parallel and perpendicular lines

Before you get started, take this readiness quiz.

Simplify: (1–4)(8−2).
If you missed this problem, review Example 7 in Fractions.

Solution

−12

Divide: 04,40.
If you missed this problem, review Example 5 in Properties of Real Numbers.

Solution

0; undefined

Simplify: 15−3,−153,−15−3.
If you missed this problem, review Example 7 in Fractions.

Solution

−5;−5;5

Find the Slope of a Line

When you graph linear equations, you may notice that some lines tilt up as they go from left to right and some lines tilt down. Some lines are very steep and some lines are flatter.

In mathematics, the measure of the steepness of a line is called the slope of the line.

The concept of slope has many applications in the real world. In construction the pitch of a roof, the slant of the plumbing pipes, and the steepness of the stairs are all applications of slope. and as you ski or jog down a hill, you definitely experience slope.

We can assign a numerical value to the slope of a line by finding the ratio of the rise and run. The rise is the amount the vertical distance changes while the run measures the horizontal change, as shown in this illustration. Slope is a rate of change. See Figure 1.

This figure has a diagram of two arrows. The first arrow is vertical and pointed up and labeled “rise”. The second arrow starts at the end of the first. The second arrow is horizontal and pointed right and labeled “run”.

Slope of a Line

The slope of a line is m=riserun.

The rise measures the vertical change and the run measures the horizontal change.

To find the slope of a line, we locate two points on the line whose coordinates are integers. Then we sketch a right triangle where the two points are vertices and one side is horizontal and one side is vertical.

To find the slope of the line, we measure the distance along the vertical and horizontal sides of the triangle. The vertical distance is called the rise and the horizontal distance is called the run,

Find the slope of a line from its graph using m=riserun.

  1. Locate two points on the line whose coordinates are integers.
  2. Starting with one point, sketch a right triangle, going from the first point to the second point.
  3. Count the rise and the run on the legs of the triangle.
  4. Take the ratio of rise to run to find the slope: m=riserun.

Find the slope of the line shown.

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 9. The y-axis runs from negative 1 to 7. The line goes through the points (0, 5), (3, 3), and (6, 1).
Solution
Locate two points on the graph whose
coordinates are integers.
(0,5) and (3,3)
Starting at (0,5), sketch a right triangle to
(3,3) as shown in this graph.
A graph illustrating the concept of slope with a downward-sloping line on a coordinate plane, showing labeled 'rise' and 'run' segments forming a right triangle.
Count the rise— since it goes down, it is negative. The rise is −2.
Count the run. The run is 3.
Use the slope formula. m=riserun
Substitute the values of the rise and run. m=−23
Simplify. m=−23
The slope of the line is −23.
So y decreases by 2 units as x increases by 3 units.

Find the slope of the line shown.

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 5. The y-axis runs from negative 6 to 1. The line goes through the points (0, negative 2) and (3, negative 6).
Solution

−43

Find the slope of the line shown.

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 3 to 6. The y-axis runs from negative 3 to 2. The line goes through the points (0, 1) and (5, negative 2).
Solution

−35

How do we find the slope of horizontal and vertical lines? To find the slope of the horizontal line, y=4, we could graph the line, find two points on it, and count the rise and the run. Let’s see what happens when we do this, as shown in the graph below.

The figure then shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 6. The y-axis runs from negative 1 to 8. The line goes through the points (0, 4) and (3, 4). What is the rise? The rise is 0. What is the run? The run is 3. What is the slope? m equals rise divided by run. m equals 0 divided by 3. m equals 0. The slope of the horizontal line y equals 4 is 0.
This table demonstrates how to calculate the slope of a horizontal line by showing that a rise of 0 over a run of 3 results in a slope of 0.
What is the rise? The rise is 0.
What is the run? The run is 3.
What is the slope? m=riserun
m=03
m=0
The slope of the horizontal line y=4is 0.

Let’s also consider a vertical line, the line x=3, as shown in the graph.

The figure then shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 2 to 6. The y-axis runs from negative 3 to 3. The line goes through the points (3, 0) and (3, 2). What is the rise? The rise is 2. What is the run? The run is 0. What is the slope? m equals rise divided by run. m equals 2 divided by 0.
This table illustrates the concepts of rise, run, and slope, including a case where the run is zero, resulting in an undefined slope.
What is the rise? The rise is 2.
What is the run? The run is 0.
What is the slope? m=riserun
m=20

The slope is undefined since division by zero is undefined. So we say that the slope of the vertical line x=3 is undefined.

All horizontal lines have slope 0. When the y-coordinates are the same, the rise is 0.

The slope of any vertical line is undefined. When the x-coordinates of a line are all the same, the run is 0.

Slope of a Horizontal and Vertical Line

The slope of a horizontal line, y=b, is 0.

The slope of a vertical line, x=a, is undefined.

Find the slope of each line: ⓐ x=8 ⓑ y=−5.

Solution

ⓐ x=8
This is a vertical line. Its slope is undefined.
ⓑ y=−5
This is a horizontal line. It has slope 0.

Find the slope of the line: x=−4.

Solution

undefined

Find the slope of the line: y=7.

Solution

0

Quick Guide to the Slopes of Lines

The image shows four arrows. The first arrow is slanted and pointing up and to the right and is labeled “positive”. The second arrow is slanted and pointing down and to the right and labeled “negative”. The third arrow is horizontal and labeled “zero”. The fourth arrow is vertical and labeled “undefined”.

Sometimes we’ll need to find the slope of a line between two points when we don’t have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but as we’ll see, there is a way to find the slope without graphing. Before we get to it, we need to introduce some algebraic notation.

We have seen that an ordered pair (x,y) gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol (x,y) be used to represent two different points? Mathematicians use subscripts to distinguish the points.

(x1,y1)read “xsub 1,ysub 1” (x2,y2)read “xsub 2,ysub 2”

We will use (x1,y1) to identify the first point and (x2,y2) to identify the second point.

If we had more than two points, we could use (x3,y3),(x4,y4), and so on.

Let’s see how the rise and run relate to the coordinates of the two points by taking another look at the slope of the line between the points (2,3) and (7,6), as shown in this graph.

The figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 7. The y-axis runs from negative 1 to 7. The line goes through the points (2, 3) and (7, 6). A right triangle is drawn by connecting the three points (2, 3), (2, 6), and (7, 6). The point (2, 3) is labeled (x 1, y 1). The point (7, 6) is labeled (x 2, y 2). The vertical side of the triangle has labels y 2 minus y 1, 6 minus 3, and 3. The horizontal side of the triangle has labels x 2 minus x 1, 7 minus 2, and 5.
Steps demonstrating the derivation of the slope formula m = (y2-y1)/(x2-x1) from two points using rise-over-run.
Since we have two points, we will use subscript notation. (2,x1,3y1)(7,6x2,y2)
m=riserun
On the graph, we counted the rise of 3 and the run of 5. m=35
Notice that the rise of 3 can be found by subtracting the
y-coordinates, 6 and 3, and the run of 5 can be found by
subtracting the x-coordinates 7 and 2.
We rewrite the rise and run by putting in the coordinates. m=6−37−2
But 6 is y2, the y-coordinate of the second point and 3 is y1, the y-coordinate
of the first point. So we can rewrite the slope using subscript notation.
m=y2−y17−2
Also 7 is the x-coordinate of the second point and 2 is the x-coordinate
of the first point. So again we rewrite the slope using subscript notation.
m=y2−y1x2−x1

We’ve shown that m=y2−y1x2−x1 is really another version of m=riserun. We can use this formula to find the slope of a line when we have two points on the line.

Slope of a line between two points

The slope of the line between two points (x1,y1) and (x2,y2) is:

m=y2−y1x2−x1.

The slope is:

yof the second point minusyof the first point over xof the second point minusxof the first point.

Use the slope formula to find the slope of the line through the points (−2,−3) and (−7,4).

Solution
This table demonstrates the step-by-step calculation of the slope between two points, including formula application, substitution, and simplification.
We’ll call (−2,−3) point #1 and (−7,4) point #2. (x1,y1−2,−3)(x2,y2−7,4)
Use the slope formula. m=y2−y1x2−x1
Substitute the values.
y of the second point minus y of the first point
x of the second point minus x of the first point m=4−(−3)−7−(−2)
Simplify. m=7−5m=−75
Let’s verify this slope on the graph shown.
The figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 2. The y-axis runs from negative 6 to 6. The line goes through the points (negative 7, 4) and (negative 2, negative 3). A right triangle is drawn by connecting the three points (negative 7, 4), (negative 7, negative 3), and (negative 2, negative 3). The vertical side of the triangle is labeled “rise”. The horizontal side of the triangle is labeled “run”.
m=riserun m=7−5 m=−75

Use the slope formula to find the slope of the line through the pair of points: (−3,4) and (2,−1).

Solution

−1

Use the slope formula to find the slope of the line through the pair of points: (−2,6) and (−3,−4).

Solution

10

Graph a Line Given a Point and the Slope

Up to now, in this chapter, we have graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.

We can also graph a line when we know one point and the slope of the line. We will start by plotting the point and then use the definition of slope to draw the graph of the line.

How to graph a Line Given a Point and the Slope

Graph the line passing through the point (1,−1) whose slope is m=34.

Solution
Step 1 is to plot the given point. Plot (1, negative 1). The figure then shows the graph of a point on the x y-coordinate plane. The x-axis runs from negative 2 to 6. The y-axis runs from negative 2 to 4. The point (1, negative 1) is plotted and labeled with its coordinates. Step 2 is to use the slope formula m equals rise divided by run to identify the rise and the run. Identify the rise and the run. m equals 3 divided by 4. Rise divided by run equals 3 divided by 4. Rise equals 3. Run equals 4. Step 3 is to start at the given point and count out the rise and run to mark the second point. Start at (1, negative 1) and count the rise and the run. Up 3 units, right 4 units. The figure then shows the graph of three points connected by two straight line segments on the x y-coordinate plane. The x-axis runs from negative 2 to 6. The y-axis runs from negative 2 to 4. The points (1, negative 1), (1, 2), and (5, 2) are plotted. A vertical line segment connects (1, negative 1) to (1, 2) and is labeled 3. A horizontal line segment connects (1, 2) to (5, 2) and is labeled 4. Step 4 is to connect the points with a line. Connect the two points with a line. . The figure then shows the graph of a straight line, three points, and two line segments on the x y-coordinate plane. The x-axis runs from negative 2 to 6. The y-axis runs from negative 2 to 4. The points (1, negative 1), (1, 2), and (5, 2) are plotted. A vertical line segment connects (1, negative 1) to (1, 2). A horizontal line segment connects (1, 2) to (5, 2) and is labeled 4. A straight line is drawn through the points (1, negative 1) and (5, 2) with arrows on both ends.

You can check your work by finding a third point. Since the slope is m=34, it can also be written as m=−3−4 (negative divided by negative is positive!). Go back to (1,−1) and count out the rise, −3, and the run, −4.

Graph the line passing through the point (2,−2) with the slopem=43.

Solution


This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (2, negative 2) and (5, 2).

Graph the line passing through the point (−2,3) with the slope m=14.

Solution


This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (negative 2, 3) and (2, 4).

Graph a line given a point and the slope.

  1. Plot the given point.
  2. Use the slope formula m=riserun to identify the rise and the run.
  3. Starting at the given point, count out the rise and run to mark the second point.
  4. Connect the points with a line.

Graph a Line Using its Slope and Intercept

We have graphed linear equations by plotting points, using intercepts, recognizing horizontal and vertical lines, and using one point and the slope of the line. Once we see how an equation in slope–intercept form and its graph are related, we’ll have one more method we can use to graph lines.

See Figure 2. Let’s look at the graph of the equation y=12x+3 and find its slope and y-intercept.

The figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, 3), (2, 4), and (4, 5). A right triangle is drawn by connecting the three points (2, 4), (2, 5), and (4, 5). The vertical side of the triangle is labeled “Rise equals 1”. The horizontal side of the triangle is labeled “Run equals 2”. The line is labeled y equals 1 divided by 2 x plus 3.

The red lines in the graph show us the rise is 1 and the run is 2. Substituting into the slope formula:

m=riserun m=12

The y-intercept is (0,3).

Look at the equation of this line.

The figure shows the equation y equals 1 divided by 2 x plus 3. The 1 divided by 2 is emphasized in red. The 3 is emphasized in blue.

Look at the slope and y-intercept.

Slope m equals 1 divided by 2 and y-intercept (0, 3). The 1 divided by 2 is emphasized in red. The 3 is emphasized in blue.

When a linear equation is solved for y, the coefficient of the x term is the slope and the constant term is the y-coordinate of the y-intercept. We say that the equation y=12x+3 is in slope–intercept form. Sometimes the slope–intercept form is called the “y-form.”

m equals 1 divided by 2; y-intercept is (0, 3). y equals 1 divided by 2 x plus 3. y equals m x plus b. The m and 1 divided by 2 are emphasized in red. The b and 3 are emphasized in blue.

Slope Intercept Form of an Equation of a Line

The slope–intercept form of an equation of a line with slope m and y-intercept, (0,b) is y=mx+b.

Let’s practice finding the values of the slope and y-intercept from the equation of a line.

Identify the slope and y-intercept of the line from the equation:

ⓐ y=−47x−2 ⓑ x+3y=9

Solution
ⓐ We compare our equation to the slope–intercept form of the equation.
Write the slope–intercept form of the equation of the line. The equation for a straight line, y = mx + b, is displayed on a white background, with 'm' in red and 'b' in blue to highlight the slope and y-intercept.
Write the equation of the line. A mathematical equation is displayed against a white background, reading y = -4/7x - 2. The fraction 4/7 is in red, and the number 2 is in light blue.
Identify the slope. The image shows the mathematical equation m = -4/7, where m represents a variable equal to a negative fraction.
Identify the y-intercept. The image displays the text 'y-intercept is (0, -2)', which represents a coordinate point on a graph where a line or curve crosses the y-axis, indicating that when x is 0, y is -2.


ⓑ When an equation of a line is not given in slope–intercept form, our first step will be to solve the equation for y.
Solve for y. x+3y=9
Subtract x from each side. A mathematical equation is displayed, reading '3y = -x + 9' on a white background.
Divide both sides by 3. A mathematical equation shows '3y divided by 3 equals the quantity of negative x plus 9, all divided by 3'. The equation simplifies to y = (-x+9)/3, demonstrating a step in solving for y.
Simplify. A mathematical equation is displayed against a white background: y = -1/3x + 3. This is the slope-intercept form of a linear equation, showing a negative slope and a positive y-intercept.
Write the slope–intercept form of the equation of the line. The equation y = mx + b, representing the slope-intercept form of a linear equation, is displayed. The 'm' is highlighted in red, indicating the slope, and 'b' is highlighted in blue, indicating the y-intercept.
Write the equation of the line. A mathematical equation, 'y = -1/3x + 3', is displayed in black text with the fraction and the number 3 highlighted in red and light blue, respectively.
Identify the slope. A mathematical equation shows 'm = -1/3', with the '1' and '3' in red font.
Identify the y-intercept. The image displays the mathematical expression 'y-intercept is (0, 3)', indicating the point where a graph crosses the y-axis.

Identify the slope and y-intercept from the equation of the line.

ⓐ y=25x−1 ⓑ x+4y=8

Solution

ⓐ m=25;(0,−1)
ⓑ m=−14;(0,2)

Identify the slope and y-intercept from the equation of the line.

ⓐ y=−43x+1 ⓑ 3x+2y=12

Solution

ⓐ m=−43;(0,1)
ⓑ m=−32;(0,6)

We have graphed a line using the slope and a point. Now that we know how to find the slope and y-intercept of a line from its equation, we can use the y-intercept as the point, and then count out the slope from there.

Graph the line of the equation y=−x+4 using its slope and y-intercept.

Solution
y=mx+b
The equation is in slope–intercept form. y=−x+4
Identify the slope and y-intercept. m=−1
y-intercept is (0,4)
Plot the y-intercept. See the graph.
Identify the rise over the run. m=−11
Count out the rise and run to mark the second point. rise −1, run 1
A coordinate plane illustrating a downward-sloping line that intersects the y-axis at (0,4) and passes through (1,3).

Draw the line as shown in the graph.

Graph the line of the equation y=−x−3 using its slope and y-intercept.

Solution


This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, negative 3) and (1, negative 4).

Graph the line of the equationy=−x−1 using its slope and y-intercept.

Solution


This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, negative 1) and (1, negative 2).

Now that we have graphed lines by using the slope and y-intercept, let’s summarize all the methods we have used to graph lines.

The table has a title row that reads “Methods to Graph Lines”. Below this are four columns. The first column contains the following: Point Plotting. A blank table with two columns and four rows. The first row is a header row with the headers “x” and “y”. Find three points. Plot the points, make sure they line up, them draw the line. The second column contains: Slope-Intercept. Y equals m x plus b. Find the slope and y-intercept, then count the slope to get a second point. The third column: Intercepts. A table with two columns and four rows. The first row is a header row with the headers “x” and “y”. In the first row there is a 0 in the x column. In the second row there is a 0 in the y column. The remaining spaces are blank. Fourth column. Recognize vertical and horizontal lines. The equation has only one variable. X equals a vertical. Y equals b horizontal.

Choose the Most Convenient Method to Graph a Line

Now that we have seen several methods we can use to graph lines, how do we know which method to use for a given equation?

While we could plot points, use the slope–intercept form, or find the intercepts for any equation, if we recognize the most convenient way to graph a certain type of equation, our work will be easier.

Generally, plotting points is not the most efficient way to graph a line. Let’s look for some patterns to help determine the most convenient method to graph a line.

Here are five equations we graphed in this chapter, and the method we used to graph each of them.

EquationMethod#1x=2Vertical line#2y=−1Horizontal line#3−x+2y=6Intercepts#44x−3y=12Intercepts#5y=−x+4Slope–intercept

Equations #1 and #2 each have just one variable. Remember, in equations of this form the value of that one variable is constant; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.

In equations #3 and #4, both x and y are on the same side of the equation. These two equations are of the form Ax+By=C. We substituted y=0 to find the x- intercept and x=0 to find the y-intercept, and then found a third point by choosing another value for x or y.

Equation #5 is written in slope–intercept form. After identifying the slope and y-intercept from the equation we used them to graph the line.

This leads to the following strategy.

Strategy for Choosing the Most Convenient Method to Graph a Line

Consider the form of the equation.

  • If it only has one variable, it is a vertical or horizontal line.
    • x=a is a vertical line passing through the x-axis at a.
    • y=b is a horizontal line passing through the y-axis at b.
  • If y is isolated on one side of the equation, in the form y=mx+b, graph by using the slope and y-intercept.
    • Identify the slope and y-intercept and then graph.
  • If the equation is of the form Ax+By=C, find the intercepts.
    • Find the x- and y-intercepts, a third point, and then graph.

Determine the most convenient method to graph each line:

ⓐ y=5 ⓑ 4x−5y=20 ⓒ x=−3 ⓓ y=−59x+8

Solution

ⓐ y=5
This equation has only one variable, y. Its graph is a horizontal line crossing the y-axis at 5.
ⓑ 4x−5y=20
This equation is of the form Ax+By=C. The easiest way to graph it will be to find the intercepts and one more point.
ⓒ x=−3
There is only one variable, x. The graph is a vertical line crossing the x-axis at−3.
ⓓ y=−59x+8
Since this equation is in y=mx+b form, it will be easiest to graph this line by using the slope and y-intercepts.

Determine the most convenient method to graph each line:

ⓐ 3x+2y=12 ⓑ y=4 ⓒ y=15x−4 ⓓ x=−7.

Solution

ⓐ intercepts ⓑ horizontal line ⓒ slope-intercept ⓓ vertical line

Determine the most convenient method to graph each line:

ⓐ x=6 ⓑ y=−34x+1 ⓒ y=−8 ⓓ 4x−3y=−1.

Solution

ⓐ vertical line ⓑ slope-intercept ⓒ horizontal line
ⓓ intercepts

Graph and Interpret Applications of Slope–Intercept

Many real-world applications are modeled by linear equations. We will take a look at a few applications here so you can see how equations written in slope–intercept form relate to real world situations.

Usually, when a linear equation models uses real-world data, different letters are used for the variables, instead of using only x and y. The variable names remind us of what quantities are being measured.

Also, we often will need to extend the axes in our rectangular coordinate system to bigger positive and negative numbers to accommodate the data in the application.

The equation F=95C+32 is used to convert temperatures, C, on the Celsius scale to temperatures, F, on the Fahrenheit scale.

ⓐ Find the Fahrenheit temperature for a Celsius temperature of 0.

ⓑ Find the Fahrenheit temperature for a Celsius temperature of 20.

ⓒ Interpret the slope and F-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ

Steps and calculations for converting 0 degrees Celsius to Fahrenheit using the formula F = (9/5)C + 32.
Find the Fahrenheit temperature for a Celsius temperature of 0. F=95C+32
Find F when C=0. F=95(0)+32
Simplify. F=32

ⓑ

Step-by-step calculation showing the conversion of 20 degrees Celsius to Fahrenheit using the temperature conversion formula.
Find the Fahrenheit temperature for a Celsius temperature of 20. F=95C+32
Find F when C=20. F=95(20)+32
Simplify. F=36+32
Simplify. F=68

ⓒ
Interpret the slope and F-intercept of the equation.
Even though this equation uses F and C, it is still in slope–intercept form.
y equals m x plus b. F equals m C plus b. The y and F are emphasized in red. The x and C are emphasized in blue. F equals 9 divided by 5 C plus 32. The slope, 95, means that the temperature Fahrenheit (F) increases 9 degrees when the temperature Celsius (C) increases 5 degrees.
The F-intercept means that when the temperature is 0° on the Celsius scale, it is 32° on the Fahrenheit scale.
ⓓ Graph the equation.
We’ll need to use a larger scale than our usual. Start at the F-intercept (0,32), and then count out the rise of 9 and the run of 5 to get a second point as shown in the graph.

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 40 to 80. The y-axis runs from negative 40 to 80. The line goes through the points (0, 32) and (5, 41).

The equation h=2s+50 is used to estimate a woman’s height in inches, h, based on her shoe size, s.

ⓐ Estimate the height of a child who wears women’s shoe size 0.

ⓑ Estimate the height of a woman with shoe size 8.

ⓒ Interpret the slope and h-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ 50 inches
ⓑ 66 inches
ⓒ The slope, 2, means that the height, h, increases by 2 inches when the shoe size, s, increases by 1. The h-intercept means that when the shoe size is 0, the height is 50 inches.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 14. The y-axis runs from negative 1 to 80. The line goes through the points (0, 50) and (10, 70).

The equation T=14n+40 is used to estimate the temperature in degrees Fahrenheit, T, based on the number of cricket chirps, n, in one minute.

ⓐ Estimate the temperature when there are no chirps.

ⓑ Estimate the temperature when the number of chirps in one minute is 100.

ⓒ Interpret the slope and T-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ 40 degrees
ⓑ 65 degrees
ⓒ The slope, 14, means that the temperature Fahrenheit (F) increases 1 degree when the number of chirps, n, increases by 4. The T-intercept means that when the number of chirps is 0, the temperature is 40°.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 140. The y-axis runs from negative 1 to 80. The line goes through the points (0, 40) and (40, 50).

The cost of running some types of business has two components—a fixed cost and a variable cost. The fixed cost is always the same regardless of how many units are produced. This is the cost of rent, insurance, equipment, advertising, and other items that must be paid regularly. The variable cost depends on the number of units produced. It is for the material and labor needed to produce each item.

Sam drives a delivery van. The equation C=0.5m+60 models the relation between his weekly cost, C, in dollars and the number of miles, m, that he drives.

ⓐ Find Sam’s cost for a week when he drives 0 miles.

ⓑ Find the cost for a week when he drives 250 miles.

ⓒ Interpret the slope and C-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ

This table demonstrates the step-by-step calculation of Sam's weekly driving cost when 0 miles are driven, applying the formula C = 0.5m + 60.
Find Sam’s cost for a week when he drives 0 miles. C=0.5m+60
Find C when m=0. C=0.5(0)+60
Simplify. C=60
Sam’s costs are $60 when he drives 0 miles.

ⓑ

Step-by-step calculation of driving costs using a given formula. It determines Sam's total cost for driving 250 miles in a week.
Find the cost for a week when he drives 250 miles. C=0.5m+60
Find C when m=250. C=0.5(250)+60
Simplify. C=185
Sam’s costs are $185 when he drives 250 miles.

ⓒ Interpret the slope and C-intercept of the equation.
y equals m x plus b. C equals 0.5 p plus 60. The y and C are emphasized in red. The x and p are emphasized in blue.
The slope, 0.5, means that the weekly cost, C, increases by $0.50 when the number of miles driven, n, increases by 1.
The C-intercept means that when the number of miles driven is 0, the weekly cost is $60.
ⓓ Graph the equation.
We’ll need to use a larger scale than our usual. Start at the C-intercept (0,60).

To count out the slope m = 0.5, we rewrite it as an equivalent fraction that will make our graphing easier.

Steps demonstrating the conversion of the decimal 0.5 to an equivalent fraction.
m=0.5
Rewrite as a fraction. m=0.51
Multiply numerator and denominator by 100. m=0.5(100)1(100)
Simplify. m=50100

So to graph the next point go up 50 from the intercept of 60 and then to the right 100. The second point will be (100, 110).

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 350. The y-axis runs from negative 1 to 350. The line goes through the points (0, 60) and (200, 160).

Stella has a home business selling gourmet pizzas. The equation C=4p+25 models the relation between her weekly cost, C, in dollars and the number of pizzas, p, that she sells.

ⓐ Find Stella’s cost for a week when she sells no pizzas.

ⓑ Find the cost for a week when she sells 15 pizzas.

ⓒ Interpret the slope and C-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ $25
ⓑ $85
ⓒ The slope, 4, means that the weekly cost, C, increases by $4 when the number of pizzas sold, p, increases by 1. The C-intercept means that when the number of pizzas sold is 0, the weekly cost is $25.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 2 to 20. The y-axis runs from negative 10 to `00. The line goes through the points (0, 25) and (1, 29).

Loreen has a calligraphy business. The equation C=1.8n+35 models the relation between her weekly cost, C, in dollars and the number of wedding invitations, n, that she writes.

ⓐ Find Loreen’s cost for a week when she writes no invitations.

ⓑ Find the cost for a week when she writes 75 invitations.

ⓒ Interpret the slope and C-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ $35
ⓑ $170
ⓒ The slope, 1.8, means that the weekly cost, C, increases by $1.80 when the number of invitations, n, increases by 1.
The C-intercept means that when the number of invitations is 0, the weekly cost is $35.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 350. The y-axis runs from negative 1 to 350. The line goes through the points (0, 35) and (75, 170).

Use Slopes to Identify Parallel and Perpendicular Lines

Two lines that have the same slope are called parallel lines. Parallel lines have the same steepness and never intersect.

We say this more formally in terms of the rectangular coordinate system. Two lines that have the same slope and different y-intercepts are called parallel lines. See Figure 3.

This figure shows the graph of a two straight lines on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The first line goes through the points (0, 3) and (5, 5). The second line goes through the points (0, negative 2) and (5, 0). The lines are parallel meaning they will always be the same distance apart and never intersect. They are slanted by the same angle.

Verify that both lines have the same slope, m=25, and different y-intercepts.

What about vertical lines? The slope of a vertical line is undefined, so vertical lines don’t fit in the definition above. We say that vertical lines that have different x-intercepts are parallel, like the lines shown in this graph.

This figure shows the graph of a two straight vertical line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The first line goes through the points (2, 0) and (2, 1). The second line goes through the points (5, 0) and (5, 1). The lines are parallel meaning they will always be the same distance apart and never intersect.

Parallel Lines

Parallel lines are lines in the same plane that do not intersect.

  • Parallel lines have the same slope and different y-intercepts.
  • If m1 and m2 are the slopes of two parallel lines then m1=m2.
  • Parallel vertical lines have different x-intercepts.

Since parallel lines have the same slope and different y-intercepts, we can now just look at the slope–intercept form of the equations of lines and decide if the lines are parallel.

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ 3x−2y=6 and y=32x+1 ⓑ y=2x-3 and −6x+3y=−9.

Solution

ⓐ

This table demonstrates converting two linear equations to slope-intercept form and identifying their slopes and y-intercepts.
3x−2y=6 and y=32x+1
Solve the first equation for y. −2y=−3x+6−2y−2=−3x+6−2
The equation is now in slope–intercept form. y=32x−3
The equation of the second line is already in slope–intercept form. y=32x+1
Identify the slope and y-intercept of both lines. y=32x−3y=mx+bm=32 y=32x+1y=mx+by=32
y-intercept is(0,−3) y-intercept is(0,1)

The lines have the same slope and different y-intercepts and so they are parallel.

You may want to graph the lines to confirm whether they are parallel.

ⓑ

This table illustrates the process of converting two linear equations into slope-intercept form to compare their properties and demonstrate they are equivalent.
y=2x−3 and −6x+3y=−9
The first equation is already in slope–intercept form. y=2x−3
Solve the second equation for y. −6x+3y=−93y=6x−93y3=6x−93y=2x−3
The second equation is now in slope–intercept form. y=2x−3
Identify the slope and y-intercept of both lines. y=2x−3y=mx+bm=2 y=2x−3y=mx+bm=2
y-intercept is(0,−3) y-intercept is(0,−3)

The lines have the same slope, but they also have the same y-intercepts. Their equations represent the same line and we say the lines are coincident. They are not parallel; they are the same line.

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ 2x+5y=5 and y=−25x−4 ⓑ y=−12x−1 and x+2y=−2.

Solution

ⓐ parallel ⓑ not parallel; same line

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ 4x−3y=6 and y=43x−1 ⓑ y=34x−3 and 3x−4y=12.

Solution

ⓐ parallel ⓑ not parallel; same line

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ y=−4 and y=3 ⓑ x=−2 and x=−5.

Solution

ⓐ y=−4 and y=3
We recognize right away from the equations that these are horizontal lines, and so we know their slopes are both 0.
Since the horizontal lines cross the y-axis at y=−4 and at y=3, we know the y-intercepts are (0,−4) and (0,3).
The lines have the same slope and different y-intercepts and so they are parallel.
ⓑ x=−2 and x=−5
We recognize right away from the equations that these are vertical lines, and so we know their slopes are undefined.
Since the vertical lines cross the x-axis at x=−2 and x=−5, we know the y-intercepts are (−2,0) and (−5,0).
The lines are vertical and have different x-intercepts and so they are parallel.

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ y=8 and y=−6 ⓑ x=1 and x=−5.

Solution

ⓐ parallel ⓑ parallel

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ y=1 and y=−5 ⓑ x=8 and x=−6.

Solution

ⓐ parallel ⓑ parallel

Let’s look at the lines whose equations are y=14x−1 and y=−4x+2, shown in Figure 5.

This figure shows the graph of a two perpendicular straight lines on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The first line goes through the points (0, negative 1) and (4, 0). The first line is labeled y equals 1 divided by 4 x minus 1. The second line goes through the points (0, 2) and (1, negative 2). The second line is labeled y equals negative 4 x plus 2. The lines are perpendicular meaning they form a right angle where they intersect.

These lines lie in the same plane and intersect in right angles. We call these lines perpendicular.

If we look at the slope of the first line, m1=14, and the slope of the second line, m2=−4, we can see that they are negative reciprocals of each other. If we multiply them, their product is −1.

m1·m214(−4)−1

This is always true for perpendicular lines and leads us to this definition.

Perpendicular Lines

Perpendicular lines are lines in the same plane that form a right angle.

  • If m1 and m2 are the slopes of two perpendicular lines, then:
    • their slopes are negative reciprocals of each other, m1=−1m2.
    • the product of their slopes is −1, m1·m2=−1.
  • A vertical line and a horizontal line are always perpendicular to each other.

We were able to look at the slope–intercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.

We find the slope–intercept form of the equation, and then see if the slopes are opposite reciprocals. If the product of the slopes is −1, the lines are perpendicular.

Use slopes to determine if the lines are perpendicular:

ⓐ y=−5x−4 and x−5y=5 ⓑ 7x+2y=3 and 2x+7y=5

Solution

ⓐ

Step-by-step analysis of two linear equations, including solving for y and identifying their slopes.
The first equation is in slope–intercept form. y=−5x−4
Solve the second equation for y. x−5y=5−5y=−x+5−5y−5=−x+5−5y=15x−1
Identify the slope of each line. y=−5x−4y=mx+bm1=−5 y=15x−1y=mx+bm2=15

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes, Since −5(15)=−1, it checks.

ⓑ

Steps to solve linear equations for 'y' and identify their slopes, illustrated with two distinct examples.
Solve the equations for y. 7x+2y=32y=−7x+32y2=−7x+32y=−72x+32 2x+7y=57y=−2x+57y7=−2x+57y=−27x+57
Identify the slope of each line. y=mx+bm1=−72 y=mx+bm1=−27

The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.

Use slopes to determine if the lines are perpendicular:

ⓐ y=−3x+2 and x−3y=4 ⓑ 5x+4y=1 and 4x+5y=3.

Solution

ⓐ perpendicular ⓑ not perpendicular

Use slopes to determine if the lines are perpendicular:

ⓐ y=2x−5 and x+2y=−6 ⓑ 2x−9y=3 and 9x−2y=1.

Solution

ⓐ perpendicular ⓑ not perpendicular

Key Concepts

  • Slope of a Line
    • The slope of a line is m=riserun.
    • The rise measures the vertical change and the run measures the horizontal change.




  • How to find the slope of a line from its graph using m=riserun.
    1. Locate two points on the line whose coordinates are integers.
    2. Starting with one point, sketch a right triangle, going from the first point to the second point.
    3. Count the rise and the run on the legs of the triangle.
    4. Take the ratio of rise to run to find the slope: m=riserun.
  • Slope of a line between two points.
    • The slope of the line between two points (x1,y1) and (x2,y2) is:
      m=y2−y1x2−x1.
  • How to graph a line given a point and the slope.
    1. Plot the given point.
    2. Use the slope formula m=riserun to identify the rise and the run.
    3. Starting at the given point, count out the rise and run to mark the second point.
    4. Connect the points with a line.
  • Slope Intercept Form of an Equation of a Line
    • The slope–intercept form of an equation of a line with slope m and y-intercept, (0,b) is y=mx+b The table has a title row that reads “Methods to Graph Lines”. Below this are four columns. The first column contains the following: Point Plotting. A blank table with two columns and four rows. The first row is a header row with the headers “x” and “y”. Find three points. Plot the points, make sure they line up, them draw the line. The second column contains: Slope-Intercept. Y equals m x plus b. Find the slope and y-intercept, then count the slope to get a second point. The third column: Intercepts. A table with two columns and four rows. The first row is a header row with the headers “x” and “y”. In the first row there is a 0 in the x column. In the second row there is a 0 in the y column. The remaining spaces are blank. Fourth column. Recognize vertical and horizontal lines. The equation has only one variable. X equals a vertical. Y equals b horizontal.
  • Parallel Lines
    • Parallel lines are lines in the same plane that do not intersect.
      Parallel lines have the same slope and different y-intercepts.
      If m1 and m2 are the slopes of two parallel lines then m1=m2.
      Parallel vertical lines have different x-intercepts.
  • Perpendicular Lines
    • Perpendicular lines are lines in the same plane that form a right angle.
    • If m1 and m2 are the slopes of two perpendicular lines, then:
      their slopes are negative reciprocals of each other, m1=−1m2.
      the product of their slopes is −1,m1·m2=−1.
    • A vertical line and a horizontal line are always perpendicular to each other.

Practice Makes Perfect

Find the Slope of a Line

In the following exercises, find the slope of each line shown.

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, negative 4) and (5, negative 2).
Solution

25

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, negative 5) and (2, negative 2).
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, negative 1) and (4, 4).
Solution

54

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, negative 2) and (3, 3).
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, 2) and (3, 1).
Solution

−13

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, negative 1) and (3, negative 3).
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, 4) and (2, negative 1).
Solution

−52

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, 2) and (4, negative 1).

In the following exercises, find the slope of each line.

y=3

Solution

0

y=−2

x=−5

Solution

undefined

x=4

In the following exercises, use the slope formula to find the slope of the line between each pair of points.

(2,5),(4,0)

Solution

−52

(3,6),(8,0)

(−3,3),(4,−5)

Solution

−87

(−2,4),(3,−1)

(−1,−2),(2,5)

Solution

73

(−2,−1),(6,5)

(4,−5),(1,−2)

Solution

−1

(3,−6),(2,−2)

Graph a Line Given a Point and the Slope

In the following exercises, graph each line with the given point and slope.

(2,5);m=−13

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (2, 5) and (5, 4).

(1,4); m=−12

(−1,−4); m=43

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (negative 1, negative 4) and (2, 0).

(−3,−5); m=32

y-intercept 3; m=−25

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (0, 3) and (5, 1).

x-intercept −2; m=34

(−4,2); m=4

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (negative 4, 2) and (negative 3, 6).

(1,5); m=−3

Graph a Line Using Its Slope and Intercept

In the following exercises, identify the slope and y-intercept of each line.

y=−7x+3

Solution

m=−7;(0,3)

y=4x−10

3x+y=5

Solution

m=−3;(0,5)

4x+y=8

6x+4y=12

Solution

m=−32;(0,3)

8x+3y=12

5x−2y=6

Solution

m=52;(0,−3)

7x−3y=9

In the following exercises, graph the line of each equation using its slope and y-intercept.

y=3x−1

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, negative 1) and (1, 2).

y=2x−3

y=−x+3

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, 3) and (1, 2).

y=−x−4

y=−25x−3

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, negative 3) and (5, negative 5).

y=−35x+2

3x−2y=4

Solution
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, negative 2) and (2, 1).

3x−4y=8

Choose the Most Convenient Method to Graph a Line

In the following exercises, determine the most convenient method to graph each line.

x=2

Solution

vertical line

y=5

y=−3x+4

Solution

slope-intercept

x−y=5

x−y=1

Solution

intercepts

y=23x−1

3x−2y=−12

Solution

intercepts

2x−5y=−10

Graph and Interpret Applications of Slope–Intercept

The equation P=31+1.75w models the relation between the amount of Tuyet’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

ⓐ Find Tuyet’s payment for a month when 0 units of water are used.

ⓑ Find Tuyet’s payment for a month when 12 units of water are used.

ⓒ Interpret the slope and P-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ $31
ⓑ $52
ⓒ The slope, 1.75, means that the payment, P, increases by $1.75 when the number of units of water used, w, increases by 1. The P-intercept means that when the number units of water Tuyet used is 0, the payment is $31.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 1 to 21. The y-axis runs from negative 1 to 80. The line goes through the points (0, 31) and (12, 52).

The equation P=28+2.54w models the relation between the amount of R and y’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

ⓐ Find the payment for a month when R and y used 0 units of water.

ⓑ Find the payment for a month when R and y used 15 units of water.

ⓒ Interpret the slope and P-intercept of the equation.

ⓓ Graph the equation.

Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day.

ⓐ Find the amount Bruce is reimbursed on a day when he drives 0 miles.

ⓑ Find the amount Bruce is reimbursed on a day when he drives 220 miles.

ⓒ Interpret the slope and R-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ $42
ⓑ $168.50
ⓒ The slope, 0.575 means that the amount he is reimbursed, R, increases by $0.575 when the number of miles driven, m, increases by 1. The R-intercept means that when the number miles driven is 0, the amount reimbursed is $42.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 50 to 250. The y-axis runs from negative 50 to 300. The line goes through the points (0, 42) and (220, 168.5).

Janelle is planning to rent a car while on vacation. The equation C=0.32m+15 models the relation between the cost in dollars, C, per day and the number of miles, m, she drives in one day.

ⓐ Find the cost if Janelle drives the car 0 miles one day.

ⓑ Find the cost on a day when Janelle drives the car 400 miles.

ⓒ Interpret the slope and C-intercept of the equation.

ⓓ Graph the equation.

Cherie works in retail and her weekly salary includes commission for the amount she sells. The equation S=400+0.15c models the relation between her weekly salary, S, in dollars and the amount of her sales, c, in dollars.

ⓐ Find Cherie’s salary for a week when her sales were $0.

ⓑ Find Cherie’s salary for a week when her sales were $3,600.

ⓒ Interpret the slope and S-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ $400
ⓑ $940
ⓒ The slope, 0.15, means that Cherie’s salary, S, increases by $0.15 for every $1 increase in her sales. The S-intercept means that when her sales are $0, her salary is $400.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 500 to 3500. The y-axis runs from negative 200 to 1000. The line goes through the points (0, 400) and (3600, 940).

Patel’s weekly salary includes a base pay plus commission on his sales. The equation S=750+0.09c models the relation between his weekly salary, S, in dollars and the amount of his sales, c, in dollars.

ⓐ Find Patel’s salary for a week when his sales were 0.

ⓑ Find Patel’s salary for a week when his sales were 18,540.

ⓒ Interpret the slope and S-intercept of the equation.

ⓓ Graph the equation.

Costa is planning a lunch banquet. The equation C=450+28g models the relation between the cost in dollars, C, of the banquet and the number of guests, g.

ⓐ Find the cost if the number of guests is 40.

ⓑ Find the cost if the number of guests is 80.

ⓒ Interpret the slope and C-intercept of the equation.

ⓓ Graph the equation.

Solution

ⓐ $1570
ⓑ $2690
ⓒ The slope gives the cost per guest. The slope, 28, means that the cost, C, increases by $28 when the number of guests increases by 1. The C-intercept means that if the number of guests was 0, the cost would be $450.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 20 to 100. The y-axis runs from negative 1000 to 7000. The line goes through the points (0, 450) and (40, 1570).

Margie is planning a dinner banquet. The equation C=750+42g models the relation between the cost in dollars, C of the banquet and the number of guests, g.

ⓐ Find the cost if the number of guests is 50.

ⓑ Find the cost if the number of guests is 100.

ⓒ Interpret the slope and C-intercept of the equation.

ⓓ Graph the equation.

Use Slopes to Identify Parallel and Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel, perpendicular, or neither.

y=34x−3;3x−4y=−2

Solution

parallel

3x+4y=−2;y=34x−3

2x−4y=6;x−2y=3

Solution

neither

8x+6y=6;12x+9y=12

x=5;x=−6

Solution

parallel

x=−3;x=−2

4x−2y=5;3x+6y=8

Solution

perpendicular

8x−2y=7;3x+12y=9

3x−6y=12;6x−3y=3

Solution

neither

9x−5y=4;5x+9y=−1

7x−4y=8;4x+7y=14

Solution

perpendicular

5x−2y=11;5x−y=7

3x−2y=8;2x+3y=6

Solution

perpendicular

2x+3y=5;3x−2y=7

3x−2y=1;2x−3y=2

Solution

neither

2x+4y=3;6x+3y=2

y=2;y=6

Solution

parallel

y=−1;y=2

Writing Exercises

How does the graph of a line with slope m=-12 differ from the graph of a line with slope m=2?

Solution

It is perpendicular to the second line.

Why is the slope of a vertical line “undefined”?

Explain how you can graph a line given a point and its slope.

Solution

Answers will vary.

Explain in your own words how to decide which method to use to graph a line.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 7 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “find the slope of a line”, “graph a line given a point and the slope”, “graph a line using its slope and intercept”, “choose the most convenient method to graph a line”, “graph and interpret applications of slope-intercept”, and “use slopes to identify parallel and perpendicular lines”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

parallel lines
Parallel lines are lines in the same plane that do not intersect.
perpendicular lines
Perpendicular lines are lines in the same plane that form a right angle.

Find the Equation of a Line

Learning Objectives

By the end of this section, you will be able to:

  • Find an equation of the line given the slope and y-intercept
  • Find an equation of the line given the slope and a point
  • Find an equation of the line given two points
  • Find an equation of a line parallel to a given line
  • Find an equation of a line perpendicular to a given line

Before you get started, take this readiness quiz.

Simplify: 25(x+15).
If you missed this problem, review Example 6 in Properties of Real Numbers.

Solution

25x+6

Simplify: −3(x−(−2)).
If you missed this problem, review Example 9 in Properties of Real Numbers.

Solution

−3x−6

Solve for y: y−3=−2(x+1).
If you missed this problem, review Example 4 in Solve a Formula for a Specific Variable.

Solution

y=−2x+1

How do online companies know that “you may also like” a particular item based on something you just ordered? How can economists know how a rise in the minimum wage will affect the unemployment rate? How do medical researchers create drugs to target cancer cells? How can traffic engineers predict the effect on your commuting time of an increase or decrease in gas prices? It’s all mathematics.

The physical sciences, social sciences, and the business world are full of situations that can be modeled with linear equations relating two variables. To create a mathematical model of a linear relation between two variables, we must be able to find the equation of the line. In this section, we will look at several ways to write the equation of a line. The specific method we use will be determined by what information we are given.

Find an Equation of the Line Given the Slope and y-Intercept

We can easily determine the slope and intercept of a line if the equation is written in slope-intercept form, y=mx+b. Now we will do the reverse—we will start with the slope and y-intercept and use them to find the equation of the line.

Find the equation of a line with slope −9 and y-intercept (0,−4).

Solution

Since we are given the slope and y-intercept of the line, we can substitute the needed values into the slope-intercept form, y=mx+b.

Name the slope. The equation 'm = -9' is displayed on a white background. The letter 'm' and the equals sign are in gray, while the number '-9' is prominently shown in red.
Name the y-intercept. The text
Substitute the values into y=mx+b. The equation for a straight line, y = mx + b, is shown with 'm' (slope) in red and 'b' (y-intercept) in blue.
A mathematical equation is displayed, reading y equals negative 9x plus parentheses negative 4 parentheses. The -9 is in red and the -4 is in light blue.
The image displays a linear equation in slope-intercept form, y = -9x - 4, which represents a straight line with a slope of -9 and a y-intercept of -4.

Find the equation of a line with slope 25 and y-intercept (0,4).

Solution

y=25x+4

Find the equation of a line with slope −1 and y-intercept (0,−3).

Solution

y=−x−3

Sometimes, the slope and intercept need to be determined from the graph.

Find the equation of the line shown in the graph.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 3, negative 6), (0, negative 4), (3, negative 2), and (6, 0).
Solution

We need to find the slope and y-intercept of the line from the graph so we can substitute the needed values into the slope-intercept form, y=mx+b.

To find the slope, we choose two points on the graph.

The y-intercept is (0,−4) and the graph passes through (3,−2).

Find the slope, by counting the rise and run. The formula for calculating the slope (m) of a line, represented as the ratio of its vertical change (rise) to its horizontal change (run): m = rise / run.
The equation m equals the fraction 2 over 3 is displayed on a white background.
Find the y-intercept. The text 'y-intercept (0, -4)' is displayed in the image, indicating the point where a line or curve crosses the y-axis.
Substitute the values into y=mx+b. The image shows the mathematical equation for a straight line in slope-intercept form, which is 'y = mx + b'. The variable 'm' is colored red, and 'b' is colored blue.
The image shows the linear equation y = (2/3)x - 4. The slope (2/3) is colored red, and the y-intercept (-4) is colored light blue, highlighting key components of the equation.

Find the equation of the line shown in the graph.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 5, negative 2), (0, 1), and (5, 4).
Solution

y=35x+1

Find the equation of the line shown in the graph.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (0, negative 5), (3, negative 1), and (6, 3).
Solution

y=43x−5

Find an Equation of the Line Given the Slope and a Point

Finding an equation of a line using the slope-intercept form of the equation works well when you are given the slope and y-intercept or when you read them off a graph. But what happens when you have another point instead of the y-intercept?

We are going to use the slope formula to derive another form of an equation of the line.

Suppose we have a line that has slope m and that contains some specific point (x1,y1) and some other point, which we will just call (x,y). We can write the slope of this line and then change it to a different form.

This table demonstrates the step-by-step derivation of the point-slope form of a linear equation from the slope formula.
m=y−y1x−x1
Multiply both sides of the equation by x−x1. m(x−x1)=(y−y1x−x1)(x−x1)
Simplify. m(x−x1)=y−y1
Rewrite the equation with the y terms on the left. y−y1=m(x−x1)

This format is called the point-slope form of an equation of a line.

Point-slope Form of an Equation of a Line

The point-slope form of an equation of a line with slope m and containing the point (x1,y1) is:

y−y1=m(x−x1)

We can use the point-slope form of an equation to find an equation of a line when we know the slope and at least one point. Then, we will rewrite the equation in slope-intercept form. Most applications of linear equations use the the slope-intercept form.

How to Find an Equation of a Line Given a Point and the Slope

Find an equation of a line with slope m=−13 that contains the point (6,−4). Write the equation in slope-intercept form.

Solution
Step 1 is to identify the slope. The slope is given. m equals negative 1 divided by 3. Step 2 is to identify the point. The point is given. x 1 is 6 and y 1 is negative 4. Step 3 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus negative 4 equals negative 1 divided by 3 times the quantity x minus 6 in parentheses. This simplifies to y plus 4 equals negative 1 divided by 3 times x plus 2. Step 4 is to write the equation in slope-intercept form. y equals negative 1 divided by 3 times x minus 2.

Find the equation of a line with slope m=−25 and containing the point (10,−5).

Solution

y=−25x−1

Find the equation of a line with slope m=−34, and containing the point (4,−7).

Solution

y=−34x−4

We list the steps for easy reference.

To find an equation of a line given the slope and a point.

  1. Identify the slope.
  2. Identify the point.
  3. Substitute the values into the point-slope form, y−y1=m(x−x1).
  4. Write the equation in slope-intercept form.

Find an equation of a horizontal line that contains the point (−2,−6). Write the equation in slope-intercept form.

Solution

Every horizontal line has slope 0. We can substitute the slope and points into the point-slope form, y−y1=m(x−x1).

Identify the slope. The mathematical equation 'm = ø' is displayed in black and light blue text on a plain white background, symbolizing the variable 'm' equaling the empty set or null value.
Identify the point. Coordinate pair (-2, -6), displayed as x₁ = -2 and y₁ = -6.
Substitute the values into y−y1=m(x−x1). The point-slope form of a linear equation: y - y1 = m(x - x1), where m represents the slope and (x1, y1) is a known point on the line.
A mathematical equation is displayed: y - (-6) = 0(x - (-2)). The numbers -6 and -2 are highlighted in red, and the 0 is highlighted in light blue.
Simplify. The image displays the equation y + 6 = 0 against a plain white background.
The image displays the algebraic equation y = -6 in a simple, clear font against a white background.
Write in slope-intercept form. It is in y-form, but could be written y=0x−6.

Did we end up with the form of a horizontal line, y=b?

Find the equation of a horizontal line containing the point (−3,8).

Solution

y=8

Find the equation of a horizontal line containing the point (−1,4).

Solution

y=4

Find an Equation of the Line Given Two Points

When real-world data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when just two points are given.

So far, we have two options for finding an equation of a line: slope-intercept or point-slope. When we start with two points, it makes more sense to use the point-slope form.

But then we need the slope. Can we find the slope with just two points? Yes. Then, once we have the slope, we can use it and one of the given points to find the equation.

How to Find the Equation of a Line Given Two Points

Find an equation of a line that contains the points (−3,−1) and (2,−2) Write the equation in slope-intercept form.

Solution
Step 1 is to find the slope using the given points. Find the slope of the line through (negative 3, negative 1) and (2, and negative 2). m equals the quotient of y 2 minus y 1 in parentheses and x 2 minus x 1 in parentheses. m equals the quotient of negative 2 minus negative 1 in parentheses and 2 minus negative 3 in parentheses. m equals negative 1 divided by 5. Step 2 is to identify the point. Choose either point. x 1 is 2 and y 1 is negative 2. Step 3 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus negative 2 equals negative 1 divided by 5 times the quantity x minus 2 in parentheses. This simplifies to y plus 2 equals negative 1 divided by 5 times x plus 2 divided by 5. Step 4 is to write the equation in slope-intercept form. y equals negative 1 divided by 5 times x minus 8 divided by 5.

Find the equation of a line containing the points (−2,−4) and (1,−3).

Solution

y=13x−103

Find the equation of a line containing the points (−4,−3) and (1,−5).

Solution

y=−25x−235

The steps are summarized here.

To find an equation of a line given two points.

  1. Find the slope using the given points. m=y2−y1x2−x1
  2. Choose one point.
  3. Substitute the values into the point-slope form: y−y1=m(x−x1).
  4. Write the equation in slope-intercept form.

Find an equation of a line that contains the points (−3,5) and (−3,4). Write the equation in slope-intercept form.

Solution

Again, the first step will be to find the slope.

Calculation of the slope of a line between two points, specifically demonstrating a case where the slope is undefined.
Find the slope of the line through (−3,5) and (−3,4). m=y2−y1x2−x1
m=4−5−3−(−3)
m=−10
The slope is undefined.

This tells us it is a vertical line. Both of our points have an x-coordinate of −3. So our equation of the line is x=−3. Since there is no y, we cannot write it in slope-intercept form.

You may want to sketch a graph using the two given points. Does your graph agree with our conclusion that this is a vertical line?

Find the equation of a line containing the points (5,1) and (5,−4).

Solution

x=5

Find the equation of a line containing the points (−4,4) and (−4,3).

Solution

x=−4

We have seen that we can use either the slope-intercept form or the point-slope form to find an equation of a line. Which form we use will depend on the information we are given.

To Write an Equation of a Line
If given: Use: Form:
Slope and y-intercept slope-intercept y=mx+b
Slope and a point point-slope y−y1=m(x−x1)
Two points point-slope y−y1=m(x−x1)

Find an Equation of a Line Parallel to a Given Line

Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point-slope equation.

First, let’s look at this graphically.

This graph shows y=2x−3. We want to graph a line parallel to this line and passing through the point (−2,1).

This figure has a graph of a straight line and a point on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The point (negative 2, 1) is plotted. The line does not go through the point (negative 2, 1).

We know that parallel lines have the same slope. So the second line will have the same slope as y=2x−3. That slope is m∥=2. We’ll use the notation m∥ to represent the slope of a line parallel to a line with slope m. (Notice that the subscript || looks like two parallel lines.)

The second line will pass through (−2,1) and have m=2.

To graph the line, we start at(−2,1) and count out the rise and run.

With m=2 (or m=21), we count out the rise 2 and the run 1. We draw the line, as shown in the graph.

This figure has a graph of a two straight lines on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The first line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The points (negative 2, 1) and (negative 1, 3) are plotted. The second line goes through the points (negative 2, 1) and (negative 1, 3).

Do the lines appear parallel? Does the second line pass through(−2,1)?

We were asked to graph the line, now let’s see how to do this algebraically.

We can use either the slope-intercept form or the point-slope form to find an equation of a line. Here we know one point and can find the slope. So we will use the point-slope form.

How to Find the Equation of a Line Parallel to a Given Line and a Point

Find an equation of a line parallel to y=2x−3 that contains the point (−2,1). Write the equation in slope-intercept form.

Solution
Step 1 is to find the slope of the given line. The line is in slope-intercept form, y equals 2 x minus 3. m equals 2. Step 2 is to find the slope of the parallel line. Parallel lines have the same slope. m equals 2. Step 3 is to identify the point. The given point is (negative 2, 1). x 1 is negative 2 and y 1 is 1. Step 4 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus 1 equals 2 times the quantity x minus negative 2 in parentheses. This simplifies to y minus 1 equals 2 x plus 4. Step 5 is to write the equation in slope-intercept form. y equals 2 x plus 5.

Look at graph with the parallel lines shown previously. Does this equation make sense? What is the y-intercept of the line? What is the slope?

Find an equation of a line parallel to the line y=3x+1 that contains the point (4,2). Write the equation in slope-intercept form.

Solution

y=3x−10

Find an equation of a line parallel to the line y=12x−3 that contains the point (6,4).

Write the equation in slope-intercept form.

Solution

y=12x+1

Find an equation of a line parallel to a given line.

  1. Find the slope of the given line.
  2. Find the slope of the parallel line.
  3. Identify the point.
  4. Substitute the values into the point-slope form: y−y1=m(x−x1).
  5. Write the equation in slope-intercept form.

Find an Equation of a Line Perpendicular to a Given Line

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point-slope equation, like we did with parallel lines.

This graph shows y=2x−3. Now, we want to graph a line perpendicular to this line and passing through (−2,1).

This figure has a graph of a straight line and a point on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The point (negative 2, 1) is plotted. The line does not go through the point (negative 2, 1).

We know that perpendicular lines have slopes that are negative reciprocals.

We’ll use the notation m⊥ to represent the slope of a line perpendicular to a line with slope m. (Notice that the subscript ⊥ looks like the right angles made by two perpendicular lines.)

y=2x−3perpendicular line m=2m⊥=−12

We now know the perpendicular line will pass through (−2,1) with m⊥=−12.

To graph the line, we will start at (−2,1) and count out the rise −1 and the run 2. Then we draw the line.

This figure has a graph of two perpendicular straight lines on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The first line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The points (negative 2, 1) and (0, 0) are plotted. A right triangle is drawn connecting the points (negative 2, 1), (negative 2, 0), and (0, 0). The second line goes through the points (negative 2, 1) and (0, 0).

Do the lines appear perpendicular? Does the second line pass through(−2,1)?

We were asked to graph the line, now, let’s see how to do this algebraically.

We can use either the slope-intercept form or the point-slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point-slope form.

How to Find the Equation of a Line Perpendicular to a Given Line and a Point

Find an equation of a line perpendicular to y=2x−3 that contains the point (−2,1). Write the equation in slope-intercept form.

Solution
Step 1 is to find the slope of the given line. The line is in slope-intercept form, y equals 2 x minus 3. m equals 2. Step 2 is to find the slope of the perpendicular line. The slopes of perpendicular lines are negative reciprocals. m equals negative 1 divided by 2 Step 3 is to identify the point. The given point is (negative 2, 1). x 1 is negative 2 and y 1 is 1. Step 4 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus 1 equals negative 1 divided by 2 times the quantity x minus negative 2 in parentheses. This simplifies to y minus 1 equals negative 1 divided by 2 times the quantity x plus 2 in parentheses. This further simplifies to y minus 1 equals negative 1 divided by 2 times x minus 1. Step 5 is to write the equation in slope-intercept form. y equals negative 1 divided by 2 times x.

Find an equation of a line perpendicular to the line y=3x+1 that contains the point (4,2). Write the equation in slope-intercept form.

Solution

y=−13x+103

Find an equation of a line perpendicular to the line y=12x−3 that contains the point (6,4). Write the equation in slope-intercept form.

Solution

y=−2x+16

Find an equation of a line perpendicular to a given line.

  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line.
  3. Identify the point.
  4. Substitute the values into the point-slope form, y−y1=m(x−x1).
  5. Write the equation in slope-intercept form.

Find an equation of a line perpendicular to x=5 that contains the point (3,−2). Write the equation in slope-intercept form.

Solution

Again, since we know one point, the point-slope option seems more promising than the slope-intercept option. We need the slope to use this form, and we know the new line will be perpendicular to x=5. This line is vertical, so its perpendicular will be horizontal. This tells us the m⊥=0.

Step-by-step derivation of a line's equation (y = -2) using a point and perpendicular slope.
Identify the point. (3,−2)
Identify the slope of the perpendicular line. m⊥=0
Substitute the values into y−y1=m(x−x1). y−y1=m(x−x1)
y−(−2)=0(x−3)
Simplify. y+2=0
y=−2

Sketch the graph of both lines. On your graph, do the lines appear to be perpendicular?

Find an equation of a line that is perpendicular to the line x=4 that contains the point (4,−5).. Write the equation in slope-intercept form.

Solution

y=−5

Find an equation of a line that is perpendicular to the line x=2 that contains the point (2,−1). Write the equation in slope-intercept form.

Solution

y=−1

In Example 9, we used the point-slope form to find the equation. We could have looked at this in a different way.

We want to find a line that is perpendicular to x=5 that contains the point (3,−2). This graph shows us the linex=5 and the point (3,−2).

This figure has a graph of a straight vertical line and a point on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (5, 0), (5, 1), and (5, 2). The point (3, negative 2) is plotted. The line does not go through the point (3, negative 2).

We know every line perpendicular to a vertical line is horizontal, so we will sketch the horizontal line through (3,−2).

This figure has a graph of a straight vertical line and a straight horizontal line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The vertical line goes through the points (5, 0), (5, 1), and (5, 2). The horizontal line goes through the points (negative 2, negative 2), (0, negative 2), (3, negative 2), and (6, negative 2).

Do the lines appear perpendicular?

If we look at a few points on this horizontal line, we notice they all have y-coordinates of −2. So, the equation of the line perpendicular to the vertical line x=5 is y=−2.

Find an equation of a line that is perpendicular to y=−3 that contains the point (−3,5). Write the equation in slope-intercept form.

Solution

The line y=−3 is a horizontal line. Any line perpendicular to it must be vertical, in the form x=a. Since the perpendicular line is vertical and passes through (−3,5), every point on it has an x-coordinate of −3. The equation of the perpendicular line is x=−3

You may want to sketch the lines. Do they appear perpendicular?

Find an equation of a line that is perpendicular to the line y=1 that contains the point (−5,1). Write the equation in slope-intercept form.

Solution

x=−5

Find an equation of a line that is perpendicular to the line y=−5 that contains the point (−4,−5). Write the equation in slope-intercept form.

Solution

x=−4

Access these online resources for additional instruction and practice with finding the equation of a line.

  • Write an Equation of Line Given its slope and Y-Intercept
  • Find the equation given two points
  • Find the equation of perpendicular and parallel lines

Key Concepts

  • How to find an equation of a line given the slope and a point.
    1. Identify the slope.
    2. Identify the point.
    3. Substitute the values into the point-slope form, y−y1=m(x−x1).
    4. Write the equation in slope-intercept form.



  • How to find an equation of a line given two points.
    1. Find the slope using the given points. m=y2−y1x2−x1
    2. Choose one point.
    3. Substitute the values into the point-slope form: y−y1=m(x−x1).
    4. Write the equation in slope-intercept form.
      To Write an Equation of a Line
      If given: Use: Form:
      Slope and y-intercept slope-intercept y=mx+b
      Slope and a point point-slope y−y1=m(x−x1)
      Two points point-slope y−y1=m(x−x1)
  • How to find an equation of a line parallel to a given line.
    1. Find the slope of the given line.
    2. Find the slope of the parallel line.
    3. Identify the point.
    4. Substitute the values into the point-slope form: y−y1=m(x−x1).
    5. Write the equation in slope-intercept form
  • How to find an equation of a line perpendicular to a given line.
    1. Find the slope of the given line.
    2. Find the slope of the perpendicular line.
    3. Identify the point.
    4. Substitute the values into the point-slope form, y−y1=m(x−x1)
    5. Write the equation in slope-intercept form.

Practice Makes Perfect

Find an Equation of the Line Given the Slope and y-Intercept

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope-intercept form.

slope 3 and
y-intercept (0,5)

Solution

y=3x+5

slope 8 and
y-intercept (0,−6)

slope −3 and
y-intercept (0,−1)

Solution

y=−3x−1

slope −1 and
y-intercept (0,3)

slope 15 and
y-intercept (0,−5)

Solution

y=15x−5

slope −34 and
y-intercept (0,−2)

slope 0 and
y-intercept (0,−1)

Solution

y=−1

slope −4 and
y-intercept (0,0)

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope-intercept form.


This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 5), (1, negative 2), and (2, 1).

Solution

y=3x−5


This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 4), (1, 2), and (2, 0).


This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 3), (2, negative 2), and (6, 0).

Solution

y=12x−3


This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 2), (4, 5), and (8, 8).


This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 3), (3, negative 1), and (6, negative 5).

Solution

y=−43x+3


This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 1), (2, negative 4), and (4, negative 7).


This figure has a graph of a horizontal straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 2), (1, negative 2), and (2, negative 2).

Solution

y=−2


This figure has a graph of a horizontal straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 6), (1, 6), and (2, 6).

Find an Equation of the Line Given the Slope and a Point

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope-intercept form.

m=58, point (8,3)

Solution

y=58x−2

m=56, point (6,7)

m=−35, point (10,−5)

Solution

y=−35x+1

m=−34, point (8,−5)

m=−32, point (−4,−3)

Solution

y=−32x−9

m=−52, point (−8,−2)

m=−7, point (−1,−3)

Solution

y=−7x−10

m=−4, point (−2,−3)

Horizontal line containing (−2,5)

Solution

y=5

Horizontal line containing (−2,−3)

Horizontal line containing (−1,−7)

Solution

y=−7

Horizontal line containing (4,−8)

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope-intercept form.

(2,6) and (5,3)

Solution

y=−x+8

(4,3) and (8,1)

(−3,−4) and (5,−2)

Solution

y=14x−134

(−5,−3) and (4,−6)

(−1,3) and (−6,−7)

Solution

y=2x+5

(−2,8) and (−4,−6)

(0,4) and (2,−3)

Solution

y=−72x+4

(0,−2) and (−5,−3)

(7,2) and (7,−2)

Solution

x=7

(−2,1) and (−2,−4)

(3,−4) and (5,−4)

Solution

y=−4

(−6,−3) and (−1,−3)

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope-intercept form.

line y=4x+2,
point (1,2)

Solution

y=4x−2

line y=−3x−1,
point (2,−3).

line 2x−y=6,
point (3,0).

Solution

y=2x−6

line 2x+3y=6,
point (0,5).

line x=−4,
point (−3,−5).

Solution

x=−3

line x−2=0,
point (1,−2)

line y=5,
point (2,−2)

Solution

y=−2

line y+2=0,
point (3,−3)

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope-intercept form.

line y=−2x+3,
point (2,2)

Solution

y=12x+1

line y=−x+5,
point (3,3)

line y=34x−2,
point (−3,4)

Solution

y=−43x

line y=23x−4,
point (2,−4)

line 2x−3y=8,
point (4,−1)

Solution

y=−32x+5

line 4x−3y=5,
point (−3,2)

line 2x+5y=6,
point (0,0)

Solution

y=52x

line 4x+5y=−3,
point (0,0)

line x=3,
point (3,4)

Solution

y=4

line x=−5,
point (1,−2)

line x=7,
point (−3,−4)

Solution

y=−4

line x=−1,
point (−4,0)

line y−3=0,
point (−2,−4)

Solution

x=−2

line y−6=0,
point (−5,−3)

line y-axis,
point (3,4)

Solution

y=4

line y-axis,
point (2,1)

Mixed Practice

In the following exercises, find the equation of each line. Write the equation in slope-intercept form.

Containing the points (4,3) and (8,1)

Solution

y=−12x+5

Containing the points (−2,0) and (−3,−2)

m=16, containing point (6,1)

Solution

y=16x

m=56, containing point (6,7)

Parallel to the line 4x+3y=6, containing point (0,−3)

Solution

y=−43x−3

Parallel to the line 2x+3y=6, containing point (0,5)

m=−34, containing point (8,−5)

Solution

y=−34x+1

m=−35, containing point (10,−5)

Perpendicular to the line y−1=0, point (−2,6)

Solution

x=−2

Perpendicular to the line y-axis, point (−6,2)

Parallel to the line x=−3, containing point (−2,−1)

Solution

x=−2

Parallel to the line x=−4, containing point (−3,−5)

Containing the points (−3,−4) and (2,−5)

Solution

y=−15x−235

Containing the points (−5,−3) and (4,−6)

Perpendicular to the line x−2y=5, point (−2,2)

Solution

y=−2x−2

Perpendicular to the line 4x+3y=1, point (0,0)

Writing Exercises

Why are all horizontal lines parallel?

Solution

Answers will vary.

Explain in your own words why the slopes of two perpendicular lines must have opposite signs.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The figure shows a table with six rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “confidently”, the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “find the equation of the line given the slope and y-intercept”, “find an equation of the line given the slope and a point”, “find an equation of the line given two points”, “find an equation of a line parallel to a given line”, and “find an equation of a line perpendicular to a given line”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

point-slope form
The point-slope form of an equation of a line with slope m and containing the point (x1,y1) is y−y1=m(x−x1).

Graph Linear Inequalities in Two Variables

Learning Objectives

By the end of this section, you will be able to:

  • Verify solutions to an inequality in two variables.
  • Recognize the relation between the solutions of an inequality and its graph.
  • Graph linear inequalities in two variables
  • Solve applications using linear inequalities in two variables

Before you get started, take this readiness quiz.

Graph x>2 on a number line.
If you missed this problem, review Example 1 in Solve Linear Inequalities.

Solution

A number line displays the inequality x > 2. It shows an open circle at 2 and a thick arrow extending to the right, indicating all real numbers greater than 2. The number line ranges from -5 to 5.

Solve: 4x+3>23.
If you missed this problem, review Example 5 in Solve Linear Inequalities.

Solution

x>5

Translate: 8>x>3.
If you missed this problem, review Example 9 in Solve Linear Inequalities.

Solution

x is between 3 and 8, not inclusive

Verify Solutions to an Inequality in Two Variables

Previously we learned to solve inequalities with only one variable. We will now learn about inequalities containing two variables. In particular we will look at linear inequalities in two variables which are very similar to linear equations in two variables.

Linear inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs—so that your business made a profit.

Linear Inequality

A linear inequality is an inequality that can be written in one of the following forms:

Ax+By>CAx+By≥CAx+By<CAx+By≤C

Where A and B are not both zero.

Recall that an inequality with one variable had many solutions. For example, the solution to the inequality x>3 is any number greater than 3. We showed this on the number line by shading in the number line to the right of 3, and putting an open parenthesis at 3. See Figure 1.


Image of the number line with the integers from negative 5 to 5. The part of the number line to the right of 3 is marked with a blue line. The number 3 is marked with a blue open parenthesis.

Similarly, linear inequalities in two variables have many solutions. Any ordered pair (x,y) that makes an inequality true when we substitute in the values is a solution to a linear inequality.

Solution to a Linear Inequality

An ordered pair (x,y) is a solution to a linear inequality if the inequality is true when we substitute the values of x and y.

Determine whether each ordered pair is a solution to the inequality y>x+4:

ⓐ (0,0) ⓑ (1,6) ⓒ (2,6) ⓓ (−5,−15) ⓔ (−8,12)

Solution
ⓐ
(0,0) A mathematical inequality is displayed, showing 'y > x + 4' in black text against a white background.
Text instruction: 'Substitute 0 for x and 0 for y.', with the first '0' in red and the second '0' in light blue. The expression '0 >? 0 + 4' questions whether zero is greater than four, with a question mark above the inequality symbol.
Simplify. The mathematical expression '0 not greater than or equal to 4' is displayed on a white background.
So, (0,0) is not a solution to y>x+4.
ⓑ
(1,6) The image shows the mathematical inequality y > x + 4.
The text 'Substitute 1 for x and 6 for y' is displayed on a white background. The number 1 is highlighted in red, and the number 6 is highlighted in light blue. An inequality 6 > 1 + 4 with a question mark, prompting evaluation of whether six is indeed greater than the sum of one and four.
Simplify. A simple mathematical inequality showing that 6 is greater than 5, represented as '6 > 5' in a bold, dark gray font on a white background.
So, (1,6) is a solution to y>x+4.
ⓒ
(2,6) The mathematical inequality y > x + 4 is displayed in black text on a white background, representing a region above the line y = x + 4 in a coordinate plane.
Substitute 2 for x and 6 for y. A math problem featuring the inequality '6 > 2 + 4' with a question mark over the greater-than sign, asking if the statement is true. The number 6 is blue and 2 is red.
Simplify. "6 is not greater than 6"
So, (2,6) is not a solution to y>x+4.
ⓓ
(−5,−15) The mathematical inequality 'y > x + 4' is displayed on a white background.
Substitute -5 for x and -15 for y. A mathematical inequality problem asks to determine if -15 is greater than -5 + 4, with a question mark above the greater than symbol, implying verification of the statement.
Simplify. The image displays the mathematical inequality -15 not greater than or equal to -1, which is equivalent to -15 < -1.
So, (−5,−15) is not a solution to y>x+4.
ⓔ
(−8,12) The mathematical inequality y > x + 4 is shown in black text on a white background.
The image displays text instructions to 'Substitute -8 for x and 12 for y.', with -8 highlighted in red and 12 highlighted in light blue. An inequality expression showing '12 > ? -8 + 4' on a white background. The number '12' is light blue, '-8' is red, and the question mark is above the greater than symbol.
Simplify. The mathematical inequality '12 > -4' is displayed in a simple, clear font against a white background.
So, (−8,12) is a solution to y>x+4.

Determine whether each ordered pair is a solution to the inequality y>x−3:

ⓐ (0,0) ⓑ (4,9) ⓒ (−2,1) ⓓ (−5,−3) ⓔ (5,1)

Solution

ⓐ yes ⓑ yes ⓒ yes ⓓ yes ⓔ no

Determine whether each ordered pair is a solution to the inequality y<x+1:

ⓐ (0,0) ⓑ (8,6) ⓒ (−2,−1) ⓓ (3,4) ⓔ (−1,−4)

Solution

ⓐ yes ⓑ yes ⓒ no ⓓ no
ⓔ yes

Recognize the Relation Between the Solutions of an Inequality and its Graph

Now, we will look at how the solutions of an inequality relate to its graph.

Let’s think about the number line in shown previously again. The point x=3 separated that number line into two parts. On one side of 3 are all the numbers less than 3. On the other side of 3 all the numbers are greater than 3. See Figure 2.

Image of the number line with the integers from negative 5 to 5. The part of the number line to the right of 3 is marked with a blue line. The number 3 is marked with a blue open parenthesis. The part of the number line to the right of 3 is labeled “numbers greater than 3”. The part of the number line to the left of 3 is labeled “numbers less than 3”.
The solution to x>3 is the shaded part of the number line to the right of x=3.

Similarly, the line y=x+4 separates the plane into two regions. On one side of the line are points with y<x+4. On the other side of the line are the points with y>x+4. We call the line y=x+4 a boundary line.

Boundary Line

The line with equation Ax+By=C is the boundary line that separates the region where Ax+By>C from the region where Ax+By<C.

For an inequality in one variable, the endpoint is shown with a parenthesis or a bracket depending on whether or not a is included in the solution:

Two number lines are shown with the middle labeled with the number “a”. In both number lines, the part to the left of the number a is marked with red. The first number line is labeled “x is less than a” and the number a is marked with an open parenthesis. The second number line is labeled “x is less than or equal to a” and the number a is marked with an open bracket.

Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to show whether or not it the line is included in the solution.

Ax+By<CAx+By≤C Ax+By>CAx+By≥C Boundary line isAx+By=CBoundary line isAx+By=C Boundary line is not included in solution.Boundary line is included in solution. Boundary line is dashed.Boundary line is solid.

Now, let’s take a look at what we found in Example 1. We’ll start by graphing the line y=x+4, and then we’ll plot the five points we tested, as shown in the graph. See Figure 3.

This figure has the graph of some points and a straight line on the x y-coordinate plane. The x and y axes run from negative 16 to 16. The points (negative 8, 12), (negative 5, negative 15), (0, 0), (1, 6), and (2, 6) are plotted and labeled with their coordinates. A straight line is drawn through the points (negative 4, 0), (0, 4), and (2, 6).

In Example 1 we found that some of the points were solutions to the inequality y>x+4 and some were not.

Which of the points we plotted are solutions to the inequality y>x+4?

The points (1,6) and (−8,12) are solutions to the inequality y>x+4. Notice that they are both on the same side of the boundary line y=x+4.

The two points (0,0) and (−5,−15) are on the other side of the boundary line y=x+4, and they are not solutions to the inequality y>x+4. For those two points, y<x+4.

What about the point (2,6)? Because 6=2+4, the point is a solution to the equation y=x+4, but not a solution to the inequality y>x+4. So the point (2,6) is on the boundary line.

Let’s take another point above the boundary line and test whether or not it is a solution to the inequality y>x+4. The point (0,10) clearly looks to above the boundary line, doesn’t it? Is it a solution to the inequality?

y>x+410>?0+410>4

So, (0,10) is a solution to y>x+4.

Any point you choose above the boundary line is a solution to the inequality y>x+4. All points above the boundary line are solutions.

Similarly, all points below the boundary line, the side with (0,0) and (−5,−15), are not solutions to y>x+4, as shown in Figure 4.

This figure has the graph of some points and a straight line on the x y-coordinate plane. The x and y axes run from negative 16 to 16. The points (negative 8, 12), (negative 5, negative 15), (0, 0), (1, 6), and (2, 6) are plotted and labeled with their coordinates. A straight line is drawn through the points (negative 4, 0), (0, 4), and (2, 6). The line divides the x y-coordinate plane into two halves. The top left half is labeled y is greater than x plus 4. The bottom right half is labeled y is less than x plus 4.

The graph of the inequality y>x+4 is shown in below.

The line y=x+4 divides the plane into two regions. The shaded side shows the solutions to the inequality y>x+4.

The points on the boundary line, those where y=x+4, are not solutions to the inequality y>x+4, so the line itself is not part of the solution. We show that by making the line dashed, not solid.

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight dashed line is drawn through the points (negative 4, 0), (0, 4), and (2, 6). The line divides the x y-coordinate plane into two halves. The top left half is colored red to indicate that this is where the solutions of the inequality are.

The boundary line shown in this graph is y=2x−1. Write the inequality shown by the graph.

 This figure has the graph of a straight solid line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight solid line is drawn through the points (0, negative 1), (1, 1), and (2, 3). The line divides the x y-coordinate plane into two halves. The top left half is colored red to indicate that this is where the solutions of the inequality are.
Solution

The line y=2x−1 is the boundary line. On one side of the line are the points with y>2x−1 and on the other side of the line are the points with y<2x−1.

Let’s test the point (0,0) and see which inequality describes its position relative to the boundary line.

At (0,0), which inequality is true: y>2x−1 or y<2x−1?

y>2x−1y<2x−1 0>?2·0−10<?2·0−1 0>−1True0<−1False

Since, y>2x−1 is true, the side of the line with (0,0), is the solution. The shaded region shows the solution of the inequality y>2x−1.

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality y≥2x−1.

We could use any point as a test point, provided it is not on the line. Why did we choose (0,0)? Because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that y<2x−1.

Write the inequality shown by the graph with the boundary line y=−2x+3.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight line is drawn through the points (0, 3), (1, 1), and (3, negative 3). The line divides the x y-coordinate plane into two halves. The line itself and the top right half are colored red to indicate that this is where the solutions of the inequality are.
Solution

y≥−2x+3

Write the inequality shown by the graph with the boundary line y=12x−4.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight line is drawn through the points (0, negative 4), (2, negative 3), and (4, negative 2). The line divides the x y-coordinate plane into two halves. The line itself and the bottom right half are colored red to indicate that this is where the solutions of the inequality are.
Solution

y≤12x−4

The boundary line shown in this graph is 2x+3y=6. Write the inequality shown by the graph.

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight dashed line is drawn through the points (0, 2), (3, 0), and (6, negative 2). The line divides the x y-coordinate plane into two halves. The bottom left half is colored red to indicate that this is where the solutions of the inequality are.
Solution

The line 2x+3y=6 is the boundary line. On one side of the line are the points with 2x+3y>6 and on the other side of the line are the points with 2x+3y<6.

Let’s test the point (0,0) and see which inequality describes its side of the boundary line.

At (0,0), which inequality is true: 2x+3y>6 or 2x+3y<6?

2x+3y>62(0)+3(0)>?60>6False2x+3y<62(0)+3(0)<?60<6True

So the side with (0,0) is the side where 2x+3y<6.

(You may want to pick a point on the other side of the boundary line and check that 2x+3y>6.)

Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.

The shaded region shows the solution to the inequality 2x+3y<6.

Write the inequality shown by the shaded region in the graph with the boundary line x−4y=8.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight line is drawn through the points (0, negative 2), (4, negative 1), and (8, 0). The line divides the x y-coordinate plane into two halves. The line itself and the top left half are colored red to indicate that this is where the solutions of the inequality are.
Solution

x−4y≤8

Write the inequality shown by the shaded region in the graph with the boundary line 3x−y=6.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight line is drawn through the points (0, negative 6), (1, negative 3), and (2, 0). The line divides the x y-coordinate plane into two halves. The line itself and the bottom right half are colored red to indicate that this is where the solutions of the inequality are.
Solution

3x−y≥6

Graph Linear Inequalities in Two Variables

Now that we know what the graph of a linear inequality looks like and how it relates to a boundary equation we can use this knowledge to graph a given linear inequality.

How to Graph a Linear Equation in Two Variables

Graph the linear inequality y≥34x−2.

Solution
Step 1 is to Identify and graph the boundary line. If the inequality is less than or equal or greater than or equal, the boundary line is solid. If the inequality is less than or greater than, the boundary line is dashed. In this example the inequality sign is greater than or equal, so we draw a solid line. Replace the inequality sign with an equal sign to find the boundary line. Graph the boundary line y = 3 divided by 4 times x minus 2. The figure then shows the graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (0, negative 2), (4, 1), and (8, 4). Step 2 is to test a point that is not on the boundary line. Is it a solution of the inequality? We will test (0, 0). At (0, 0) is y greater than or equal to 3 divided by 4 times x minus 2? Is 0 greater than or equal to 3 divided by 4 times 0 minus 2? 0 is greater than or equal to negative 2 so (0, 0) is a solution. Step 3 is to shade in one side of the boundary line. If the test point is a solution, shade in the side that includes the point. If the test point is not a solution, shade in the opposite side. The test point (0, 0), is a solution to y greater than or equal to 3 divided by 4 times x minus 2. So we shade in the side that contains (0, 0). The figure then shows the graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (0, negative 2), (4, 1), and (8, 4). The top left half of the coordinate plane is shaded to indicate that this is where the solution set is located.

Graph the linear inequality y≥52x−4.

Solution


This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 4), (2, 1), and (4, 6). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region and on the boundary line, represent the solutions to y>52x−4.

Graph the linear inequality y<23x−5.

Solution


This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 5), (3, negative 3), and (5, negative 1). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region, but not those on the boundary line, represent the solutions to y<23x−5.

The steps we take to graph a linear inequality are summarized here.

Graph a linear inequality in two variables.

  1. Identify and graph the boundary line.
    • If the inequality is ≤or≥, the boundary line is solid.
    • If the inequality is <or>, the boundary line is dashed.
  2. Test a point that is not on the boundary line. Is it a solution of the inequality?
  3. Shade in one side of the boundary line.
    • If the test point is a solution, shade in the side that includes the point.
    • If the test point is not a solution, shade in the opposite side.

Graph the linear inequality x−2y<5.

Solution

First, we graph the boundary line x−2y=5. The inequality is < so we draw a dashed line.

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight dashed line is drawn through the points (negative 3, negative 4), (1, negative 2), and (5, 0).

Then, we test a point. We’ll use (0,0) again because it is easy to evaluate and it is not on the boundary line.

Is (0,0) a solution of x−2y<5?

Is 0 minus 2 times 0 less than 5? Is 0 minus 0 less than 5? 0 is less than 5.

The point (0,0) is a solution of x−2y<5, so we shade in that side of the boundary line.

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight dashed line is drawn through the points (negative 3, negative 4), (1, negative 2), and (5, 0). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.

All points in the shaded region, but not those on the boundary line, represent the solutions to x−2y<5.

Graph the linear inequality: 2x−3y<6.

Solution


This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 2), (3, 0), and (6, 2). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region, but not those on the boundary line, represent the solutions to 2x−3y<6.

Graph the linear inequality: 2x−y>3.

Solution


This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 3), (1, negative 1), and (2, 1). The line divides the x y-coordinate plane into two halves. The bottom right half is shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region, but not those on the boundary line, represent the solutions to 2x−y>3.

What if the boundary line goes through the origin? Then, we won’t be able to use (0,0) as a test point. No problem—we’ll just choose some other point that is not on the boundary line.

Graph the linear inequality: y≤​−4x.

Solution

First, we graph the boundary line y=−4x. It is in slope–intercept form, with m=−4 and b=0. The inequality is ≤ so we draw a solid line.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A straight is drawn through the points (0, 0), (1, negative 4), and (negative 1, 4).

Now we need a test point. We can see that the point (1,0) is not on the boundary line.

Is (1,0) a solution of y≤−4x?

Mathematical expressions demonstrating 0 is less than or equal to -4(1) with a red question mark, which simplifies to the false statement 0 <= -4, followed by the true statement 0 != -4.

The point (1,0) is not a solution to y≤​−4x, so we shade in the opposite side of the boundary line.

Is 0 less than or equal to negative 4 times 1? 0 is not less than or equal to negative 4.

All points in the shaded region and on the boundary line represent the solutions to y≤​−4x.

Graph the linear inequality: y>−3x.

Solution


This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (negative 1, 3), (0, 0), and (1, negative 3). The line divides the x y-coordinate plane into two halves. The top right half is shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region, but not those on the boundary line, represent the solutions to y>−3x.

Graph the linear inequality: y≥−2x.

Solution


This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (negative 1, 2), (0, 0), and (1, negative 2). The line divides the x y-coordinate plane into two halves. The top right half is shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region and on the boundary line, represent the solutions to y≥−2x.

Some linear inequalities have only one variable. They may have an x but no y, or a y but no x. In these cases, the boundary line will be either a vertical or a horizontal line.

Recall that:

x=avertical line y=bhorizontal line

Graph the linear inequality: y>3.

Solution

First, we graph the boundary line y=3. It is a horizontal line. The inequality is > so we draw a dashed line.

We test the point (0,0).

y>30>3

So, (0,0) is not a solution to y>3.

So we shade the side that does not include (0,0) as shown in this graph.

This figure has the graph of a straight horizontal dashed line on the x y-coordinate plane. The x and y axes run from negative 8 to 8. A horizontal dashed line is drawn through the points (negative 1, 3), (0, 3), and (1, 3). The line divides the x y-coordinate plane into two halves. The top half is shaded red to indicate that this is where the solutions of the inequality are.

All points in the shaded region, but not those on the boundary line, represent the solutions to y>3.

Graph the linear inequality: y<5.

Solution


This figure has the graph of a straight horizontal dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A horizontal dashed line is drawn through the points (negative 1, 5), (0, 5), and (1, 5). The line divides the x y-coordinate plane into two halves. The bottom half is shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region, but not those on the boundary line, represent the solutions to y<5.

Graph the linear inequality: y≤−1.

Solution


This figure has the graph of a straight horizontal line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A horizontal line is drawn through the points (negative 1, negative 1), (0, negative 1), and (1, negative 1). The line divides the x y-coordinate plane into two halves. The line and the bottom half are shaded red to indicate that this is where the solutions of the inequality are.
All points in the shaded region and on the boundary line represent the solutions to y≤−1.

Solve Applications using Linear Inequalities in Two Variables

Many fields use linear inequalities to model a problem. While our examples may be about simple situations, they give us an opportunity to build our skills and to get a feel for how they might be used.

Hilaria works two part time jobs in order to earn enough money to meet her obligations of at least $240 a week. Her job in food service pays $10 an hour and her tutoring job on campus pays $15 an hour. How many hours does Hilaria need to work at each job to earn at least $240?

ⓐ Let x be the number of hours she works at the job in food service and let y be the number of hours she works tutoring. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x,y) that would be solutions to the inequality. Then, explain what that means for Hilaria.

Solution

ⓐ We let x be the number of hours she works at the job in food service and let y be the number of hours she works tutoring.

She earns $10 per hour at the job in food service and $15 an hour tutoring. At each job, the number of hours multiplied by the hourly wage will gives the amount earned at that job.

10 x plus 15 y is greater than 240. The “10 x” is labeled “Amount earned at the food service job”. The “15 y” is labeled “the amount earned tutoring”. The “is greater than 240” is labeled “is at least 240”.

ⓑ To graph the inequality, we put it in slope–intercept form.

10x+15y≥24015y≥−10x+240y≥−23x+16

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from 0 to 25. A line is drawn through the points (0, 16), (15, 6), and (24, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.

ⓒ From the graph, we see that the ordered pairs (15,10),(0,16),(24,0) represent three of infinitely many solutions. Check the values in the inequality.

First we test the point (15, 10) in the inequality 10 x plus 15 y greater than or equal to 240. Is 10 times 15 plus 15 times 10 greater than or equal to 240? Since 300 is greater than or equal to 240 (15, 10) is a solution. Next we test the point (0, 16) in the inequality 10 x plus 15 y greater than or equal to 240. Is 10 times 0 plus 15 times 16 greater than or equal to 240? Since 240 is greater than or equal to 240 (0, 16) is a solution. Then we test the point (24, 0) in the inequality 10 x plus 15 y greater than or equal to 240. Is 10 times 24 plus 15 times 0 greater than or equal to 240? Since 240 is greater than or equal to 240 (24, 0) is a solution.

For Hilaria, it means that to earn at least $240, she can work 15 hours tutoring and 10 hours at her fast-food job, earn all her money tutoring for 16 hours, or earn all her money while working 24 hours at the job in food service.

Hugh works two part time jobs. One at a grocery store that pays $10 an hour and the other is babysitting for $13 hour. Between the two jobs, Hugh wants to earn at least $260 a week. How many hours does Hugh need to work at each job to earn at least $260?

ⓐ Let x be the number of hours he works at the grocery store and let y be the number of hours he works babysitting. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x, y) that would be solutions to the inequality. Then, explain what that means for Hugh.

Solution

ⓐ 10x+13y≥260
ⓑ
This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from 0 to 30. A line is drawn through the points (0, 20), (13, 10), and (26, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.
ⓒ Answers will vary.

Veronica works two part time jobs in order to earn enough money to meet her obligations of at least $280 a week. Her job at the day spa pays $10 an hour and her administrative assistant job on campus pays $17.50 an hour. How many hours does Veronica need to work at each job to earn at least $280?

ⓐ Let x be the number of hours she works at the day spa and let y be the number of hours she works as administrative assistant. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x, y) that would be solutions to the inequality. Then, explain what that means for Veronica

Solution

ⓐ 10x+17.5y≥280
ⓑ
This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from 0 to 25. A line is drawn through the points (0, 16) and (28, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.
ⓒ Answers will vary.

Access this online resource for additional instruction and practice with graphing linear inequalities in two variables.

  • Graphing Linear Inequalities in Two Variables

Key Concepts

  • How to graph a linear inequality in two variables.
    1. Identify and graph the boundary line.
      If the inequality is ≤or≥, the boundary line is solid.
      If the inequality is <or>, the boundary line is dashed.
    2. Test a point that is not on the boundary line. Is it a solution of the inequality?
    3. Shade in one side of the boundary line.
      If the test point is a solution, shade in the side that includes the point.
      If the test point is not a solution, shade in the opposite side.

Practice Makes Perfect

Verify Solutions to an Inequality in Two Variables

In the following exercises, determine whether each ordered pair is a solution to the given inequality.

Determine whether each ordered pair is a solution to the inequality y>x−1:


ⓐ (0,1)
ⓑ (−4,−1)
ⓒ (4,2)
ⓓ (3,0)
ⓔ (−2,−3)

Solution

ⓐ yes ⓑ yes ⓒ no ⓓ no ⓔ no

Determine whether each ordered pair is a solution to the inequality y>x−3:


ⓐ (0,0)
ⓑ (2,1)
ⓒ (−1,−5)
ⓓ (−6,−3)
ⓔ (1,0)

Determine whether each ordered pair is a solution to the inequality y<3x+2:


ⓐ (0,3)
ⓑ (−3,−2)
ⓒ (−2,0)
ⓓ (0,0)
ⓔ (−1,4)

Solution

ⓐ no ⓑ no ⓒ no ⓓ yes ⓔ no

Determine whether each ordered pair is a solution to the inequality y<−2x+5:


ⓐ (−3,0)
ⓑ (1,6)
ⓒ (−6,−2)
ⓓ (0,1)
ⓔ (5,−4)

Determine whether each ordered pair is a solution to the inequality 3x−4y>4:


ⓐ (5,1)
ⓑ (−2,6)
ⓒ (3,2)
ⓓ (10,−5)
ⓔ (0,0)

Solution

ⓐ yes ⓑ no ⓒ no ⓓ yes ⓔ no

Determine whether each ordered pair is a solution to the inequality 2x+3y>2:


ⓐ (1,1)
ⓑ (4,−3)
ⓒ (0,0)
ⓓ (−8,12)
ⓔ (3,0)

Recognize the Relation Between the Solutions of an Inequality and its Graph

In the following exercises, write the inequality shown by the shaded region.

Write the inequality shown by the graph with the boundary line y=3x−4.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 4), (1, negative 1), and (2, 2). The line divides the x y-coordinate plane into two halves. The line and the bottom right half are shaded red to indicate that this is where the solutions of the inequality are.
Solution

y≤3x−4

Write the inequality shown by the graph with the boundary line y=2x−4.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 4), (1, negative 2), and (2, 0). The line divides the x y-coordinate plane into two halves. The line and the bottom right half are shaded red to indicate that this is where the solutions of the inequality are.

Write the inequality shown by the graph with the boundary line y=12x+1.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, 1), (2, 0), and (4, negative 1). The line divides the x y-coordinate plane into two halves. The line and the bottom right half are shaded red to indicate that this is where the solutions of the inequality are.
Solution

y≤12x+1

Write the inequality shown by the graph with the boundary line y=−13x−2.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 2), (3, negative 3), and (6, negative 4). The line divides the x y-coordinate plane into two halves. The line and the bottom left half are shaded red to indicate that this is where the solutions of the inequality are.

Write the inequality shown by the shaded region in the graph with the boundary line x+y=5.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, 5), (1, 4), and (5, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.
Solution

x+y≥5

Write the inequality shown by the shaded region in the graph with the boundary line x+y=3.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, 3), (1, 2), and (3, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.

Write the inequality shown by the shaded region in the graph with the boundary line 3x−y=6.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 6), (1, negative 3), and (2, 0). The line divides the x y-coordinate plane into two halves. The line and the top left half are shaded red to indicate that this is where the solutions of the inequality are.
Solution

3x−y≤6

Write the inequality shown by the shaded region in the graph with the boundary line 2x−y=4.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 4), (1, negative 2), and (2, 0). The line divides the x y-coordinate plane into two halves. The line and the top left half are shaded red to indicate that this is where the solutions of the inequality are.

Graph Linear Inequalities in Two Variables

In the following exercises, graph each linear inequality.

Graph the linear inequality: y>23x−1.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 1), (3, 1), and (6, 3). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: y<35x+2.

Graph the linear inequality: y≤−12x+4.

Solution

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, 4), (2, 3), and (4, 2). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: y≥−13x−2.

Graph the linear inequality: x−y≤3.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 3), (1, negative 2), and (3, 0). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: x−y≥−2.

Graph the linear inequality: 4x+y>−4.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 4), (negative 1, 0), and (1, negative 8). The line divides the x y-coordinate plane into two halves. The top right half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: x+5y<−5.

Graph the linear inequality: 3x+2y≥−6.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 3), (3, negative 5), and (negative 2, 0). The line divides the x y-coordinate plane into two halves. The top right half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: 4x+2y≥−8.

Graph the linear inequality: y>4x.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, 0), (negative 1, negative 4), and (1, 4). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: y≤−3x.

Graph the linear inequality: y<−10.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, 0), (negative 1, 3), and (1, negative 3). The line divides the x y-coordinate plane into two halves. The bottom left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: y≥2.

Graph the linear inequality: x≤5.

Solution

This figure has the graph of a straight vertical dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A vertical dashed line is drawn through the points (5, negative 1), (5, 0), and (5, 1). The line divides the x y-coordinate plane into two halves. The left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: x≥0.

Graph the linear inequality: x−y<4.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, negative 4), (1, negative 3), and (4, 0). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: x−y<−3.

Graph the linear inequality: y≥32x.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, 0), (2, 3), and (negative 2, negative 3). The line divides the x y-coordinate plane into two halves. The top left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: y≤54x.

Graph the linear inequality: y>−2x+1.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, 1), (1, negative 1), and (2, negative 3). The line divides the x y-coordinate plane into two halves. The top right half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: y<−3x−4.

Graph the linear inequality: 2x+y≥−4.

Solution

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 4), (1, negative 6), and (negative 2, 0). The line divides the x y-coordinate plane into two halves. The line and the bottom left half are shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: x+2y≤−2.

Graph the linear inequality: 2x−5y>10.

Solution

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 2), (5, 0), and (negative 5, negative 4). The line divides the x y-coordinate plane into two halves. The line and the bottom right half are shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality: 4x−3y>12.

Solve Applications using Linear Inequalities in Two Variables

Harrison works two part time jobs. One at a gas station that pays $11 an hour and the other is IT troubleshooting for $16.50 an hour. Between the two jobs, Harrison wants to earn at least $330 a week. How many hours does Harrison need to work at each job to earn at least $330?

ⓐ Let x be the number of hours he works at the gas station and let y be the number of (hours he works troubleshooting. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x,y) that would be solutions to the inequality. Then, explain what that means for Harrison.

Solution

ⓐ 11x+16.5y≥330
ⓑ
This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from 0 to 35. A line is drawn through the points (0, 20), (15, 10), and (30, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.
ⓒ Answers will vary.

Elena needs to earn at least $450 a week during her summer break to pay for college. She works two jobs. One as a swimming instructor that pays $9 an hour and the other as an intern in a genetics lab for $22.50 per hour. How many hours does Elena need to work at each job to earn at least $450 per week?

ⓐ Let x be the number of hours she works teaching swimming and let y be the number of hours she works as an intern. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x,y) that would be solutions to the inequality. Then, explain what that means for Elena.

The doctor tells Laura she needs to exercise enough to burn 500 calories each day. She prefers to either run or bike and burns 15 calories per minute while running and 10 calories a minute while biking.

ⓐ If x is the number of minutes that Laura runs and y is the number minutes she bikes, find the inequality that models the situation.

ⓑ Graph the inequality.

ⓒ List three solutions to the inequality. What options do the solutions provide Laura?

Solution

ⓐ 15x+10y≥500
ⓑ
This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from 0 to 70. A line is drawn through the points (0, 50) and (34, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.
ⓒ Answers will vary.

Armando’s workouts consist of kickboxing and swimming. While kickboxing, he burns 10 calories per minute and he burns 7 calories a minute while swimming. He wants to burn 600 calories each day.

ⓐ If x is the number of minutes that Armando will kickbox and y is the number minutes he will swim, find the inequality that will help Armando create a workout for today.

ⓑ Graph the inequality.

ⓒ List three solutions to the inequality. What options do the solutions provide Armando?

Writing Exercises

Lester thinks that the solution of any inequality with a > sign is the region above the line and the solution of any inequality with a < sign is the region below the line. Is Lester correct? Explain why or why not.

Solution

Answers will vary.

Explain why, in some graphs of linear inequalities, the boundary line is solid but in other graphs it is dashed.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “verify solutions to an inequality in two variables.”, “recognize the relation between the solutions of an inequality and its graph”, and “graph linear inequalities”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

boundary line
The line with equation Ax+By=C is the boundary line that separates the region where Ax+By>C from the region where Ax+By<C.
linear inequality
A linear inequality is an inequality that can be written in one of the following forms: Ax+By>C,Ax+By≥C,Ax+By<C, or Ax+By≤C, where A and B are not both zero.
solution to a linear inequality
An ordered pair (x,y) is a solution to a linear inequality if the inequality is true when we substitute the values of x and y.

Relations and Functions

Learning Objectives

By the end of this section, you will be able to:

  • Find the domain and range of a relation
  • Determine if a relation is a function
  • Find the value of a function

Before you get started, take this readiness quiz.

Evaluate 3x−5 when x=−2.
If you missed this problem, review Example 6 in Use the Language of Algebra.

Solution

−11

Evaluate 2x2−x−3 when x=a.
If you missed this problem, review Example 6 in Use the Language of Algebra.

Solution

2a2−a−3

Simplify: 7x−1−4x+5.
If you missed this problem, review Example 7 in Use the Language of Algebra.

Solution

3x+4

Find the Domain and Range of a Relation

As we go about our daily lives, we have many data items or quantities that are paired to our names. Our social security number, student ID number, email address, phone number and our birthday are matched to our name. There is a relationship between our name and each of those items.

When your professor gets her class roster, the names of all the students in the class are listed in one column and then the student ID number is likely to be in the next column. If we think of the correspondence as a set of ordered pairs, where the first element is a student name and the second element is that student’s ID number, we call this a relation.

(Student name, Student ID #)

The set of all the names of the students in the class is called the domain of the relation and the set of all student ID numbers paired with these students is the range of the relation.

There are many similar situations where one variable is paired or matched with another. The set of ordered pairs that records this matching is a relation.

Relation

A relation is any set of ordered pairs,(x,y). All the x-values in the ordered pairs together make up the domain. All the y-values in the ordered pairs together make up the range.

For the relation {(1,1),(2,4),(3,9),(4,16),(5,25)}:

ⓐ Find the domain of the relation.

ⓑ Find the range of the relation.

Solution

{(1,1),(2,4),(3,9),(4,16),(5,25)}

ⓐ The domain is the set of all x-values of the relation. {1,2,3,4,5}

ⓑ The range is the set of all y-values of the relation. {1,4,9,16,25}

For the relation {(1,1),(2,8),(3,27),(4,64),(5,125)}:

ⓐ Find the domain of the relation.

ⓑ Find the range of the relation.

Solution

ⓐ {1,2,3,4,5}
ⓑ {1,8,27,64,125}

For the relation {(1,3),(2,6),(3,9),(4,12),(5,15)}:

ⓐ Find the domain of the relation.

ⓑ Find the range of the relation.

Solution

ⓐ {1,2,3,4,5}
ⓑ {3,6,9,12,15}

Mapping

A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range.

Use the mapping of the relation shown to ⓐ list the ordered pairs of the relation, ⓑ find the domain of the relation, and ⓒ find the range of the relation.

This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Alison”, “Penelope”, “June”, “Gregory”, “Geoffrey”, “Lauren”, “Stephen”, “Alice”, “Liz”, “Danny”. The table on the right has the header “Birthday” and lists the dates “January 12”, “February 3”, “April 25”, “May 10”, “May 23”, “July 24”, “August 2”, and “September 15”. There is one arrow for each name in the Name table that starts at the name and points toward a date in the Birthday table. While most dates have only one arrow pointing to them, there are two arrows pointing to July 24: one from Stephen and one from Liz.
Solution

ⓐ The arrow shows the matching of the person to their birthday. We create ordered pairs with the person’s name as the x-value and their birthday as the y-value.

{(Alison, April 25), (Penelope, May 23), (June, August 2), (Gregory, September 15), (Geoffrey, January 12), (Lauren, May 10), (Stephen, July 24), (Alice, February 3), (Liz, August 2), (Danny, July 24)}

ⓑ The domain is the set of all x-values of the relation.

{Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, Danny}

ⓒ The range is the set of all y-values of the relation.

{January 12, February 3, April 25, May 10, May 23, July 24, August 2, September 15}

Use the mapping of the relation shown to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Khanh Nguyen”, “Abigail Brown”, “Sumantha Mishal”, and “Jose Hern and ez”. The table on the right has the header “Student ID #” and lists the codes “a b 56781”, “j h 47983”, “k n 68413”, and “s m 32479”. There is one arrow for each name in the Name table that starts at the name and points toward a code in the student ID table. The first arrow goes from Khanh Nguyen to k n 68413. The second arrow goes from Abigail Brown to a b 56781. The third arrow goes from Sumantha Mishal to s m 32479. The fourth arrow goes from Jose Hern and ez to j h 47983.
Solution

ⓐ (Khanh Nguyen, kn68413), (Abigail Brown, ab56781), (Sumantha Mishal, sm32479), (Jose Hern and ez, jh47983) ⓑ {Khanh Nguyen, Abigail Brown, Sumantha Mishal, Jose Hern and ez} ⓒ {kn68413, ab56781, sm32479, jh47983}

Use the mapping of the relation shown to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Maria”, “Arm and o”, “Cynthia”, “Kelly”, and “Rachel”. The table on the right has the header “Birthday” and lists the dates “January 18”, “March 15”, “November 6”, and “December 8”. There is one arrow for each name in the Name table that starts at the name and points toward a date in the Birthday table. The first arrow goes from Maria to November 6. The second arrow goes from Arm and o to a January 18. The third arrow goes from Cynthia to December 8. The fourth arrow goes from Kelly to March 15. The fifth arrow goes from Rachel to November 6.
Solution

ⓐ (Maria, November 6), (Arm and o, January 18), (Cynthia, December 8), (Kelly, March 15), (Rachel, November 6) ⓑ {Maria, Arm and o, Cynthia, Kelly, Rachel} ⓒ {November 6, January 18, December 8, March 15}

A graph is yet another way that a relation can be represented. The set of ordered pairs of all the points plotted is the relation. The set of all x-coordinates is the domain of the relation and the set of all y-coordinates is the range. Generally we write the numbers in ascending order for both the domain and range.

Use the graph of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The points (negative 3, 4), (negative 3, negative 1), (0, 3), (1, 5), (2, negative 2), and (4, negative 2).
Solution

ⓐ The ordered pairs of the relation are: {(1,5),(−3,−1),(4,−2),(0,3),(2,−2),(−3,4)}.

ⓑ The domain is the set of all x-values of the relation: {−3,0,1,2,4}.

Notice that while −3 repeats, it is only listed once.

ⓒ The range is the set of all y-values of the relation: {−2,−1,3,4,5}.

Notice that while −2 repeats, it is only listed once.

Use the graph of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The points (negative 3, 3), (negative 2, 2), (negative 1, 0), (0, negative 1), (2, negative 2), and (4, negative 4).
Solution

ⓐ (−3,3),(−2,2),(−1,0),
(0,−1),(2,−2),(4,−4)
ⓑ {−3,−2,−1,0,2,4}
ⓒ {3,2,0,−1,−2,−4}

Use the graph of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The points (negative 3, 5), (negative 3, 0), (negative 3, negative 6), (negative 1, negative 2), (1, 2), and (4, negative 4).
Solution

ⓐ (−3,0),(−3,5),(−3,−6),
(−1,−2),(1,2),(4,−4)
ⓑ {−3,−1,1,4}
ⓒ {−6,0,5,−2,2,−4}

Determine if a Relation is a Function

A special type of relation, called a function, occurs extensively in mathematics. A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x-value is matched with only one y-value.

Function

A function is a relation that assigns to each element in its domain exactly one element in the range.

The birthday example from Example 2 helps us understand this definition. Every person has a birthday but no one has two birthdays. It is okay for two people to share a birthday. It is okay that Danny and Stephen share July 24th as their birthday and that June and Liz share August 2nd. Since each person has exactly one birthday, the relation in Example 2 is a function.

The relation shown by the graph in Example 3 includes the ordered pairs (−3,−1) and (−3,4). Is that okay in a function? No, as this is like one person having two different birthdays.

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the relation.

ⓐ {(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}

ⓑ {(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}

Solution

ⓐ {(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}

(i) Each x-value is matched with only one y-value. So this relation is a function.

(ii) The domain is the set of all x-values in the relation.
The domain is: {−3,−2,−1,0,1,2,3}.

(iii) The range is the set of all y-values in the relation. Notice we do not list range values twice.
The range is: {27,8,1,0}.

ⓑ {(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}

(i) The x-value 9 is matched with two y-values, both 3 and −3. So this relation is not a function.

(ii) The domain is the set of all x-values in the relation. Notice we do not list domain values twice.
The domain is: {0,1,4,9}.

(iii) The range is the set of all y-values in the relation.
The range is: {−3,−2,−1,0,1,2,3}.

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the function.

ⓐ {(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}

ⓑ {(8,−4),(4,−2),(2,−1),(0,0),(2,1),(4,2),(8,4)}

Solution

ⓐ Yes; {−3,−2,−1,0,1,2,3};
{−6,−4,−2,0,2,4,6}
ⓑ No; {0,2,4,8};
{−4,−2,−1,0,1,2,4}

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the relation.

ⓐ {(27,−3),(8,−2),(1,−1),(0,0),(1,1),(8,2),(27,3)}

ⓑ {(7,−3),(−5,−4),(8,0),(0,0),(−6,4),(−2,2),(−1,3)}

Solution

ⓐ No; {0,1,8,27};
{−3,−2,−1,0,2,2,3}
ⓑ Yes; {7,−5,8,0,−6,−2,−1};
{−3,−4,0,4,2,3}

Use the mapping to ⓐ determine whether the relation is a function ⓑ find the domain of the relation ⓒ find the range of the relation.

This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Lydia”, “Eugene”, “Janet”, “Rick”, and “Marty”. The table on the right has the header “Phone number” and lists the numbers “321-549-3327 home”, “427-658-2314 cell”, “321-964-7324 cell”, “684-358-7961 home”, “684-369-7231 cell”, and “798-367-8541 cell”. There are arrows that start at a name and points toward a number in the phone number table. The first arrow goes from Lydia to 321-549-3327 home. The second arrow goes from Lydia to a 321-964-7324 cell. The third arrow goes from Eugene to 427-658-2314 cell. The fourth arrow goes from Janet to 427-658-2314 cell. The fifth arrow goes from Rick to 798-367-8541 cell. The sixth arrow goes from Marty to 684-358-7961 home. The seventh arrow goes from Marty to 684-369-7231 cell.
Solution

ⓐ Both Lydia and Marty have two phone numbers. So each x-value is not matched with only one y-value. So this relation is not a function.

ⓑ The domain is the set of all x-values in the relation. The domain is: {Lydia, Eugene, Janet, Rick, Marty}

ⓒ The range is the set of all y-values in the relation. The range is:

{321-549-3327, 427-658-2314, 321-964-7324, 684-358-7961, 684-369-7231, 798-367-8541}

Use the mapping to ⓐ determine whether the relation is a function ⓑ find the domain of the relation ⓒ find the range of the relation.

This figure shows two table that each have one column. The table on the left has the header “Network” and lists the television stations “NBC”, “HGTV”, and “HBO”. The table on the right has the header “Program” and lists the television shows “Ellen Degeneres Show”, “Law and Order”, “Tonight Show”, “Property Brothers”, “House Hunters”, “Love it or List it”, “Game of Thrones”, “True Detective”, and “Sesame Street”. There are arrows that start at a network in the first table and point toward a program in the second table. The first arrow goes from NBC to Ellen Degeneres Show. The second arrow goes from NBC to Law and Order. The third arrow goes from NBC to Tonight Show. The fourth arrow goes from HGTV to Property Brothers. The fifth arrow goes from HGTV to House Hunters. The sixth arrow goes from HGTV to Love it or List it. The seventh arrow goes from HBO to Game of Thrones. The eighth arrow goes from HBO to True Detective. The ninth arrow goes from HBO to Sesame Street.
Solution

ⓐ no ⓑ {NBC, HGTV, HBO} ⓒ {Ellen Degeneres Show, Law and Order, Tonight Show, Property Brothers, House Hunters, Love it or List it, Game of Thrones, True Detective, Sesame Street}

Use the mapping to ⓐ determine whether the relation is a function ⓑ find the domain of the relation ⓒ find the range of the relation.

This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Neal”, “Krystal”, “Kelvin”, “George”, “Christa”, and “Mike”. The table on the right has the header “Phone number” and lists the numbers “123-567-4389 work”, “231-378-5941 cell”, “753-469-9731 cell”, “567-534-2970 work”, “684-369-7231 cell”, “798-367-8541 cell”, and “639-847-6971 cell”. There are arrows that start at a name and points toward a number in the phone number table. The first arrow goes from Neal to 753-469-9731 cell. The second arrow goes from Krystal to a 684-369-7231 cell. The third arrow goes from Kelvin to 231-378-5941 cell. The fourth arrow goes from George to 123-567-4389 work. The fifth arrow goes from George to 639-847-6971 cell. The sixth arrow goes from Christa to 567-534-2970 work. The seventh arrow goes from Mike to 567-534-2970 work. The eighth arrow goes from Mike to 798-367-8541 cell.
Solution

ⓐ No ⓑ {Neal, Krystal, Kelvin, George, Christa, Mike} ⓒ {123-567-4839 work, 231-378-5941 cell, 743-469-9731 cell, 567-534-2970 work, 684-369-7231 cell, 798-367-8541 cell, 639-847-6971 cell}

In algebra, more often than not, functions will be represented by an equation. It is easiest to see if the equation is a function when it is solved for y. If each value of x results in only one value of y, then the equation defines a function.

Determine whether each equation is a function. Assume x is the independent variable.

ⓐ 2x+y=7 ⓑ y=x2+1 ⓒ x+y2=3

Solution

ⓐ 2x+y=7

For each value of x, we multiply it by −2 and then add 7 to get the y-value

The image displays the linear equation y = -2x + 7, shown in a simple, clear font against a white background.
For example, if x=3: The equation y = -2 '.' 3 + 7 is shown, with the number 3 highlighted in red.
The image shows the mathematical equation y = 1 in a bold, sans-serif font on a white background. It represents a horizontal line where the y-coordinate is constant at 1.

We have that when x=3, then y=1. It would work similarly for any value of x. Since each value of x, corresponds to only one value of y the equation defines a function.

ⓑ y=x2+1

For each value of x, we square it and then add 1 to get the y-value.

The equation y = x^2 + 1 is displayed in a mathematical expression. This represents a parabola shifted upwards by one unit on the y-axis, centered at x=0.
For example, if x=2: A mathematical equation is displayed on a white background, which reads 'y = 2^2 + 1'. The number 2 in '2^2' is highlighted in red, suggesting it might be the focus of the equation.
The image displays the mathematical equation 'y = 5' in a clean, straightforward manner on a white background, highlighting a constant value for the variable 'y'.

We have that when x=2, then y=5. It would work similarly for any value of x. Since each value of x, corresponds to only one value of y the equation defines a function.

ⓒ
A mathematical equation is displayed, showing 'x + y^2 = 3' in a clear, dark font against a white background.
Isolate the y term. The image displays the mathematical equation y^2 = -x + 3, presented in a clear, digital format on a plain white background.
Let’s substitute x=2. A mathematical equation is displayed, showing 'y² = -2 + 3' with the number 2 highlighted in red, indicating a specific part of the calculation.
The mathematical equation y^2 = 1 is displayed in black text on a white background.
This give us two values for y. y=1y=−1

We have shown that when x=2, then y=1 and y=−1. It would work similarly for any value of x. Since each value of x does not corresponds to only one value of y the equation does not define a function.

Determine whether each equation is a function.

ⓐ 4x+y=−3 ⓑ x+y2=1 ⓒ y−x2=2

Solution

ⓐ yes ⓑ no ⓒ yes

Determine whether each equation is a function.

ⓐ x+y2=4 ⓑ y=x2−7 ⓒ y=5x−4

Solution

ⓐ no ⓑ yes ⓒ yes

Find the Value of a Function

It is very convenient to name a function and most often we name it f, g, h, F, G, or H. In any function, for each x-value from the domain we get a corresponding y-value in the range. For the function f, we write this range value y as f(x). This is called function notation and is read f of x or the value of f at x. In this case the parentheses does not indicate multiplication.

Function Notation

For the function y=f(x)

fis the name of the function xis the domain value f(x)is the range valueycorresponding to the valuex

We read f(x) as f of x or the value of f at x.

We call x the independent variable as it can be any value in the domain. We call y the dependent variable as its value depends on x.

Independent and Dependent Variables

For the function y=f(x),

xis the independent variable as it can be any value in the domain ythe dependent variable as its value depends onx

Much as when you first encountered the variable x, function notation may be rather unsettling. It seems strange because it is new. You will feel more comfortable with the notation as you use it.

Let’s look at the equation y=4x−5. To find the value of y when x=2, we know to substitute x=2 into the equation and then simplify.

A linear equation is displayed as y = 4x - 5, representing a straight line with a slope of 4 and a y-intercept of -5.
Let x=2. A mathematical equation reads y = 4 ×7 2 ×1 5, where the number 2 is highlighted in red. This shows the initial setup for solving for y.
The equation y = 3 is displayed on a white background, representing a horizontal line in a Cartesian coordinate system where the y-coordinate for any point on the line is always 3.

The value of the function at x=2 is 3.

We do the same thing using function notation, the equation y=4x−5 can be written as f(x)=4x−5. To find the value when x=2, we write:

The image displays the function f(x) = 4x - 5, a linear equation often encountered in algebra and pre-calculus.
Let x=2. A mathematical equation showing the function f evaluated at 2, expressed as f(2) = 4 * 2 - 5. The number 2 in '4 * 2' is highlighted in red.
The mathematical equation f(2) = 3 is displayed on a white background.

The value of the function at x=2 is 3.

This process of finding the value of f(x) for a given value of x is called evaluating the function.

For the function f(x)=2x2+3x−1, evaluate the function.

ⓐ f(3) ⓑ f(−2) ⓒ f(a)

Solution
ⓐ
A mathematical equation for a quadratic function, f(x) = 2x^2 + 3x - 1, is displayed on a white background. The equation is rendered clearly in black text, showing the function definition and its polynomial expression.
To evaluate f(3), substitute 3 for x. An equation evaluating f(3) as 2(3)^2 + 3 * 3 - 1, with the substituted value of '3' highlighted in red.
Simplify. A mathematical equation shows 'f(3) = 2 multiplied by 9 plus 3 multiplied by 3 minus 1'.
A mathematical equation is displayed, showing 'f(3) = 18 + 9 - 1' in a horizontal layout against a white background.
A mathematical equation reads 'f(3) = 26' in a clear, sans-serif font against a plain white background.
ⓑ
A mathematical equation displays the quadratic function f(x) = 2x^2 + 3x - 1 on a white background, representing a parabola in a clear, straightforward algebraic form.
The image displays the mathematical instruction: 'To evaluate f(-2), substitute -2 for x.' An image displays the mathematical expression for function f, where f(-2) is calculated as 2 multiplied by -2 squared, plus 3 multiplied by -2, minus 1. The value -2 is highlighted in red.
Simplify. A mathematical equation f(-2) = 2 * 4 + (-6) - 1 is shown on a white background.
The image shows a mathematical equation in black text on a white background. The equation reads as 'f(-2) = 8 + (-6) - 1'.
A mathematical equation is displayed on a white background, reading 'f(-2) = 1' in black text.
ⓒ
A mathematical equation is displayed, showing a quadratic function: f(x) = 2x^2 + 3x - 1. The text is in a dark gray font against a white background.
To evaluate f(a), substitute a for x. A mathematical equation is displayed, showing f(a) = 2(a)^2 + 3 * a - 1. The variable 'a' is highlighted in red within the function and the expression.
Simplify. A mathematical equation for a quadratic function, shown as f(a) = 2a^2 + 3a - 1, against a white background.

For the function f(x)=3x2−2x+1, evaluate the function.

ⓐ f(3) ⓑ f(−1) ⓒ f(t)

Solution

ⓐ f(3)=22 ⓑ f(−1)=6 ⓒ f(t)=3t2−2t+1

For the function f(x)=2x2+4x−3, evaluate the function.

ⓐ f(2) ⓑ f(−3) ⓒ f(h)

Solution

ⓐ (2)=13 ⓑ f(−3)=3
ⓒ f(h)=2h2+4h−3

In the last example, we found f(x) for a constant value of x. In the next example, we are asked to find g(x) with values of x that are variables. We still follow the same procedure and substitute the variables in for the x.

For the function g(x)=3x−5, evaluate the function.

ⓐ g(h2) ⓑ g(x+2) ⓒ g(x)+g(2)

Solution
ⓐ
A mathematical equation is displayed on a white background, reading g(x) = 3x - 5, representing a linear function.
To evaluate g(h2), substitute h2 for x. A mathematical equation is displayed on a white background: g(h^2) = 3h^2 - 5. The variable 'h' is highlighted in red, indicating its specific role within the function.
A mathematical equation shows g(h squared) equals 3h squared minus 5, representing a function where the input and the variable in the expression are both h squared.
ⓑ
A mathematical equation for a linear function, g(x) = 3x - 5, is displayed in black text on a white background.
To evaluate g(x+2), substitute x+2 for x. The equation g(x + 2) = 3(x + 2) - 5 is displayed, showing a function transformation where the input (x) is replaced by (x + 2) in a linear function.
Simplify. The image displays a mathematical equation, g(x + 2) = 3x + 6 - 5, presented in a black font on a white background.
A mathematical equation is displayed: g(x + 2) = 3x + 1, against a white background.
ⓒ
The mathematical function g(x) = 3x - 5 is displayed.
To evaluate g(x)+g(2), first find g(2). A mathematical equation shows 'g(2) = 3 * 2 - 5', with the number 2 highlighted in red in both instances.
The image displays a mathematical equation, g(2) = 1, presented in a clear, digital format on a plain white background.
A mathematical expression asks to 'Now find g(x) + g(2)'. Equation g(x) + g(2) = 3x - 5 + 1, where 3x - 5 is identified as g(x) and 1 as g(2) by blue underbraces.
Simplify. A mathematical equation is displayed, showing 'g(x) + g(2) = 3x - 5 + 1' in black text on a white background.
The image shows a mathematical equation in black text on a white background. The equation is 'g(x) + g(2) = 3x - 4'.

Notice the difference between part ⓑ and ⓒ. We get g(x+2)=3x+1 and g(x)+g(2)=3x−4. So we see that g(x+2)≠g(x)+g(2).

For the function g(x)=4x−7, evaluate the function.

ⓐ g(m2) ⓑ g(x−3) ⓒ g(x)−g(3)

Solution

ⓐ 4m2−7 ⓑ 4x−19
ⓒ 4x−12

For the function h(x)=2x+1, evaluate the function.

ⓐ h(k2) ⓑ h(x+1) ⓒ h(x)+h(1)

Solution

ⓐ 2k2+1 ⓑ 2x+3
ⓒ 2x+4

Many everyday situations can be modeled using functions.

The number of unread emails in Sylvia’s account is 75. This number grows by 10 unread emails a day. The function N(t)=75+10t represents the relation between the number of emails, N, and the time, t, measured in days.

ⓐ Determine the independent and dependent variable.

ⓑ Find N(5). Explain what this result means.

Solution

ⓐ The number of unread emails is a function of the number of days. The number of unread emails, N, depends on the number of days, t. Therefore, the variable N, is the dependent variable and the variable t is the independent variable.

ⓑ Find N(5). Explain what this result means.

A mathematical equation is displayed on a white background: N(t) = 75 + 10t, with 'N' and 't' in italicized font.
Substitute in t=5. A mathematical expression shows N(5) = 75 + 10 * 5, representing an equation with multiplication and addition.
Simplify. A mathematical equation shows 'N(5) = 75 + 50' in a bold, sans-serif font against a white background.
A mathematical expression 'N(5) = 125' is displayed, indicating that the value of the function N at 5 is 125.

Since 5 is the number of days, N(5), is the number of unread emails after 5 days. After 5 days, there are 125 unread emails in the account.

The number of unread emails in Bryan’s account is 100. This number grows by 15 unread emails a day. The function N(t)=100+15t represents the relation between the number of emails, N, and the time, t, measured in days.

ⓐ Determine the independent and dependent variable.

ⓑ Find N(7). Explain what this result means.

Solution

ⓐ t IND; N DEP ⓑ 205; the number of unread emails in Bryan’s account on the seventh day.

The number of unread emails in Anthony’s account is 110. This number grows by 25 unread emails a day. The function N(t)=110+25t represents the relation between the number of emails, N, and the time, t, measured in days.

ⓐ Determine the independent and dependent variable.

ⓑ Find N(14). Explain what this result means.

Solution

ⓐ t IND; N DEP ⓑ 460; the number of unread emails in Anthony’s account on the fourteenth day

Access this online resource for additional instruction and practice with relations and functions.

  • Introduction to Functions

Key Concepts

  • Function Notation: For the function y=f(x)
    • f is the name of the function
    • x is the domain value
    • f(x) is the range value y corresponding to the value x
      We read f(x) as f of x or the value of f at x.
  • Independent and Dependent Variables: For the function y=f(x),
    • x is the independent variable as it can be any value in the domain
    • y is the dependent variable as its value depends on x

Practice Makes Perfect

Find the Domain and Range of a Relation

In the following exercises, for each relation ⓐ find the domain of the relation ⓑ find the range of the relation.

{(1,4),(2,8),(3,12),(4,16),(5,20)}

Solution

ⓐ {1, 2, 3, 4, 5} ⓑ {4, 8, 12, 16, 20}

{(1,−2),(2,−4),(3,−6),(4,−8),(5,−10)}

{(1,7),(5,3),(7,9),(−2,−3),(−2,8)}

Solution

ⓐ {1, 5, 7, −2} ⓑ {7, 3, 9, −3, 8}

{(11,3),(−2,−7),(4,−8),(4,17),(−6,9)}

In the following exercises, use the mapping of the relation to ⓐ list the ordered pairs of the relation, ⓑ find the domain of the relation, and ⓒ find the range of the relation.


This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Rebecca”, “Jennifer”, “John”, “Hector”, “Luis”, “Ebony”, “Raphael”, “Meredith”, “Karen”, and “Joseph”. The table on the right has the header “Birthday” and lists the dates “January 18”, “February 15”, “April 1”, “April 7”, “June 23”, “July 30”, “August 19”, and “November 6”. There are arrows starting at names in the Name table and pointing towards dates in the Birthday table. The first arrow goes from Rebecca to January 18. The second arrow goes from Jennifer to April 1. The third arrow goes from John to January 18. The fourth arrow goes from Hector to June 23. The fifth arrow goes from Luis to February 15. The sixth arrow goes from Ebony to April 7. The seventh arrow goes from Raphael to November 6. The eighth arrow goes from Meredith to August 19. The ninth arrow goes from Karen to August 19. The tenth arrow goes from Joseph to July 30.

Solution

ⓐ (Rebecca, January 18), (Jennifer, April 1), (John, January 18), (Hector, June 23), (Luis, February 15), (Ebony, April 7), (Raphael, November 6), (Meredith, August 19), (Karen, August 19), (Joseph, July 30)
ⓑ {Rebecca, Jennifer, John, Hector, Luis, Ebony, Raphael, Meredith, Karen, Joseph}
ⓒ {January 18, April 1, June 23, February 15, April 7, November 6, August 19, July 30}


This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Amy”, “Carol”, “Devon”, “Harrison”, “Jackson”, “Labron”, “Mason”, “Natalie”, “Paul”, and “Sylvester”. The table on the right has the header “Birthday” and lists the dates “January 5”, “January 7”, “February 14”, “March 1”, “April 7”, “May 30”, “July 20”, “August 1”, “November 13”, and “November 26”. There are arrows starting at names in the Name table and pointing towards dates in the Birthday table. The first arrow goes from Amy to February 14. The second arrow goes from Carol to May 30. The third arrow goes from Devon to January 5. The fourth arrow goes from Harrison to January 7. The fifth arrow goes from Jackson to November 26. The sixth arrow goes from Labron to April 7. The seventh arrow goes from Mason to July 20. The eighth arrow goes from Natalie to March 1. The ninth arrow goes from Paul to August 1. The tenth arrow goes from Sylvester to November 13.

For a woman of height 5′4″ the mapping below shows the corresponding Body Mass Index (BMI). The body mass index is a measurement of body fat based on height and weight. A BMI of 18.5–24.9 is considered healthy.

This figure shows two table that each have one column. The table on the left has the header “Weight (lbs)” and lists the numbers plus 100, 110, 120, 130, 140, 150, and 160. The table on the right has the header “BMI” and lists the numbers 18. 9, 22. 3, 17. 2, 24. 0, 25. 7, 20. 6, and 27. 5. There are arrows starting at numbers in the weight table and pointing towards numbers in the BMI table. The first arrow goes from plus 100 to 17. 2. The second arrow goes from 110 to 18. 9. The third arrow goes from 120 to 20. 6. The fourth arrow goes from 130 to 22. 3. The fifth arrow goes from 140 to 24. 0. The sixth arrow goes from 150 to 25. 7. The seventh arrow goes from 160 to 27. 5.
Solution

ⓐ (+100, 17. 2), (110, 18.9), (120, 20.6), (130, 22.3), (140, 24.0), (150, 25.7), (160, 27.5) ⓑ {+100, 110, 120, 130, 140, 150, 160,} ⓒ {17.2, 18.9, 20.6, 22.3, 24.0, 25.7, 27.5}

For a man of height 5′11′′ the mapping below shows the corresponding Body Mass Index (BMI). The body mass index is a measurement of body fat based on height and weight. A BMI of 18.5–24.9 is considered healthy.
This figure shows two table that each have one column. The table on the left has the header “Weight (lbs)” and lists the numbers 130, 140, 150, 160, 170, 180, 190, and 200. The table on the right has the header “BMI” and lists the numbers 22. 3, 19. 5, 20. 9, 27. 9, 25. 1, 26. 5, 23. 7, and 18. 1. There are arrows starting at numbers in the weight table and pointing towards numbers in the BMI table. The first arrow goes from 130 to 18. 1. The second arrow goes from 140 to 19. 5. The third arrow goes from 150 to 20. 9. The fourth arrow goes from 160 to 22. 3. The fifth arrow goes from 170 to 23. 7. The sixth arrow goes from 180 to 25. 1. The seventh arrow goes from 190 to 26. 5. The eighth arrow goes from 200 to 27. 9.

In the following exercises, use the graph of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.


The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The points (negative 3, 4), (negative 3, negative 1), (0, negative 3), (2, 3), (4, negative 1), and (4, negative 3).

Solution

ⓐ (2, 3), (4, −3), (−2, −1), (−3, 4), (4, −1), (0, −3) ⓑ {−3, −2, 0, 2, 4}
ⓒ {−3, −1, 3, 4}


The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The points (negative 3, 4), (negative 3, negative 4), (negative 2, 0), (negative 1, 3), (1, 5), and (4, negative 2).


The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The points (negative 1, 4), (negative 1, negative 4), (0, 3), (0, negative 3), (1, 4), and (1, negative 4).

Solution

ⓐ (1, 4), (1, −4), (−1, 4), (−1, −4), (0, 3), (0, −3) ⓑ {−1, 0, 1} ⓒ {−4, −3, 3,4}


The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The points (negative 2, negative 6), (negative 2, negative 3), (0, 0), (0. 5, 1. 5), (1, 3), and (3, 6).

Determine if a Relation is a Function

In the following exercises, use the set of ordered pairs to ⓐ determine whether the relation is a function, ⓑ find the domain of the relation, and ⓒ find the range of the relation.

{(−3,9),(−2,4),(−1,1),
(0,0),(1,1),(2,4),(3,9)}

Solution

ⓐ yes ⓑ {−3, −2, −1, 0, 1, 2, 3} ⓒ {9, 4, 1, 0}

{(9,−3),(4,−2),(1,−1),
(0,0),(1,1),(4,2),(9,3)}

{(−3,27),(−2,8),(−1,1),
(0,0),(1,1),(2,8),(3,27)}

Solution

ⓐ yes ⓑ {−3, −2, −1, 0, 1, 2, 3} ⓒ 0, 1, 8, 27}

{(−3,−27),(−2,−8),(−1,−1),
(0,0),(1,1),(2,8),(3,27)}

In the following exercises, use the mapping to ⓐ determine whether the relation is a function, ⓑ find the domain of the function, and ⓒ find the range of the function.


This figure shows two table that each have one column. The table on the left has the header “Number” and lists the numbers negative 3, negative 2, negative 1, 0, 1, 2, and 3. The table on the right has the header “Absolute Value” and lists the numbers 0, 1, 2, and 3. There are arrows starting at numbers in the number table and pointing towards numbers in the absolute value table. The first arrow goes from negative 3 to 3. The second arrow goes from negative 2 to 2. The third arrow goes from negative 1 to 1. The fourth arrow goes from 0 to 0. The fifth arrow goes from 1 to 1. The sixth arrow goes from 2 to 2. The seventh arrow goes from 3 to 3.

Solution

ⓐ yes ⓑ {−3, −2, −1, 0, 1, 2, 3} ⓒ {0, 1, 2, 3}


This figure shows two table that each have one column. The table on the left has the header “Number” and lists the numbers negative 3, negative 2, negative 1, 0, 1, 2, and 3. The table on the right has the header “Square” and lists the numbers 0, 1, 4, and 9. There are arrows starting at numbers in the number table and pointing towards numbers in the square table. The first arrow goes from negative 3 to 9. The second arrow goes from negative 2 to 4. The third arrow goes from negative 1 to 1. The fourth arrow goes from 0 to 0. The fifth arrow goes from 1 to 1. The sixth arrow goes from 2 to 4. The seventh arrow goes from 3 to 9.


This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Jenny”, “R and y”, “Dennis”, “Emily”, and “Raul”. The table on the right has the header “Email” and lists the email addresses RHern and ez@state. edu, JKim@gmail.com, Raul@gmail.com, ESmith@state. edu, DBrown@aol.com, jenny@aol.com, and R and y@gmail.com. There are arrows starting at names in the name table and pointing towards addresses in the email table. The first arrow goes from Jenny to JKim@gmail.com. The second arrow goes from Jenny to jenny@aol.com. The third arrow goes from R and y to R and y@gmail.com. The fourth arrow goes from Dennis to DBrown@aol.com. The fifth arrow goes from Emily to ESmith@state. edu. The sixth arrow goes from Raul to RHern and ez@state. edu. The seventh arrow goes from Raul to Raul@gmail.com.

Solution

ⓐ no ⓑ {Jenny, R and y, Dennis, Emily, Raul} ⓒ {RHern and ez@state.edu, JKim@gmail.com, Raul@gmail.com, ESmith@state.edu, DBroen@aol.com, jenny@aol.cvom, R and y@gmail.com}


This figure shows two table that each have one column. The table on the left has the header “Name” and lists the names “Jon”, “Rachel”, “Matt”, “Leslie”, “Chris”, “Beth”, and “Liz”. The table on the right has the header “Email” and lists the email addresses chrisg@gmail.com, lizzie@aol.com, jong@gmail.com, mattg@gmail.com, Rachel@state. edu, leslie@aol.com, and bethc@gmail.com. There are arrows starting at names in the name table and pointing towards addresses in the email table. The first arrow goes from Jon to jong@gmail.com. The second arrow goes from Rachel to Rachel@state. edu. The third arrow goes from Matt to mattg@gmail.com. The fourth arrow goes from Leslie to leslie@aol.com. The fifth arrow goes from Chris to chrisg@gmail.com. The sixth arrow goes from Beth to bethc@gmail.com. The seventh arrow goes from Liz to lizzie@aol.com.

In the following exercises, determine whether each equation is a function.


ⓐ 2x+y=−3
ⓑ y=x2
ⓒ x+y2=−5

Solution

ⓐ yes ⓑ yes ⓒ no


ⓐ y=3x−5
ⓑ y=x3
ⓒ 2x+y2=4


ⓐ y−3x3=2
ⓑ x+y2=3
ⓒ 3x−2y=6

Solution

ⓐ yes ⓑ no ⓒ yes


ⓐ 2x−4y=8
ⓑ −4=x2−y
ⓒ y2=−x+5

Find the Value of a Function

In the following exercises, evaluate the function: ⓐ f(2) ⓑ f(−1) ⓒ f(a).

f(x)=5x−3

Solution

ⓐ f(2)=7 ⓑ f(−1)=−8 ⓒ f(a)=5a−3

f(x)=3x+4

f(x)=−4x+2

Solution

ⓐ f(2)=−6 ⓑ f(−1)=6 ⓒ f(a)=−4a+2

f(x)=−6x−3

f(x)=x2−x+3

Solution

ⓐ f(2)=5 ⓑ f(−1)=5
ⓒ f(a)=a2−a+3

f(x)=x2+x−2

f(x)=2x2−x+3

Solution

ⓐ f(2)=9 ⓑ f(−1)=6
ⓒ f(a)=2a2−a+3

f(x)=3x2+x−2

In the following exercises, evaluate the function: ⓐ g(h2) ⓑ g(x+2) ⓒ g(x)+g(2).

g(x)=2x+1

Solution

ⓐ g(h2)=2h2+1
ⓑ g(x+2)=2x+5
ⓒ g(x)+g(2)=2x+6

g(x)=5x−8

g(x)=−3x−2

Solution

ⓐ g(h2)=−3h2−2
ⓑ g(x+2)=−3x−8
ⓒ g(x)+g(2)=−3x−10

g(x)=−8x+2

g(x)=3−x

Solution

ⓐ g(h2)=3−h2
ⓑ g(x+2)=1−x
ⓒ g(x)+g(2)=4−x

g(x)=7−5x

In the following exercises, evaluate the function.

f(x)=3x2−5x; f(2)

Solution

2

g(x)=4x2−3x; g(3)

F(x)=2x2−3x+1;
F(−1)

Solution

6

G(x)=3x2−5x+2;
G(−2)

h(t)=2|t−5|+4; h(−4)

Solution

22

h(y)=3|y−1|−3; h(−4)

f(x)=x+2x−1; f(2)

Solution

4

g(x)=x−2x+2; g(4)

In the following exercises, solve.

The number of unwatched shows in Sylvia’s DVR is 85. This number grows by 20 unwatched shows per week. The function N(t)=85+20t represents the relation between the number of unwatched shows, N, and the time, t, measured in weeks.

ⓐ Determine the independent and dependent variable.

ⓑ Find N(4). Explain what this result means

Solution

ⓐ t IND; N DEP
ⓑ N(4)=165 the number of unwatched shows in Sylvia’s DVR at the fourth week.

Every day a new puzzle is downloaded into Ken’s account. Right now he has 43 puzzles in his account. The function N(t)=43+t represents the relation between the number of puzzles, N, and the time, t, measured in days.

ⓐ Determine the independent and dependent variable.

ⓑ Find N(30). Explain what this result means.

The daily cost to the printing company to print a book is modeled by the function C(x)=3.25x+1500 where C is the total daily cost in dollars and x is the number of books printed.

ⓐ Determine the independent and dependent variable.

ⓑ Find C(0). Explain what this result means.

ⓒ Find C(1000). Explain what this result means.

Solution

ⓐ x IND; C DEP
ⓑ N(0)=1500 the daily cost if no books are printed
ⓒ N(1000)=4750 the daily cost of printing 1000 books

The daily cost to the manufacturing company is modeled by the function C(x)=7.25x+2500 where C(x) is the total daily cost and x is the number of items manufactured.

ⓐ Determine the independent and dependent variable.

ⓑ Find C(0). Explain what this result means.

ⓒ Find C(1000). Explain what this result means.

Writing Exercises

In your own words, explain the difference between a relation and a function.

In your own words, explain what is meant by domain and range.

Is every relation a function? Is every function a relation?

How do you find the value of a function?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The figure shows a table with four rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “confidently”, the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “find the domain and range of a relation”, “determine if a relation is a function”, and “find the value of a function”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

domain of a relation
The domain of a relation is all the x-values in the ordered pairs of the relation.
function
A function is a relation that assigns to each element in its domain exactly one element in the range.
mapping
A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range.
range of a relation
The range of a relation is all the y-values in the ordered pairs of the relation.
relation
A relation is any set of ordered pairs,(x,y). All the x-values in the ordered pairs together make up the domain. All the y-values in the ordered pairs together make up the range.

Graphs of Functions

Learning Objectives

By the end of this section, you will be able to:

  • Use the vertical line test
  • Identify graphs of basic functions
  • Read information from a graph of a function

Before you get started, take this readiness quiz.

Evaluate: ⓐ 23 ⓑ 32.
If you missed this problem, review Example 5 in Use the Language of Algebra.

Solution

ⓐ 8; ⓑ 9

Evaluate: ⓐ |7| ⓑ |−3|.
If you missed this problem, review Example 3 in Integers.

Solution

ⓐ 7; ⓑ 3

Evaluate: ⓐ 4 ⓑ 16.
If you missed this problem, review Example 8 in Decimals.

Solution

ⓐ 2; ⓑ 4

Use the Vertical Line Test

In the last section we learned how to determine if a relation is a function. The relations we looked at were expressed as a set of ordered pairs, a mapping or an equation. We will now look at how to tell if a graph is that of a function.

An ordered pair (x,y) is a solution of a linear equation, if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.

The graph of a linear equation is a straight line where every point on the line is a solution of the equation and every solution of this equation is a point on this line.

In Figure 1, we can see that, in graph of the equation y=2x−3, for every x-value there is only one y-value, as shown in the accompanying table.

plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The line is labeled y equals2 x minus 3. There are several vertical arrows that relate values on the x-axis to points on the line. The first arrow relates x equalsnegative 2 on the x-axis to the point (negative 2, negative 7) on the line. The second arrow relates x equalsnegative 1 on the x-axis to the point (negative 1, negative 5) on the line. The next arrow relates x equals0 on the x-axis to the point (0, negative 3) on the line. The next arrow relates x equals3 on the x-axis to the point (3, 3) on the line. The last arrow relates x equals4 on the x-axis to the point (4, 5) on the line. The table has 7 rows and 3 columns. The first row is a title row with the label y equals2 x minus 3. The second row is a header row with the headers x, y, and (x, y). The third row has the coordinates negative 2, negative 7, and (negative 2, negative 7). The fourth row has the coordinates negative 1, negative 5, and (negative 1, negative 5). The fifth row has the coordinates 0, negative 3, and (0, negative 3). The sixth row has the coordinates 3, 3, and (3, 3). The seventh row has the coordinates 4, 5, and (4, 5).

A relation is a function if every element of the domain has exactly one value in the range. So the relation defined by the equation y=2x−3 is a function.

If we look at the graph, each vertical dashed line only intersects the line at one point. This makes sense as in a function, for every x-value there is only one y-value.

If the vertical line hit the graph twice, the x-value would be mapped to two y-values, and so the graph would not represent a function.

This leads us to the vertical line test. A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. If any vertical line intersects the graph in more than one point, the graph does not represent a function.

Vertical Line Test

A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point.

If any vertical line intersects the graph in more than one point, the graph does not represent a function.

Determine whether each graph is the graph of a function.

The figure has two graphs. In graph a there is a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 2), (3, 0), and (6, negative 2). In graph b there is a parabola opening to the right graphed on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The parabola goes through the points (negative 1, 0), (0, 1), (0, negative 1), (3, 2), and (3, negative 2).
Solution

ⓐ Since any vertical line intersects the graph in at most one point, the graph is the graph of a function.
The figure has a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 2), (3, 0), and (6, negative 2). Three dashed vertical straight lines are drawn at x equalsnegative 5, x equalsnegative 3, and x equals3. Each line intersects the slanted line at exactly one point.

ⓑ One of the vertical lines shown on the graph, intersects it in two points. This graph does not represent a function.
The figure has a parabola opening to the right graphed on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The parabola goes through the points (negative 1, 0), (0, 1), (0, negative 1), (3, 2), and (3, negative 2). Three dashed vertical straight lines are drawn at x equalsnegative 2, x equalsnegative 1, and x equals2. The vertical line x – negative 2 does not intersect the parabola. The vertical line x equalsnegative 1 intersects the parabola at exactly one point. The vertical line x equals3 intersects the parabola at two separate points.

Determine whether each graph is the graph of a function.

The figure has two graphs. In graph a there is a parabola opening up graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (0, negative 1), (negative 1, 0), (1, 0), (negative 2, 3), and (2, 3). In graph b there is a circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The circle goes through the points (negative 2, 0), (2, 0), (0, negative 2), and (0, 2).
Solution

ⓐ yes ⓑ no

Determine whether each graph is the graph of a function.

The figure has two graphs. In graph a there is an ellipse graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The ellipse goes through the points (0, negative 3), (negative 2, 0), (2, 0), and (0, 3). In graph b there is a straight line graphed on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (0, negative 2), (2, 0), and (4, 2).
Solution

ⓐ no ⓑ yes

Identify Graphs of Basic Functions

We used the equation y=2x−3 and its graph as we developed the vertical line test. We said that the relation defined by the equation y=2x−3 is a function.

We can write this as in function notation as f(x)=2x−3. It still means the same thing. The graph of the function is the graph of all ordered pairs (x,y) where y=f(x). So we can write the ordered pairs as (x,f(x)). It looks different but the graph will be the same.

Compare the graph of y=2x−3 previously shown in Figure 1 with the graph of f(x)=2x−3 shown in Figure 2. Nothing has changed but the notation.

This figure has a graph next to a table. The graph has a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The line is labeled f of x equals2 x minus 3. There are several vertical arrows that relate values on the x-axis to points on the line. The first arrow relates x equalsnegative 2 on the x-axis to the point (negative 2, negative 7) on the line. The second arrow relates x equalsnegative 1 on the x-axis to the point (negative 1, negative 5) on the line. The next arrow relates x equals0 on the x-axis to the point (0, negative 3) on the line. The next arrow relates x equals3 on the x-axis to the point (3, 3) on the line. The last arrow relates x equals4 on the x-axis to the point (4, 5) on the line. The table has 7 rows and 3 columns. The first row is a title row with the label f of x equals2 x minus 3. The second row is a header row with the headers x, f of x, and (x, f of x). The third row has the coordinates negative 2, negative 7, and (negative 2, negative 7). The fourth row has the coordinates negative 1, negative 5, and (negative 1, negative 5). The fifth row has the coordinates 0, negative 3, and (0, negative 3). The sixth row has the coordinates 3, 3, and (3, 3). The seventh row has the coordinates 4, 5, and (4, 5).

Graph of a Function

The graph of a function is the graph of all its ordered pairs, (x,y) or using function notation, (x,f(x)) where y=f(x).

fname of functionxx-coordinate of the ordered pairf(x)y-coordinate of the ordered pair

As we move forward in our study, it is helpful to be familiar with the graphs of several basic functions and be able to identify them.

Through our earlier work, we are familiar with the graphs of linear equations. The process we used to decide if y=2x−3 is a function would apply to all linear equations. All non-vertical linear equations are functions. Vertical lines are not functions as the x-value has infinitely many y-values.

We wrote linear equations in several forms, but it will be most helpful for us here to use the slope-intercept form of the linear equation. The slope-intercept form of a linear equation is y=mx+b. In function notation, this linear function becomes f(x)=mx+b where m is the slope of the line and b is the y-intercept.

The domain is the set of all real numbers, and the range is also the set of all real numbers.

Linear Function

This figure has a graph of a straight line on the x y-coordinate plane. The line goes through the point (0, b). Next to the graph are the following: “f of x equalsm x plus b”, “m, b: all real numbers”, “m: slope of the line”, “b: y-intercept”, “Domain: (negative infinity, infinity)”, and “Range: (negative infinity, infinity)”.

We will use the graphing techniques we used earlier, to graph the basic functions.

Graph: f(x)=−2x−4.

Solution
f(x)=−2x−4
We recognize this as a linear function.
Find the slope and y-intercept. m=−2
b=−4
Graph using the slope intercept. A coordinate plane with a straight line passing through (-2, 0) and (0, -4).

Graph: f(x)=−3x−1

Solution


The figure has the graph of a linear function on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The line goes through the points (1, negative 4), (0, negative 1), and (negative 1, 2).

Graph: f(x)=−4x−5

Solution


The figure has the graph of a linear function on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The line goes through the points (negative 2, 3), (0, negative 5), and (negative 1, negative 1).

The next function whose graph we will look at is called the constant function and its equation is of the form f(x)=b, where b is any real number. If we replace the f(x) with y, we get y=b. We recognize this as the horizontal line whose y-intercept is b. The graph of the function f(x)=b, is also the horizontal line whose y-intercept is b.

Notice that for any real number we put in the function, the function value will be b. This tells us the range has only one value, b.

Constant Function

This figure has a graph of a straight horizontal line on the x y-coordinate plane. The line goes through the point (0, b). Next to the graph are the following: “f of x equalsb”, “b: any real number”, “b: y-intercept”, “Domain: (negative infinity, infinity)”, and “Range: b”.

Graph: f(x)=4.

Solution
f(x)=4
We recognize this as a constant function.
The graph will be a horizontal line through (0,4). A graph displays a horizontal line at y=4, extending across the x-axis, on a Cartesian coordinate plane with labeled axes from -7 to 7 for x, and -1 to 11 for y, marked with grid lines.

Graph: f(x)=−2.

Solution


The figure has the graph of a constant function on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (0, negative 2), (1, negative 2), and (2, negative 2).

Graph: f(x)=3.

Solution


The figure has the graph of a constant function on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (0, 3), (1, 3), and (2, 3).

The identity function, f(x)=x is a special case of the linear function. If we write it in linear function form, f(x)=1x+0, we see the slope is 1 and the y-intercept is 0.

Identity Function

This figure has a graph of a straight line on the x y-coordinate plane. The line goes through the points (0, 0), (1, 1), and (2, 2). Next to the graph are the following: “f of x equalsx”, “m: 1”, “b: 0”, “Domain: (negative infinity, infinity)”, and “Range: (negative infinity, infinity)”.

The next function we will look at is not a linear function. So the graph will not be a line. The only method we have to graph this function is point plotting. Because this is an unfamiliar function, we make sure to choose several positive and negative values as well as 0 for our x-values.

Graph: f(x)=x2.

Solution

We choose x-values. We substitute them in and then create a chart as shown.
This figure has a graph next to a table. In the graph there is a parabola opening up graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 2 to 6. The parabola goes through the points (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9). The table has 8 rows and 3 columns. The first row is a header row with the headers x, f of x equalsx squared, and (x, f of x). The second row has the coordinates negative 3, 9, and (negative 3, 9). The third row has the coordinates negative 2, 4, and (negative 2, 4). The fourth row has the coordinates negative 1, 1, and (negative 1, 1). The fifth row has the coordinates 0, 0, and (0, 0). The sixth row has the coordinates 1, 1, and (1, 1). The seventh row has the coordinates 2, 4, and (2, 4). The seventh row has the coordinates 3, 9, and (3, 9).

Graph: f(x)=x2.

Solution


This figure has a graph next to a table. In the graph there is a parabola opening up graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 4 to 8. The parabola goes through the points (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4).

f(x)=−x2

Solution


This figure has a graph next to a table. In the graph there is a parabola opening up graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 4 to 8. The parabola goes through the points (negative 2, negative 4), (negative 1, negative 1), (0, 0), (1, negative 1), and (2, negative 4).

Looking at the result in Example 4, we can summarize the features of the square function. We call this graph a parabola. As we consider the domain, notice any real number can be used as an x-value. The domain is all real numbers.

The range is not all real numbers. Notice the graph consists of values of y never go below zero. This makes sense as the square of any number cannot be negative. So, the range of the square function is all non-negative real numbers.

Square Function

This figure has a graph of a parabola opening up graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 2 to 6. The parabola goes through the points (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4). Next to the graph are the following: “f of x equalsx squared”, “Domain: (negative infinity, infinity)”, and “Range: [0, infinity)”.

The next function we will look at is also not a linear function so the graph will not be a line. Again we will use point plotting, and make sure to choose several positive and negative values as well as 0 for our x-values.

Graph: f(x)=x3.

Solution

We choose x-values. We substitute them in and then create a chart.
This figure has a curved line graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The curved line goes through the points (negative 2, negative 8), (negative 1, negative 1), (0, 0), (1, 1), and (2, 8). Next to the graph is a table. The table has 6 rows and 3 columns. The first row is a header row with the headers x, f of x equalsx cubed, and (x, f of x). The second row has the coordinates negative 2, negative 8, and (negative 2, negative 8). The third row has the coordinates negative 1, negative 1, and (negative 1, negative 1). The fourth row has the coordinates 0, 0, and (0, 0). The fifth row has the coordinates 1, 1, and (1, 1). The sixth row has the coordinates 2, 8, and (2, 8).

Graph: f(x)=x3.

Solution


This figure has a curved line graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 2, negative 8), (negative 1, negative 1), (0, 0), (1, 1), and (2, 8).

Graph: f(x)=−x3.

Solution


This figure has a curved line graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 2, 8), (negative 1, 1), (0, 0), (1, negative 1), and (2, negative 8).

Looking at the result in Example 5, we can summarize the features of the cube function. As we consider the domain, notice any real number can be used as an x-value. The domain is all real numbers.

The range is all real numbers. This makes sense as the cube of any non-zero number can be positive or negative. So, the range of the cube function is all real numbers.

Cube Function

This figure has a curved line graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The curved line goes through the points (negative 2, negative 8), (negative 1, negative 1), (0, 0), (1, 1), and (2, 8).). Next to the graph are the following: “f of x equalsx cubed”, “Domain: (negative infinity, infinity)”, and “Range: (negative infinity, infinity)”.

The next function we will look at does not square or cube the input values, but rather takes the square root of those values.

Let’s graph the function f(x)=x and then summarize the features of the function. Remember, we can only take the square root of non-negative real numbers, so our domain will be the non-negative real numbers.

f(x)=x

Solution

We choose x-values. Since we will be taking the square root, we choose numbers that are perfect squares, to make our work easier. We substitute them in and then create a chart.
This figure has a curved half-line graphed on the x y-coordinate plane. The x-axis runs from 0 to 8. The y-axis runs from 0 to 8. The curved half-line starts at the point (0, 0) and then goes up and to the right. The curved half line goes through the points (1, 1) and (4, 2). Next to the graph is a table. The table has 5 rows and 3 columns. The first row is a header row with the headers x, f of x equalssquare root of x, and (x, f of x). The second row has the coordinates 0, 0, and (0, 0). The third row has the coordinates 1, 1, and (1, 1). The fourth row has the coordinates 4, 2, and (4, 2). The fifth row has the coordinates 9, 3, and (9, 3).

Graph: f(x)=x.

Solution


This figure has a curved half-line graphed on the x y-coordinate plane. The x-axis runs from 0 to 10. The y-axis runs from 0 to 10. The curved half-line starts at the point (0, 0) and then goes up and to the right. The curved half line goes through the points (1, 1), (4, 2), and (9, 3).

Graph: f(x)=−x.

Solution


This figure has a curved half-line graphed on the x y-coordinate plane. The x-axis runs from 0 to 10. The y-axis runs from negative 10 to 0. The curved half-line starts at the point (0, 0) and then goes down and to the right. The curved half line goes through the points (1, negative 1), (4, negative 2), and (9, negative 3).

Square Root Function

This figure has a curved half-line graphed on the x y-coordinate plane. The x-axis runs from 0 to 8. The y-axis runs from 0 to 8. The curved half-line starts at the point (0, 0) and then goes up and to the right. The curved half line goes through the points (1, 1) and (4, 2). Next to the graph are the following: “f of x equalssquare root of x”, “Domain: [0, infinity)”, and “Range: [0, infinity)”.

Our last basic function is the absolute value function, f(x)=|x|. Keep in mind that the absolute value of a number is its distance from zero. Since we never measure distance as a negative number, we will never get a negative number in the range.

Graph: f(x)=|x|.

Solution

We choose x-values. We substitute them in and then create a chart.
This figure has a v-shaped line graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 1 to 6. The v-shaped line goes through the points (negative 3, 3), (negative 2, 2), (negative 1, 1), (0, 0), (1, 1), (2, 2), and (3, 3). Next to the graph is a table. The table has 8 rows and 3 columns. The first row is a header row with the headers x, f of x equalsabsolute value of x, and (x, f of x). The second row has the coordinates negative 3, 3, and (negative 3, 3). The third row has the coordinates negative 2, 2, and (negative 2, 2). The fourth row has the coordinates negative 1, 1, and (negative 1, 1). The fifth row has the coordinates 0, 0, and (0, 0). The sixth row has the coordinates 1, 1, and (1, 1). The seventh row has the coordinates 2, 2, and (2, 2). The eighth row has the coordinates 3, 3, and (3, 3).

Graph: f(x)=|x|.

Solution


This figure has a v-shaped line graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The v-shaped line goes through the points (negative 3, 3), (negative 2, 2), (negative 1, 1), (0, 0), (1, 1), (2, 2), and (3, 3).

Graph: f(x)=−|x|.

Solution


This figure has a v-shaped line graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 8 to 4. The v-shaped line goes through the points (negative 3, negative 3), (negative 2, negative 2), (negative 1, negative 1), (0, 0), (1, negative 1), (2, negative 2), and (3, negative 3).

Absolute Value Function

This figure has a v-shaped line graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 1 to 6. The v-shaped line goes through the points (negative 3, 3), (negative 2, 2), (negative 1, 1), (0, 0), (1, 1), (2, 2), and (3, 3). The point (0, 0) where the line changes slope is called the vertex. Next to the graph are the following: “f of x equalsabsolute value of x”, “Domain: (negative infinity, infinity)”, and “Range: [0, infinity)”.

Read Information from a Graph of a Function

In the sciences and business, data is often collected and then graphed. The graph is analyzed, information is obtained from the graph and then often predictions are made from the data.

We will start by reading the domain and range of a function from its graph.

Remember the domain is the set of all the x-values in the ordered pairs in the function. To find the domain we look at the graph and find all the values of x that have a corresponding value on the graph. Follow the value x up or down vertically. If you hit the graph of the function then x is in the domain.

Remember the range is the set of all the y-values in the ordered pairs in the function. To find the range we look at the graph and find all the values of y that have a corresponding value on the graph. Follow the value y left or right horizontally. If you hit the graph of the function then y is in the range.

Use the graph of the function to find its domain and range. Write the domain and range in interval notation.

This figure has a curved line segment graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The curved line segment goes through the points (negative 3, negative 1), (1.5, 3), and (3, 1). The interval [negative 3, 3] is marked on the horizontal axis. The interval [negative 1, 3] is marked on the vertical axis.
Solution

To find the domain we look at the graph and find all the values of x that correspond to a point on the graph. The domain is highlighted in red on the graph. The domain is [−3,3].

To find the range we look at the graph and find all the values of y that correspond to a point on the graph. The range is highlighted in blue on the graph. The range is [−1,3].

Use the graph of the function to find its domain and range. Write the domain and range in interval notation.

This figure has a curved line segment graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line segment goes through the points (negative 5, negative 4), (0, negative 3), and (1, 2). The interval [negative 5, 1] is marked on the horizontal axis. The interval [negative 4, 2] is marked on the vertical axis.
Solution

The domain is [−5,1]. The range is [−4,2].

Use the graph of the function to find its domain and range. Write the domain and range in interval notation.

This figure has a curved line segment graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 5. The y-axis runs from negative 6 to 4. The curved line segment goes through the points (negative 2, 1), (0, 3), and (4, negative 5). The interval [negative 2, 4] is marked on the horizontal axis. The interval [negative 5, 3] is marked on the vertical axis.
Solution

The domain is [−2,4]. The range is [−5,3].

We are now going to read information from the graph that you may see in future math classes.

Use the graph of the function to find the indicated values.

This figure has a wavy curved line graphed on the x y-coordinate plane. The x-axis runs from negative 2 times pi to 2 times pi. The y-axis runs from negative 4 to 4. The curved line segment goes through the points (negative 2 times pi, 0), (negative 3 divided by 2 times pi, 1), (negative pi, 0), (negative 1 divided by 2 times pi, negative 1), (0, 0), (1 divided by 2 times pi, 1), (pi, 0), (3 divided by 2 times pi, negative 1), and (2 times pi, 0). The points (negative 3 divided by 2 times pi, 1) and (1 divided by 2 times pi, 1) are the highest points on the graph. The points (negative 1 divided by 2 times pi, negative 1) and (3 divided by 2 times pi, negative 1) are the lowest points on the graph. The pattern extends infinitely to the left and right.

ⓐ Find: f(0).
ⓑ Find: f(32π).
ⓒ Find: f(−12π).
ⓓ Find the values for x when f(x)=0.
ⓔ Find the x-intercepts.
ⓕ Find the y-intercepts.
ⓖ Find the domain. Write it in interval notation.
ⓗ Find the range. Write it in interval notation.

Solution

ⓐ When x=0, the function crosses the y-axis at 0. So, f(0)=0.
ⓑ When x=32π, the y-value of the function is −1. So, f(32π)=−1.
ⓒ When x=−12π, the y-value of the function is −1. So, f(−12π)=−1.
ⓓ The function is 0 at the points, (−2π,0),(−π,0),(0,0),(π,0),(2π,0). The x-values when f(x)=0 are −2π,−π,0,π,2π.
ⓔ The x-intercepts occur when y=0. So the x-intercepts occur when f(x)=0. The x-intercepts are (−2π,0),(−π,0),(0,0),(π,0),(2π,0).
ⓕ The y-intercepts occur when x=0. So the y-intercepts occur at f(0). The y-intercept is (0,0).
ⓖ This function has a value for all values of x. Therefore, the domain in interval notation is (−∞,∞)
ⓗ This function values, or y-values go from −1 to 1. Therefore, the range, in interval notation, is [−1,1].

Use the graph of the function to find the indicated values.

This figure has a wavy curved line graphed on the x y-coordinate plane. The x-axis runs from negative 2 times pi to 2 times pi. The y-axis runs from negative 6 to 6. The curved line segment goes through the points (negative 2 times pi, 0), (negative 3 divided by 2 times pi, 2), (negative pi, 0), (negative 1 divided by 2 times pi, negative 2), (0, 0), (1 divided by 2 times pi, 2), (pi, 0), (3 divided by 2 times pi, negative 2), and (2 times pi, 0). The points (negative 3 divided by 2 times pi, 2) and (1 divided by 2 times pi, 2) are the highest points on the graph. The points (negative 1 divided by 2 times pi, negative 2) and (3 divided by 2 times pi, negative 2) are the lowest points on the graph. The line extends infinitely to the left and right.

ⓐ Find: f(0).
ⓑ Find: f(12π).
ⓒ Find: f(−32π).
ⓓ Find the values for x in [-2π,2π] when f(x)=0.
ⓔ Find the x-intercepts.
ⓕ Find the y-intercepts.
ⓖ Find the domain. Write it in interval notation.
ⓗ Find the range. Write it in interval notation.

Solution

ⓐ f(0)=0 ⓑ f=(π2)=2 ⓒ f=(−3π2)=2 ⓓ f(x)=0 for x=−2π,−π,0,π,2π ⓔ (−2π,0),(−π,0),(0,0),(π,0),(2π,0) ⓕ (0,0) ⓖ (−∞,∞) ⓗ [−2,2]

Use the graph of the function to find the indicated values.

This figure has a wavy curved line graphed on the x y-coordinate plane. The x-axis runs from negative 2 times pi to 2 times pi. The y-axis runs from negative 6 to 6. The curved line segment goes through the points (negative 2 times pi, 1), (negative 3 divided by 2 times pi, 0), (negative pi, negative 1), (negative 1 divided by 2 times pi, 0), (0, 1), (1 divided by 2 times pi, 0), (pi, negative 1), (3 divided by 2 times pi, 0), and (2 times pi, 1). The points (negative 2 times pi, 1), (0, 1), and (2 times pi, 1) are the highest points on the graph. The points (negative pi, negative 1) and (pi, negative 1) are the lowest points on the graph. The pattern extends infinitely to the left and right.

ⓐ Find: f(0).
ⓑ Find: f(π).
ⓒ Find: f(−π).
ⓓ Find the values for x in [-2π,2π] when f(x)=0.
ⓔ Find the x-intercepts.
ⓕ Find the y-intercepts.
ⓖ Find the domain. Write it in interval notation.
ⓗ Find the range. Write it in interval notation.

Solution

ⓐ f(0)=1 ⓑ f(π)=−1 ⓒ f(−π)=−1 ⓓ f(x)=0 for x=−3π2,−π2,π2,3π2 ⓔ (−3π2,0),(−π2,0),(π2,0),(3π2,0) ⓕ (0,1) ⓖ (−∞,∞) ⓗ [−1,1]

Access this online resource for additional instruction and practice with graphs of functions.

  • Find Domain and Range

Key Concepts

  • Vertical Line Test
    • A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point.
    • If any vertical line intersects the graph in more than one point, the graph does not represent a function.
  • Graph of a Function
    • The graph of a function is the graph of all its ordered pairs, (x,y) or using function notation, (x,f(x)) where y=f(x).
      fname of functionxx-coordinate of the ordered pairf(x)y-coordinate of the ordered pair
  • Linear Function
    This figure has a graph of a straight line on the x y-coordinate plane. The line goes through the point (0, b). Next to the graph are the following: “f of x equalsm x plus b”, “m, b: all real numbers”, “m: slope of the line”, “b: y-intercept”, “Domain: (negative infinity, infinity)”, and “Range: (negative infinity, infinity)”.
  • Constant Function
    This figure has a graph of a straight horizontal line on the x y-coordinate plane. The line goes through the point (0, b). Next to the graph are the following: “f of x equalsb”, “b: any real number”, “b: y-intercept”, “Domain: (negative infinity, infinity)”, and “Range: b”.
  • Identity Function
    This figure has a graph of a straight line on the x y-coordinate plane. The line goes through the points (0, 0), (1, 1), and (2, 2). Next to the graph are the following: “f of x equalsx”, “m: 1”, “b: 0”, “Domain: (negative infinity, infinity)”, and “Range: (negative infinity, infinity)”.
  • Square Function
    This figure has a graph of a parabola opening up graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 2 to 6. The parabola goes through the points (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4). Next to the graph are the following: “f of x equalsx squared”, “Domain: (negative infinity, infinity)”, and “Range: [0, infinity)”.
  • Cube Function
    This figure has a curved line graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The curved line goes through the points (negative 2, negative 8), (negative 1, negative 1), (0, 0), (1, 1), and (2, 8).). Next to the graph are the following: “f of x equalsx cubed”, “Domain: (negative infinity, infinity)”, and “Range: (negative infinity, infinity)”.
  • Square Root Function
    This figure has a curved half-line graphed on the x y-coordinate plane. The x-axis runs from 0 to 8. The y-axis runs from 0 to 8. The curved half-line starts at the point (0, 0) and then goes up and to the right. The curved half line goes through the points (1, 1) and (4, 2). Next to the graph are the following: “f of x equalssquare root of x”, “Domain: [0, infinity)”, and “Range: [0, infinity)”.
  • Absolute Value Function
    This figure has a v-shaped line graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 4. The y-axis runs from negative 1 to 6. The v-shaped line goes through the points (negative 3, 3), (negative 2, 2), (negative 1, 1), (0, 0), (1, 1), (2, 2), and (3, 3). The point (0, 0) where the line changes slope is called the vertex. Next to the graph are the following: “f of x equalsabsolute value of x”, “Domain: (negative infinity, infinity)”, and “Range: [0, infinity)”.

Section Exercises

Practice Makes Perfect

Use the Vertical Line Test

In the following exercises, determine whether each graph is the graph of a function.

ⓐ
The figure has a circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The circle goes through the points (negative 3, 0), (3, 0), (0, negative 3), and (0, 3).
ⓑ
The figure has a parabola opening up graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 4 to 8. The parabola goes through the points (negative 2, 6), (1, 3), (0, 2), (1, 3), and (2, 6).

Solution

ⓐ no ⓑ yes

ⓐ
The figure has an s-shaped curved line graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The s-shaped curved line goes through the points (negative 1, 1), (0, 0), and (1, 1).
ⓑ
The figure has a circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The circle goes through the points (negative 4, 0), (4, 0), (0, negative 4), and (0, 4).

ⓐ
The figure has a parabola opening right graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The parabola goes through the points (negative 2, 0), (negative 1, 1), (negative 1, negative 1), (negative 2, 2), and (2, 2).
ⓑ
The figure has a cube function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 1, negative 1), (0, 0), and (1, 1).

Solution

ⓐ no ⓑ yes

ⓐ
The figure has two curved lines graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line on the left goes through the points (negative 2, 0), (negative 4, 5), and (negative 4, negative 5). The curved line on the right goes through the points (2, 0), (4, 5), and (4, negative 5).
ⓑ
The figure has a sideways absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line bends at the point (0, 2) and goes to the right. The line goes through the points (1, 3), (2, 4), (1, 1), and (2, 0).

Identify Graphs of Basic Functions

In the following exercises, ⓐ graph each function ⓑ state its domain and range. Write the domain and range in interval notation.

f(x)=3x+4

Solution

ⓐ
The figure has a linear function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (negative 2, negative 2), (negative 1, 1), and (0, 4).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=2x+5

f(x)=−x−2

Solution

ⓐ
The figure has a linear function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (negative 2, 0), (0, negative 2), and (2, negative 4).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=−4x−3

f(x)=−2x+2

Solution

ⓐ
The figure has a linear function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (negative 2, 2), (negative 1, 0), and (0, negative 2).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=−3x+3

f(x)=12x+1

Solution

ⓐ
The figure has a linear function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (negative 2, 0), (0, 1), and (2, 2).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=23x−2

f(x)=5

Solution

ⓐ
The figure has a constant function graphed on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (negative 2, 5), (negative 1, 5), and (0, 5).

ⓑ D:(-∞,∞), R:{5}

f(x)=2

f(x)=−3

Solution

ⓐ
The figure has a constant function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (0, negative 3), (1, negative 3), and (2, negative 3).

ⓑ D:(-∞,∞), R: {−3}

f(x)=−1

f(x)=2x

Solution

ⓐ
The figure has a linear function graphed on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, 0), (2, 4), and (negative 2, negative 4).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=3x

f(x)=−2x

Solution

ⓐ
The figure has a linear function graphed on the x y-coordinate plane. The x-axis runs from negative 12 to 12. The y-axis runs from negative 12 to 12. The line goes through the points (0, 0), (1, negative 2), and (negative 1, 2).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=−3x

f(x)=3x2

Solution

ⓐ
The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (negative 1, 3), (0, 0), and (1, 3). The lowest point on the graph is (0, 0).

ⓑ D:(-∞,∞), R:[0,∞)

f(x)=2x2

f(x)=−3x2

Solution

ⓐ
The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 10 to 2. The parabola goes through the points (negative 1, negative 3), (0, 0), and (1, negative 3). The highest point on the graph is (0, 0).

ⓑ (-∞,∞), R:(-∞,0]

f(x)=−2x2

f(x)=12x2

Solution

ⓐ
The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (negative 4, 8), (negative 2, 2), (0, 0), (2, 2), and (4, 8). The lowest point on the graph is (0, 0).

ⓑ (-∞,∞), R:[0,∞)

f(x)=13x2

f(x)=x2−1

Solution

ⓐ
The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (negative 2, 3), (negative 1, 0), (0, negative 1), (1, 0), and (2, 3). The lowest point on the graph is (0, negative 1).

ⓑ (-∞,∞), R:[−1, ∞)

f(x)=x2+1

f(x)=−2x3

Solution

ⓐ
The figure has a cube function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 1, 2), (0, 0), and (1, negative 2).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=2x3

f(x)=x3+2

Solution

ⓐ
The figure has a cube function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 1, 1), (0, 2), and (1, 3).

ⓑ D:(-∞,∞), R:(-∞,∞)

f(x)=x3−2

f(x)=2x

Solution

ⓐ
The figure has a square root function graphed on the x y-coordinate plane. The x-axis runs from 0 to 10. The y-axis runs from 0 to 10. The half-line starts at the point (0, 0) and goes through the points (1, 2) and (4, 4).

ⓑ D:[0,∞), R:[0,∞)

f(x)=−2x

f(x)=x−1

Solution

ⓐ
The figure has a square root function graphed on the x y-coordinate plane. The x-axis runs from 0 to 10. The y-axis runs from 0 to 10. The half-line starts at the point (1, 0) and goes through the points (2, 1) and (5, 2).

ⓑ D:[1,∞), R:[0,∞)

f(x)=x+1

f(x)=3|x|

Solution

ⓐ
The figure has an absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The vertex is at the point (0, 0). The line goes through the points (negative 1, 3) and (1, 3).

ⓑ D:−∞,∞,R:[0,∞)

f(x)=−2|x|

f(x)=|x|+1

Solution

ⓐ
The figure has an absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The vertex is at the point (0, 1). The line goes through the points (negative 1, 2) and (1, 2).

ⓑ D:(-∞,∞), R:[1,∞)

f(x)=|x|−1

Read Information from a Graph of a Function

In the following exercises, use the graph of the function to find its domain and range. Write the domain and range in interval notation.

The figure has a square root function graphed on the x y-coordinate plane. The x-axis runs from negative 2 to 8. The y-axis runs from negative 2 to 8. The half-line starts at the point (2, 0) and goes through the points (3, 1) and (6, 2).
Solution

D: [2,∞), R: [0,∞)

The figure has a square root function graphed on the x y-coordinate plane. The x-axis runs from negative 2 to 8. The y-axis runs from negative 2 to 10. The half-line starts at the point (negative 3, 0) and goes through the points (negative 2, 1) and (1, 2).
The figure has an absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from 0 to 12. The vertex is at the point (0, 4). The line goes through the points (negative 2, 6) and (2, 6).
Solution

D: (-∞,∞), R: [4,∞)

The figure has an absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 4 to 8. The vertex is at the point (0, negative 1). The line goes through the points (negative 1, 0) and (1, 0).
The figure has a half-circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line segment starts at the point (negative 2, 0). The line goes through the point (0, 2) and ends at the point (2, 0).
Solution

D: [−2,2], R: [0, 2]

The figure has a half-circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The curved line segment starts at the point (negative 3, 3). The line goes through the point (0, 6) and ends at the point (3, 3).

In the following exercises, use the graph of the function to find the indicated values.

This figure has a wavy curved line graphed on the x y-coordinate plane. The x-axis runs from negative 2 times pi to 2 times pi. The y-axis runs from negative 6 to 6. The curved line segment goes through the points (negative 2 times pi, 0), (negative 3 divided by 2 times pi, negative 1), (negative pi, 0), (negative 1 divided by 2 times pi, 1), (0, 0), (1 divided by 2 times pi, negative 1), (pi, 0), (3 divided by 2 times pi, 1), and (2 times pi, 0). The points (negative 3 divided by 2 times pi, negative 1) and (1 divided by 2 times pi, negative 1) are the lowest points on the graph. The points (negative 1 divided by 2 times pi, 1) and (3 divided by 2 times pi, 1) are the highest points on the graph. The pattern extends infinitely to the left and right.

ⓐ Find: f(0).
ⓑ Find: f(12π).
ⓒ Find: f(−32π).
ⓓ Find the values for x when f(x)=0.
ⓔ Find the x-intercepts.
ⓕ Find the y-intercepts.
ⓖ Find the domain. Write it in interval notation.
ⓗ Find the range. Write it in interval notation.

Solution

ⓐ f(0)=0 ⓑ fπ2=−1
ⓒ f−3π2=−1 ⓓ f(x)=0 for x=−2π,−π,0,π,2π
ⓔ (−2π,0),(−π,0), (0,0),(π,0),(2π,0)
ⓕ 0,0 ⓖ (−∞,∞)
ⓗ [−1,1]

This figure has a wavy curved line graphed on the x y-coordinate plane. The x-axis runs from negative 2 times pi to 2 times pi. The y-axis runs from negative 6 to 6. The curved line segment goes through the points (negative 2 times pi, negative 1), (negative 3 divided by 2 times pi, 0), (negative pi, 1), (negative 1 divided by 2 times pi, 0), (0, negative 1), (1 divided by 2 times pi, 0), (pi, 1), (3 divided by 2 times pi, 0), and (2 times pi, negative 1). The points (negative 2 times pi, negative 1) and (2 times pi, negative 1) are the lowest points on the graph. The points (negative pi, 1) and (pi, 1) are the highest points on the graph. The pattern extends infinitely to the left and right.

ⓐ Find: f(0).
ⓑ Find: f(π).
ⓒ Find: f(−π).
ⓓ Find the values for x when f(x)=0.
ⓔ Find the x-intercepts.
ⓕ Find the y-intercepts.
ⓖ Find the domain. Write it in interval notation.
ⓗ Find the range. Write it in interval notation

The figure has the top half of a circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 4 to 8. The curved line segment starts at the point (negative 3, 2). The line goes through the point (0, 5) and ends at the point (3, 2). The point (0, 5) is the highest point on the graph. The points (negative 3, 2) and (3, 2) are the lowest points on the graph.

ⓐ Find: f(0).
ⓑ Find: f(−3).
ⓒ Find: f(3).
ⓓ Find the values for x when f(x)=0.
ⓔ Find the x-intercepts.
ⓕ Find the y-intercepts.
ⓖ Find the domain. Write it in interval notation.
ⓗ Find the range. Write it in interval notation.

Solution

ⓐ 5 ⓑ 2 ⓒ 2 ⓓ f(x)=0 for no x ⓔ none ⓕ 0,5 ⓖ [−3,3]
ⓗ [2,5]

The figure has the top half of a circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 4 to 8. The curved line segment starts at the point (negative 4, 0). The line goes through the point (0, 4) and ends at the point (4, 0). The point (0, 4) is the highest point on the graph. The points (negative 4, 0) and (4, 0) are the lowest points on the graph.

ⓐ Find: f(0).
ⓑ Find the values for x when f(x)=0.
ⓒ Find the x-intercepts.
ⓓ Find the y-intercepts.
ⓔ Find the domain. Write it in interval notation.
ⓕ Find the range. Write it in interval notation

Writing Exercises

Explain in your own words how to find the domain from a graph.

Explain in your own words how to find the range from a graph.

Explain in your own words how to use the vertical line test.

Draw a sketch of the square and cube functions. What are the similarities and differences in the graphs?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The figure shows a table with four rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is "confidently", the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “use the vertical line test”, “identify graphs of basic functions”, and “read information from a graph”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Chapter Review Exercises

Graph Linear Equations in Two Variables

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system.


ⓐ (−1,−5)
ⓑ (−3,4)
ⓒ (2,−3)
ⓓ (1,52)

Solution

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 5 to 5. The point labeled a is 1 units to the left of the origin and 5 units below the origin and is located in quadrant III. The point labeled b is 3 units to the left of the origin and 4 units above the origin and is located in quadrant II. The point labeled c is 2 units to the right of the origin and 3 units below the origin and is located in quadrant IV. The point labeled d is 1 unit to the right of the origin and 2.5 units above the origin and is located in quadrant I.


ⓐ (−2,0)
ⓑ (0,−4)
ⓒ (0,5)
ⓓ (3,0)

In the following exercises, determine which ordered pairs are solutions to the given equations.

5x+y=10;
ⓐ (5,1)
ⓑ (2,0)
ⓒ (4,−10)

Solution

ⓑ , ⓒ

y=6x−2;
ⓐ (1,4)
ⓑ (13,0)
ⓒ (6,−2)

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

y=4x−3

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 1, negative 7), (0, negative 3), (1, negative 1), and (2, 3).

y=−3x

y=12x+3

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 6, 0), (0, 3), (2, 4), and (4, 5).

y=−45x−1

x−y=6

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 1, negative 7), (0, negative 6), (3, negative 3), and (6, 0).

2x+y=7

3x−2y=6

Solution

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 2, negative 6), (0, negative 3), (2, 0), and (4, 3).

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

y=−2

x=3

Solution

This figure shows a vertical straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (3, negative 1), (3, 0), and (3, 1).

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

y=−2x and y=−2

y=43x and y=43

Solution

The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 5 to 5. The horizontal line goes through the points (0, 4 divided by 3), (1, 4 divided by 3), and (2, 4 divided by 3). The slanted line goes through the points (0, 0), (1, 4 divided by 3), and (2, 8 divided by 3).

Find x- and y-Intercepts

In the following exercises, find the x- and y-intercepts.

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 6, negative 2), (negative 4, 0), (negative 2, 2), (0, 4), (2, 6), and (4, 8).
The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 2, 5), (negative 1, 4), (0, 3), (3, 0), and (6, negative 3).
Solution

(0,3)(3,0)

In the following exercises, find the intercepts of each equation.

x−y=−1

x+2y=6

Solution

(6,0),(0,3)

2x+3y=12

y=34x−12

Solution

(16,0),(0,−12)

y=3x

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

−x+3y=3

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 3, 0), (0, 1), (3, 2), and (6, 3).

x−y=4

2x−y=5

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (0, negative 5), (1, negative 3), (2, negative 1), and (3, 1).

2x−4y=8

y=4x

Solution

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 1, 4), (0, 0), and (1, negative 4).

Slope of a Line

Find the Slope of a Line

In the following exercises, find the slope of each line shown.

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (0, 0) and (1, negative 3).
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (negative 4, 0) and (0, 4).
Solution

1

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (negative 4, negative 4) and (2, negative 2).
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (1, 4) and (5, 2).
Solution

−12

In the following exercises, find the slope of each line.

y=2

x=5

Solution

undefined

x=−3

y=−1

Solution

0

Use the Slope Formula to find the Slope of a Line between Two Points

In the following exercises, use the slope formula to find the slope of the line between each pair of points.

(−1,−1),(0,5)

(3,5),(4,−1)

Solution

−6

(−5,−2),(3,2)

(2,1),(4,6)

Solution

52

Graph a Line Given a Point and the Slope

In the following exercises, graph each line with the given point and slope.

(2,−2); m=52

(−3,4); m=−13

Solution

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (negative 3, 4) and (0, 3).

x-intercept −4; m=3

y-intercept 1; m=−34

Solution

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (0, 1) and (4, negative 2).

Graph a Line Using Its Slope and Intercept

In the following exercises, identify the slope and y-intercept of each line.

y=−4x+9

y=53x−6

Solution

m=53;(0,−6)

5x+y=10

4x−5y=8

Solution

m=45;(0,−85)

In the following exercises, graph the line of each equation using its slope and y-intercept.

y=2x+3

y=−x−1

Solution

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, negative 1) and (1, negative 2).

y=−25x+3

4x−3y=12

Solution

This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (0, negative 4) and (3, 0).

In the following exercises, determine the most convenient method to graph each line.

x=5

y=−3

Solution

horizontal line

2x+y=5

x−y=2

Solution

intercepts

y=22x+2

y=34x−1

Solution

plotting points

Graph and Interpret Applications of Slope-Intercept

Katherine is a private chef. The equation C=6.5m+42 models the relation between her weekly cost, C, in dollars and the number of meals, m, that she serves.

ⓐ Find Katherine’s cost for a week when she serves no meals.
ⓑ Find the cost for a week when she serves 14 meals.
ⓒ Interpret the slope and C-intercept of the equation.
ⓓ Graph the equation.

Marjorie teaches piano. The equation P=35s−250 models the relation between her weekly profit, P, in dollars and the number of student lessons, s, that she teaches.

ⓐ Find Marjorie’s profit for a week when she teaches no student lessons.
ⓑ Find the profit for a week when she teaches 20 student lessons.
ⓒ Interpret the slope and P-intercept of the equation.
ⓓ Graph the equation.

Solution

ⓐ −$250
ⓑ $450
ⓒ The slope, 35, means that Marjorie’s weekly profit, P, increases by $35 for each additional student lesson she teaches.
The P-intercept means that when the number of lessons is 0, Marjorie loses $250.
ⓓ
This figure shows the graph of a straight line on the x y-coordinate plane. The x-axis runs from negative 4 to 28. The y-axis runs from negative 250 to 450. The line goes through the points (0, negative 250) and (20, 450).

Use Slopes to Identify Parallel and Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel, perpendicular, or neither.

4x−3y=−1;y=43x−3

y=5x−1;10x+2y=0

Solution

neither

3x−2y=5;2x+3y=6

2x−y=8;x−2y=4

Solution

neither

Find the Equation of a Line

Find an Equation of the Line Given the Slope and y-Intercept

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope–intercept form.

slope 13 and y-intercept (0,−6)

slope −5 and y-intercept (0,−3)

Solution

y=−5x−3

slope 0 and y-intercept (0,4)

slope −2 and y-intercept (0,0)

Solution

y=−2x

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope–intercept form.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 1), (1, 3), and (2, 5).
This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 5), (1, 2), and (2, negative 1).
Solution

y=−3x+5

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 2), (4, 1), and (8, 4).
This figure has a graph of a horizontal straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 4), (1, negative 4), and (2, negative 4).
Solution

y=−4

Find an Equation of the Line Given the Slope and a Point

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope–intercept form.

m=−14, point (−8,3)

m=35, point (10,6)

Solution

y=35x

Horizontal line containing (−2,7)

m=−2, point (−1,−3)

Solution

y=−2x−5

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope–intercept form.

(2,10) and (−2,−2)

(7,1) and (5,0)

Solution

y=12x−52

(3,8) and (3,−4)

(5,2) and (−1,2)

Solution

y=2

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope–intercept form.

line y=−3x+6, point (1,−5)

line 2x+5y=−10, point (10,4)

Solution

y=−25x+8

line x=4, point (−2,−1)

line y=−5, point (−4,3)

Solution

y=3

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope–intercept form.

line y=−45x+2, point (8,9)

line 2x−3y=9, point (−4,0)

Solution

y=−32x−6

line y=3, point (−1,−3)

line x=−5 point (2,1)

Solution

y=1

Graph Linear Inequalities in Two Variables

Verify Solutions to an Inequality in Two Variables

In the following exercises, determine whether each ordered pair is a solution to the given inequality.

Determine whether each ordered pair is a solution to the inequality y<x−3:

ⓐ (0,1) ⓑ (−2,−4) ⓒ (5,2) ⓓ (3,−1)
ⓔ (−1,−5)

Determine whether each ordered pair is a solution to the inequality x+y>4:

ⓐ (6,1) ⓑ (−3,6) ⓒ (3,2) ⓓ (−5,10) ⓔ (0,0)

Solution

ⓐ yes ⓑ no ⓒ yes ⓓ yes; ⓔ no

Recognize the Relation Between the Solutions of an Inequality and its Graph

In the following exercises, write the inequality shown by the shaded region.

Write the inequality shown by the graph with the boundary line y=−x+2.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, 2), (1, 1), and (2, 0). The line divides the x y-coordinate plane into two halves. The line and the bottom left half are shaded red to indicate that this is where the solutions of the inequality are.

Write the inequality shown by the graph with the boundary line y=23x−3.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 3), (3, negative 1), and (6, 1). The line divides the x y-coordinate plane into two halves. The line and the top left half are shaded red to indicate that this is where the solutions of the inequality are.
Solution

y≥23x−3

Write the inequality shown by the shaded region in the graph with the boundary line x+y=−4.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 4), (negative 2, negative 2), and (negative 4, 0). The line divides the x y-coordinate plane into two halves. The line and the top right half are shaded red to indicate that this is where the solutions of the inequality are.

Write the inequality shown by the shaded region in the graph with the boundary line x−2y=6.

This figure has the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A line is drawn through the points (0, negative 3), (2, negative 2), and (6, 0). The line divides the x y-coordinate plane into two halves. The line and the bottom right half are shaded red to indicate that this is where the solutions of the inequality are.
Solution

x−2y≥6

Graph Linear Inequalities in Two Variables

In the following exercises, graph each linear inequality.

Graph the linear inequality y>25x−4.

Graph the linear inequality y≤−14x+3.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, 3), (4, 2), and (8, 1). The line divides the x y-coordinate plane into two halves. The bottom left half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality x−y≤5.

Graph the linear inequality 3x+2y>10.

Solution

This figure has the graph of a straight dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, 5), (2, 2), and (4, negative 1). The line divides the x y-coordinate plane into two halves. The top right half is shaded red to indicate that this is where the solutions of the inequality are.

Graph the linear inequality y≤−3x.

Graph the linear inequality y<6.

Solution

This figure has the graph of a straight horizontal dashed line on the x y-coordinate plane. The x and y axes run from negative 10 to 10. A straight dashed line is drawn through the points (0, 6), (1, 6), and (2, 6). The line divides the x y-coordinate plane into two halves. The bottom half is shaded red to indicate that this is where the solutions of the inequality are.

Solve Applications using Linear Inequalities in Two Variables

Shanthie needs to earn at least $500 a week during her summer break to pay for college. She works two jobs. One as a swimming instructor that pays $10 an hour and the other as an intern in a law office for $25 hour. How many hours does Shanthie need to work at each job to earn at least $500 per week?

ⓐ Let x be the number of hours she works teaching swimming and let y be the number of hours she works as an intern. Write an inequality that would model this situation.
ⓑ Graph the inequality.
ⓒ Find three ordered pairs (x,y) that would be solutions to the inequality. Then, explain what that means for Shanthie.

Atsushi he needs to exercise enough to burn 600 calories each day. He prefers to either run or bike and burns 20 calories per minute while running and 15 calories a minute while biking.

ⓐ If x is the number of minutes that Atsushi runs and y is the number minutes he bikes, find the inequality that models the situation.
ⓑ Graph the inequality.
ⓒ List three solutions to the inequality. What options do the solutions provide Atsushi?

Solution

ⓐ 20x+15y≥600
ⓑ
The figure has a straight line graphed on the x y-coordinate plane. The x-axis runs from 0 to 50. The y-axis runs from 0 to 50. The line goes through the points (0, 40) and (30, 0). The line divides the coordinate plane into two halves. The top right half and the line are colored red to indicate that this is the solution set.

ⓒ Answers will vary.

Relations and Functions

Find the Domain and Range of a Relation

In the following exercises, for each relation, ⓐ find the domain of the relation ⓑ find the range of the relation.

{(5,−2),(5,−4),(7,−6),
(8,−8),(9,−10)}

{(−3,7),(−2,3),(−1,9),
(0,−3),(−1,8)}

Solution

ⓐ D: {−3, −2, −1, 0}
ⓑ R: {7, 3, 9, −3, 8}

In the following exercise, use the mapping of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

The mapping below shows the average weight of a child according to age.

This figure shows two table that each have one column. The table on the left has the header “Age (yrs)” and lists the numbers 1, 2, 3, 4, 5, 6, and 7. The table on the right has the header “Weight (pounds)” and lists the numbers 20, 35, 30, 45, 40, 25, and 50. There are arrows starting at numbers in the age table and pointing towards numbers in the weight table. The first arrow goes from 1 to 20. The second arrow goes from 2 to 25. The third arrow goes from 3 to 30. The fourth arrow goes from 4 to 35. The fifth arrow goes from 5 to 40. The sixth arrow goes from 6 to 45. The seventh arrow goes from 7 to 50.

In the following exercise, use the graph of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

The figure shows the graph of some points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The points (negative 3, 1), (negative 2, negative 1), (negative 2, negative 3), (0, negative 1), (0, 4), and (4, 3).
Solution

ⓐ (4, 3), (−2, −3), (−2, −1), (−3, 1), (0, −1), (0, 4),
ⓑ D: {−3, −2, 0, 4}
ⓒ R: {−3, −1, 1, 3, 4}

Determine if a Relation is a Function

In the following exercises, use the set of ordered pairs to ⓐ determine whether the relation is a function ⓑ find the domain of the relation ⓒ find the range of the relation.

{(9,−5),(4,−3),(1,−1),
(0,0),(1,1),(4,3),(9,5)}

{(−3,27),(−2,8),(−1,1),
(0,0),(1,1),(2,8),(3,27)}

Solution

ⓐ yes ⓑ {−3, −2, −1, 0, 1, 2, 3}
ⓒ {0, 1, 8, 27}

In the following exercises, use the mapping to ⓐ determine whether the relation is a function ⓑ find the domain of the function ⓒ find the range of the function.

This figure shows two table that each have one column. The table on the left has the header “x” and lists the numbers negative 3, negative 2, negative 1, 0, 1, 2, and 3. The table on the right has the header “x to the fourth power” and lists the numbers 0, 1, 16, and 81. There are arrows starting at numbers in the x table and pointing towards numbers in the x to the fourth power table. The first arrow goes from negative 3 to 81. The second arrow goes from negative 2 to 16. The third arrow goes from negative 1 to 1. The fourth arrow goes from 0 to 0. The fifth arrow goes from 1 to 1. The sixth arrow goes from 2 to 16. The seventh arrow goes from 3 to 81.
This figure shows two table that each have one column. The table on the left has the header “x” and lists the numbers negative 3, negative 2, negative 1, 0, 1, 2, and 3. The table on the right has the header “x to the fifth power” and lists the numbers 0, 1, 32, 243, negative 1, negative 32, and negative 243. There are arrows starting at numbers in the x table and pointing towards numbers in the x to the fifth power table. The first arrow goes from negative 3 to negative 243. The second arrow goes from negative 2 to negative 32. The third arrow goes from negative 1 to 1. The fourth arrow goes from 0 to 0. The fifth arrow goes from 1 to 1. The sixth arrow goes from 2 to 32. The seventh arrow goes from 3 to 243.
Solution

ⓐ yes
ⓑ {−3, −2, −1, 0, 1, 2, 3}
ⓒ {−243, −32, −1, 0, 1, 32, 243}

In the following exercises, determine whether each equation is a function.

2x+y=−3

y=x2

Solution

yes

y=3x−5

y=x3

Solution

yes

2x+y2=4

Find the Value of a Function

In the following exercises, evaluate the function:

ⓐ f(−2) ⓑ f(3) ⓒ f(a).

f(x)=3x−4

Solution

ⓐ f(−2)=−10 ⓑ f(3)=5 ⓒ f(a)=3a−4

f(x)=−2x+5

f(x)=x2−5x+6

Solution

ⓐ f(−2)=20 ⓑ f(3)=0 ⓒ f(a)=a2−5a+6

f(x)=3x2−2x+1

In the following exercises, evaluate the function.

g(x)=3x2−5x; g(2)

Solution

2

F(x)=2x2−3x+1;
F(−1)

h(t)=4|t−1|+2; h(−3)

Solution

18

f(x)=x+2x−1; f(3)

Graphs of Functions

Use the Vertical line Test

In the following exercises, determine whether each graph is the graph of a function.

The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (negative 2, 5), (negative 1, 2), (0, 1), (1, 2), and (2, 5). The lowest point on the graph is (0, 1).
Solution

yes

The figure has an s-shaped function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curve goes through the points (negative 1, negative 1), (0, 0), and (1, 1).
The figure has a circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The circle goes through the points (negative 5, 0), (5, 0), (0, negative 5), and (0, 5).
Solution

no

The figure has a parabola opening to the right graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (negative 2, 0), (negative 1, 1), (negative 1, negative 1), (2, 2), and (2, negative 2). The left-most point on the graph is (negative 2, 0).
The figure has a cube function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 1, negative 1), (0, 0), and (1, 1).
Solution

yes

The figure has two curved lines graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line on the left goes through the points (negative 3, 0), (negative 4, 2), and (negative 4, negative 2). The curved line on the right goes through the points (3, 0), (4, 2), and (4, negative 2).
The figure has a sideways absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line bends at the point (0, negative 1) and goes to the right. The line goes through the points (1, 0), (1, negative 2), (2, 1), and (2, negative 3).
Solution

no

Identify Graphs of Basic Functions

In the following exercises, ⓐ graph each function ⓑ state its domain and range. Write the domain and range in interval notation.

f(x)=5x+1

f(x)=−4x−2

Solution

ⓐ
The figure has a linear function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The line goes through the points (negative 2, 6), (negative 1, 2), and (0, negative 2).

ⓑ D: (-∞,∞), R: (-∞,∞)

f(x)=23x−1

f(x)=−6

Solution

ⓐ
The figure has a constant function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 8 to 4. The line goes through the points (0, negative 6), (1, negative 6), and (2, negative 6).

ⓑ D: (-∞,∞), R: (6)

f(x)=2x

f(x)=3x2

Solution

ⓐ
The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (negative 1, 3), (0, 0), and (1, 3). The lowest point on the graph is (0, 0).

ⓑ D: (-∞,∞), R: [0,∞)

f(x)=−12x2

f(x)=x2+2

Solution

ⓐ
The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 4 to 8. The parabola goes through the points (negative 2, 6), (negative 1, 3), (0, 2), (1, 3), and (2, 6). The lowest point on the graph is (0, 2).

ⓑ D: (-∞,∞), R: [2,∞)

f(x)=x3−2

f(x)=x+2

Solution

ⓐ
The figure has a square root function graphed on the x y-coordinate plane. The x-axis runs from negative 4 to 8. The y-axis runs from negative 2 to 10. The half-line starts at the point (negative 2, 0) and goes through the points (negative 1, 1) and (2, 2).

ⓑ D: [−2, ∞), R: [0,∞)

f(x)=−|x|

f(x)=|x|+1

Solution

ⓐ
The figure has an absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The vertex is at the point (0, 1). The line goes through the points (negative 1, 2) and (1, 2).

ⓑ D: (-∞,∞), R: [1,∞)

Read Information from a Graph of a Function

In the following exercises, use the graph of the function to find its domain and range. Write the domain and range in interval notation

The figure has a square root function graphed on the x y-coordinate plane. The x-axis runs from 0 to 10. The y-axis runs from 0 to 10. The half-line starts at the point (1, 0) and goes through the points (2, 1) and (5, 2).
The figure has an absolute value function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The vertex is at the point (0, 2). The line goes through the points (negative 1, 3) and (1, 3).
Solution

D: (-∞,∞), R: [2,∞)

The figure has a cubic function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 2, negative 4), (0, 0), and (2, 4).

In the following exercises, use the graph of the function to find the indicated values.

This figure has a wavy curved line graphed on the x y-coordinate plane. The x-axis runs from negative 2 times pi to 2 times pi. The y-axis runs from negative 6 to 6. The curved line segment goes through the points (negative 2 times pi, 0), (negative 3 divided by 2 times pi, 1), (negative pi, 0), (negative 1 divided by 2 times pi, negative 1), (0, 0), (1 divided by 2 times pi, 1), (pi, 0), (3 divided by 2 times pi, negative 1), and (2 times pi, 0). The points (negative 3 divided by 2 times pi, 1) and (1 divided by 2 times pi, 1) are the highest points on the graph. The points (negative 1 divided by 2 times pi, negative 1) and (3 divided by 2 times pi, negative 1) are the lowest points on the graph. The pattern extends infinitely to the left and right.

ⓐ Find f(0).
ⓑ Find f(12π).
ⓒ Find f(−32π).
ⓓ Find the values for x when f(x)=0.
ⓔ Find the x-intercepts.
ⓕ Find the y-intercepts.
ⓖ Find the domain. Write it in interval notation.
ⓗ Find the range. Write it in interval notation.

Solution

ⓐ f(x)=0 ⓑ fπ2=1
ⓒ f−3π2=1 ⓓ f(x)=0 for x=−2π,−π,0,π,2π
ⓔ (−2π,0), (−π,0), (0,0), (π,0), (2π,0) ⓕ 0,0
ⓖ −∞,∞ ⓗ [−1,1]

The figure has a half-circle graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line segment starts at the point (negative 2, 0). The line goes through the point (0, 2) and ends at the point (2, 0). The point (0, 2) is the highest point on the graph.

ⓐ Find f(0).
ⓑ Find the values for x when f(x)=0.
ⓒ Find the x-intercepts.
ⓓ Find the y-intercepts.
ⓔ Find the domain. Write it in interval notation.
ⓕ Find the range. Write it in interval notation.

Practice Test

Plot each point in a rectangular coordinate system.

ⓐ (2,5)
ⓑ (−1,−3)
ⓒ (0,2)
ⓓ (−4,32)
ⓔ (5,0)

Solution

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 10 to 10. The point labeled a is 2 units to the right of the origin and 5 units above the origin and is located in quadrant I. The point labeled b is 1 unit to the left of the origin and 3 units below the origin and is located in quadrant III. The point labeled c is 2 units above the origin and is located on the y-axis. The point labeled d is 4 units to the left of the origin and 1.5 units above the origin and is located in quadrant II. The point labeled e is 5 units to the right of the origin and is located on the x-axis.

Which of the given ordered pairs are solutions to the equation 3x−y=6?

ⓐ (3,3) ⓑ (2,0) ⓒ (4,−6)

Find the slope of each line shown.

ⓐ
The figure has a straight line graphed on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (negative 5, 2) (0, negative 1), and (5, negative 4).

ⓑ
The figure has a straight vertical line graphed on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (2, 0) (2, negative 1), and (2, 1).

Solution

ⓐ −35 ⓑ undefined

Find the slope of the line between the points (5,2) and (−1,−4).

Graph the line with slope 12 containing the point (−3,−4).

Solution

The figure has a straight line graphed on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (negative 3, negative 4) (negative 1, negative 3), and (1, negative 2).

Find the intercepts of 4x+2y=−8 and graph.

Graph the line for each of the following equations.

y=53x−1

Solution

The figure has a straight line graphed on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (negative 3, negative 6) (0, negative 1), and (3, 4).

y=−x

y=2

Solution

The figure has a straight horizontal line graphed on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (negative 1, 2) (0, 2), and (1, 2).

Find the equation of each line. Write the equation in slope-intercept form.

slope −34 and y-intercept (0,−2)

m=2, point (−3,−1)

Solution

y=2x+5

containing (10,1) and (6,−1)

perpendicular to the line y=54x+2, containing the point (−10,3)

Solution

y=−45x−5

Write the inequality shown by the graph with the boundary line y=−x−3.

The figure has a straight line graphed on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (negative 3, 0), (0, negative 3), and (1, negative 4). The line divides the coordinate plane into two halves. The bottom left half and the line are colored red to indicate that this is the solution set.

Graph each linear inequality.

y>32x+5

Solution

The figure has a straight dashed line graphed on the x y-coordinate plane. The x-axis runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The line goes through the points (negative 2, 2), (0, 5), and (2, 8). The line divides the coordinate plane into two halves. The top left half is colored red to indicate that this is the solution set.

x−y≥−4

y≤−5x

Solution

The figure has a straight line graphed on the x y-coordinate plane. The x-axis runs from negative 8 to 8. The y-axis runs from negative 8 to 8. The line goes through the points (negative 1, 5), (0, 0), and (1, negative 5). The line divides the coordinate plane into two halves. The bottom left half and the line are colored red to indicate that this is the solution set.

Hiro works two part time jobs in order to earn enough money to meet her obligations of at least $450 a week. Her job at the mall pays $10 an hour and her administrative assistant job on campus pays $15 an hour. How many hours does Hiro need to work at each job to earn at least $450?

ⓐ Let x be the number of hours she works at the mall and let y be the number of hours she works as administrative assistant. Write an inequality that would model this situation.
ⓑ Graph the inequality .
ⓒ Find three ordered pairs(x,y) that would be solutions to the inequality. Then explain what that means for Hiro.

Use the set of ordered pairs to ⓐ determine whether the relation is a function, ⓑ find the domain of the relation, and ⓒ find the range of the relation.

{(−3,27),(−2,8),(−1,1),(0,0),
(1,1),(2,8),(3,27)}

Solution

ⓐ yes ⓑ {−3,−2,−1,0,1,2,3} ⓒ {0, 1, 8, 27}

Evaluate the function: ⓐ f(−1) ⓑ f(2) ⓒ f(c).

f(x)=4x2−2x−3

For h(y)=3|y−1|−3, evaluate h(−4).

Solution

12

Determine whether the graph is the graph of a function. Explain your answer.

The figure has a cube function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The curved line goes through the points (negative 1, 1), (0, 2), and (1, 3).

In the following exercises, ⓐ graph each function ⓑ state its domain and range.
Write the domain and range in interval notation.

f(x)=x2+1

Solution

ⓐ
The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 2 to 10. The parabola goes through the points (negative 2, 5), (negative 1, 2), (0, 1), (1, 2), and (2, 5). The lowest point on the graph is (0, 1).

ⓑ D: (-∞,∞), R: [1,∞)

f(x)=x+1

The figure has a square function graphed on the x y-coordinate plane. The x-axis runs from negative 6 to 6. The y-axis runs from negative 6 to 6. The parabola goes through the points (negative 2, 0), (negative 1, negative 3), (0, negative 4), (1, negative 3), and (2, 0). The lowest point on the graph is (0, negative 4).

ⓐ Find the x-intercepts.
ⓑ Find the y-intercepts.
ⓒ Find f(−1).
ⓓ Find f(1).
ⓔ Find the domain. Write it in interval notation.
ⓕ Find the range. Write it in interval notation.

Solution

ⓐ x=−2,2 ⓑ y=−4
ⓒ f(−1)=−3 ⓓ f(1)=−3
ⓔ D: (-∞,∞) ⓕ R: [−4, ∞)

Introduction

A close up photo of a bright orange Lamborghini Aventador.
In the future, car drivers may become passengers because cars will be able to drive themselves. (credit: jingoba/Pixabay)

Climb into your car. Put on your seatbelt. Choose your destination and then…relax. That’s right. You don’t have to do anything else because you are in an autonomous car, or one that navigates its way to your destination! No cars are fully autonomous at the moment and so you theoretically still need to have your hands on the wheel. Self-driving cars may help ease traffic congestion, prevent accidents, and lower pollution. The technology is thanks to computer programmers who are developing software to control the navigation of the car. These programmers rely on their understanding of mathematics, including relationships between equations. In this chapter, you will learn how to solve systems of linear equations in different ways and use them to analyze real-world situations.

Solve Systems of Linear Equations with Two Variables

Learning Objectives

By the end of this section, you will be able to:

  • Determine whether an ordered pair is a solution of a system of equations
  • Solve a system of linear equations by graphing
  • Solve a system of equations by substitution
  • Solve a system of equations by elimination
  • Choose the most convenient method to solve a system of linear equations

Before you get started, take this readiness quiz.

For the equation y=23x−4,
ⓐ Is (6,0) a solution? ⓑ Is (−3,−2) a solution?
If you missed this problem, review Example 2 in Graph Linear Equations in Two Variables.

Solution

ⓐ yes; ⓑ no

Find the slope and y-intercept of the line 3x−y=12.
If you missed this problem, review Example 5 in Slope of a Line.

Solution

m=3;b=−12

Find the x- and y-intercepts of the line 2x−3y=12.
If you missed this problem, review Example 8 in Graph Linear Equations in Two Variables.

Solution

6,0,0,−4

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations, we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations.

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations.

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

{2x+y=7x−2y=6

A linear equation in two variables, such as 2x+y=7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs (x,y) that make both equations true. These are called the solutions of a system of equations.

Solutions of a System of Equations

The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x,y).

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Determine whether the ordered pair is a solution to the system {x−y=−12x−y=−5.

ⓐ (−2,−1) ⓑ (−4,−3)

Solution

ⓐ
The equations are x minus y equals minus 1 and 2 x minus y equals minus 5. We substitute x equal to minus 2 and y equal to minus 1 into both equations. So, x minus y equals minus 1 becomes minus 2 minus open parentheses minus 1 close parentheses equal to or not equal to minus 1. Simplifying, we get minus 1 equals minus 1 which is correct. The equation 2 x minus y equals minus 5 becomes 2 times minus 2 minus open parentheses minus 1 close parentheses equal to or not equal to minus 5. Simplifying, we get minus 3 not equal to minus 5. Hence, the ordered pair minus 2, minus 1 does not make both equations true. So, it is not a solution.

ⓑ
We substitute x equal to minus 4 and y equal to minus 3 into both equations. So, x minus y equals minus 1 becomes minus 4 minus open parentheses minus 3 close parentheses equal to or not equal to minus 1. Simplifying, we get minus 1 equals minus 1, which is correct. The equation 2 x minus y equals minus 5 becomes 2 times minus 4 minus open parentheses minus 3 close parentheses equal to or not equal to minus 5. Simplifying, we get minus 5 equals minus 5, which is correct. The ordered pair minus 4, minus 3 does make both equations true. Hence, it is a solution.

Determine whether the ordered pair is a solution to the system {3x+y=0x+2y=−5.

ⓐ (1,−3) ⓑ (0,0)

Solution

ⓐ yes ⓑ no

Determine whether the ordered pair is a solution to the system {x−3y=−8−3x−y=4.

ⓐ (2,−2) ⓑ (−2,2)

Solution

ⓐ no ⓑ yes

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Figure shows three graphs. In the first, the lines intersect at point 3, minus 1. The intersecting lines have one point in common. There is one solution to the system. In the second graph, the lines are parallel. Parallel lines have no points in common. There is no solution to the system. The third graph has only one line. Here, both equations give the same line. Because we have only one line, there are infinite many solutions.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

How to Solve a System of Equations by Graphing

Solve the system by graphing {2x+y=7x−2y=6.

Solution
Step 1 is to graph the first equation. To graph the first line, write the equation in slope intercept form. So, 2 x plus y equals 7 becomes y equal to minus 2 x plus 7. Here, m is minus 2 and b is 7. So the graph will be a line with slope equal to minus 2 and y intercept equal to 7. Step 2 is to graph the second equation on the same rectangular coordinate system. To graph the second line, use intercepts. For x minus 2y equals 6, the intercepts are 0, minus 3 and 6, 0. Step 3 is to determine whether the lines intersect, are parallel, or are the same line. Here, they intersect. Step 4 is to identify the solution to the system. If the lines intersect, identify the point of intersection. The lines intersect at 4, minus 1. Now, check to make sure it is a solution to both equations. When x and y are substituted with 4 and minus 1 respectively, both equations hold true. This is the solution to the system. In step 4, if the lines are parallel, the system has no solution and if the lines are the same, the system has an infinite number of solutions.

Solve the system by graphing: {x−3y=−3x+y=5.

Solution

(3,2)

Solve the system by graphing: {−x+y=13x+2y=12.

Solution

(2,3)

The steps to use to solve a system of linear equations by graphing are shown here.

Solve a system of linear equations by graphing.

  1. Graph the first equation.
  2. Graph the second equation on the same rectangular coordinate system.
  3. Determine whether the lines intersect, are parallel, or are the same line.
  4. Identify the solution to the system.
    • If the lines intersect, identify the point of intersection. This is the solution to the system.
    • If the lines are parallel, the system has no solution.
    • If the lines are the same, the system has an infinite number of solutions.
  5. Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Solve the system by graphing: {3x+y=−12x+y=0.

Solution

We’ll solve both of these equations for y so that we can easily graph them using their slopes and y-intercepts.

A system of two linear equations is displayed, featuring 3x + y = -1 and 2x + y = 0, indicating a mathematical problem for solving x and y.
Solve the first equation for y. Two lines of algebraic equations are displayed. The first is 3x + y = -1, and the second is its rearranged form solving for y, shown as y = -3x - 1.
Find the slope and y-intercept. A white background displays two lines of mathematical notation in black text. The top line reads 'm = -3', and the bottom line reads 'b = -1'.
Solve the second equation for y. The image displays the algebraic steps to rearrange the linear equation 2x + y = 0 into its slope-intercept form, resulting in y = -2x.
Find the slope and y-intercept. The image displays mathematical equations showing m = -2 and b = 0, likely representing the slope and y-intercept in a linear equation.
Graph the lines. A Cartesian coordinate system with two lines, blue and red, intersecting at the point (-1, 2). The axes range from -7 to 7.
Determine the point of intersection. The lines intersect at (−1,2).
Check the solution in both equations.

The image shows the verification of the solution (-1, 2) for two linear equations: 3x + y = -1 and 2x + y = 0. The substitution confirms the equalities for both equations.
The solution is (−1,2).

Solve the system by graphing: {−x+y=12x+y=10.

Solution

(3,4)

Solve the system by graphing: {2x+y=6x+y=1.

Solution

(5,−4)

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Solve the system by graphing: {y=12x−3x−2y=4.

Solution
A system of two linear equations is presented, with the first equation being y = (1/2)x - 3 and the second equation being x - 2y = 4, enclosed by a blue brace on the left.
To graph the first equation, we will use its
slope and y-intercept.
The image displays the linear equation y = (1/2)x - 3, with its slope m = 1/2 and y-intercept b = -3 explicitly stated below.
To graph the second equation, we will use
the intercepts.
A light blue mathematical equation 'x - 2y = 4' is displayed on a plain white background, appearing as if a student has written it out on a blank sheet or digital canvas.
A table showing two (x,y) coordinate pairs: (0, -2) and (4, 0).
Graph the lines. A graph displaying two parallel lines in a Cartesian coordinate system. The blue line has a y-intercept of -2, and the red line has a y-intercept of -4.
Determine the points of intersection. The lines are parallel.
Since no point is on both lines, there is no
ordered pair that makes both equations
true. There is no solution to this system.

Solve the system by graphing: {y=−14x+2x+4y=−8.

Solution

no solution

Solve the system by graphing: {y=3x−16x−2y=6.

Solution

no solution

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

Solve the system by graphing: {y=2x−3−6x+3y=−9.

Solution
A system of two linear equations is shown, enclosed by a blue brace on the left. The first equation is y = 2x - 3, and the second equation is -6x + 3y = -9.
Find the slope and y-intercept of the first equation. Mathematical equations illustrating the slope-intercept form y = 2x - 3, where m (slope) equals 2 and b (y-intercept) equals -3, detailing key components of a linear function.
Find the intercepts of the second equation. A mathematical equation is displayed on a white background, which reads '-6x + 3y = -9'.
A table displays two columns labeled 'x' and 'y'. The first row shows x=0 and y=-3. The second row shows x=3/2 and y=0.
Graph the lines. A graph showing a straight line with a positive slope, passing through the points (0, -3) and (3, 3) on a Cartesian coordinate plane.
The lines are the same!
Since every point on the line makes both
equations true, there are infinitely many
ordered pairs that make both equations true.
There are infinitely many solutions to this system.

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y-intercept.

Solve the system by graphing: { y=−3x−66x+2y=−12.

Solution

infinitely many solutions

Solve the system by graphing: {y=12x−42x−4y=16.

Solution

infinitely many solutions

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident. Coincident lines have the same slope and same y-intercept.

Coincident Lines

Coincident lines have the same slope and same y-intercept.

The systems of equations in Example 2 and Example 3 each had two intersecting lines. Each system had one solution.

In Example 5, the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like Example 4, has no solution. We call a system of equations like this inconsistent. It has no solution.

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.

The figure shows three graphs. The first one has two intersecting line. The second one has two parallel lines. The third one has only one line. This is labeled coincident.
Lines Intersecting Parallel Coincident
Number of solutions 1 point No solution Infinitely many
Consistent/inconsistent Consistent Inconsistent Consistent
Dependent/ independent Independent Independent Dependent

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ {y=3x−16x−2y=12 ⓑ {2x+y=−3x−5y=5

Solution

ⓐ We will compare the slopes and intercepts of the two lines.

Analyzing a system of linear equations to determine if lines are parallel by comparing their slope-intercept forms.
The first equation is already in slope-intercept form. {y=3x−16x−2y=12y=3x−1
Write the second equation in slope-intercept form. 6x−2y=12−2y=−6x+12−2y−2=−6x+12−2y=3x−6
Find the slope and intercept of each line. y=3x−1y=3x−6m=3m=3b=−1b=−6
Since the slopes are the same and y-intercepts are different, the lines are parallel.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

This table demonstrates the step-by-step process of solving a system of linear equations by converting to slope-intercept form and analyzing their slopes.
{2x+y=−3x−5y=5
Write both equations in slope–intercept form. 2x+y=−3x−5y=5y=−2x−3−5y=−x+5−5y−5=−x+5−5y=15x−1
Find the slope and intercept of each line. y=−2x−3y=15x−1m=−2m=15b=−3b=−1
Since the slopes are different, the lines intersect.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ {y=−2x−44x+2y=9 ⓑ {3x+2y=22x+y=1

Solution

ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ {y=13x−5x−3y=6 ⓑ {x+4y=12−x+y=3

Solution

ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

{2x+y=7x−2y=6

We will first solve one of the equations for either x or y. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

How to Solve a System of Equations by Substitution

Solve the system by substitution: {2x+y=7x−2y=6.

Solution
The equations are 2 x plus y equals 7 and x minus 2y equals 6. Step 1 is to solve one of the equations for either variable. We’ll solve the first equation for y. We get y equals 7 minus 2 x. In step 2, substitute the expression from step 1 into the other equation. We replace y in the second equation with the expression 7 minus 2 x. So, we get x minus 2 open parentheses 7 minus 2 x close parentheses equals 6. Step 3 is to solve the resulting equation. Now we have an equation with just 1 variable. We solve it to get x equal to 4. Step 4 is to substitute the solution in step 3 into one of the original equations to find the other variable. We’ll use the first equation and replace x with 4. We get, 2 times 4 plus y equals 7. Simplifying, we get y equal to minus 1. Step 5 is to write the solution as an ordered pair. The ordered pair is 4, minus 1. Step 6 is to check that the ordered pair is a solution to both original equations. To do that we Substitute x equal to 4 and y equal to minus 1 into both equations and make sure they are both true.

Solve the system by substitution: {−2x+y=−11x+3y=9.

Solution

(6,1)

Solve the system by substitution: {2x+y=−14x+3y=3.

Solution

(−3,5)

Solve a system of equations by substitution.

  1. Solve one of the equations for either variable.
  2. Substitute the expression from Step 1 into the other equation.
  3. Solve the resulting equation.
  4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
  5. Write the solution as an ordered pair.
  6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

Solve the system by substitution: {4x+2y=46x−y=8.

Solution

We need to solve one equation for one variable. We will solve the first equation for y.

A system of two linear equations is presented. The first equation is 4x + 2y = 4, and the second equation is 6x - y = 8. A left curly brace encloses both equations, indicating they form a system.
Solve the first equation for y.
Substitute −2x+2 for y in the second equation.
Steps to solve a system of linear equations by substitution, showing how the first equation 4x + 2y = 4 is rearranged to y = -2x + 2, which will then be substituted into the second equation 6x - y = 8.
Replace the y with −2x+2. A mathematical equation is displayed: 6x - (-2x + 2) = 8. The terms inside the parenthesis, -2x + 2, are highlighted in red, indicating a focus on that part of the expression.
Solve the equation for x. A step-by-step solution for the algebraic equation 6x + 2x - 2 = 8, simplifying to 8x - 2 = 8, then 8x = 10, and finally revealing the value of x as 5/4, highlighted by a red oval.
Substitute x=54 into 4x+2y=4 to find y. Step-by-step solution to find 'y' in the equation 4x + 2y = 4 after substituting x with 5/4, resulting in y = -1/2.
The ordered pair is (54,−12).
Check the ordered pair in both equations.

This image illustrates the verification process for a solution to a system of two linear equations. It shows x=5/4 and y=-1/2 being substituted into 4x + 2y = 4 and 6x - y = 8, confirming both equations hold true.
The solution is (54,−12).

Solve the system by substitution: {x−4y=−4−3x+4y=0.

Solution

(2,32)

Solve the system by substitution: {4x−y=02x−3y=5.

Solution

(−12,−2)

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d.

ifa=bandc=dthena+c=b+d.

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

{3x+y=52x−y=0—————5x=5

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

{x+4y=22x+5y=−2

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

Minus 2 open parentheses x plus 4y close parentheses is minus 2 times 2. And, 2 x plus 5y is minus 2.

Then rewrite the system of equations.

Minus 2 x minus 8y is minus 4 and 2 x plus 5y is minus 2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Minus 2 x minus 8y is minus 4 and 2 x plus 5y is minus 2. Adding these, we get minus 3y equals minus 6.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

How to Solve a System of Equations by Elimination

Solve the system by elimination: {2x+y=7x−2y=6.

Solution
The equations are 2 x plus y equals 7 and x minus 2y equals 6. Step 1 is to write both equations in standard form. Both equations are in standard form, Ax plus By equals C. If any coefficients are fractions, clear them. There are no fractions. Step 2 is to make the coefficients of one variable opposites. First decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites. We can eliminate the y’s by multiplying the first equation by 2. We get 4x plus 2y equals 14. Step 3 is to add the equations resulting from step 2 to eliminate one variable. Adding, we get 5x equals 20. Step 4 is to solve for the remaining variable. Solving for x, we get x equals 4. Step 5 is to substitute the solution from step 4 into one of the original equations. Then solve for the other variable. Substituting x equal to 4 into the second equation, we get 4 minus 2y equals 6. Solving for y, we get y equal to minus 1. Step 6 is to write the solution as an ordered pair. Here, the ordered pair is 4, minus 1. Step 7 is to check that the ordered pair is a solution to both original equations. The ordered pair makes both original equations true.

Solve the system by elimination: {3x+y=52x−3y=7.

Solution

(2,−1)

Solve the system by elimination: { 4x+y=−5−2x−2y=−2.

Solution

(−2,3)

The steps are listed here for easy reference.

Solve a system of equations by elimination.

  1. Write both equations in standard form. If any coefficients are fractions, clear them.
  2. Make the coefficients of one variable opposites.
    • Decide which variable you will eliminate.
    • Multiply one or both equations so that the coefficients of that variable are opposites.
  3. Add the equations resulting from Step 2 to eliminate one variable.
  4. Solve for the remaining variable.
  5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
  6. Write the solution as an ordered pair.
  7. Check that the ordered pair is a solution to both original equations.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Solve the system by elimination: {4x−3y=97x+2y=−6.

Solution

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

A system of two linear equations is shown with a brace on the left side. The first equation is 4x - 3y = 9, and the second equation is 7x + 2y = -6.
Both equations are in standard form.
To get opposite coefficients of y, we will
multiply the first equation by 2 and the
second equation by 3.
A system of linear equations being prepared for solution using the elimination method. The top equation is multiplied by 2, and the bottom equation is multiplied by 3 on both sides.
Simplify. A system of two linear equations is displayed. The first equation is 8x - 6y = 18, and the second equation is 21x + 6y = -18. A left brace encloses both equations.
Add the two equations to eliminate y. A system of linear equations is solved using the elimination method, where 8x - 6y = 18 and 21x + 6y = -18 are added to eliminate the 'y' term, resulting in 29x = 0.
Solve for x. A red arrow points from 'x = 0' to the '7x' term in the equation '7x + 2y = -6', indicating the substitution of x = 0 into the equation.
Substitute x=0 into one of the original equations. A mathematical equation reads '7 multiplied by 0 plus 2y equals -6' on a white background. The number 0 is highlighted in red.
Solve for y. A mathematical solution showing '2y = -6' simplified to 'y = -3' by dividing both sides by 2.
Write the solution as an ordered pair. The ordered pair is (0,−3).
Check that the ordered pair is a solution to
both original equations.

This image demonstrates checking if the point (0, -3) satisfies two linear equations. Substitutions show 4(0)-3(-3)=9 and 7(0)+2(-3)=-6 are both true, confirming it's a solution to the system.
The solution is (0,−3).

Solve the system by elimination: { 3x−4y=−95x+3y=14.

Solution

(1,3)

Solve each system by elimination: { 7x+8y=43x−5y=27.

Solution

(4,−3)

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Solve the system by elimination: {x+12y=632x+23y=172.

Solution

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

A system of two linear equations is shown, enclosed by a blue brace. The first equation is x + (1/2)y = 6. The second equation is (3/2)x + (2/3)y = 17/2.
To clear the fractions, multiply each
equation by its LCD.
Two linear equations being scaled. The first equation is multiplied by 2, and the second by 6, to eliminate fractions and simplify the system before solving.
Simplify. A system of two linear equations is shown, with the first equation being 2x + y = 12 and the second equation being 9x + 4y = 51, enclosed by a left curly brace.
Now we are ready to eliminate one
of the variables. Notice that both equations are in
standard form.
We can eliminate y by multiplying the top equation by −4. A system of two linear equations is shown, with the top equation being prepared for solving by multiplication. The equations are -4(2x + y) = -4(12) and 9x + 4y = 51.
Simplify and add.

Substitute x=3 into one of the original equations.
Solving a system of linear equations by elimination. The equations -8x - 4y = -48 and 9x + 4y = 51 are added to eliminate 'y', resulting in x=3. This value is then applied to the equation x + 1/2y = 6.
Solve for y. A mathematical equation is displayed against a white background, showing '3 + 1/2y = 6'. The number '3' is highlighted in red.
A mathematical equation shows '1/2y = 3'.
The image displays the mathematical equation 'y = 6' in black text on a plain white background, showing a simple algebraic expression.
Write the solution as an ordered pair. The ordered pair is (3,6).
Check that the ordered pair is a solution to
both original equations.

Verification of the solution (x=3, y=6) for two linear equations, demonstrating how to check if a given point satisfies each equation through substitution and simplification.
The solution is (3,6).

Solve each system by elimination: { 13x−12y=134x−y=52 .

Solution

(6,2)

Solve each system by elimination: { x+35y=−15 −12x−23y=56 .

Solution

(1,−2)

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Solve the system by elimination: {3x+4y=12y=3−34x.

Solution
Step-by-step solution of a system of linear equations, showing the algebraic manipulations that lead to an identity, indicating infinitely many solutions.
{3x+4y=12y=3−34x
Write the second equation in standard form. {3x+4y=1234x+y=3
Clear the fractions by multiplying the second equation by 4. {3x+4y=124(34x+y)=4(3)
Simplify. {3x+4y=123x+4y=12
To eliminate a variable, we multiply the second equation by −1. Simplify and add. {3x+4y=12−3x−4y=−12______________0=0

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Solve the system by elimination: { 5x−3y=15y=−5+53x.

Solution

infinitely many solutions

Solve the system by elimination: { x+2y=6y=−12x+3.

Solution

infinitely many solutions

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Choose the Most Convenient Method to Solve a System of Linear EquationsGraphing————Substitution—————Elimination—————Use when you need aUse when one equation isUse when the equations are picture of the situation.already solved or can bein standard form. easily solved for onevariable.

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ {3x+8y=407x−4y=−32 ⓑ {5x+6y=12y=23x−1

Solution

ⓐ

{3x+8y=407x−4y=−32

Since both equations are in standard form, using elimination will be most convenient.

ⓑ

{5x+6y=12y=23x−1

Since one equation is already solved for y, using substitution will be most convenient.

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ {4x−5y=−323x+2y=−1 ⓑ {x=2y−13x−5y=−7

Solution

ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for x, using substitution will be most convenient.

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ {y=2x−13x−4y=−6 ⓑ {6x−2y=123x+7y=−13

Solution

ⓐ Since one equation is already solved for y, using substitution will be most convenient. ⓑ Since both equations are in standard form, using elimination will be most convenient.

Key Concepts

  • How to solve a system of linear equations by graphing.
    1. Graph the first equation.
    2. Graph the second equation on the same rectangular coordinate system.
    3. Determine whether the lines intersect, are parallel, or are the same line.
    4. Identify the solution to the system.
      If the lines intersect, identify the point of intersection. This is the solution to the system.
      If the lines are parallel, the system has no solution.
      If the lines are the same, the system has an infinite number of solutions.
    5. Check the solution in both equations.
  • How to solve a system of equations by substitution.
    1. Solve one of the equations for either variable.
    2. Substitute the expression from Step 1 into the other equation.
    3. Solve the resulting equation.
    4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
    5. Write the solution as an ordered pair.
    6. Check that the ordered pair is a solution to both original equations.
  • How to solve a system of equations by elimination.
    1. Write both equations in standard form. If any coefficients are fractions, clear them.
    2. Make the coefficients of one variable opposites.
      Decide which variable you will eliminate.
      Multiply one or both equations so that the coefficients of that variable are opposites.
    3. Add the equations resulting from Step 2 to eliminate one variable.
    4. Solve for the remaining variable.
    5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
    6. Write the solution as an ordered pair.
    7. Check that the ordered pair is a solution to both original equations.
      Choose the Most Convenient Method to Solve a System of Linear EquationsGraphing—————Substitution———————Elimination———————Use when you need apicture of the situation.Use when one equation isalready solved or can beeasily solved for onevariable.Use when the equations arein standard form.

Practice Makes Perfect

Determine Whether an Ordered Pair is a Solution of a System of Equations

In the following exercises, determine if the following points are solutions to the given system of equations.

{2x−6y=03x−4y=5

ⓐ (3,1) ⓑ (−3,4)

Solution

ⓐ yes ⓑ no

{−3x+y=8−x+2y=−9

ⓐ (−5,−7) ⓑ (−5,7)

{x+y=2y=34x

ⓐ (87,67) ⓑ (1,34)

Solution

ⓐ yes ⓑ no

{2x+3y=6y=23x+2
ⓐ (−6,2) ⓑ (−3,4)

Solve a System of Linear Equations by Graphing

In the following exercises, solve the following systems of equations by graphing.

{3x+y=−32x+3y=5

Solution

(−2,3)

{−x+y=22x+y=−4

{y=x+2y=−2x+2

Solution

(0,2)

{y=x−2y=−3x+2

{y=32x+1y=−12x+5

Solution

(2,4)

{y=23x−2y=−13x−5

{x+y=−4−x+2y=−2

Solution

(−2,−2)

{−x+3y=3x+3y=3

{−2x+3y=3x+3y=12

Solution

(3,3)

{2x−y=42x+3y=12

{x+3y=−6y=−43x+4

Solution

(6,−4)

{−x+2y=−6y=−12x−1

{−2x+4y=4y=12x

Solution

no solution

{3x+5y=10y=−35x+1

{4x−3y=88x−6y=14

Solution

no solution

{x+3y=4−2x−6y=3

{x=−3y+42x+6y=8

Solution

infinite solutions

{4x=3y+78x−6y=14

{2x+y=6−8x−4y=−24

Solution

infinite solutions

{5x+2y=7−10x−4y=−14

Without graphing, determine the number of solutions and then classify the system of equations.

{y=23x+1−2x+3y=5

Solution

No solution, Inconsistent, Independent

{y=32x+12x−3y=7

{5x+3y=42x−3y=5

Solution

1 point, consistent and independent

{y=−12x+5x+2y=10

{5x−2y=10y=52x−5

Solution

infinite solutions, consistent, dependent

Solve a System of Equations by Substitution

In the following exercises, solve the systems of equations by substitution.

{2x+y=−43x−2y=−6

{2x+y=−23x−y=7

Solution

(1,−4)

{x−2y=−52x−3y=−4

{x−3y=−92x+5y=4

Solution

(−3,2)

{5x−2y=−6y=3x+3

{−2x+2y=6y=−3x+1

Solution

(−12,52)

{2x+5y=1y=13x−2

{3x+4y=1y=−25x+2

Solution

(−5,4)

{2x+y=5x−2y=−15

{4x+y=10x−2y=−20

Solution

(0,10)

{y=−2x−1y=−13x+4

{y=x−6y=−32x+4

Solution

(4,−2)

{x=2y4x−8y=0

{2x−16y=8−x−8y=−4

Solution

(4,0)

{y=78x+4−7x+8y=6

{y=−23x+52x+3y=11

Solution

no solution

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

{5x+2y=2−3x−y=0

{6x−5y=−12x+y=13

Solution

(4,5)

{2x−5y=73x−y=17

{5x−3y=−12x−y=2

Solution

(7,12)

{3x−5y=−95x+2y=16

{4x−3y=32x+5y=−31

Solution

(−3,−5)

{3x+8y=−32x+5y=−3

{11x+9y=−57x+5y=−1

Solution

(2,−3)

{3x+8y=675x+3y=60

{2x+9y=−43x+13y=−7

Solution

(−11,2)

{13x−y=−3x+52y=2

{x+12y=3215x−15y=3

Solution

(6,−9)

{x+13y=−113x+12y=1

{13x−y=−323x+52y=3

Solution

(−3,2)

{2x+y=36x+3y=9

{x−4y=−1−3x+12y=3

Solution

infinitely many

{−3x−y=86x+2y=−16

{4x+3y=220x+15y=10

Solution

infinitely many

Choose the Most Convenient Method to Solve a System of Linear Equations

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.


ⓐ {8x−15y=−326x+3y=−5
ⓑ {x=4y−34x−2y=−6


ⓐ {y=7x−53x−2y=16
ⓑ {12x−5y=−423x+7y=−15

Solution

ⓐ substitution ⓑ elimination


ⓐ {y=4x+95x−2y=−21
ⓑ {9x−4y=243x+5y=−14


ⓐ {14x−15y=−307x+2y=10
ⓑ {x=9y−112x−7y=−27

Solution

ⓐ elimination ⓑ substituion

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: {3x+y=12x=y−8.

Solution

Answers will vary.

Solve the system of equations by elimination and explain all your steps in words: {5x+4y=102x=3y+27.

Solve the system of equations {x+y=10x−y=6

ⓐ by graphing ⓑ by substitution
ⓒ Which method do you prefer? Why?

Solution

Answers will vary.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns 5 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: determine whether an ordered pair is a solution of a system of equations, solve a system of linear equations by graphing, solve a system of equations by substitution, solve a system of equations by elimination, choose the most convenient method to solve a system of linear equations. The remaining columns are blank.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

coincident lines
Coincident lines have the same slope and same y-intercept.
consistent and inconsistent systems
Consistent system of equations is a system of equations with at least one solution; inconsistent system of equations is a system of equations with no solution.
solutions of a system of equations
Solutions of a system of equations are the values of the variables that make all the equations true; solution is represented by an ordered pair (x,y).
system of linear equations
When two or more linear equations are grouped together, they form a system of linear equations.

Solve Applications with Systems of Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve direct translation applications
  • Solve geometry applications
  • Solve uniform motion applications

Before you get started, take this readiness quiz.

The sum of twice a number and nine is 31. Find the number.
If you missed this problem, review Example 2 in Use a Problem Solving Strategy.

Solution

11

Twins Jon and Ron together earned $96,000 last year. Ron earned $8000 more than three times what Jon earned. How much did each of the twins earn?
If you missed this problem, review Example 6 in Use a Problem Solving Strategy.

Solution

Jon earned $22,000 and Ron earned $74,000.

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.
If you missed this problem, review Example 5 in Solve Mixture and Uniform Motion Applications.

Solution

The speed of the local train is 48 mph and the speed of the express train is 60 mph.

Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

Solve applications with systems of equations.

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are looking for.
  3. Name what we are looking for. Choose variables to represent those quantities.
  4. Translate into a system of equations.
  5. Solve the system of equations using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for two numbers.
Step 3. Name what we are looking for. Let n= the first number.
m= the second number
Step 4. Translate into a system of equations. The sum of two numbers is zero.
A clear, concise image displays the algebraic equation 'n + m = 0' written in black characters on a plain white background, presenting a fundamental mathematical concept.
One number is nine less than the other.
A mathematical equation is displayed on a white background, showing 'n = m - 9' in black text, representing a relationship between three variables.
The system is: A system of two linear equations is presented, enclosed by a large left curly brace. The first equation is 'n + m = 0', and the second equation is 'n = m - 9'.
Step 5. Solve the system of
equations. We will use substitution
since the second equation is solved
for n.
Substitute m − 9 for n in the first equation. Two mathematical equations are displayed: n = m - 9 is circled, and an arrow points from it to the second equation, n + m = 0.
Solve for m. A mathematical equation, m - 9 + m = 0, is displayed on a white background. The 'm' and '-9' are in a reddish hue, while the second 'm' and '= 0' are in gray.
A mathematical equation is displayed, reading '2m - 9 = 0'. The text is rendered in a dark gray font against a clean white background.
A simple algebraic equation is displayed, showing '2m = 9' in bold, grey text against a plain white background. This equation involves a variable 'm' being multiplied by 2 and equaling 9.
Substitute m=92 into the second equation
and then solve for n.
A mathematical diagram displays two equations: m = 9/2 and n = m - 9. The value 9/2 for 'm' is circled in red, with an arrow indicating its substitution into the second equation to solve for 'n'.
The image displays two steps of an algebraic equation. The first line is m = 9/2 - 9. The second line converts the integer 9 into a fraction with a common denominator, showing m = 9/2 - 18/2.
A mathematical equation is displayed on a white background, showing 'n = -9/2'.
Step 6. Check the answer in the problem. Do these numbers make sense in
the problem? We will leave this to
you!
Step 7. Answer the question. The numbers are 92 and −92.

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Solution

3, 7

The sum of two numbers is −6. One number is 10 less than the other. Find the numbers.

Solution

2, −8

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000+$40 for each training session. How many training sessions would make the salary options equal?

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for the number of
training sessions that would make
the pay equal.
Step 3. Name what we are looking for. Let s= Heather’s salary.
n= the number of training sessions
Step 4. Translate into a system of equations. Option A would pay her $25,000
plus $15 for each training
session.
A mathematical equation is displayed on a white background: s = 25,000 + 15n.
Option B would pay her $10,000
+ $40 for each training session.
A mathematical equation is displayed against a white background, reading 's = 10,000 + 40n'. The equation uses a sans-serif font and appears to be a formula for calculating a value 's' based on a base number 10,000 and a variable 'n' multiplied by 40.
The system is shown. A system of two linear equations is displayed. The first equation is s = 25,000 + 15n, and the second equation is s = 10,000 + 40n, where 's' and 'n' are variables.
Step 5. Solve the system of equations.
We will use substitution.
Two linear equations are shown: s = 25,000 + 15n (circled in red) and s = 10,000 + 40n. An arrow points from the first to the second, possibly indicating a transformation or choice between the two.
Substitute 25,000 +15n for s in the second
equation.
A mathematical equation reads '25,000 + 15n = 10,000 + 40n' with the numbers 25,000 and 15n in red, and the rest of the equation in gray.
Solve for n. A step-by-step algebraic solution for 'n' is presented. The initial equation, 25,000 = 10,000 + 25n, is simplified to 15,000 = 25n, and finally solved to show that n = 600.
Step 6. Check the answer. Are 600 training sessions a year reasonable?
Are the two options equal when n = 600?
Step 7. Answer the question. The salary options would be equal for 600 training
sessions.

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

Solution

160 policies

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Solution

1000 suits

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for the number of
calories burned each minute on the
elliptical trainer and each minute of
circuit training.
Step 3. Name what we are looking for. Let e= number of calories burned per
minute on the elliptical trainer.
c= number of calories burned per
minute while circuit training
Step 4. Translate into a system of equations. 10 minutes on the elliptical and circuit
training for 20 minutes, burned
278 calories
A mathematical equation is displayed on a white background, which reads '10e + 20c = 278' in a dark gray sans-serif font.
20 minutes on the elliptical and
30 minutes of circuit training burned
473 calories
A mathematical equation is displayed on a white background, reading 20e + 30c = 473.
The system is: A system of two linear equations is displayed. The first equation is 10e + 20c = 278, and the second equation is 20e + 30c = 473. A curly brace groups the two equations together.
Step 5. Solve the system of equations.
Multiply the first equation by −2 to get
opposite coefficients of e.
The image shows a system of two linear equations. The first equation is given as -2(10e + 20c) = -2(278), indicating that the original equation has been multiplied by -2 on both sides. The second equation in the system is 20e + 30c = 473.
Simplify and add the equations.
Solve for c.
A system of two linear equations, -20e - 40c = -556 and 20e + 30c = 473, is solved using the elimination method, resulting in -10c = -83, which simplifies to c = 8.3.
Substitute c = 8.3 into one of the
original equations to solve for e.
A step-by-step solution to an algebraic equation, starting with 10e + 20c = 278, substituting c with 8.3, and solving for e, resulting in e = 11.2.
Step 6. Check the answer in the problem. Check the math on your own.
Two mathematical expressions are displayed with question marks questioning their equality: 10(11.2) + 20(8.3) ?= 278 and 20(11.2) + 30(8.3) ?= 473. A brace indicates they are grouped together.
Step 7. Answer the question. Jenna burns 8.3 calories per minute
circuit training and 11.2 calories per
minute while on the elliptical trainer.

Translate to a system of equations and then solve:

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Solution

Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

Translate to a system of equations and then solve:

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

Solution

Erin burned 11 calories for each minute on the rowing machine and 5 calories for each minute of weight lifting.

Solve Geometry Applications

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

Complementary and Supplementary Angles

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Translate to a system of equations and then solve.

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

Solution
Step-by-step solution for finding two complementary angles given their difference, using a system of linear equations.
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for the measure of each angle.
Step 3. Name what we are looking for. Letx=the measure of the first angle.  y=the measure of the second angle
Step 4. Translate into a system of equations. The angles are complementary.
x+y=90
The difference of the two angles is 26 degrees.
x−y=26
The system is shown. {x+y=90x−y=26
Step 5. Solve the system of equations by elimination. {x+y=90x−y=26___________ 2x=116
Substitute x=58 into the first equation. x=58x+y=9058+y=90y=32
Step 6. Check the answer in the problem.
58+32=90✓58−32=26✓
Step 7. Answer the question. The angle measures are 58 and 32 degrees.

Translate to a system of equations and then solve:

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

Solution

The angle measures are 55 and 35.

Translate to a system of equations and then solve:

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

Solution

The angle measures are 5 and 85.

In the next example, we remember that the measures of supplementary angles add to 180.

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for measure of each
angle.
Step 3. Name what we are looking for. Let x= the measure of the first angle.
y= the measure of the second angle
Step 4. Translate into a system of equations. The angles are supplementary.
The equation 'x + y = 180' is displayed in black text on a plain white background.
The larger angle is twelve less than five
times the smaller angle.
A mathematical equation is displayed on a white background, which reads 'y = 5x - 12'.
The system is shown:
Step 5. Solve the system of equations substitution.
A system of equations: x+y=180 and y=5x-12. An arrow shows substituting '5x-12' for 'y' into the equation x+y=180, illustrating the substitution method for solving linear equations.
Substitute 5x − 12 for y in the first equation.
Solve for x.
Two lines of an algebraic equation are shown. The first line is x + 5x - 12 = 180. The second line, which simplifies the first, reads 6x - 12 = 180.


Substitute 32 for x in the second
equation, then solve for y.
A mathematical problem showing how to solve for 'x' in the equation 6x = 192, yielding x = 32. An arrow indicates this value should be substituted into the equation y = 5x - 12 to find 'y'.
A step-by-step mathematical calculation is shown where y = 5 * 32 - 12 simplifies to y = 160 - 12, resulting in the final answer of y = 148.
Step 6. Check the answer in the problem. Two mathematical equations with checkmarks indicating correctness are shown: 32 + 148 = 180 and 5 * 32 - 12 = 148.
Step 7. Answer the question. The angle measures are 148 and 32 degrees.

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

Solution

The angle measures are 42 and 138.

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

Solution

The angle measures are 66 and 114.

Recall that the angles of a triangle add up to 180 degrees. A right triangle has one angle that is 90 degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

Solution

We will draw and label a figure.

Step 1. Read the problem. A right-angled triangle with a vertical side labeled 'a' and an acute angle labeled 'b'. A square symbol indicates the 90-degree angle.
Step 2. Identify what you are looking for. We are looking for the measures of the angles.
Step 3. Name what we are looking for. Let a= the measure of the first angle.
b= the measure of the second angle
Step 4. Translate into a system of equations. The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle.
A mathematical equation is displayed on a white background, which reads 'a = 3b + 10' in black text.
The sum of the measures of the angles of a triangle is 180.
The image displays the algebraic equation a + b + 90 = 180, suggesting a problem involving angles or geometric figures where variables a and b, along with a 90-degree angle, sum up to 180 degrees.

The system is shown. A system of two linear equations is presented. The first equation is a = 3b + 10. The second equation is a + b + 90 = 180.
Step 5. Solve the system of equations. We will use substitution since the first equation is solved for a. An algebraic problem with two equations: a = 3b + 10 and a + b + 90 = 180. The first equation, circled, points to the second, indicating a substitution scenario.
Substitute 3b+10 for a in the second equation. A mathematical equation is displayed: (3b + 10) + b + 90 = 180, featuring variables and numerical values.
Solve for b. A mathematical solution showing the steps to find the value of 'b' from the equation 4b + 100 = 180, resulting in b = 20, and then showing the next equation a = 3b + 10.
Substitute b=20 into the first equation and then solve for a. A mathematical equation is displayed on a white background: a = 3 * 20 + 10, which resolves to a = 70. The number '20' in the first line is highlighted in red.
Step 6. Check the answer in the problem. We will leave this to you!
Step 7. Answer the question. The measures of the small angles are 20 and 70 degrees.

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Solution

22, 68

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Solution

36, 54

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

Translate to a system of equations and then solve:

Randall has 125 feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Solution
Step 1. Read the problem.
Step 2. Identify what you are looking for. We are looking for the length and width.
A brown paper bag with a textured handle is depicted against a plain white background in a simple, top-down perspective illustration.
Step 3. Name what we are looking for. Let L= the length of the fenced yard.
W= the width of the fenced yard
Step 4. Translate into a system of equations. One length and two widths equal 125.
A mathematical equation is displayed on a white background: L + 2W = 125.
The length will be 5 feet more than
four times the width.
The image shows the mathematical equation L = 4W + 5, which defines the variable L in terms of the variable W.
The system is shown.

Step 5. Solve The system of equations
by substitution.
A system of two linear equations is shown with the second equation, L = 4W + 5, prepared for substitution into the first equation, L + 2W = 125. The expression '4W + 5' is highlighted and indicated to replace 'L'.
Substitute L = 4W + 5 into the first
equation, then solve for W.
A mathematical equation is displayed on a white background: 4W + 5 + 2W = 125. The numbers '4' and '5' are in red, while the rest of the equation is in gray.
Algebraic equation 6W + 5 = 125 is solved step-by-step, resulting in W = 20, which is circled.
Substitute 20 for W in the second
equation, then solve for L.
An algebraic calculation shows L = 4W + 5. Substituting W with 20, the equation becomes L = 4 * 20 + 5, which simplifies to L = 80 + 5, resulting in L = 85.
Step 6. Check the answer in the
problem.
Arithmetic problem: 20 + 85 + 20 = 125. Below, 85 is shown as 4 * 20 + 5, followed by a checkmark confirming 85 = 85. This illustrates a numerical calculation and its verification.
Step 7. Answer the equation. The length is 85 feet and the width is 20 feet.

Translate to a system of equations and then solve:

Mario wants to put a fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

Solution

The length is 60 feet and the width is 35 feet.

Translate to a system of equations and then solve:

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

Solution

The length is 60 feet and the width is 38 feet.

Solve uniform motion applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D=rt where D is the distance traveled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Translate to a system of equations and then solve:

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

Solution

A diagram is useful in helping us visualize the situation.


A diagram illustrating a word problem with two people, Joni and Kelly, traveling from St. Louis towards Denver. Joni travels at 65 mph, while Kelly travels at 78 mph, starting 0.5 hours later.

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k, and Joni, j, will each drive.


A table titled 'Rate * Time = Distance' shows calculations for Joni and Kelly. Joni has a rate of 65, time 'j', and distance '65j'. Kelly has a rate of 78, time 'k', and distance '78k'.

Since D=r·t we can fill in the Distance column.

Translate into a system of equations.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

65j=78k

Also, since Kelly left later, her time will be 12 hour less than Joni’s time. So,

Step-by-step solution for a word problem using a system of equations. Calculates when Kelly catches up to Joni, considering their speeds and travel times.
k=j−12
Now we have the system. {k=j−1265j=78k
Solve the system of equations by substitution.
Substitute k=j−12 into the second equation, then solve for j.
65j=78k65j=78(j−12)65j=78j−39−13j=−39j=3
To find Kelly’s time, substitute j=3 into the first equation, then solve for k. k=j−12
k=3−12
k=52ork=212
Check the answer in the problem.
Joni 3hours(65mph)=195miles
Kelly 212hours(78mph)=195miles
Yes, they will have traveled the same distance when they meet.
Answer the question. Kelly will catch up to Joni in 212 hours. By then, Joni will have traveled 3 hours.

Translate to a system of equations and then solve:

Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

Solution

It will take Clark 4 hours to catch Mitchell.

Translate to a system of equations and then solve:

Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes (14hour) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

Solution

It will take Sally 112 hours to catch up to Charlie.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c.

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b+c.

Figure shows a boat and two horizontal arrows, both pointing left. The one to the left of the boat is b and the one to the right is c.

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is b−c.

Figure shows a boat and two horizontal arrows to its left. One, labeled b, points left and the other, labeled c, points right.

We’ll put some numbers to this situation in the next example.

Translate to a system of equations and then solve.

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Solution
Read the problem. This is a uniform motion problem and a
picture will help us visualize the situation.
A diagram illustrates a 60-mile river journey with a current 'c'. It takes 4 hours to travel downstream (with the current) and 5 hours to travel upstream (against the current).
Identify what we are looking for. We are looking for the speed of the ship
in still water and the speed of the current.
Name what we are looking for. Let s= the rate of the ship in still water.
c= the rate of the current
A chart will help us organize the information.
The ship goes downstream and then upstream.
Going downstream, the current helps the
ship and so the ship's actual rate is s + c.
Going upstream, the current slows the ship
and so the actual rate is s − c.
A table showing rate, time, and distance for downstream and upstream travel. Downstream: rate s+c, time 4, distance 60. Upstream: rate s-c, time 5, distance 60.
Downstream it takes 4 hours.
Upstream it takes 5 hours.
Each way the distance is 60 miles.
Translate into a system of equations.
Since rate times time is distance, we can
write the system of equations.
A system of two linear equations is presented. The first equation is 4(s + c) = 60, and the second equation is 5(s - c) = 60. The equations are enclosed within a brace on the left.
Solve the system of equations.
Distribute to put both equations in standard
form, then solve by elimination.
A system of two linear equations is presented. The first equation is 4s + 4c = 60, and the second equation is 5s - 5c = 60. A curly brace indicates they are part of a system.
Multiply the top equation by 5 and the
bottom equation by 4.
Add the equations, then solve for s.
A step-by-step solution demonstrating the elimination method for a system of linear equations: 20s + 20c = 300 and 20s - 20c = 240, yielding s = 13.5, with an arrow pointing to 4(s + c) = 60.
Substitute s = 13.5 into of the original
equations.
An algebraic equation showing the steps to solve for 'c'. The equation 4(13.5 + c) = 60 is solved, leading to c = 1.5 through distribution, subtraction, and division.
Check the answer in the problem.
The downstream rate would be
 13.5+1.5=15 mph.
In 4 hours the ship would travel
  15·4=60 miles.
The upstream rate would be
 13.5−1.5=12 mph.
In 5 hours the ship would travel
  12·5=60 miles.
Answer the question. The rate of the ship is 13.5 mph and
the rate of the current is 1.5 mph.

Translate to a system of equations and then solve:

A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

Solution

The rate of the boat is 11 mph and the rate of the current is 1 mph.

Translate to a system of equations and then solve:

Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

Solution

The speed of the canoe is 7 mph and the speed of the current is 1 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind. A wind current blowing against the direction of the plane is called a headwind.

Translate to a system of equations and then solve:

A private jet can fly 1,095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution
Read the problem. This is a uniform motion problem and a
picture will help us visualize.
This diagram compares distances traveled over 3 hours with and without wind assistance: 1,095 miles with a tailwind vs. 987 miles with a headwind, demonstrating wind's effect on travel.
Identify what we are looking for. We are looking for the speed of the jet
in still air and the speed of the wind.
Name what we are looking for. Let j= the speed of the jet in still air.
w= the speed of the wind.
A chart will help us organize the information.
The jet makes two trips—one in a tailwind
and one in a headwind.
In a tailwind, the wind helps the jet and so
the rate is j + w.
In a headwind, the wind slows the jet and
so the rate is j ‒ w.
A table illustrating Rate, Time, and Distance for travel with tailwind and headwind. Tailwind values are (j+w), 3, and 1095. Headwind values are (j-w), 3, and 987, respectively.
Each trip takes 3 hours.
In a tailwind the jet flies 1,095 miles.
In a headwind the jet flies 987 miles.
Translate into a system of equations.
Since rate times time is distance, we get the
system of equations.
A system of two linear equations is presented. The first equation is 3 multiplied by the sum of j and w, which equals 1095. The second equation is 3 multiplied by the difference of j and w, which equals 987.
Solve the system of equations.
Distribute, then solve by elimination.
Add, and solve for j.
A system of linear equations, {3j + 3w = 1095, 3j - 3w = 987}, is solved by adding the equations to eliminate 'w', resulting in 6j = 2082 and j = 347.
Substitute j = 347 into one of the original
equations, then solve for w.
A step-by-step algebraic solution showing how to solve for 'w' in the equation 3(347 + w) = 1095, resulting in w = 18. The steps include distribution, subtraction, and division to isolate the variable.
Check the answer in the problem.
With the tailwind, the actual rate of the
jet would be
 347+18=365 mph.
In 3 hours the jet would travel
  365·3=1,095 miles
Going into the headwind, the jet’s actual
rate would be
 347−18=329 mph.
In 3 hours the jet would travel
  329·3=987 miles.
Answer the question. The rate of the jet is 347 mph and the
rate of the wind is 18 mph.

Translate to a system of equations and then solve:

A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1,035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

The speed of the jet is 236 mph and the speed of the wind is 29 mph.

Translate to a system of equations and then solve:

A commercial jet can fly 1,728 miles in 4 hours with a tailwind but only 1,536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

The speed of the jet is 408 mph and the speed of the wind is 24 mph.

Access this online resource for additional instruction and practice with systems of equations.

  • Systems of Equations

Key Concepts

  • How To Solve Applications with Systems of Equations
    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what we are looking for.
    3. Name what we are looking for. Choose variables to represent those quantities.
    4. Translate into a system of equations.
    5. Solve the system of equations using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.

Practice Makes Perfect

Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

The sum of two number is 15. One number is 3 less than the other. Find the numbers.

The sum of two number is 30. One number is 4 less than the other. Find the numbers.

Solution

13 and 17

The sum of two number is −16. One number is 20 less than the other. Find the numbers.

The sum of two number is −26. One number is 12 less than the other. Find the numbers.

Solution

−7 and −19

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

Solution

14 and 23

The sum of two numbers is −27. Their difference is −59. Find the numbers.

The sum of two numbers is −45. Their difference is −89. Find the numbers.

Solution

22 and −67

Maxim has been offered positions by two car companies. The first company pays a salary of $10,000 plus a commission of $1000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Solution

Eighty cable packages would need to be sold to make the total pay the same.

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Solution

Mitchell would need to sell 120 stoves for the companies to be equal.

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

Solution

8 and 40 gallons

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball? How many calories did he burn per hour when canoeing?

Solution

1000 calories playing basketball and 400 calories canoeing

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.

Nancy bought seven pounds of oranges and three pounds of bananas for $17. Her husband later bought three pounds of oranges and six pounds of bananas for $12. What was the cost per pound of the oranges and the bananas?

Solution

Oranges cost $2 per pound and bananas cost $1 per pound

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

Solution

Package of paper $4, stapler $7

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

Solution

Hot dog 150 calories, cup of cottage cheese 220 calories

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water as juice. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

Solution

Owen will need 80 quarts of water and 20 quarts of concentrate to make 100 quarts of lemonade.

Solve Geometry Applications

In the following exercises, translate to a system of equations and solve.

The difference of two complementary angles is 55 degrees. Find the measures of the angles.

The difference of two complementary angles is 17 degrees. Find the measures of the angles.

Solution

53.5 degrees and 36.5 degrees

Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

Solution

16 degrees and 74 degrees

The difference of two supplementary angles is 8 degrees. Find the measures of the angles.

The difference of two supplementary angles is 88 degrees. Find the measures of the angles.

Solution

134 degrees and 46 degrees

Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.

Solution

37 degrees and 143 degrees

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

Solution

16 degrees and 74 degrees

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Solution

45 degrees and 45 degrees

Wayne is hanging a string of lights 45 feet long around the three sides of his patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his front yard. The length is five feet less than three times the width. Find the length and width of the fencing.

Solution

Width is 41 feet and length is 118 feet.

A frame around a family portrait has a perimeter of 90 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

The perimeter of a toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.

Solution

Width is 10 feet and length is 40 feet.

Solve Uniform Motion Applications

In the following exercises, translate to a system of equations and solve.

Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?

College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?

Solution

12 hours

At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?

Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

Solution

1.83 hour

The Jones family took a 12-mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

Solution

Boat rate is 16 mph and current rate is 4 mph.

A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current.

A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current.

Solution

Boat rate is 18 mph and current rate is 2 mph.

A small jet can fly 1072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A small jet can fly 1435 miles in 5 hours with a tailwind but only 1,215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

Jet rate is 265 mph and wind speed is 22 mph.

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

Jet rate is 415 mph and wind speed is 25 mph.

Writing Exercises

Write an application problem similar to Example 1. Then translate to a system of equations and solve it.

Write a uniform motion problem similar to Example 2 that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve direct translation applications, solve geometry applications, solve uniform motion applications. The remaining columns are empty.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

complementary angles
Two angles are complementary if the sum of the measures of their angles is 90 degrees.
supplementary angles
Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

Solve Mixture Applications with Systems of Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve mixture applications
  • Solve interest applications
  • Solve applications of cost and revenue functions

Before you get started, take this readiness quiz.

Multiply: 4.025(1,562).
If you missed this problem, review Example 3 in Decimals.

Solution

6,287.05

Write 8.2% as a decimal.
If you missed this problem, review Example 7 in Decimals.

Solution

0.082

Earl’s dinner bill came to $32.50 and he wanted to leave an 18% tip. How much should the tip be?
If you missed this problem, review Example 7 in Use a Problem Solving Strategy.

Solution

$5.85

Solve Mixture Applications

Mixture application involve combining two or more quantities. When we solved mixture applications with coins and tickets earlier, we started by creating a table so we could organize the information. For a coin example with nickels and dimes, the table looked like this:

This table has 4 columns and two rows. The first column labels each row nickels and dimes. The header labels the columns number times value equals total value.

Using one variable meant that we had to relate the number of nickels and the number of dimes. We had to decide if we were going to let n be the number of nickels and then write the number of dimes in terms of n, or if we would let d be the number of dimes and write the number of nickels in terms of d.

Now that we know how to solve systems of equations with two variables, we’ll just let n be the number of nickels and d be the number of dimes. We’ll write one equation based on the total value column, like we did before, and the other equation will come from the number column.

For the first example, we’ll do a ticket problem where the ticket prices are in whole dollars, so we won’t need to use decimals just yet.

Translate to a system of equations and solve:

A science center sold 1,363 tickets on a busy weekend. The receipts totaled $12,146. How many $12 adult tickets and how many $7 child tickets were sold?

Solution
Step 1. Read the problem. We will create a table to organize the information.
Step 2. Identify what we are looking for. We are looking for the number of adult tickets
and the number of child tickets sold.
Step 3. Name what we are looking for. Let a= the number of adult tickets.
c= the number of child tickets
A table will help us organize the data.
We have two types of tickets, adult and child.
Write in a and c for the number of tickets.
Write the total number of tickets sold at the bottom
of the Number column.
Altogether 1,363 were sold.
Write the value of each type of ticket in the
Value column.
The value of each adult ticket is $12.
The value of each child tickets is $7.
The number times the value gives the total value,
so the total value of adult tickets is a·12=12a,
and the total value of child tickets is c·7=7c.
Fill in the Total Value column.
Altogether the total value of the tickets was $12,146. A two-row table with four columns titled 'Type', 'Number', 'Value', and 'Total Value'. Rows represent 'adult' and 'child' categories. For adults: Number is 'a', Value is '12', Total Value is '12a'. For children: Number is 'c', Value is '7', Total Value is '7c'. The total number is '1,363' and the total value is '12,146'.
Step 4. Translate into a system of equations.
The Number column and the Total value column
give us the system of equations.
A system of two linear equations is presented, with the first equation being a + c = 1,363 and the second equation being 12a + 7c = 12,146.
We will use the elimination method to solve
this system. Multiply the first equation by −7.
A system of two linear equations is presented. The first equation is -7(a + c) = -7(1,363), and the second is 12a + 7c = 12,146.
Simplify and add, then solve for a. Solving a system of linear equations by elimination, demonstrating the addition of two equations to eliminate 'c' and solve for 'a', resulting in a = 521.
Substitute a=521 into the first equation, then
solve for c.
A math problem demonstrating the solution for 'c' where a + c = 1,363. By substituting a with 521, the value of 'c' is derived as 842 using simple arithmetic.
Step 6. Check the answer in the
problem.
521 adult at $12  per ticket makes  $  6,252
 842 child at $7 per ticket makes  $5,894
      The total receipts are $12,146✓
Step 7. Answer the question. The science center sold 521 adult tickets and
842 child tickets.

Translate to a system of equations and solve:

The ticket office at the zoo sold 553 tickets one day. The receipts totaled $3,936. How many $9 adult tickets and how many $6 child tickets were sold?

Solution

206 adults, 347 children

Translate to a system of equations and solve:

The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled $1,302. How many $11 adult and how many $8 child tickets were sold?

Solution

42 adults, 105 children

In the next example, we’ll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the ‘number’ column will be easy.

Translate to a system of equations and solve:

Juan has a pocketful of nickels and dimes. The total value of the coins is $8.10. The number of dimes is 9 less than twice the number of nickels. How many nickels and how many dimes does Juan have?

Solution
Step 1. Read the problem.
We will create a table to organize the information.
Step 2. Identify what we are looking for. We are looking for the number of
nickels and the number of dimes.
Step 3. Name what we are looking for. Let n= the number of nickels.
  d= the number of dimes
A table will help us organize the data.
We have two types of coins, nickels and dimes.
Write n and d for the number of
each type of coin.
Fill in the Value column with the value of each
type of coin.
The value of each nickel is $0.05.
The value of each dime is $0.10.
The number times the value gives the total
value, so, the total value of the nickels is
n(0.05)=0.05n and the total value of dimes is
d(0.10)=0.10d.
Altogether the total value of the coins is $8.10.
An algebraic table for solving coin problems, showing the number (n for nickels, d for dimes), individual value (0.05, 0.10), and total value for each coin type, summing to $8.10.
Step 4. Translate into a system of equations.
The Total Value column gives one equation. A mathematical equation is displayed: 0.05n + 0.10d = 8.10. This equation represents a monetary sum involving two variables, likely nickels (n) and dimes (d), totaling eight dollars and ten cents.
We also know the number of dimes is 9 less than
twice the number of nickels.
Translate to get the second equation. The image displays the mathematical equation d = 2n - 9 on a white background, representing a linear relationship between the variables d and n.
Now we have the system to solve. A system of two linear equations is presented on a white background. The first equation is 0.05n + 0.10d = 8.10, and the second equation is d = 2n - 9.
Step 5. Solve the system of equations
We will use the substitution method.
Substitute d=2n−9 into the first equation. A mathematical problem showing the substitution method for solving a system of linear equations. The variable 'd' is substituted with '2n - 9' in the second equation.
Simplify and solve for n. A step-by-step solution showing the algebraic process to solve the equation 0.05n + 0.2n - 0.90 = 8.10, combining like terms to get 0.25n = 9.00, and ultimately finding n = 36.
To find the number of dimes, substitute
n=36 into the second equation.
A mathematical calculation is shown on a white background, starting with the equation d = 2n - 9. This is followed by d = 2 multiplied by 36 (where 36 is highlighted in red) minus 9. The final line shows the result as d = 63.
Step 6. Check the answer in the problem
 63 dimes at  $0.10=$6.30
 36 nickels at $0.05=$1.80
       Total =$8.10✓
Step 7. Answer the question. Juan has 36 nickels and 63 dimes.

Translate to a system of equations and solve:

Matilda has a handful of quarters and dimes, with a total value of $8.55. The number of quarters is 3 more than twice the number of dimes. How many dimes and how many quarters does she have?

Solution

13 dimes and 29 quarters

Translate to a system of equations and solve:

Priam has a collection of nickels and quarters, with a total value of $7.30. The number of nickels is six less than three times the number of quarters. How many nickels and how many quarters does he have?

Solution

19 quarters and 51 nickels

Some mixture applications involve combining foods or drinks. Example situations might include combining raisins and nuts to make a trail mix or using two types of coffee beans to make a blend.

Translate to a system of equations and solve:

Carson wants to make 20 pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix costs him $7.60. per pound. Nuts cost $9.00 per pound and chocolate chips cost $2.00 per pound. How many pounds of nuts and how many pounds of chocolate chips should he use?

Solution
Step 1. Read the problem.
We will create a table to organize the information.
Step 2. Identify what we are looking for. We are looking for the number of pounds of
nuts and the number of pounds of chocolate
chips.
Step 3. Name what we are looking for. Let n= the number of pound of nuts.
  c= the number of pounds of chips
Carson will mix nuts and chocolate chips to get
trail mix.
Write in n and c for the number of pounds of
nuts and chocolate chips.
A table showing the number of pounds, value per pound, and total value for nuts, chocolate chips, and trail mix. Variables 'n' and 'c' are used for the pounds of nuts and chocolate chips.
There will be 20 pounds of trail mix.
Put the price per pound of each item in
the Value column.
Fill in the last column using
 Number•Value=Total Value
Step 4. Translate into a system of equations.
We get the equations from the Number
and Total Value columns.
A system of two linear equations is displayed, showing 'n + c = 20' and '9n + 2c = 152' enclosed by a light blue curly bracket on the left, indicating they form a connected set.
Step 5. Solve the system of equations
We will use elimination to solve the system.
Multiply the first equation by −2 to eliminate c.
A system of two linear equations is displayed, with the first equation being -2(n+c) = -2(20) and the second being 9n + 2c = 152, enclosed by a left curly brace.
Simplify and add.
Solve for n.
A system of two linear equations, -2n - 2c = -40 and 9n + 2c = 152, is solved by elimination, resulting in 7n = 112 and ultimately n = 16.
To find the number of pounds of chocolate
chips, substitute n=16 into the first equation,
then solve for c.
An algebraic problem is displayed with three lines of equations. The first line is n + c = 20, the second line shows 16 + c = 20, and the third line presents the solution as c = 4.
Step 6. Check the answer in the problem.
16+4=20✓9·16+2·4=152✓
Step 7. Answer the question. Carson should mix 16 pounds of nuts with 4
pounds of chocolate chips to create the trail
mix.

Translate to a system of equations and solve:

Greta wants to make 5 pounds of a nut mix using peanuts and cashews. Her budget requires the mixture to cost her $6 per pound. Peanuts are $4 per pound and cashews are $9 per pound. How many pounds of peanuts and how many pounds of cashews should she use?

Solution

3 pounds peanuts and 2 pounds cashews

Translate to a system of equations and solve:

Sammy has most of the ingredients he needs to make a large batch of chili. The only items he lacks are beans and ground beef. He needs a total of 20 pounds combined of beans and ground beef and has a budget of $3 per pound. The price of beans is $1 per pound and the price of ground beef is $5 per pound. How many pounds of beans and how many pounds of ground beef should he purchase?

Solution

10 pounds of beans, 10 pounds of ground beef

Another application of mixture problems relates to concentrated cleaning supplies, other chemicals, and mixed drinks. The concentration is given as a percent. For example, a 20% concentrated household cleanser means that 20% of the total amount is cleanser, and the rest is water. To make 35 ounces of a 20% concentration, you mix 7 ounces (20% of 35) of the cleanser with 28 ounces of water.

For these kinds of mixture problems, we’ll use “percent” instead of “value” for one of the columns in our table.

Translate to a system of equations and solve:

Sasheena is lab assistant at her community college. She needs to make 200 milliliters of a 40% solution of sulfuric acid for a lab experiment. The lab has only 25% and 50% solutions in the storeroom. How much should she mix of the 25% and the 50% solutions to make the 40% solution?

Solution
Step 1. Read the problem.
A figure may help us visualize the
situation, then we will create a table to
organize the information.
Sasheena must mix some of the 25% solution and
some of the 50% solution together to get 200 ml of
the 40% solution.
A diagram illustrates two solutions, 'x' at 25% and 'y' at 50% concentration, being mixed together to form a resulting solution of 40% concentration and a total volume of 200 ml.
Step 2. Identify what we are looking for. We are looking for how much of each solution she
needs.
Step 3. Name what we are looking for. Let x= number of ml of 25% solution.
  y= number of ml of 50% solution
A table will help us organize the data. She will
mix x ml of 25% with y ml of 50% to get 200 ml
of 40% solution. We write the percents as decimals
in the chart.
We multiply the number of units times the
concentration to get the total amount of
sulfuric acid in each solution.
A table illustrating how to calculate the amount based on the number of units and their concentration percentage, with examples for 25%, 50%, and 40% types using variables x, y, and a value of 200.
Step 4. Translate into a system of
equations.
We get the equations from the Number
column and the Amount column.
Now we have the system.
A system of two linear equations is presented, with the first equation being x + y = 200, and the second equation being 0.25x + 0.50y = 0.40(200), enclosed by a left brace.
Step 5. Solve the system of equations
We will solve the system by elimination.
Multiply the first equation by −0.5 to
eliminate y.
A system of two linear equations is presented, with the first equation in red showing -0.5(x + y) = -0.5(200), and the second equation in black as 0.25x + 0.50y = 80.
Simplify and add to solve for x. A mathematical problem showing a system of two linear equations solved using the elimination method. The equations are -0.5x - 0.5y = -100 and 0.25x + 0.5y = 80, resulting in x = 80.
To solve for y, substitute x=80 into the first
equation.
Solving the equation x + y = 200 by substituting x with 80, which then results in y = 120. The steps demonstrate basic algebraic substitution to find the value of 'y'.
Step 6. Check the answer in the problem.
80+120=200✓0.25(80)+0.50(120)=200✓Yes!
Step 7. Answer the question. Sasheena should mix 80 ml of the 25% solution with
120 ml of the 50% solution to get the 200 ml of the
40% solution.

Translate to a system of equations and solve:

LeBron needs 150 milliliters of a 30% solution of sulfuric acid for a lab experiment but only has access to a 25% and a 50% solution. How much of the 25% and how much of the 50% solution should he mix to make the 30% solution?

Solution

120 ml of 25% solution and 30 ml of 50% solution

Translate to a system of equations and solve:

Anatole needs to make 250 milliliters of a 25% solution of hydrochloric acid for a lab experiment. The lab only has a 10% solution and a 40% solution in the storeroom. How much of the 10% and how much of the 40% solutions should he mix to make the 25% solution?

Solution

125 ml of 10% solution and 125 ml of 40% solution

Solve Interest Applications

The formula to model simple interest applications is I=Prt. Interest, I, is the product of the principal, P, the rate, r, and the time, t. In our work here, we will calculate the interest earned in one year, so t will be 1.

We modify the column titles in the mixture table to show the formula for interest, as you’ll see in the next example.

Translate to a system of equations and solve:

Adnan has $40,000 to invest and hopes to earn 7.1% interest per year. He will put some of the money into a stock fund that earns 8% per year and the rest into bonds that earns 3% per year. How much money should he put into each fund?

Solution
Step 1. Read the problem. A chart will help us organize the information.
Step 2. Identify what we are looking for. We are looking for the amount to invest in each fund.
Step 3. Name what we are looking for. Let s= the amount invested in stocks.
b= the amount invested in stocks
Write the interest rate as a decimal for
each fund.
Multiply: Principal · Rate · Time
A table outlining investment details for a stock fund and bonds. It shows principal, rate, time, and calculated interest for each account, with a total investment of $40,000 at a 0.071 total rate.
Step 4. Translate into a system of
equations.
We get our system of equations from
the Principal column and the
Interest column.
A mathematical problem displaying a system of two linear equations: s + b = 40,000 and 0.08s + 0.03b = 0.071(40,000). A brace groups the equations together.
Step 5. Solve the system of equations
by elimination.
Multiply the top equation by −0.03.
A system of two linear equations is shown: -0.03(s + b) = -0.03(40,000) and 0.08s + 0.03b = 2,840.
Simplify and add to solve for s. A system of two linear equations, -0.03s - 0.03b = -1,200 and 0.08s + 0.03b = 2,840, being solved by adding them together to eliminate 'b', resulting in 0.05s = 1,640, which simplifies to s = 32,800.
To find b, substitute s = 32,800 into
the first equation.
This image illustrates an algebraic solution where 's' is substituted with 32,800 into 's + b = 40,000', leading to '32,800 + b = 40,000' and finally 'b = 7,200'.
Step 6. Check the answer in the
problem.
We leave the check to you.
Step 7. Answer the question. Adnan should invest $32,800 in stock and
$7,200 in bonds.

Did you notice that the Principal column represents the total amount of money invested while the Interest column represents only the interest earned? Likewise, the first equation in our system, s+b=40,000, represents the total amount of money invested and the second equation, 0.08s+0.03b=0.071(40,000), represents the interest earned.

Translate to a system of equations and solve:

Leon had $50,000 to invest and hopes to earn 6.2% interest per year. He will put some of the money into a stock fund that earns 7% per year and the rest in to a savings account that earns 2% per year. How much money should he put into each fund?

Solution

$42,000 in the stock fund and $8000 in the savings account

Translate to a system of equations and solve:

Julius invested $7000 into two stock investments. One stock paid 11% interest and the other stock paid 13% interest. He earned 12.5% interest on the total investment. How much money did he put in each stock?

Solution

$1750 at 11% and $5250 at 13%

The next example requires that we find the principal given the amount of interest earned.

Translate to a system of equations and solve:

Rosie owes $21,540 on her two student loans. The interest rate on her bank loan is 10.5% and the interest rate on the federal loan is 5.9%. The total amount of interest she paid last year was $1,669.68. What was the principal for each loan?

Solution
Step 1. Read the problem. A chart will help us organize the information.
Step 2. Identify what we are looking for. We are looking for the principal of each loan.
Step 3. Name what we are looking for. Let b= the principal for the bank loan.
f= the principal on the federal loan
The total loans are $21,540.
Record the interest rates as decimals
in the chart.
Multiply using the formula I = Prt to
get the Interest.
A table detailing simple interest calculations for 'Bank' and 'Federal' accounts, showing principal (b, f), rates (0.105, 0.059), time (1 year), and calculated interest (0.105b, 0.059f). The total principal is 21,540, and total interest is 1669.68.
Step 4. Translate into a system of
equations.
The system of equations comes from
the Principal column and the Interest
column.
A system of two linear equations is displayed, featuring b + f = 21,540 and 0.105b + 0.059f = 1669.68. This represents a typical algebraic problem with two variables.
Step 5. Solve the system of equations
We will use substitution to solve.
Solve the first equation for b.
Two lines of equations are centered on a white background. The first line reads b + f = 21,540. The second line reads b = -f + 21,540.
A mathematical equation is displayed with the text '0.105b + 0.059f = 1669.68' against a white background.
Substitute b = −f + 21.540 into
the second equation.
Two lines of a math problem simplifying an equation, showing the distribution of a decimal factor.
Simplify and solve for f. A series of algebraic equations are displayed, showing the solution for 'f'. The first equation is -0.046f + 2261.70 = 1669.68, which simplifies to -0.046f = -592.02, leading to f = 12,870.
To find b, substitute f = 12,870 into the first equation. Three lines of algebraic calculation show the equation b + f = 21,540, followed by the substitution of f with 12,870, leading to b + 12,870 = 21,540, and finally the solution b = 8,670.
Step 6. Check the answer in the
problem.
We leave the check to you.
Step 7. Answer the question. The principal of the federal loan was $12,870 and
the principal for the bank loan was $8,670.

Translate to a system of equations and solve:

Laura owes $18,000 on her student loans. The interest rate on the bank loan is 2.5% and the interest rate on the federal loan is 6.9%. The total amount of interest she paid last year was $1,066. What was the principal for each loan?

Solution

Bank $4,000; Federal $14,000

Translate to a system of equations and solve:

Jill’s Sandwich Shoppe owes $65,200 on two business loans, one at 4.5% interest and the other at 7.2% interest. The total amount of interest owed last year was $3,582. What was the principal for each loan?

Solution

$41,200 at 4.5%, $24,000 at 7.2%

Solve applications of cost and revenue functions

Suppose a company makes and sells x units of a product. The cost to the company is the total costs to produce x units. This is the cost to manufacture for each unit times x, the number of units manufactured, plus the fixed costs.

The revenue is the money the company brings in as a result of selling x units. This is the selling price of each unit times the number of units sold.

When the costs equal the revenue we say the business has reached the break-even point.

Cost and Revenue Functions

The cost function is the cost to manufacture each unit times x, the number of units manufactured, plus the fixed costs.

C(x)=(cost per unit)·x+fixed costs

The revenue function is the selling price of each unit times x, the number of units sold.

R(x)=(selling price per unit)·x

The break-even point is when the revenue equals the costs.

C(x)=R(x)

The manufacturer of a weight training bench spends $105 to build each bench and sells them for $245. The manufacturer also has fixed costs each month of $7,000.

ⓐ Find the cost function C when x benches are manufactured.

ⓑ Find the revenue function R when x benches are sold.

ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid.

ⓓ Find the break-even point. Interpret what the break-even point means.

Solution

ⓐ The manufacturer has $7,000 of fixed costs no matter how many weight training benches it produces. In addition to the fixed costs, the manufacturer also spends $105 to produce each bench. Suppose x benches are sold.

Derivation of a cost function, illustrating the general formula and its application with specific cost values.
Write the general Cost function formula. C(x)=(cost per unit)·x+fixed costs
Substitute in the cost values. C(x)=105x+7000

ⓑ The manufacturer sells each weight training bench for $245. We get the total revenue by multiplying the revenue per unit times the number of units sold.

Illustrates the derivation of a revenue function, moving from a general formula to a specific instance with substituted values.
Write the general Revenue function. R(x)=(sellingpriceperunit)·x
Substitute in the revenue per unit. R(x)=245x

ⓒ Essentially we have a system of linear equations. We will show the graph of the system as this helps make the idea of a break-even point more visual.

{C(x)=105x+7000R(x)=245xor{y=105x+7000y=245x

Figure shows a graph with two intersecting lines. One of them passes through the origin.

ⓓ To find the actual value, we remember the break-even point occurs when costs equal revenue.

This table illustrates the steps involved in setting up and solving a break-even formula.
Write the break-even formula. C(x)=R(x)105x+7000=245x
Solve. 7000=140x50=x

When 50 benches are sold, the costs equal the revenue.

C of x is 105x plus 7000. C of 50 is 105 times 50 plus 7000, which is equal to 12250. R of x is 245x. R of 50 is 245 times 50, which is 12250.

When 50 benches are sold, the revenue and costs are both $12,250. Notice this corresponds to the ordered pair (50,12,250).

The manufacturer of a weight training bench spends $15 to build each bench and sells them for $32. The manufacturer also has fixed costs each month of $25,500.

ⓐ Find the cost function C when x benches are manufactured.

ⓑ Find the revenue function R when x benches are sold.

ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid.

ⓓ Find the break-even point. Interpret what the break-even point means.

Solution

ⓐ C(x)=15x+25,500

ⓑ R(x)=32x

ⓒ
Figure shows a graph with two intersecting lines. One of them passes through the origin. The other crosses the y axis at point 25,687.

ⓓ 1,500; when 1,500 benches are sold, the cost and revenue will be both 48,000

The manufacturer of a weight training bench spends $120 to build each bench and sells them for $170. The manufacturer also has fixed costs each month of $150,000.

ⓐ Find the cost function C when x benches are manufactured.

ⓑ Find the revenue function R when x benches are sold.

ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid.

ⓓ Find the break-even point. Interpret what the break-even point means.

Solution

ⓐ C(x)=120x+150,000

ⓑ R(x)=170x

ⓒ
Figure shows a graph with two intersecting lines. One of them passes through the origin.

ⓓ 3,000; when 3,000 benches are sold, the revenue and costs are both $510,000

Access this online resource for additional instruction and practice with interest and mixtures.

  • Interest and Mixtures

Key Concepts

  • Cost function: The cost function is the cost to manufacture each unit times x, the number of units manufactured, plus the fixed costs.
    C(x)=(cost per unit)·x+fixed costs
  • Revenue: The revenue function is the selling price of each unit times x, the number of units sold.
    R(x)=(sellingpriceperunit)·x
  • Break-even point: The break-even point is when the revenue equals the costs.
    C(x)=R(x)

Practice Makes Perfect

Solve Mixture Applications

In the following exercises, translate to a system of equations and solve.

Tickets to a Broadway show cost $35 for adults and $15 for children. The total receipts for 1650 tickets at one performance were $47,150. How many adult and how many child tickets were sold?

Tickets for the Cirque du Soleil show are $70 for adults and $50 for children. One evening performance had a total of 300 tickets sold and the receipts totaled $17,200. How many adult and how many child tickets were sold?

Solution

110 adult tickets, 190 child tickets

Tickets for an Amtrak train cost $10 for children and $22 for adults. Josie paid $1200 for a total of 72 tickets. How many children tickets and how many adult tickets did Josie buy?

Tickets for a Minnesota Twins baseball game are $69 for Main Level seats and $39 for Terrace Level seats. A group of sixteen friends went to the game and spent a total of $804 for the tickets. How many of Main Level and how many Terrace Level tickets did they buy?

Solution

6 good seats, 10 cheap seats

Tickets for a dance recital cost $15 for adults and $7 dollars for children. The dance company sold 253 tickets and the total receipts were $2771. How many adult tickets and how many child tickets were sold?

Tickets for the community fair cost $12 for adults and $5 dollars for children. On the first day of the fair, 312 tickets were sold for a total of $2204. How many adult tickets and how many child tickets were sold?

Solution

92 adult tickets, 220 children tickets

Brandon has a cup of quarters and dimes with a total value of $3.80. The number of quarters is four less than twice the number of dimes. How many quarters and how many dimes does Brandon have?

Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is $0.95. The number of nickels is two less than five times the number of dimes. How many nickels and how many dimes are in the coin purse?

Solution

13 nickels, 3 dimes

Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value $13.10. The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have?

Lucinda had a pocketful of dimes and quarters with a value of $6.20. The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have?

Solution

42 dimes, 8 quarters

A cashier has 30 bills, all of which are $10 or $20 bills. The total value of the money is $460. How many of each type of bill does the cashier have?

A cashier has 54 bills, all of which are $10 or $20 bills. The total value of the money is $910. How many of each type of bill does the cashier have?

Solution

17 $10 bills, 37 $20 bills

Marissa wants to blend candy selling for $1.80 per pound with candy costing $1.20 per pound to get a mixture that costs her $1.40 per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use?

How many pounds of nuts selling for $6 per pound and raisins selling for $3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him $5 per pound?

Solution

80 pounds nuts and 40 pounds raisins

Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is $1.79 per gallon and the cost of the fruit drink is $2.49 per gallon. Hannah’s budget requires that the punch cost $2.21 per gallon. How many gallons of soda and how many gallons of fruit drink does she need?

Joseph would like to make twelve pounds of a coffee blend at a cost of $6 per pound. He blends Ground Chicory at $5 a pound with Jamaican Blue Mountain at $9 per pound. How much of each type of coffee should he use?

Solution

9 pounds of Chicory coffee, 3 pounds of Jamaican Blue Mountain coffee

Julia and her husband own a coffee shop. They experimented with mixing a City Roast Columbian coffee that cost $7.80 per pound with French Roast Columbian coffee that cost $8.10 per pound to make a twenty-pound blend. Their blend should cost them $7.92 per pound. How much of each type of coffee should they buy?

Twelve-year old Melody wants to sell bags of mixed candy at her lemonade stand. She will mix M&M’s that cost $4.89 per bag and Reese’s Pieces that cost $3.79 per bag to get a total of twenty-five bags of mixed candy. Melody wants the bags of mixed candy to cost her $4.23 a bag to make. How many bags of M&M’s and how many bags of Reese’s Pieces should she use?

Solution

10 bags of M&M’s, 15 bags of Reese’s Pieces

Jotham needs 70 liters of a 50% solution of an alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution?

Joy is preparing 15 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution?

Solution

7.5 liters of each solution

A scientist needs 65 liters of a 15% alcohol solution. She has available a 25% and a 12% solution. How many liters of the 25% and how many liters of the 12% solutions should she mix to make the 15% solution?

A scientist needs 120 milliliters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many liters of the 25% and how many liters of the 10% solutions should the scientist mix to make the 20% solution?

Solution

80 liters of the 25% solution and 40 liters of the 10% solution

A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used?

A 90% antifreeze solution is to be mixed with a 75% antifreeze solution to get 360 liters of an 85% solution. How many liters of the 90% and how many liters of the 75% solutions will be used?

Solution

240 liters of the 90% solution and 120 liters of the 75% solution

Solve Interest Applications

In the following exercises, translate to a system of equations and solve.

Hattie had $3000 to invest and wants to earn 10.6% interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year. How much money should she put into each account?

Carol invested $2560 into two accounts. One account paid 8% interest and the other paid 6% interest. She earned 7.25% interest on the total investment. How much money did she put in each account?

Solution

$1600 at 8%, 960 at 6%

Sam invested $48,000, some at 6% interest and the rest at 10%. How much did he invest at each rate if he received $4000 in interest in one year?

Arnold invested $64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received $4500 in interest in one year?

Solution

$28,000 at 9%, $36,000 at 5.5%

After four years in college, Josie owes $65, 800 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was $2878.50. What is the amount of each loan?

Mark wants to invest $10,000 to pay for his daughter’s wedding next year. He will invest some of the money in a short term CD that pays 12% interest and the rest in a money market savings account that pays 5% interest. How much should he invest at each rate if he wants to earn $1095 in interest in one year?

Solution

$8500 CD, $1500 savings account

A trust fund worth $25,000 is invested in two different portfolios. This year, one portfolio is expected to earn 5.25% interest and the other is expected to earn 4%. Plans are for the total interest on the fund to be $1150 in one year. How much money should be invested at each rate?

A business has two loans totaling $85,000. One loan has a rate of 6% and the other has a rate of 4.5% This year, the business expects to pay $4,650 in interest on the two loans. How much is each loan?

Solution

$55,000 on loan at 6% and $30,000 on loan at 4.5%

Solve Applications of Cost and Revenue Functions

The manufacturer of an energy drink spends $1.20 to make each drink and sells them for $2. The manufacturer also has fixed costs each month of $8,000.

ⓐ Find the cost function C when x energy drinks are manufactured.

ⓑ Find the revenue function R when x drinks are sold.

ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid.

ⓓ Find the break-even point. Interpret what the break-even point means.

The manufacturer of a water bottle spends $5 to build each bottle and sells them for $10. The manufacturer also has fixed costs each month of $6500. ⓐ Find the cost function C when x bottles are manufactured. ⓑ Find the revenue function R when x bottles are sold. ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. ⓓ Find the break-even point. Interpret what the break-even point means.

Solution

ⓐ C(x)=5x+6500

ⓑ R(x)=10x

ⓒ
Figure shows a graph with two intersecting lines. One of them passes through the origin. The other crosses the y axis at point 6560.

ⓓ 1,300; when 1,300 water bottles are sold, the cost and the revenue equal $13,000

Writing Exercises

Take a handful of two types of coins, and write a problem similar to Example 2 relating the total number of coins and their total value. Set up a system of equations to describe your situation and then solve it.

In Example 5, we used elimination to solve the system of equations
{s+b=40,0000.08s+0.03b=0.071(40,000).

Could you have used substitution or elimination to solve this system? Why?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 2 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve mixture applications, solve interest applications. The remaining columns are blank.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

cost function
The cost function is the cost to manufacture each unit times x, the number of units manufactured, plus the fixed costs; C(x) = (cost per unit)x + fixed costs.
revenue
The revenue is the selling price of each unit times x, the number of units sold; R(x) = (selling price per unit)x.
break-even point
The point at which the revenue equals the costs is the break-even point; C(x)=R(x).

Solve Systems of Equations with Three Variables

Learning Objectives

By the end of this section, you will be able to:

  • Determine whether an ordered triple is a solution of a system of three linear equations with three variables
  • Solve a system of linear equations with three variables
  • Solve applications using systems of linear equations with three variables

Before you get started, take this readiness quiz.

Evaluate 5x−2y+3z when x=−2, y=−4, and z=3.
If you missed this problem, review Example 10 in Integers.

Solution

7

Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. {−2x+y=−11x+3y=9.
If you missed this problem, review Example 6 in Use a General Strategy to Solve Linear Equations.

Solution

conditional; one solution

Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. {7x+8y=43x−5y=27.
If you missed this problem, review Example 8 in Use a General Strategy to Solve Linear Equations.

Solution

conditional; one solution

Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables

In this section, we will extend our work of solving a system of linear equations. So far we have worked with systems of equations with two equations and two variables. Now we will work with systems of three equations with three variables. But first let's review what we already know about solving equations and systems involving up to two variables.

We learned earlier that the graph of a linear equation, ax+by=c, is a line. Each point on the line, an ordered pair (x,y), is a solution to the equation. For a system of two equations with two variables, we graph two lines. Then we can see that all the points that are solutions to each equation form a line. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions

We know when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Figure shows three graphs. In the first one, two lines intersect. Intersecting lines have one point in common. There is one solution to this system. The graph is labeled Consistent Independent. In the second graph, two lines are parallel. Parallel lines have no points in common. There is no solution to this system. The graph is labeled inconsistent. In the third graph, there is just one line. Both equations give the same line. Because we have just one line, there are infinitely many solutions. It is labeled consistent dependent.

Similarly, for a linear equation with three variables ax+by+cz=d, every solution to the equation is an ordered triple, (x,y,z), that makes the equation true.

Linear Equation in Three Variables

A linear equation with three variables, where a, b, c, and d are real numbers and a, b, and c are not all 0, is of the form

ax+by+cz=d

Every solution to the equation is an ordered triple, (x,y,z) that makes the equation true.

All the points that are solutions to one equation form a plane in three-dimensional space. And, by finding what the planes have in common, we’ll find the solution to the system.

When we solve a system of three linear equations represented by a graph of three planes in space, there are three possible cases.

Eight figures are shown. The first one shows three intersecting planes with one point in common. It is labeled Consistent system and Independent equations. The second figure has three parallel planes with no points in common. It is labeled Inconsistent system. In the third figure two planes are coincident and parallel to the third plane. The planes have no points in common. In the fourth figure, two planes are parallel and each intersects the third plane. The planes have no points in common. In the fifth figure, each plane intersects the other two, but all three share no points. The planes have no points in common. In the sixth figure, three planes intersect in one line. There is just one line, so there are infinitely many solutions. In the seventh figure, two planes are coincident and intersect the third plane in a line. There is just one line, so there are infinitely many solutions. In the last figure, three planes are coincident. There is just one plane, so there are infinitely many solutions. Three parallel planes demonstrate an inconsistent system with no solution. As parallel planes never intersect, they have no points in common. Text states 'Two planes are coincident and parallel to the third plane. The planes have no points in common.' An illustration shows two distinct parallel planes, highlighting the contradictory nature of the statements. Two parallel planes (blue) are shown, each intersected by a third plane (orange). The illustration emphasizes that the two parallel planes themselves have no points in common. Three planes are depicted, intersecting pairwise with each other. Notably, there is no single point where all three planes converge, illustrating a unique geometric relationship. Three planes intersect along a single line, illustrating a consistent system with dependent equations and infinitely many solutions, as there are endless points along the shared line. Depiction of two planes intersecting, illustrating a scenario where two coincident planes intersect a third in a line, resulting in infinitely many solutions as described by the text. Three coincident planes are represented by a single blue plane, illustrating that there is only one distinct plane and thus infinitely many solutions.

To solve a system of three linear equations, we want to find the values of the variables that are solutions to all three equations. In other words, we are looking for the ordered triple (x,y,z) that makes all three equations true. These are called the solutions of the system of three linear equations with three variables.

Solutions of a System of Linear Equations with Three Variables

Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple (x,y,z).

To determine if an ordered triple is a solution to a system of three equations, we substitute the values of the variables into each equation. If the ordered triple makes all three equations true, it is a solution to the system.

Determine whether the ordered triple is a solution to the system: {x−y+z=22x−y−z=−62x+2y+z=−3.

ⓐ (−2,−1,3) ⓑ (−4,−3,4)

Solution

ⓐ
The equations are x minus y plus z equals 2, 2x minus y minus z equals minus 6 and 2x plus 2y plus z equals minus 3. Substituting minus 2 for x, minus 1 for y and 3 for z into all three equations, we find that all three hold true. Hence, minus 2, minus 1, 3 is a solution.

ⓑ
The equations are x minus y plus z equals 2, 2x minus y minus z equals minus 6 and 2x plus 2y plus z equals minus 3. Substituting minus minus 4 for x, minus 3 for y and 4 for z into all three equations, we find that all three hold true. Hence, minus 4, minus 3, 4 is not a solution.

Determine whether the ordered triple is a solution to the system: {3x+y+z=2x+2y+z=−33x+y+2z=4.

ⓐ (1,−3,2) ⓑ (4,−1,−5)

Solution

ⓐ yes ⓑ no

Determine whether the ordered triple is a solution to the system: {x−3y+z=−5−3x−y−z=12x−2y+3z=1.

ⓐ (2,−2,3) ⓑ (−2,2,3)

Solution

ⓐ no ⓑ yes

Solve a System of Linear Equations with Three Variables

To solve a system of linear equations with three variables, we basically use the same techniques we used with systems that had two variables. We start with two pairs of equations and in each pair we eliminate the same variable. This will then give us a system of equations with only two variables and then we know how to solve that system!

Next, we use the values of the two variables we just found to go back to the original equation and find the third variable. We write our answer as an ordered triple and then check our results.

How to Solve a System of Equations With Three Variables by Elimination

Solve the system by elimination: {x−2y+z=32x+y+z=43x+4y+3z=−1.

Solution

The equations are x minus 2y plus z equals 3, 2x plus y plus z equals 4 and 3x plus 4y plus 3z equals minus 1. Step 1 is to write the equations in standard form. They are. If any coefficients are fractions, clear them. There are none. Step 2 is to eliminate the same variable from two equations. Decide which variable you will eliminate. We can eliminate the y’s from equations 1 and 2 by multiplying equation 2 by 2. Work with a pair of equations to eliminate the chosen variable. Multiply one or both equations so that the coefficients of that variable are opposites. Add the equations resulting from Step 2 to eliminate one variable. The new equation we get is 5x plus 3z equals 11. Step 3 is to repeat step 2 using two other equations and eliminate the same variable as in step 2. We can again eliminate the y’s using the equations 1, 3 by multiplying equation 1 by 2. Add the new equations and the result will be 5x plus 5z equals 5. Step 4. The two new equations form a system of two equations with two variables. Solve this system. Eliminating x, we get z equal to minus 3. Substituting this in one of the new equations, we get x equal to 4. Step 5 is to use the values of the two variables found in step 4 to find the third variable. Substituting values of x and z in one of the original equations, we get y equal to minus 1. Step 6 is to write the solution as an ordered triple 4, minus 1, minus 3. Step 7 is to check that the ordered triple is a solution to all three original equations. It makes all three equations true.

Solve the system by elimination: {3x+y−z=22x−3y−2z=14x−y−3z=0.

Solution

(2,−1,3)

Solve the system by elimination: {4x+y+z=−1−2x−2y+z=22x+3y−z=1.

Solution

(−2,3,4)

The steps are summarized here.

Solve a system of linear equations with three variables.

  1. Write the equations in standard form
    • If any coefficients are fractions, clear them.
  2. Eliminate the same variable from two equations.
    • Decide which variable you will eliminate.
    • Work with a pair of equations to eliminate the chosen variable.
    • Multiply one or both equations so that the coefficients of that variable are opposites.
    • Add the equations resulting from Step 2 to eliminate one variable
  3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
  4. The two new equations form a system of two equations with two variables. Solve this system.
  5. Use the values of the two variables found in Step 4 to find the third variable.
  6. Write the solution as an ordered triple.
  7. Check that the ordered triple is a solution to all three original equations.

Solve: {3x−4z=03y+2z=−32x+3y=−5.

Solution
{3x−4z=0(1)3y+2z=−3(2)2x+3y=−5(3)

We can eliminate z from equations (1) and (2) by multiplying equation (2) by 2 and then adding the resulting equations.

The equations are 3 x minus 4 equals 0, 3y plus 2 z equals minus 3 and 2 x plus 3 y equals minus 5. Multiply equation 2 by 2 and add to equation 1. We get 3 x plus 6 y equals minus 6.

Notice that equations (3) and (4) both have the variables x and y. We will solve this new system for x and y.

Multiply equation 3 by minus 2 and add that to equation 4. We get x equal to minus 4.

To solve for y, we substitute x=−4 into equation (3).

Substitute minus 4 into equation 3 and solve for y. We get y equal to 1.

We now have x=−4 and y=1. We need to solve for z. We can substitute x=−4 into equation (1) to find z.

Substituting minus 4 into equation 1 for x, we get z equal to minus 3.

We write the solution as an ordered triple. (−4,1,−3)

We check that the solution makes all three equations true.

3x−4z=0(1)3(−4)−4(−3)=?00=0✓ 3y+2z=−3(2)3(1)+2(−3)=?−3−3=−3✓ 2x+3y=−5(3)2(−4)+3(1)=?−5−5=−5✓The solution is(−4,1,−3).

Solve: {3x−4z=−12y+3z=22x+3y=6.

Solution

(−3,4,−2)

Solve: {4x−3z=−53y+2z=73x+4y=6.

Solution

(−2,3,−1)

When we solve a system and end up with no variables and a false statement, we know there are no solutions and that the system is inconsistent. The next example shows a system of equations that is inconsistent.

Solve the system of equations: {x+2y−3z=−1x−3y+z=12x−y−2z=2.

Solution
{x+2y−3z=−1(1)x−3y+z=1(2)2x−y−2z=2(3)

Use equation (1) and (2) to eliminate z.

The equations are x plus 2y minus 3z equals minus 1, x minus 3y plus z equals 1 and 2x minus y minus 2z equals 2.

Use (2) and (3) to eliminate z again.

Multiplying equation 2 by 3 and adding it to equation 1, we get equation 4, 4x minus 7y equals 2. Multiplying equation 2 by 2 and adding it to equation 3, we get equation 5, 4x minus 7y equals 4.

Use (4) and (5) to eliminate a variable.

Equations 4 and 5 both have 2 variables. Multiply equation 5 by minus 1 and add it to equation 4. We get 0 equal to minus 2, which is false.

There is no solution.

We are left with a false statement and this tells us the system is inconsistent and has no solution.

Solve the system of equations: {x+2y+6z=5−x+y−2z=3x−4y−2z=1.

Solution

no solution

Solve the system of equations: {2x−2y+3z=64x−3y+2z=0−2x+3y−7z=1.

Solution

no solution

When we solve a system and end up with no variables but a true statement, we know there are infinitely many solutions. The system is consistent with dependent equations. Our solution will show how two of the variables depend on the third.

Solve the system of equations: {x+2y−z=12x+7y+4z=11x+3y+z=4.

Solution
{x+2y−z=1(1)2x+7y+4z=11(2)x+3y+z=4(3)

Use equation (1) and (3) to eliminate x.

The equations are x plus 2y minus z equals 1, 2x plus 7y plus 4z equals 11 and x plus 3y plus z equals 4. Multiply equation 1 with minus 1 and add it to equation 3. We get equation 4, y plus 2z equals 3.

Use equation (1) and (2) to eliminate x again.

Multiply equation 1 with minus 2 and add it to equation 2. We get equation 5, 3y plus 6z equals 9.

Use equation (4) and (5) to eliminate y.

Multiply equation 4 with minus 3 and add it to equation 5. We get 0 equal to 0. There are infinite many solutions. Solving equation 4 for y, we get y equal to minus 2z plus 3. Substituting this into equation 1, we get x equal to 5z minus 5. The true statement 0 equal to 0 tells us that this is a dependent system that has infinitely many solutions. The solutions are of the form x, y, z where x is 5z minus 5, y is minus 2z plus 3 and z is any real number.
There are infinitely many solutions.
Solve equation (4) for y. Represent the solution showing how x and y are dependent on z.
y+2z=3y=−2z+3
Use equation (1) to solve for x. x+2y−z=1
Substitute y=−2z+3. x+2(−2z+3)−z=1x−4z+6−z=1x−5z+6=1x=5z−5

The true statement 0=0 tells us that this is a dependent system that has infinitely many solutions. The solutions are of the form (x,y,z) where x=5z−5;y=−2z+3and z is any real number.

Solve the system by equations: {x+y−z=02x+4y−2z=63x+6y−3z=9.

Solution

infinitely many solutions(x,3,z) where x=z−3;y=3;z is any real number

Solve the system by equations: {x−y−z=1−x+2y−3z=−43x−2y−7z=0.

Solution

infinitely many solutions (x,y,z) wherex=5z−2;y=4z−3;z is any real number

Solve Applications using Systems of Linear Equations with Three Variables

Applications that are modeled by a systems of equations can be solved using the same techniques we used to solve the systems. Many of the application are just extensions to three variables of the types we have solved earlier.

The community college theater department sold three kinds of tickets to its latest play production. The adult tickets sold for $15, the student tickets for $10 and the child tickets for $8. The theater department was thrilled to have sold 250 tickets and brought in $2,825 in one night. The number of student tickets sold is twice the number of adult tickets sold. How many of each type did the department sell?

Solution
We will use a chart to organize the information. A table displays ticket pricing details for adults, students, and children. It shows the number of each type (x, y, z), their individual values ($15, $10, $8), and their total values (15x, 10y, 8z). The total number of tickets is 250, and the total value is $2825.
Number of students is twice number of adults.
Rewrite the equation in standard form. y=2x2x−y=0
The image displays a mathematical problem with the prompt 'Write the system of equations.' To the right, a system of three linear equations with variables x, y, and z is presented.
Use equations (1) and (2) to eliminate z.
Solving a system of linear equations by multiplying the first equation by -8 and adding it to the second equation to eliminate 'z', resulting in 7x + 2y = 825.
Use (3) and (4) to eliminate y.
A step-by-step solution demonstrating how to solve a system of linear equations by elimination. The first equation (-2x + y = 0) is multiplied by -2 to yield 4x - 2y = 0, which is then added to the second equation (7x + 2y = 825) to eliminate 'y', resulting in 11x = 825.
Solve for x.  x=75 adult tickets
Use equation (3) to find y. −2x+y=0
Substitute x=75. −2(75)+y=0−150+y=0y=150student tickets
Use equation (1) to find z. x+y+z=250
Substitute in the values
x=75,y=150.

75+150+z=250225+z=250z=25child tickets
Write the solution. The theater department sold 75 adult tickets,
150 student tickets, and 25 child tickets.

The community college fine arts department sold three kinds of tickets to its latest dance presentation. The adult tickets sold for $20, the student tickets for $12 and the child tickets for $10.The fine arts department was thrilled to have sold 350 tickets and brought in $4,650 in one night. The number of child tickets sold is the same as the number of adult tickets sold. How many of each type did the department sell?

Solution

The fine arts department sold 75 adult tickets, 200 student tickets, and 75 child tickets.

The community college soccer team sold three kinds of tickets to its latest game. The adult tickets sold for $10, the student tickets for $8 and the child tickets for $5. The soccer team was thrilled to have sold 600 tickets and brought in $4,900 for one game. The number of adult tickets is twice the number of child tickets. How many of each type did the soccer team sell?

Solution

The soccer team sold 200 adult tickets, 300 student tickets, and 100 child tickets.

Access this online resource for additional instruction and practice with solving a linear system in three variables with no or infinite solutions.

  • Solving a Linear System in Three Variables with No or Infinite Solutions
  • 3 Variable Application

Key Concepts

  • Linear Equation in Three Variables: A linear equation with three variables, where a, b, c, and d are real numbers and a, b, and c are not all 0, is of the form
    ax+by+cz=d

    Every solution to the equation is an ordered triple, (x,y,z) that makes the equation true.
  • How to solve a system of linear equations with three variables.
    1. Write the equations in standard form
      If any coefficients are fractions, clear them.
    2. Eliminate the same variable from two equations.
      Decide which variable you will eliminate.
      Work with a pair of equations to eliminate the chosen variable.
      Multiply one or both equations so that the coefficients of that variable are opposites.
      Add the equations resulting from Step 2 to eliminate one variable
    3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
    4. The two new equations form a system of two equations with two variables. Solve this system.
    5. Use the values of the two variables found in Step 4 to find the third variable.
    6. Write the solution as an ordered triple.
    7. Check that the ordered triple is a solution to all three original equations.

Practice Makes Perfect

Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables

In the following exercises, determine whether the ordered triple is a solution to the system.

{2x−6y+z=33x−4y−3z=22x+3y−2z=3

ⓐ (3,1,3) ⓑ (4,3,7)

{−3x+y+z=−4−x+2y−2z=12x−y−z=−1

ⓐ (−5,−7,4) ⓑ (5,7,4)

Solution

ⓐ no ⓑ yes

{y−10z=−82x−y=2x−5z=3

ⓐ (7,12,2) ⓑ (2,2,1)

{x+3y−z=15y=23x−2x−3y+z=−2

ⓐ (−6,5,12) ⓑ (5,43,−3)

Solution

ⓐ no ⓑ no

Solve a System of Linear Equations with Three Variables

In the following exercises, solve the system of equations.

{5x+2y+z=5−3x−y+2z=62x+3y−3z=5

{6x−5y+2z=32x+y−4z=53x−3y+z=−1

Solution

(4,5,2)

{2x−5y+3z=83x−y+4z=7x+3y+2z=−3

{5x−3y+2z=−52x−y−z=43x−2y+2z=−7

Solution

(7,12,−2)

{3x−5y+4z=55x+2y+z=02x+3y−2z=3

{4x−3y+z=72x−5y−4z=33x−2y−2z=−7

Solution

(−3,−5,4)

{3x+8y+2z=−52x+5y−3z=0x+2y−2z=−1

{11x+9y+2z=−97x+5y+3z=−74x+3y+z=−3

Solution

(2,−3,−2)

{13x−y−z=1x+52y+z=−22x+2y+12z=−4

{x+12y+12z=015x−15y+z=013x−13y+2z=−1

Solution

(6,−9,−3)

{x+13y−2z=−113x+y+12z=012x+13y−12z=−1

{13x−y+12z=423x+52y−4z=0x−12y+32z=2

Solution

(3,−4,−2)

{x+2z=04y+3z=−22x−5y=3

{2x+5y=43y−z=34x+3z=−3

Solution

(−3,2,3)

{2y+3z=−15x+3y=−67x+z=1

{3x−z=−35y+2z=−64x+3y=−8

Solution

(−2,0,−3)

{4x−3y+2z=0−2x+3y−7z=12x−2y+3z=6

{x−2y+2z=1−2x+y−z=2x−y+z=5

Solution

no solution

{2x+3y+z=12x+y+z=93x+4y+2z=20

{x+4y+z=−84x−y+3z=92x+7y+z=0

Solution

x=20316;y=–2516;z=–23116;

{x+2y+z=4x+y−2z=3−2x−3y+z=−7

{x+y−2z=3−2x−3y+z=−7x+2y+z=4

Solution

(x,y,z) where x=5z+2;y=−3z+1;z is any real number

{x+y−3z=−1y−z=0−x+2y=1

{x−2y+3z=1x+y−3z=73x−4y+5z=7

Solution

(x,y,z) where x=5z−2;y=4z−3;z is any real number

Solve Applications using Systems of Linear Equations with Three Variables

In the following exercises, solve the given problem.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is twice the measure of the first angle. The third angle is twelve more than the second. Find the measures of the three angles.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is three times the measure of the first angle. The third angle is fifteen more than the second. Find the measures of the three angles.

Solution

45 degrees, 60 degrees, 75 degrees

After watching a major musical production at the theater, the patrons can purchase souvenirs. If a family purchases 4 t-shirts, the video, and 1 stuffed animal, their total is $135.

A couple buys 2 t-shirts, the video, and 3 stuffed animals for their nieces and spends $115. Another couple buys 2 t-shirts, the video, and 1 stuffed animal and their total is $85. What is the cost of each item?

The church youth group is selling snacks to raise money to attend their convention. Amy sold 2 pounds of candy, 3 boxes of cookies and 1 can of popcorn for a total sales of $65. Brian sold 4 pounds of candy, 6 boxes of cookies and 3 cans of popcorn for a total sales of $140. Paulina sold 8 pounds of candy, 8 boxes of cookies and 5 cans of popcorn for a total sales of $250. What is the cost of each item?

Solution

$20, $5, $10

Writing Exercises

In your own words explain the steps to solve a system of linear equations with three variables by elimination.

How can you tell when a system of three linear equations with three variables has no solution? Infinitely many solutions?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first row contains the following statements: determine whether an ordered triple is a solution of a system of three linear equations with three variables, solve a system of linear equations with three variables, solve applications using systems of linear equations with three variables. The remaining columns are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

solutions of a system of linear equations with three variables
The solutions of a system of equations are the values of the variables that make all the equations true; a solution is represented by an ordered triple (x,y,z).

Solve Systems of Equations Using Matrices

Learning Objectives

By the end of this section, you will be able to:

  • Write the augmented matrix for a system of equations
  • Use row operations on a matrix
  • Solve systems of equations using matrices

Before you get started, take this readiness quiz.

Solve: 3(x+2)+4=4(2x−1)+9.
If you missed this problem, review Example 2 in Use a General Strategy to Solve Linear Equations.

Solution

x=1

Solve: 0.25p+0.25(p+4)=5.20.
If you missed this problem, review Example 13 in Use a General Strategy to Solve Linear Equations.

Solution

p=8.4

Evaluate when x=−2 and y=3:2x2−xy+3y2.
If you missed this problem, review Example 10 in Integers.

Solution

41

Write the Augmented Matrix for a System of Equations

Solving a system of equations can be a tedious operation where a simple mistake can wreak havoc on finding the solution. An alternative method which uses the basic procedures of elimination but with notation that is simpler is available. The method involves using a matrix. A matrix is a rectangular array of numbers arranged in rows and columns.

Matrix

A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix with m rows and n columns has order m×n. The matrix on the left below has 2 rows and 3 columns and so it has order 2×3. We say it is a 2 by 3 matrix.

Figure shows two matrices. The one on the left has the numbers minus 3, minus 2 and 2 in the first row and the numbers minus 1, 4 and 5 in the second row. The rows and columns are enclosed within brackets. Thus, it has 2 rows and 3 columns. It is labeled 2 cross 3 or 2 by 3 matrix. The matrix on the right is similar but with 3 rows and 4 columns. It is labeled 3 by 4 matrix.

Each number in the matrix is called an element or entry in the matrix.

We will use a matrix to represent a system of linear equations. We write each equation in standard form and the coefficients of the variables and the constant of each equation becomes a row in the matrix. Each column then would be the coefficients of one of the variables in the system or the constants. A vertical line replaces the equal signs. We call the resulting matrix the augmented matrix for the system of equations.

The equations are 3x plus y equals minus 3 and 2x plus 3y equals 6. A 2 by 3 matrix is shown. The first row is 3, 1, minus 3. The second row is 2, 3, 6. The first column is labeled coefficients of x. The second column is labeled coefficients of y and the third is labeled constants.

Notice the first column is made up of all the coefficients of x, the second column is the all the coefficients of y, and the third column is all the constants.

Write each system of linear equations as an augmented matrix:

ⓐ {5x−3y=−1y=2x−2 ⓑ {6x−5y+2z=32x+y−4z=53x−3y+z=−1

Solution

ⓐ The second equation is not in standard form. We rewrite the second equation in standard form.

y=2x−2−2x+y=−2

We replace the second equation with its standard form. In the augmented matrix, the first equation gives us the first row and the second equation gives us the second row. The vertical line replaces the equal signs.

The equations are 3x plus y equals minus 3 and 2x plus 3y equals 6. A 2 by 3 matrix is shown. The first row is 3, 1, minus 3. The second row is 2, 3, 6. The first column is labeled coefficients of x. The second column is labeled coefficients of y and the third is labeled constants.

ⓑ All three equations are in standard form. In the augmented matrix the first equation gives us the first row, the second equation gives us the second row, and the third equation gives us the third row. The vertical line replaces the equal signs.

The equations are 6x minus 5y plus 2z equals 3, 2x plus y minus 4z equals 5 and 3x minus 3y plus z equals minus 1. A 4 by 3 matrix is shown whose first row is 6, minus 5, 2, 3. Its second row is 2, 1, minus 4, 5. Its third row is 3, minus 3, 1 and minus 1. Its first three columns are labeled x, y and z respectively.

Write each system of linear equations as an augmented matrix:

ⓐ {3x+8y=−32x=−5y−3 ⓑ {2x−5y+3z=83x−y+4z=7x+3y+2z=−3

Solution


ⓐ [38−325−3]
ⓑ [2−5383−147132−3]

Write each system of linear equations as an augmented matrix:

ⓐ {11x=−9y−57x+5y=−1 ⓑ {5x−3y+2z=−52x−y−z=43x−2y+2z=−7

Solution


ⓐ [119−575−1]
ⓑ [5−32−52−1−143−22−7]

It is important as we solve systems of equations using matrices to be able to go back and forth between the system and the matrix. The next example asks us to take the information in the matrix and write the system of equations.

Write the system of equations that corresponds to the augmented matrix:

[4−3312−1−2−13|−12−4].

Solution

We remember that each row corresponds to an equation and that each entry is a coefficient of a variable or the constant. The vertical line replaces the equal sign. Since this matrix is a 4×3, we know it will translate into a system of three equations with three variables.

A 3 by 4 matrix is shown. Its first row is 4, minus 3, 3, minus 1. Its second row is 1, 2, minus 1, 2. Its third row is minus 2, minus 1, 3, minus 4. The three equations are 4x minus 3y plus 3z equals minus 1, x plus 2y minus z equals 2 and minus 2x minus y plus 3z equals minus 4.

Write the system of equations that corresponds to the augmented matrix: [1−12321−214−120].

Solution

{x−y+2z=32x+y−2z=14x−y+2z=0

Write the system of equations that corresponds to the augmented matrix: [111423−1811−13].

Solution

{x+y+z=42x+3y−z=8x+y−z=3

Use Row Operations on a Matrix

Once a system of equations is in its augmented matrix form, we will perform operations on the rows that will lead us to the solution.

To solve by elimination, it doesn’t matter which order we place the equations in the system. Similarly, in the matrix we can interchange the rows.

When we solve by elimination, we often multiply one of the equations by a constant. Since each row represents an equation, and we can multiply each side of an equation by a constant, similarly we can multiply each entry in a row by any real number except 0.

In elimination, we often add a multiple of one row to another row. In the matrix we can replace a row with its sum with a multiple of another row.

These actions are called row operations and will help us use the matrix to solve a system of equations.

Row Operations

In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.

  1. Interchange any two rows.
  2. Multiply a row by any real number except 0.
  3. Add a nonzero multiple of one row to another row.

Performing these operations is easy to do but all the arithmetic can result in a mistake. If we use a system to record the row operation in each step, it is much easier to go back and check our work.

We use capital letters with subscripts to represent each row. We then show the operation to the left of the new matrix. To show interchanging a row:

A 2 by 3 matrix is shown. Its first row, labeled R2 is 2, minus 1, 2. Its second row, labeled R1 is 5, minus 3, minus 1.

To multiply row 2 by −3:

A 2 by 3 matrix is shown. Its first row is 5, minus 3, minus 1. Its second row is 2, minus 1, 2. An arrow point from this matrix to another one on the right. The first row of the new matrix is the same. The second row is preceded by minus 3 R2. It is minus 6, 3, minus 6.

To multiply row 2 by −3 and add it to row 1:

A 2 by 3 matrix is shown. Its first row is 5, minus 3, minus 1. Its second row is 2, minus 1, 2. An arrow point from this matrix to another one on the right. The first row of the new matrix is preceded by minus 3 R2 plus R1. It is minus 1, 0, minus 7. The second row is 2, minus 1, 2.

Perform the indicated operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.

ⓑ Multiply row 2 by 5.

ⓒ Multiply row 3 by −2 and add to row 1.

[6−5221−43−31|35−1]
Solution

ⓐ We interchange rows 2 and 3.
Two 3 by 4 matrices are shown. In the one on the left, the first row is 6, minus 5, 2, 3. The second row is 2, 1, minus 4, 5. The third row is 3, minus 3, 1, minus 1. The second matrix is similar except that rows 2 and 3 are interchanged.

ⓑ We multiply row 2 by 5.
Two 3 by 4 matrices are shown. In the one on the left, the first row is 6, minus 5, 2, 3. The second row is 2, 1, minus 4, 5. The third row is 3, minus 3, 1, minus 1. The second matrix is similar to the first except that row 2, preceded by 5 R2, is 10, 5, minus 20, 25.

ⓒ We multiply row 3 by −2 and add to row 1.
In the 3 by 4 matrix, the first row is 6, minus 5, 2, 3. The second row is 2, 1, minus 4, 5. The third row is 3, minus 3, 1, minus 1. Performing the operation minus 2 R3 plus R1 on the first row, the first row becomes 6 plus minus 2 times 3, minus 5 plus minus 2 times minus 3, 2 plus minus 2 times 1 and 3 plus minus 2 times minus 1. This becomes 0, 1, 0, 5. The remaining 2 rows of the new matrix are the same.

Perform the indicated operations sequentially on the augmented matrix:

ⓐ Interchange rows 1 and 3.

ⓑ Multiply row 3 by 3.

ⓒ Multiply row 3 by 2 and add to row 2.

[5−2−24−1−4−230|−24−1]

Solution


ⓐ [−230−14−1−445−2−2−2]
ⓑ [−23004−1−4415−6−6−6]
ⓒ [−230−134−13−16−815−6−6−6]

Perform the indicated operations on the augmented matrix:

ⓐ Interchange rows 1 and 2,

ⓑ Multiply row 1 by 2,

ⓒ Multiply row 2 by 3 and add to row 1.

[2−3−241−3504|−42−1]

Solution


ⓐ [41−322−3−2−4504−1]
ⓑ [82−642−3−2−4504−1]
ⓒ [14−7−12−82−3−2−4504−1]

Now that we have practiced the row operations, we will look at an augmented matrix and figure out what operation we will use to reach a goal. This is exactly what we did when we did elimination. We decided what number to multiply a row by in order that a variable would be eliminated when we added the rows together.

Given this system, what would you do to eliminate x?

The two equations are x minus y equals 2 and 4x minus 8y equals 0. Multiplying the first by minus 4, we get minus 4x plus 4y equals minus 8. Adding this to the second equation we get minus 4y equals minus 8.

This next example essentially does the same thing, but to the matrix.

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [1−14−8|20].

Solution

To make the 4 a 0, we could multiply row 1 by −4 and then add it to row 2.

The 2 by 3 matrix is 1, minus 1, 2 and 4, minus 8, 0. Performing the operation minus 4R1 plus R2 on row 2, the second row of the new matrix becomes 0, minus 4, minus 8. The first row remains the same.

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [1−13−6|22].

Solution

[1−120−3−4]

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [1−1−2−3|32].

Solution

[1−130−58]

Solve Systems of Equations Using Matrices

To solve a system of equations using matrices, we transform the augmented matrix into a matrix in row-echelon form using row operations. For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Row-Echelon Form

For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

A 2 by 3 matrix is shown on the left. Its first row is 1, a, b. Its second row is 0, 1, c. An arrow points diagonally down and right, overlapping both the 1s in the matrix. A 3 by 4 matrix is shown on the right. Its first row is 1, a, b, d. Its second row is 0, 1, c, e. Its third row is 0, 0, 1, f. An arrow points diagonally down and right, overlapping all the 1s in the matrix. a, b, c, d, e, f are real numbers.

Once we get the augmented matrix into row-echelon form, we can write the equivalent system of equations and read the value of at least one variable. We then substitute this value in another equation to continue to solve for the other variables. This process is illustrated in the next example.

How to Solve a System of Equations Using a Matrix

Solve the system of equations using a matrix: {3x+4y=5x+2y=1.

Solution

The equations are 3x plus 4y equals 5 and x plus 2y equals 1. Step 1. Write the augmented matrix for the system of equations. We get a 2 by 3 matrix with first row 3, 4, 5 and second row 1, 2, 1. Step 2. Using row operations get the entry in row 1, column 1 to be 1. Interchange rows R1 and R2. Step 3. Using row operations, get zeros in column 1 below the 1. Multiply row 1 by minus 3 and add it to row 2. Row 2 becomes 0, minus 2, 2. Step 4. Using row operations, get the entry in row 2, column 2 to be 1. Multiply row 2 by minus half. Row 2 becomes 0, 1, minus 1. Step 5. Continue the process until the matrix is in row-echelon form. The matrix is now in row-echelon form. Step 6. Write the corresponding system of equations. We get x plus 2y equals 1 and y equals minus 1. Step 7. Use substitution to find the remaining variables. Substitute y equals negative 1 into x plus 2y equals 1. X plus 2 times negative 1 equals 1. X minus 2 equals 1. We get x equal to 3. Step 8. Write the solution as an ordered pair or triple. Ordered pair is (3, negative 1). Step 9. Check that the solution makes the original equations true.

Solve the system of equations using a matrix: {2x+y=7x−2y=6.

Solution

The solution is (4,−1).

Solve the system of equations using a matrix: {2x+y=−4x−y=−2.

Solution

The solution is (−2,0).

The steps are summarized here.

Solve a system of equations using matrices.

  1. Write the augmented matrix for the system of equations.
  2. Using row operations get the entry in row 1, column 1 to be 1.
  3. Using row operations, get zeros in column 1 below the 1.
  4. Using row operations, get the entry in row 2, column 2 to be 1.
  5. Continue the process until the matrix is in row-echelon form.
  6. Write the corresponding system of equations.
  7. Use substitution to find the remaining variables.
  8. Write the solution as an ordered pair or triple.
  9. Check that the solution makes the original equations true.

Here is a visual to show the order for getting the 1’s and 0’s in the proper position for row-echelon form.

The figure shows 3 steps for a 2 by 3 matrix and 6 steps for a 3 by 4 matrix. For the former, step 1 is to get a 1 in row 1 column 1. Step to is to get a 0 is row 2 column 1. Step 3 is to get a 1 in row 2 column 2. For a 3 by 4 matrix, step 1 is to get a 1 in row 1 column 1. Step 2 is to get a 0 in row 2 column 1. Step 3 is to get a 0 in row 3 column 1. Step 4 is to get a 1 in row 2 column 2. Step 5 is to get a 0 in row 3 column 2. Step 6 is to get a 1 in row 3 column 3.

We use the same procedure when the system of equations has three equations.

Solve the system of equations using a matrix: {3x+8y+2z=−52x+5y−3z=0x+2y−2z=−1.

Solution
A system of three linear equations with variables x, y, and z. The equations are: 3x + 8y + 2z = -5, 2x + 5y - 3z = 0, and x + 2y - 2z = -1.
Write the augmented matrix for the equations. An augmented matrix representing a system of linear equations, with a vertical line separating the coefficient matrix from the constant terms. The matrix is 3x4.
Interchange row 1 and 3 to get the entry in
row 1, column 1 to be 1.
An augmented 3x4 matrix with numerical values, featuring curved red arrows and labels R1 and R3 indicating an elementary row operation between the first and third rows.
Using row operations, get zeros in column 1 below the 1. An augmented matrix illustrating the elementary row operation -2R_1 + R_2, with the resulting second row highlighted in red.
A matrix shown after performing the row operation -3R1 + R3 on the third row. The modified third row, (0, 2, 8, -2), is highlighted in red, indicating a step in Gaussian elimination.
The entry in row 2, column 2 is now 1.
Continue the process until the matrix
is in row-echelon form.
Augmented matrix after applying the row operation -2R_2 + R_3, with the modified third row values highlighted in red during a step in Gaussian elimination.
An augmented matrix with the elementary row operation (1/6)R3 applied, resulting in the third row elements 0, 0, 1, -1 (shown in red).
The matrix is now in row-echelon form. An augmented matrix in row echelon form with a diagonal arrow highlighting the leading ones of the coefficient matrix.
Write the corresponding system of equations. A system of three linear equations is shown with a brace on the left. The equations are x + 2y - 2z = -1, y + z = 2, and z = -1. This system can be solved using back-substitution.
Use substitution to find the remaining variables. A mathematical problem solving for 'y'. The equation 'y + z = 2' is shown, where 'z' is substituted with -1 (highlighted in red), leading to 'y + (-1) = 2', which simplifies to 'y = 3'.
The image illustrates the substitution of y=3 and z=-1 into the equation x + 2y - 2z = -1, leading to the simplified form x + 6 + 2 = -1 to solve for x.
The algebraic expression 'x = -9' is clearly shown on a white background.
Write the solution as an ordered pair or triple. The image shows the mathematical coordinates (-9, 3, -1) in black text on a white background.
Check that the solution makes the original equations true. We leave the check for you.

Solve the system of equations using a matrix: {2x−5y+3z=83x−y+4z=7x+3y+2z=−3.

Solution

(6,−1,−3)

Solve the system of equations using a matrix: {−3x+y+z=−4−x+2y−2z=12x−y−z=−1.

Solution

(5,7,4)

So far our work with matrices has only been with systems that are consistent and independent, which means they have exactly one solution. Let’s now look at what happens when we use a matrix for a dependent or inconsistent system.

Solve the system of equations using a matrix: {x+y+3z=0x+3y+5z=02x+4z=1.

Solution
A system of three linear equations with variables x, y, and z. The equations are: x + y + 3z = 0, x + 3y + 5z = 0, and 2x + 4z = 1.
Write the augmented matrix for the equations. A 3x4 augmented matrix for a system of linear equations, featuring rows [1 1 3 | 0], [1 3 5 | 0], and [2 0 4 | 1].
The entry in row 1, column 1 is 1.
Using row operations, get zeros in column 1 below the 1. A 3x4 augmented matrix is shown with the row operation -1R1 + R2 indicated. The second row, which results from this operation (0 2 2 | 0), is highlighted in red, alongside other matrix entries.
Augmented matrix displaying the result of an elementary row operation where the third row (0 -2 -2 1) and the instruction -2R1 + R3 are highlighted in red.
Continue the process until the matrix is in row-echelon form. An augmented matrix displays the result of a row operation where the second row (R2) has been scaled by 1/2. The elements of this new R2, [0, 1, 1, 0], are highlighted in red.
Multiply row 2 by 2 and add it to row 3. An augmented matrix showing the next row operation '2R2 + R3' to be performed as part of Gaussian elimination.
At this point, we have all zeros on the left of row 3.
Write the corresponding system of equations. A system of conditions including two linear equations, x+y+3z=0 and y+z=0, and the inequality 0  M 1.
Since 0≠1 we have a false statement. Just as when we solved a system using other methods, this tells us we have an inconsistent system. There is no solution.

Solve the system of equations using a matrix: {x−2y+2z=1−2x+y−z=2x−y+z=5.

Solution

no solution

Solve the system of equations using a matrix: {3x+4y−3z=−22x+3y−z=−12x+y−2z=6.

Solution

no solution

The last system was inconsistent and so had no solutions. The next example is dependent and has infinitely many solutions.

Solve the system of equations using a matrix: {x−2y+3z=1x+y−3z=73x−4y+5z=7.

Solution
A system of three linear equations with variables x, y, and z. The equations are: x - 2y + 3z = 1, x + y - 3z = 7, and 3x - 4y + 5z = 7.
Write the augmented matrix for the equations. An augmented matrix representing a system of three linear equations. The matrix is 3x4, with coefficients (1, -2, 3), (1, 1, -3), (3, -4, 5) and constants (1, 7, 7) respectively.
The entry in row 1, column 1 is 1.
Using row operations, get zeros in column 1 below the 1. Augmented matrix demonstrating the elementary row operation -1R1 + R2, where the modified second row (0, 3, -6, 6) is highlighted in red, illustrating a step in Gaussian elimination.
An augmented matrix displaying a system of linear equations. The third row, highlighted in red as [0, 2, -4, 4], is presented alongside the row operation '-3R1 + R3', indicating its derivation.
Continue the process until the matrix is in row-echelon form. An augmented matrix displays the row operation (1/3)R2. The elements of the second row, highlighted in red, show the current state of the matrix after the operation, featuring values 0, 1, -2, and 2.
Multiply row 2 by −2 and add it to row 3. An augmented matrix is shown after the row operation -2R2 + R3 was applied, resulting in the third row being all zeros. This represents a consistent system of linear equations with infinitely many solutions.
At this point, we have all zeros in the bottom row.
Write the corresponding system of equations. A system of three linear equations with three variables x, y, and z is displayed. The equations are: x - 2y + 3z = 1, y - 2z = 2, and 0 = 0, indicating infinitely many solutions.
Since 0=0 we have a true statement. Just as when we solved by substitution, this tells us we have a dependent system. There are infinitely many solutions.
Solve for y in terms of z in the second equation. Two mathematical equations are displayed: y - 2z = 2, and its rearranged form y = 2z + 2, illustrating how to isolate the variable y.
Solve the first equation for x in terms of z. A mathematical equation is displayed with a white background, showing 'x - 2y + 3z = 1' in black text.
Substitute y=2z+2. A mathematical equation is displayed on a white background: x - 2(2z + 2) + 3z = 1.
Simplify. A mathematical equation is displayed with the expression x - 4z - 4 + 3z = 1 written in black against a white background.
Simplify. An algebraic equation is shown on a white background, reading 'X - Z - 4 = 1'.
Simplify. The equation X = Z + 5 is displayed on a white background, representing a simple algebraic relationship between the variables X and Z, with a constant of 5 added to Z.
The system has infinitely many solutions (85,-425,-245)

Solve the system of equations using a matrix: {x+y−z=02x+4y−2z=63x+6y−3z=9.

Solution

infinitely many solutions (x,y,z), where x=z−3;y=3;z is any real number.

Solve the system of equations using a matrix: {x−y−z=1−x+2y−3z=−43x−2y−7z=0.

Solution

infinitely many solutions (x,y,z), where x=5z−2;y=4z−3;z is any real number.

Access this online resource for additional instruction and practice with Gaussian Elimination.

  • Gaussian Elimination

Key Concepts

  • Matrix: A matrix is a rectangular array of numbers arranged in rows and columns. A matrix with m rows and n columns has order m×n. The matrix on the left below has 2 rows and 3 columns and so it has order 2×3. We say it is a 2 by 3 matrix.
    Figure shows two matrices. The one on the left has the numbers minus 3, minus 2 and 2 in the first row and the numbers minus 1, 4 and 5 in the second row. The rows and columns are enclosed within brackets. Thus, it has 2 rows and 3 columns. It is labeled 2 cross 3 or 2 by 3 matrix. The matrix on the right is similar but with 3 rows and 4 columns. It is labeled 3 by 4 matrix.
    Each number in the matrix is called an element or entry in the matrix.
  • Row Operations: In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.
    • Interchange any two rows
    • Multiply a row by any real number except 0
    • Add a nonzero multiple of one row to another row
  • Row-Echelon Form: For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.
    Figure shows two matrices. The one on the left has the numbers minus 3, minus 2 and 2 in the first row and the numbers minus 1, 4 and 5 in the second row. The rows and columns are enclosed within brackets. Thus, it has 2 rows and 3 columns. It is labeled 2 cross 3 or 2 by 3 matrix. The matrix on the right is similar but with 3 rows and 4 columns. It is labeled 3 by 4 matrix.
  • How to solve a system of equations using matrices.
    1. Write the augmented matrix for the system of equations.
    2. Using row operations get the entry in row 1, column 1 to be 1.
    3. Using row operations, get zeros in column 1 below the 1.
    4. Using row operations, get the entry in row 2, column 2 to be 1.
    5. Continue the process until the matrix is in row-echelon form.
    6. Write the corresponding system of equations.
    7. Use substitution to find the remaining variables.
    8. Write the solution as an ordered pair or triple.
    9. Check that the solution makes the original equations true.

Practice Makes Perfect

Write the Augmented Matrix for a System of Equations

In the following exercises, write each system of linear equations as an augmented matrix.


ⓐ {3x−y=−12y=2x+5
ⓑ {4x+3y=−2x−2y−3z=72x−y+2z=−6


ⓐ {2x+4y=−53x−2y=2
ⓑ {3x−2y−z=−2−2x+y=55x+4y+z=−1

Solution


ⓐ [24−53−22]
ⓑ [3−2−1−2−2105541−1]


ⓐ {3x−y=−42x=y+2
ⓑ {x−3y−4z=−24x+2y+2z=52x−5y+7z=−8


ⓐ {2x−5y=−34x=3y−1
ⓑ {4x+3y−2z=−3−2x+y−3z=4−x−4y+5z=−2

Solution


ⓐ [2−5−34−3−1]
ⓑ [43−2−3−21−34−1−45−2]

Write the system of equations that corresponds to the augmented matrix.

[2−11−3|42]

[2−43−3|−2−1]

Solution

{2x−4y=−23x−3y=−1

[10−31−200−12|−1−23]

[2−2002−130−1|−12−2]

Solution

{2x−2y=−12y−z=23x−z=−2

Use Row Operations on a Matrix

In the following exercises, perform the indicated operations on the augmented matrices.

[6−43−2|31]

ⓐ Interchange rows 1 and 2

ⓑ Multiply row 2 by 3

ⓒ Multiply row 2 by −2 and add row 1 to it.

[4−632|−31]

ⓐ Interchange rows 1 and 2

ⓑ Multiply row 1 by 4

ⓒ Multiply row 2 by 3 and add row 1 to it.

Solution


ⓐ [3214−6−3]
ⓑ [12844−6−3]
ⓒ [128424−10−5]

[4−12−84−2−3−62−1|16−1−1]

ⓐ Interchange rows 2 and 3

ⓑ Multiply row 1 by 4

ⓒ Multiply row 2 by −2 and add to row 3.

[6−5221−43−31|35−1]

ⓐ Interchange rows 2 and 3

ⓑ Multiply row 2 by 5

ⓒ Multiply row 3 by −2 and add to row 1.

Solution


ⓐ 6−5233−31−121−45
ⓑ 6−52315−155−521−45
ⓒ 2−710−715−155−521−45

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [12−3−4|5−1].

Perform the needed row operations that will get the first entry in both row 2 and row 3 to be zero in the augmented matrix: [1−233−1−22−3−4|−45−1].

Solution

[1−23−405−111701−107]

Solve Systems of Equations Using Matrices

In the following exercises, solve each system of equations using a matrix.

{2x+y=2x−y=−2

{3x+y=2x−y=2

Solution

(1,−1)

{−x+2y=−2x+y=−4

{−2x+3y=3x+3y=12

Solution

(3,3)

In the following exercises, solve each system of equations using a matrix.

{2x−3y+z=19−3x+y−2z=−15x+y+z=0

{2x−y+3z=−3−x+2y−z=10x+y+z=5

Solution

(−2,5,2)

{2x−6y+z=33x+2y−3z=22x+3y−2z=3

{4x−3y+z=72x−5y−4z=33x−2y−2z=−7

Solution

(−3,−5,4)

{x+2z=04y+3z=−22x−5y=3

{2x+5y=43y−z=34x+3z=−3

Solution

(−3,2,3)

{2y+3z=−15x+3y=−67x+z=1

{3x−z=−35y+2z=−64x+3y=−8

Solution

(−2,0,−3)

{2x+3y+z=12x+y+z=93x+4y+2z=20

{x+2y+6z=5−x+y−2z=3x−4y−2z=1

Solution

no solution

{x+2y−3z=−1x−3y+z=12x−y−2z=2

{4x−3y+2z=0−2x+3y−7z=12x−2y+3z=6

Solution

no solution

{x−y+2z=−42x+y+3z=2−3x+3y−6z=12

{−x−3y+2z=14−x+2y−3z=−43x+y−2z=6

Solution

85,-425,-245

{x+y−3z=−1y−z=0−x+2y=1

{x+2y+z=4x+y−2z=3−2x−3y+z=−7

Solution

infinitely many solutions (x,y,z) where x=5z+2;y=−3z+1;z is any real number

Writing Exercises

Solve the system of equations {x+y=10x−y=6 ⓐ by graphing and ⓑ by substitution. ⓒ Which method do you prefer? Why?

Solve the system of equations {3x+y=12x=y−8 by substitution and explain all your steps in words.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns 5 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: Write the augmented matrix for a system of equations, Use row operations on a matrix, Solve systems of equations using matrices, Write the augmented matrix for a system of equations, Use row operations on a matrix. The remaining columns are blank.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

matrix
A matrix is a rectangular array of numbers arranged in rows and columns.
row-echelon form
A matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Solve Systems of Equations Using Determinants

Learning Objectives

By the end of this section, you will be able to:

  • Evaluate the determinant of a 2×2 matrix
  • Evaluate the determinant of a 3×3 matrix
  • Use Cramer’s Rule to solve systems of equations
  • Solve applications using determinants

Before you get started, take this readiness quiz.

Simplify: 5(−2)−(−4)(1).
If you missed this problem, review Example 9 in Integers.

Solution

−6

Simplify: −3(8−10)+(−2)(6−3)−4(−3−(−4)).
If you missed this problem, review Example 8 in Integers.

Solution

−4

Simplify: −12−8.
If you missed this problem, review Example 7 in Integers.

Solution

32

In this section we will learn of another method to solve systems of linear equations called Cramer’s rule. Before we can begin to use the rule, we need to learn some new definitions and notation.

Evaluate the Determinant of a 2×2 Matrix

If a matrix has the same number of rows and columns, we call it a square matrix. Each square matrix has a real number associated with it called its determinant. To find the determinant of the square matrix [abcd], we first write it as |abcd|. To get the real number value of the determinant we subtract the products of the diagonals, as shown.

A 2 by 2 determinant is show, with its first row being a, b and second one being c, d. These values are written between two vertical lines instead of brackets as in the case of matrices. Two arrows are shown, one from a to d, the other from c to b. This determinant is equal to ad minus bc.

Determinant

The determinant of any square matrix [abcd], where a, b, c, and d are real numbers, is

|abcd|=ad−bc

Evaluate the determinant of ⓐ [4−23−1] ⓑ [−3−4−20].

Solution
ⓐ
A 2x2 matrix is shown, containing the numbers 4, -2, 3, and -1. The first row elements are 4 and -2, while the second row elements are 3 and -1.
Write the determinant. A 2x2 matrix with values 4, -2, 3, -1, illustrating diagonal multiplication with blue arrows, often used for determinants.
Subtract the products of the diagonals. A mathematical expression showing 4 multiplied by -1, minus 3 multiplied by -2. The expression is 4(-1) - 3(-2).
Simplify. The mathematical expression -4 + 6 is displayed in a dark gray font against a white background, showing the addition of a negative number and a positive number.
Simplify. The numeral 2 is prominently displayed in the center of a plain white background, rendered in a dark gray hue with a slight gradient effect.


ⓑ
A 2x2 matrix is shown, containing the elements -3, -4 in the first row and -2, 0 in the second row, enclosed within square brackets.
Write the determinant. A mathematical diagram showing cross-multiplication with numbers -3, -4, -2, and 0 arranged in a grid, featuring diagonal blue arrows indicating pairings.
Subtract the products of the diagonals. An algebraic expression featuring multiplication and subtraction of integers: -3(0) - (-2)(-4).
Simplify. The image displays the numbers '0-8' in a simple, grayscale font on a plain white background, likely representing a range or identifier.
Simplify. A vertical representation of the number -8, with the minus sign positioned to the left of the numeral eight, against a plain white background.

Evaluate the determinant of ⓐ [5−32−4] ⓑ [−4−607].

Solution

ⓐ −14; ⓑ −28

Evaluate the determinant of ⓐ [−13−24] ⓑ [−7−3−50].

Solution

ⓐ 2 ⓑ −15

Evaluate the Determinant of a 3×3 Matrix

To evaluate the determinant of a 3×3 matrix, we have to be able to evaluate the minor of an entry in the determinant. The minor of an entry is the 2×2 determinant found by eliminating the row and column in the 3×3 determinant that contains the entry.

Minor of an entry in 3×3 a Determinant

The minor of an entry in a 3×3 determinant is the 2×2 determinant found by eliminating the row and column in the 3×3 determinant that contains the entry.

To find the minor of entry a1, we eliminate the row and column which contain it. So we eliminate the first row and first column. Then we write the 2×2 determinant that remains.

The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. a1 is highlighted. Lines strike out the first row and the first column. What remains is called minor of a1. It is shown as a separate determinant whose first row is b2, c2 and second row is b3, c3.

To find the minor of entry b2, we eliminate the row and column that contain it. So we eliminate the 2nd row and 2nd column. Then we write the 2×2 determinant that remains.

The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. b2 is highlighted. Lines strike out the second row and second column. What remains is minor of b2. It is written as a separate determinant whose first row is a1, c1 and second row is a3, c3.

For the determinant |4−2310−3−2−42|, find and then evaluate the minor of ⓐ a1 ⓑ b3 ⓒ c2.

Solution
ⓐ
A 3x3 matrix with elements [4, -2, 3; 1, 0, -3; -2, -4, 2] enclosed by vertical bars, typically denoting a determinant or the absolute value of the matrix.
Eliminate the row and column that contains a1. A 3x3 matrix showing numerical values: top row is 4, -2, 3; middle row is 1, 0, -3; and bottom row is -2, -4, 2. Vertical and horizontal lines are drawn over some of the numbers.
Write the 2×2 determinant that remains. A mathematical expression showing the minor of a1 as a 2x2 matrix with elements 0, -3 in the first row and -4, 2 in the second row, enclosed by vertical lines indicating a determinant.
Evaluate. A mathematical expression 0(2) - (-3)(-4) is displayed, requiring calculation of a product of zero and two, minus the product of negative three and negative four.
Simplify. The image displays the number -12.


ⓑ
Eliminate the row and column that contains b3. A 3x3 matrix with values: 4, -2, 3 in the first row; 1, 0, -3 in the second; and 2, 4, 2 in the third. Light brown grid lines intersect between the numbers.
Write the 2×2 determinant that remains. The image shows the mathematical notation for the minor of b_3, which is represented by a 2x2 determinant with the first row containing 4 and 3, and the second row containing 1 and -3.
Evaluate. A mathematical expression displaying 4 multiplied by -3, followed by a subtraction sign, and then the product of 1 and 3.
Simplify. The number -15 is centrally displayed in a bold, sans-serif font against a plain white background, appearing as simple numerical text.


ⓒ
A 3x3 matrix with integer entries. The first row is 4, -2, 3. The second row is 1, 0, -3. The third row is -2, -4, 2.
Eliminate the row and column that contains c2. A 3x3 mathematical matrix is shown with elements 4, -2, 3; 1, 0, -3; -2, -4, 2. Red horizontal and vertical lines highlight specific rows and columns, indicating a calculation step.
Write the 2×2 determinant that remains. The image displays the mathematical expression 'minor of c₂' next to a 2x2 matrix. The matrix contains the elements 4, -2 in the first row and -2, -4 in the second row, enclosed by vertical bars.
Evaluate. A mathematical expression showing the calculation: 4 multiplied by -4, minus the product of -2 and -2.
Simplify. The number negative twenty (-20) isolated on a white background.

For the determinant |1−1402−1−2−33|, find and then evaluate the minor of ⓐ a1 ⓑ b2 ⓒ c3.

Solution

ⓐ 3 ⓑ 11 ⓒ 2

For the determinant |−2−1030−1−1−23|, find and then evaluate the minor of ⓐ a2 ⓑ b3 ⓒ c2.

Solution

ⓐ −3 ⓑ 2 ⓒ 3

We are now ready to evaluate a 3×3 determinant. To do this we expand by minors, which allows us to evaluate the 3×3 determinant using 2×2 determinants—which we already know how to evaluate!

To evaluate a 3×3 determinant by expanding by minors along the first row, we use the following pattern:

A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.

Remember, to find the minor of an entry we eliminate the row and column that contains the entry.

Expanding by Minors along the First Row to Evaluate a 3×3 Determinant

To evaluate a 3×3 determinant by expanding by minors along the first row, the following pattern:

A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.

Evaluate the determinant |2−3−1320−1−1−2| by expanding by minors along the first row.

Solution
A 3x3 matrix with numerical entries. The first row, in red, is 2, -3, -1. The second row is 3, 2, 0. The third row is -1, -1, -2.
Expand by minors along the first row Calculation of a 3x3 matrix determinant using cofactor expansion, with minors of 2, -3, and -1 explicitly shown.
Evaluate each determinant. A mathematical expression reads 2(4-0)+3(6-0)-1(-3-(-2)), presented in a horizontal line with red-orange coloring on the leading coefficients and plus/minus signs.
Simplify. A mathematical expression displays 2 multiplied by -4, added to 3 multiplied by -6, and then subtracting 1 multiplied by -1. The calculation is 2(-4) + 3(-6) - 1(-1).
Simplify. A mathematical expression displays the numbers -8, -18, and +1, representing an arithmetic problem involving addition and subtraction of integers.
Simplify. The number -25 is displayed in a dark gray font on a plain white background.

Evaluate the determinant |3−240−1−223−1|, by expanding by minors along the first row.

Solution

37

Evaluate the determinant |3−2−22−14−10−3|, by expanding by minors along the first row.

Solution

7

To evaluate a 3×3 determinant we can expand by minors using any row or column. Choosing a row or column other than the first row sometimes makes the work easier.

When we expand by any row or column, we must be careful about the sign of the terms in the expansion. To determine the sign of the terms, we use the following sign pattern chart.

|+−+−+−+−+|

Sign Pattern

When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern.

|+−+−+−+−+|

Notice that the sign pattern in the first row matches the signs between the terms in the expansion by the first row.

A 3 by 3 determinant has row 1: plus, minus, plus, row 2: minus, plus, minus and row 3: plus, minus, plus. The three signs in the first row each point to a minor determinant in the expansion of a 3 by 3 determinant. Plus points to minor of a1, minus to the minor of b1 and plus to the minor of c1.

Since we can expand by any row or column, how do we decide which row or column to use? Usually we try to pick a row or column that will make our calculation easier. If the determinant contains a 0, using the row or column that contains the 0 will make the calculations easier.

Evaluate the determinant |4−1−33025−4−3| by expanding by minors.

Solution

To expand by minors, we look for a row or column that will make our calculations easier. Since 0 is in the second row and second column, expanding by either of those is a good choice. Since the second row has fewer negatives than the second column, we will expand by the second row.

A 3x3 determinant with elements: first row (4, -1, -3), second row (3, 0, 2) in red, and third row (5, -4, -3).
Expand using the second row.
Be careful of the signs. This image illustrates the process of calculating a 3x3 determinant using cofactor expansion. It shows the alternating sign pattern for cofactors and the terms for the expansion, where each term is a coefficient multiplied by its 2x2 minor.
Evaluate each determinant. A mathematical expression featuring negative numbers, parentheses, and arithmetic operations: -3(3-12)+0(-12-(-15))-2(-16-(-5)).
Simplify. A mathematical expression featuring negative numbers, multiplication, and addition/subtraction: -3(-9) + 0 - 2(-11).
Simplify. The image displays the mathematical expression '27 + 0 + 22' in bold, dark gray text against a plain white background.
Add. The number 49 is displayed.

Evaluate the determinant |2−1−303−43−4−3| by expanding by minors.

Solution

−11

Evaluate the determinant |−2−1−3−1224−40| by expanding by minors.

Solution

−12

Use Cramer’s Rule to Solve Systems of Equations

Cramer’s Rule is a method of solving systems of equations using determinants. It can be derived by solving the general form of the systems of equations by elimination. Here we will demonstrate the rule for both systems of two equations with two variables and for systems of three equations with three variables.

Let’s start with the systems of two equations with two variables.

Cramer’s Rule for Solving a System of Two Equations

For the system of equations {a1x+b1y=k1a2x+b2y=k2, the solution (x,y) can be determined by

x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants

Notice that to form the determinant D, we use take the coefficients of the variables.

The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is D with row 1: a1, b1 and row 2: a2, b2. Column 1 has coefficients of x and column 2 has coefficients of

Notice that to form the determinant Dx and Dy, we substitute the constants for the coefficients of the variable we are finding.

The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is Dx has row 1: k1, b1 and row 2: k2, b2. Here columns 1 and 2 re constants and coefficients of y respectively. Determinant Dy has row 1: a1, k1 and row 2: a2, k2. Here, columns 1 and 2 are coefficients of x and constants respectively.

How to Solve a System of Equations Using Cramer’s Rule

Solve using Cramer’s Rule: {2x+y=−43x−2y=−6.

Solution
The equations are 2x plus y equals minus 4 and 3x minus 2y equals minus 6. Step 1. Evaluate the determinant D, using the coefficients of the variables. Determinant D has row 1: 2, 1 and row 2: 3, minus 2. So, D is minus 7. Step 2. Evaluate the determinant Dx. Use the constants in place of the x coefficients. We replace the coefficients of x, 2 and 3, with the constants, negative 4 and negative 6. We get Dx equal to 14. Step 3. Evaluate the determinant Dy. Use the constants in place of the y coefficients. We replace the coefficients of y, 1 and 2, with the constants, negative 4 and negative 6. We get Dy equal to 0. Step 4. Find x and y. Substituting values of D, Dx and Dy in the equations x equal to Dx upon D and y equal to Dy upon D, we get x equal to minus 2 and y equal to 0. Step 5. Write the solution as an ordered pair minus 2, 0. Step 6. Check that the ordered pair is a solution to both original equations.

Solve using Cramer’s rule: {3x+y=−32x+3y=6.

Solution

(−157,247)

Solve using Cramer’s rule: {−x+y=22x+y=−4.

Solution

(−2,0)

Solve a system of two equations using Cramer’s rule.

  1. Evaluate the determinant D, using the coefficients of the variables.
  2. Evaluate the determinant Dx. Use the constants in place of the x coefficients.
  3. Evaluate the determinant Dy. Use the constants in place of the y coefficients.
  4. Find x and y. x=DxD, y=DyD
  5. Write the solution as an ordered pair.
  6. Check that the ordered pair is a solution to both original equations.

To solve a system of three equations with three variables with Cramer’s Rule, we basically do what we did for a system of two equations. However, we now have to solve for three variables to get the solution. The determinants are also going to be 3×3 which will make our work more interesting!

Cramer’s Rule for Solving a System of Three Equations

For the system of equations {a1x+b1y+c1z=k1a2x+b2y+c2z=k2a3x+b3y+c3z=k3, the solution (x,y,z) can be determined by

The image shows equations for x, y, and z expressed as determinants. Each determinant uses variables a through c and k through d. The first determinant's first column contains the variable x coefficients. The second determinant’s first column contains the variable y coefficients. The third determinant’s first column contains the variable z coefficients.

Solve the system of equations using Cramer’s Rule: {3x−5y+4z=55x+2y+z=02x+3y−2z=3.

Solution
Evaluate the determinant D. Matrix D with highlighted elements in the first column, showing their rows: (3, -5, 4), (5, 2, 1) and (2, 3, -2).
Expand by minors using column 1.
Be careful with the signs. Expansion of a 3x3 determinant by cofactors to calculate D.
Evaluate the determinants. A mathematical equation involving operations with whole numbers, including multiplication and subtraction within parentheses to be solved.
Simplify. The image shows the equation D = 3(-7) - 5(-2) + 2(-13), a mathematical operation with integers.
Simplify. A mathematical equation showing D is equal to negative 21 plus 10 minus 26.
Simplify. The image shows the equation D = -37.
Evaluate the determinant Dx. Use the
constants to replace the coefficients of x.
A 3x3 Dx matrix with elements 5, -5, 4 in the first row; 0, 2, 1 in the second; and 3, 3, -2 in the third. The first column is highlighted in red.
Expand by minors using column 1. Calculation of the determinant D_x using cofactor expansion, showing the algebraic sum of products of elements and subdeterminants.
Evaluate the determinants. A mathematical equation for Dx is shown on a white background: Dx = 5(-4-3) - 0(10-12) + 3(-5-8). The numbers 5, 0, and 3 are highlighted in red.
Simplify. Mathematical expression: Dx is equal to 5 multiplied by -7, minus 0, plus 3 multiplied by -13. Numbers in red highlight specific terms of the operation.
Simplify. The equation D subscript x equals -74 is displayed on a white background.
Evaluate the determinant Dy. Use the
constants to replace the coefficients of y.
The 3x3 matrix D_y with numeric elements, where the second column (5, 0, 3) is highlighted in red.
Math instructions: 'Expand by minors using column 2. Be careful with the signs.' A 3x3 matrix is ​​shown with the pattern of alternating signs (+ - + / - + - / + - +) to remember. Calculation of Dy through the expansion of a 3x3 determinant. The numbers in red indicate the elements used in the cofactor expansion for the first row.
Evaluate the determinants. A mathematical equation that calculates Dy: Dy = -5(-10-2) + 0(-10-12) - 3(3-20). Some numbers appear in red, possibly indicating key steps or coefficients.
Simplify. A mathematical equation, Dy = -5(-12) + 0 - 3(-17), is shown in black text on a white background, detailing a calculation with positive and negative integers.
Simplify. Equation showing Dy equal to the sum of 60, 0, and 51.
Simplify. D sub y is equal to 111.
Evaluate the determinant Dz. Use the
constants to replace the coefficients of z.
D_z matrix with the third column highlighted in red, showing the values ​​5, 0 and 3.
Expand by minors using column 3. Be careful with signs. (A sign matrix for cofactor expansion is shown) Calculation of the determinant D_x through expansion by cofactors.
Evaluate the determinants. A mathematical equation is displayed: Dx = 5(15-4) - 0(9-(-10)) + 3(6-(-25)).
Simplify. A mathematical expression for D subscript x showing the calculation as 5 multiplied by 11, minus 0, plus 3 multiplied by 31.
Simplify. Mathematical equation with D sub x equal to fifty-five minus zero plus ninety-three, showing a simple arithmetic operation.
Simplify. A close-up view of the equation Dx = 148 against a white background.
Find x, y, and z. Formulas to calculate x, y, and z using the determinants Dx, Dy, Dz, and the principal determinant D, as in Cramer's Rule.
Substitute in the values. The equations show values ​​for x, y, and z as fractions: x = -74/-37, y = 111/-37, and z = 148/-37.
Simplify. The values ​​of the variables are shown: x is equal to 2, y is equal to -3, and z is equal to -4.
Write the solution as an ordered triple. A set of three-dimensional coordinates, (2, -3, -4), is displayed in dark gray text against a plain white background.
Check that the ordered triple is a solution
to all three original equations.
We leave the check to you.
The solution is (2,−3,−4).

Solve the system of equations using Cramer’s Rule: {3x+8y+2z=−52x+5y−3z=0x+2y−2z=−1.

Solution

(−9,3,−1)

Solve the system of equations using Cramer’s Rule: {3x+y−6z=−32x+6y+3z=03x+2y−3z=−6.

Solution

(−6,3,−2)

Cramer’s rule does not work when the value of the D determinant is 0, as this would mean we would be dividing by 0. But when D=0, the system is either inconsistent or dependent.

When the value of D=0 and Dx,Dy and Dz are all zero, the system is consistent and dependent and there are infinitely many solutions.

When the value of D=0 and Dx,Dy and Dz are not all zero, the system is inconsistent and there is no solution.

Dependent and Inconsistent Systems of Equations

For any system of equations, where the value of the determinant D=0,
Value of determinantsType of systemSolutionD=0andDx,DyandDzare all zeroconsistent and dependentinfinitely many solutionsD=0andDx,DyandDzare not all zeroinconsistentno solution

In the next example, we will use the values of the determinants to find the solution of the system.

Solve the system of equations using Cramer’s rule : {x+3y=4−2x−6y=3.

Solution

{x+3y=4−2x−6y=3 Evaluate the determinantD,using thecoefficients of the variables.D=|13−2−6| D=−6−(−6)D=0

We cannot use Cramer’s Rule to solve this system. But by looking at the value of the determinants Dx and Dy, we can determine whether the system is dependent or inconsistent.

Evaluate the determinantDx.Dx=|433−6|Dx=−24−9Dx=-33

Since all the determinants are not zero, the system is inconsistent. There is no solution.

Solve the system of equations using Cramer’s rule: {4x−3y=88x−6y=14.

Solution

no solution

Solve the system of equations using Cramer’s rule: {x=−3y+42x+6y=8.

Solution

infinite solutions

Solve Applications using Determinants

An interesting application of determinants allows us to test if points are collinear. Three points (x1,y1), (x2,y2) and (x3,y3) are collinear if and only if the determinant below is zero.

|x1y11x2y21x3y31|=0

Test for Collinear Points

Three points (x1,y1), (x2,y2) and (x3,y3) are collinear if and only if

|x1y11x2y21x3y31|=0

We will use this property in the next example.

Determine whether the points (5,−5), (4,−3), and (3,−1) are collinear.

Solution
Mathematical matrix or determinant showing three points (x1, y1), (x2, y2), (x3, y3) and a column of ones, commonly used in geometry for area calculations or collinearity tests.
Substitute the values into the determinant.
(5,−5), (4,−3), and (3,−1)
A 3x3 determinant is shown with values 5, -5, 1 in the first row; 4, -3, 1 in the second; and 3, -1, 1 in the third row.
Evaluate the determinant by expanding
by minors using column 3.
A mathematical expression showing the cofactor expansion of a 3x3 determinant, represented as a sum and difference of three 2x2 determinants with coefficients 1, -1, and 1.
Evaluate the determinants. A mathematical expression involving integers and operations is shown, featuring the sum and differences of parenthesized terms: 1(-4-(-9)) - 1(-5-(-15)) + 1(-15-(-20)).
Simplify. A mathematical expression 1(5) - 1(10) + 1(5) is shown on a white background, which simplifies to 0.
Simplify. A minimal black vertical line with a slight oval curvature is centered on a plain white background, creating a stark and simple abstract design.
The value of the determinant is 0, so the
points are collinear.

Determine whether the points (3,−2), (5,−3), and (1,−1) are collinear.

Solution

yes

Determine whether the points (−4,−1), (−6,2), and (−2,−4) are collinear.

Solution

yes

Access these online resources for additional instruction and practice with solving systems of linear equations using determinants or Cramer's Rule.

  • Solving Systems of Linear Equations Using Determinants
  • Solving Systems of Linear Equations Using Cramer's Rule

Key Concepts

  • Determinant: The determinant of any square matrix [abcd], where a, b, c, and d are real numbers, is
    |abcd|=ad−bc
  • Expanding by Minors along the First Row to Evaluate a 3 × 3 Determinant: To evaluate a 3×3 determinant by expanding by minors along the first row, the following pattern:
    A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.
  • Sign Pattern: When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern.
    |+−+−+−+−+|
  • Cramer’s Rule: For the system of equations {a1x+b1y=k1a2x+b2y=k2, the solution (x,y) can be determined by
    x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants.
    Notice that to form the determinant D, we use take the coefficients of the variables.
  • How to solve a system of two equations using Cramer’s rule.
    1. Evaluate the determinant D, using the coefficients of the variables.
    2. Evaluate the determinant Dx. Use the constants in place of the x coefficients.
    3. Evaluate the determinant Dy. Use the constants in place of the y coefficients.
    4. Find x and y. x=DxD, y=DyD.
    5. Write the solution as an ordered pair.
    6. Check that the ordered pair is a solution to both original equations.
    7. Dependent and Inconsistent Systems of Equations: For any system of equations, where the value of the determinant D=0,
      Value of determinantsType of systemSolutionD=0andDx,DyandDzare all zeroconsistent and dependentinfinitely many solutionsD=0andDx,DyandDzare not all zeroinconsistentno solution
    8. Test for Collinear Points: Three points (x1,y1), (x2,y2), and (x3,y3) are collinear if and only if
      |x1y11x2y21x3y31|=0

Practice Makes Perfect

Evaluate the Determinant of a 2 × 2 Matrix

In the following exercises, evaluate the determinant of each square matrix.

[6−23−1]

[−48−35]

Solution

4

[−350−4]

[−207−5]

Solution

10

Evaluate the Determinant of a 3 × 3 Matrix

In the following exercises, find and then evaluate the indicated minors.

|3−14−10−2−415|
Find the minor ⓐ a1 ⓑ b2 ⓒ c3

|−1−324−2−1−20−3|
Find the minor ⓐ a1 ⓑ b1 ⓒ c2

Solution

ⓐ 6 ⓑ −14 ⓒ −6

|2−3−4−12−30−1−2|
Find the minor ⓐ a2 ⓑ b2 ⓒ c2

|−2−231−30−23−2|
Find the minor ⓐ a3 ⓑ b3 ⓒ c3

Solution

ⓐ 9 ⓑ −3 ⓒ 8

In the following exercises, evaluate each determinant by expanding by minors along the first row.

|−23−1−12−231−3|

|4−1−2−3−21−2−57|

Solution

−77

|−2−3−45−67−120|

|13−25−640−2−1|

Solution

49

In the following exercises, evaluate each determinant by expanding by minors.

|−5−1−440−32−26|

|4−133−22−104|

Solution

−24

|354−130−261|

|2−4−35−1−4320|

Solution

25

Use Cramer’s Rule to Solve Systems of Equations

In the following exercises, solve each system of equations using Cramer’s Rule.

{−2x+3y=3x+3y=12

{x−2y=−52x−3y=−4

Solution

(7,6)

{x−3y=−92x+5y=4

{2x+y=−43x−2y=−6

Solution

(−2,0)

{x−2y=−52x−3y=−4

{x−3y=−92x+5y=4

Solution

(−3,2)

{5x−3y=−12x−y=2

{3x+8y=−32x+5y=−3

Solution

(−9,3)

{6x−5y+2z=32x+y−4z=53x−3y+z=−1

{4x−3y+z=72x−5y−4z=33x−2y−2z=−7

Solution

(−3,−5,4)

{2x−5y+3z=83x−y+4z=7x+3y+2z=−3

{11x+9y+2z=−97x+5y+3z=−74x+3y+z=−3

Solution

(2,−3,−2)

{x+2z=04y+3z=−22x−5y=3

{2x+5y=43y−z=34x+3z=−3

Solution

(−3,2,3)

{2y+3z=−15x+3y=−67x+z=1

{3x−z=−35y+2z=−64x+3y=−8

Solution

(−2,0,−3)

{2x+y=36x+3y=9

{x−4y=−1−3x+12y=3

Solution

infinitely many solutions

{−3x−y=46x+2y=−16

{4x+3y=220x+15y=5

Solution

inconsistent

{x+y−3z=−1y−z=0−x+2y=1

{2x+3y+z=12x+y+z=93x+4y+2z=20

Solution

inconsistent

{3x+4y−3z=−22x+3y−z=−12x+y−2z=6

{x−2y+3z=1x+y−3z=73x−4y+5z=7

Solution

infinitely many solutions

Solve Applications Using Determinants

In the following exercises, determine whether the given points are collinear.

(0,1), (2,0), and (−2,2).

(0,−5), (−2,−2), and (2,−8).

Solution

yes

(4,−3), (6,−4), and (2,−2).

(−2,1), (−4,4), and (0,−2).

Solution

yes

Writing Exercises

Explain the difference between a square matrix and its determinant. Give an example of each.

Explain what is meant by the minor of an entry in a square matrix.

Solution

Answers will vary.

Explain how to decide which row or column you will use to expand a 3×3 determinant.

Explain the steps for solving a system of equations using Cramer’s rule.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 4 rows and a header row. The header row labels each column: I ca, confidently, with some help and no, I don’t get it. The first column has the following statements: Evaluate the Determinant of a 2 by 2 Matrix, Evaluate the Determinant of a 3 by 3 Matrix, Use Cramer’s Rule to Solve Systems of Equations, Solve Applications Using Determinants. The remaining columns are blank.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

determinant
Each square matrix has a real number associated with it called its determinant.
minor of an entry in a 3×3 determinant
The minor of an entry in a 3×3 determinant is the 2×2 determinant found by eliminating the row and column in the 3×3 determinant that contains the entry.
square matrix
A square matrix is a matrix with the same number of rows and columns.

Graphing Systems of Linear Inequalities

Learning Objectives

By the end of this section, you will be able to:

  • Determine whether an ordered pair is a solution of a system of linear inequalities
  • Solve a system of linear inequalities by graphing
  • Solve applications of systems of inequalities

Before you get started, take this readiness quiz.

Solve the inequality 2a<5a+12.
If you missed this problem, review Example 5 in Solve Linear Inequalities.

Solution

−4<a

Determine whether the ordered pair (3,12) is a solution to the system y>2x+3.
If you missed this problem, review Example 1 in Graph Linear Inequalities in Two Variables.

Solution

no

Determine whether an ordered pair is a solution of a system of linear inequalities

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations.

System of Linear Inequalities

Two or more linear inequalities grouped together form a system of linear inequalities.

A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here.

{x+4y≥103x−2y<12

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs (x,y) that make both inequalities true.

Solutions of a System of Linear Inequalities

Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true.

The solution of a system of linear inequalities is shown as a shaded region in the x, y coordinate system that includes all the points whose ordered pairs make the inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Determine whether the ordered pair is a solution to the system {x+4y≥103x−2y<12.

ⓐ (−2,4) ⓑ (3,1)

Solution

ⓐ Is the ordered pair (−2,4) a solution?

We substitute x equal to negative 2 and y equal to 4 into both inequalities. First inequality is x plus 4 times y greater than or equal to 10. So negative 2 plus 4 open parentheses 4 close parenthesis is greater than or equal to 10 or not. 14 is greater than or equal to 10 is true. Second inequality, 3 times x minus 2 times y is less than 12. Three open parentheses negative 2 close parentheses minus two open parentheses 4 close parentheses is less than 12 or not. Negative 14 is less than 12 is true.

The ordered pair (−2,4) made both inequalities true. Therefore (−2,4) is a solution to this system.

ⓑ Is the ordered pair (3,1) a solution?

We substitute x equal to three and y equal to one into both inequalities. First inequality is x plus four times y greater than or equal to ten. So three plus four open parentheses one close parenthesis is greater than or equal to ten or not. Seven greater than or equal to ten is false. Second inequality, three times x minus two times y is less than twelve. Three open parentheses three close parentheses minus two open parentheses one close parentheses is less than twelve or not. Seven less than 12 holds true.

The ordered pair (3,1) made one inequality true, but the other one false. Therefore (3,1) is not a solution to this system.

Determine whether the ordered pair is a solution to the system: {x−5y>102x+3y>−2.

ⓐ (3,−1) ⓑ (6,−3)

Solution

ⓐ no ⓑ yes

Determine whether the ordered pair is a solution to the system: {y>4x−24x−y<20.

ⓐ (−2,1) ⓑ (4,−1)

Solution

ⓐ yes ⓑ no

Solve a System of Linear Inequalities by Graphing

The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

How to Solve a System of Linear Inequalities by Graphing

Solve the system by graphing: {y≥2x−1y<x+1.

Solution
Step 1. Graph the first inequality. We graph y less than 2x minus 1. Graph the boundary line y equal to 2x minus 1. It is a solid line because the inequality sign is less than. Shade in the side of the boundary line where the inequality is true. We choose 0, 0 as a test point. It is a solution to the equation, so we shade in above the boundary line. Step 2. On the same grid, graph the second inequality y less than x plus 1. Graph the boundary line y equal to x plus 1. It is a dashed line because the inequality sign is less than. Shade in the side of that boundary line where the inequality is true. Again, we use 0, 0 as a test point. It is a solution so we shade below the line y equals x plus 1. Step 3. The solution is the region where the shading overlaps. The point where the boundary lines intersect is not a solution because it is not a solution to y less than x plus 1. The solution is all points in the area bounded by the lines on the bottom left. Step 4. Check by choosing a test point. We use minus 1, minus 1. Substituting in the inequality y less than 2x minus 1, we get minus 1 less than minus 3 which is true. Hence, it is a solution. Similarly, it is also true for the other inequality. The region containing minus 1, minus 1 is the solution to this system.

Solve the system by graphing: {y<3x+2y>−x−1.

Solution

The figure shows a graph plotted for the inequalities y less than three times x plus two and y greater than minus x minus one. Two lines intersect each other on the graph. An area to the right of both the lines is colored in grey. It is the solution.
The solution is the grey region.

Solve the system by graphing: {y<−12x+3y<3x−4.

Solution

The figure shows the graph plotted for the inequalities y less than minus half of x plus three and y less than three times x minus four. Two intersecting lines are shown on the graph. The area bound by the two lines to the bottom right is shown in grey. It is the solution.
The solution is the grey region.

Solve a system of linear inequalities by graphing.

  1. Graph the first inequality.
    • Graph the boundary line.
    • Shade in the side of the boundary line where the inequality is true.
  2. On the same grid, graph the second inequality.
    • Graph the boundary line.
    • Shade in the side of that boundary line where the inequality is true.
  3. The solution is the region where the shading overlaps.
  4. Check by choosing a test point.

Solve the system by graphing: {x−y>3y<−15x+4.

Solution
{x−y>3y<−15x+4
Graph x − y > 3, by graphing x − y = 3
and testing a point.

The intercepts are x = 3 and y = −3 and the
boundary line will be dashed.

Test (0, 0) which makes the inequality false so shade
(red) the side that does not contain (0, 0).
A coordinate plane displays the graph of the linear inequality y < (1/2)x - 3. A dashed line represents the boundary, and the region below it is shaded in orange.
Graph y<−15x+4 by graphing y=−15x+4
using the slope m=−15 and y-intercept b = 4.
The boundary line will be dashed

Test (0, 0) which makes the inequality true, so
shade (blue) the side that contains (0, 0).

Choose a test point in the solution and verify that it is a solution to both inequalties.
A Cartesian coordinate graph displays two linear inequalities. The area above the dashed blue line is shaded light blue, representing one solution set. The area below the dashed red line is shaded dark grey, representing another solution set.

The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region.

Solve the system by graphing: {x+y≤2y≥23x−1.

Solution

The figure shows the graph for the inequalities x plus y less than or equal to two and y greater than or equal to two by three of x minus one. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.
The solution is the grey region.

Solve the system by graphing: {3x−2y≤6y>−14x+5.

Solution

The figure shows graph for the inequalities three times x minus two times y less than or equal to six and y greater than or equal to minus one by four of x plus five. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.
The solution is the grey region.

Solve the system by graphing: {x−2y<5y>−4.

Solution
{x−2y<5y>−4
Graph x−2y<5, by graphing x−2y=5
and testing a point. The intercepts are
x = 5 and y = −2.5 and the
boundary line will be dashed.

Test (0, 0) which makes the inequality true, so shade
(red) the side that contains (0, 0).
A graph displays the solution set for the inequality y > (1/2)x - 2. A dashed red line represents y = (1/2)x - 2, and the region above the line is shaded orange.
Graph y>−4, by graphing y=−4 and
recognizing that it is a horizontal line
through y=−4. The boundary line will
be dashed.

Test (0, 0) which makes the inequality
true so shade (blue) the side that contains (0, 0).
Graph of a system of linear inequalities with two dashed lines, y=1/2x-3 and y=-4, showing three distinct shaded regions: gray, light blue, and light orange.

The point (0,0) is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed.

The solution is the area shaded twice—which appears as the darkest shaded region.

Solve the system by graphing: {y≥3x−2y<−1.

Solution

The figure shows graph for the inequalities y greater than or equal to three times x minus two and y less than minus one. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution
The solution is the grey region.

Solve the system by graphing: {x>−4x−2y≥−4.

Solution

The figure shows graph for the inequalities x greater than or equal to minus four and x minus two times y greater than minus four. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.
The solution is the grey region.

Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in the next example.

Solve the system by graphing: {4x+3y≥12y<−43x+1.

Solution
{4x+3y≥12y<−43x+1
Graph 4x+3y≥12, by graphing 4x+3y=12
and testing a point. The intercepts are x = 3
and y = 4 and the boundary line will be solid.

Test (0, 0) which makes the inequality false, so
shade (red) the side that does not contain (0, 0).
A coordinate plane displaying a linear inequality: the area below the dashed line y = -x + 4 is shaded, indicating the solution set.
Graph y<−43x+1 by graphing y=−43x+1
using the slope m=−43 and y-intercept
b = 1. The boundary line will be dashed.

Test (0, 0) which makes the inequality true, so
shade (blue) the side that contains (0, 0).
A graph showing two linear inequalities. A solid red line represents y = -x + 4, with the region y > -x + 4 shaded orange. A dashed blue line represents y = -x, with the region y < -x shaded blue.

There is no point in both shaded regions, so the system has no solution.

Solve the system by graphing: {3x−2y≥12y≥32x+1.

Solution

The graph of three times x minus two times y greater than or equal to twelve and y greater than or equal to three by two of x plus one is shown. Two intersecting lines are shown. The inequalities do not have a solution.
No solution.

Solve the system by graphing: {x+3y>8y<−13x−2.

Solution

The graph of x plus three times y greater than eight and y less than minus one by three of x minus two is shown. Two intersecting lines are shown. The inequalities do not have a solution.
No solution.

Some systems of linear inequalities where the boundary lines are parallel will have a solution. We’ll see this in the next example.

Solve the system by graphing: {y>12x−4x−2y<−4.

Solution
{y>12x−4x−2y<−4
Graph y>12x−4 by graphing y=12x−4
using the slope m=12 and the intercept
b = −4. The boundary line will be dashed.

Test (0, 0) which makes the inequality true, so
shade (red) the side that contains (0, 0).
A graph on a coordinate plane shows a dashed red line passing through (0,-4) and (8,0). The region above the line is shaded in orange, representing the solution set of a linear inequality.
Graph x−2y<−4 by graphing x−2y=−4
and testing a point. The intercepts are
x = −4 and y = 2 and the boundary line will be dashed.

Choose a test point in the solution and verify
that it is a solution to both inequalties.

Test (0, 0) which makes the inequality false, so
shade (blue) the side that does not contain (0, 0).
A coordinate plane shows two parallel dashed lines, y = 0.5x + 2 and y = 0.5x - 4. The region above the top line is dark gray, and the region between the lines is light orange.

No point on the boundary lines is included in the solution as both lines are dashed.

The solution is the region that is shaded twice which is also the solution to x−2y<−4.

Solve the system by graphing: {y≥3x+1−3x+y≥−4.

Solution

The figure shows the graph of the inequalities y greater than or equal to three times x plus one and minus three times x plus y greater than or equal to minus four. Two parallel lines are shown and the region to the left of both is colored in grey. It is the solution.
The solution is the grey region.

Solve the system by graphing: {y≤−14x+2x+4y≤4.

Solution

The figure shows the graph of the inequalities y less than or equal to minus one fourth of x plus 2 and x plus four times y less than or equal to four. Two parallel lines are shown and the region to the bottom of both is colored in grey. It is the solution.
The solution is the grey region.

Solve Applications of Systems of Inequalities

The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so we add inequalities to the system as additional requirements.

Christy sells her photographs at a booth at a street fair. At the start of the day, she wants to have at least 25 photos to display at her booth. Each small photo she displays costs her $4 and each large photo costs her $10. She doesn’t want to spend more than $200 on photos to display.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could she display 10 small and 20 large photos?
ⓓ Could she display 20 small and 10 large photos?

Solution
ⓐ
This table illustrates the process of translating a word problem about purchasing photos into a system of linear inequalities, defining variables and formulating constraints.
Let x=the number of small photos.
y=the number of large photos
To find the system of equations translate the information.
She wants to have at least 25 photos.
The number of small plus the number of large should be at least 25.
x+y≥25
$4 for each small and $10 for each large must be no more than $200
4x+10y≤200
The number of small photos must be greater than or equal to 0.
x≥0
The number of large photos must be greater than or equal to 0.
y≥0
We have our system of equations.
{x+y≥254x+10y≤200x≥0y≥0
ⓑ
Since x≥0 and y≥0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.
To graph x+y≥25, graph x+y=25 as a solid line.
Choose (0, 0) as a test point. Since it does not make the inequality true, shade (red) the side that does not include the point (0, 0).

To graph 4x+10y≤200, graph 4x+10y=200 as a solid line.
Choose (0, 0) as a test point. Since it does make the inequality true, shade (blue) the side that include the point (0, 0).
A graph on a coordinate plane shows two downward-sloping lines representing linear inequalities. Individual shaded regions highlight their solutions, with a darker overlapping area indicating the feasible region.

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly-shaded section are included in the solution as are the points on the x-axis from (25, 0) to (55, 0).

ⓒ To determine if 10 small and 20 large photos would work, we look at the graph to see if the point (10, 20) is in the solution region. We could also test the point to see if it is a solution of both equations.

It is not, Christy would not display 10 small and 20 large photos.

ⓓ To determine if 20 small and 10 large photos would work, we look at the graph to see if the point (20, 10) is in the solution region. We could also test the point to see if it is a solution of both equations.

It is, so Christy could choose to display 20 small and 10 large photos.

Notice that we could also test the possible solutions by substituting the values into each inequality.

A trailer can carry a maximum weight of 160 pounds and a maximum volume of 15 cubic feet. A microwave oven weighs 30 pounds and has 2 cubic feet of volume, while a printer weighs 20 pounds and has 3 cubic feet of space.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could 4 microwaves and 2 printers be carried on this trailer?
ⓓ Could 7 microwaves and 3 printers be carried on this trailer?

Solution

ⓐ {30m+20p≤1602m+3p≤15
ⓑ
The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.
ⓒ yes
ⓓ no

Mary needs to purchase supplies of answer sheets and pencils for a standardized test to be given to the juniors at her high school. The number of the answer sheets needed is at least 5 more than the number of pencils. The pencils cost $2 and the answer sheets cost $1. Mary’s budget for these supplies allows for a maximum cost of $400.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could Mary purchase 100 pencils and 100 answer sheets?
ⓓ Could Mary purchase 150 pencils and 150 answer sheets?

Solution

ⓐ {a≥p+5a+2p≤400
ⓑ
The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.
ⓒ no
ⓓ no

When we use variables other than x and y to define an unknown quantity, we must change the names of the axes of the graph as well.

Omar needs to eat at least 800 calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than $5. At the hamburger restaurant near his college, each hamburger has 240 calories and costs $1.40. Each cookie has 160 calories and costs $0.50.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he eat 3 hamburgers and 1 cookie?
ⓓ Could he eat 2 hamburgers and 4 cookies?

Solution
ⓐ
Demonstrates how to formulate a system of linear inequalities from a word problem by defining variables and translating calorie and cost constraints for hamburgers and cookies.
Let h=the number of hamburgers.
c=the number of cookies
To find the system of equations translate the information.
The calories from hamburgers at 240 calories each, plus the calories from cookies at 160 calories each
must be more that 800.
240h+160c≥800
The amount spent on hamburgers at $1.40 each, plus the amount spent on cookies at $0.50 each
must be no more than $5.00.
1.40h+0.50c≤5
The number of hamburgers must be greater than or equal to 0.
h≥0
The number of cookies must be greater than or equal to 0.
c≥0
We have our system of equations. {240h+160c≥8001.40h+0.50c≤5h≥0c≥0
ⓑ
Since h>=0 and c>=0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.
To graph 240h+160c≥800, graph 240h+160c=800 as a solid line.


Choose (0, 0) as a test point. Since it does not make the inequality true, shade (red) the side that does not include the point (0, 0).
A graph on an h-c coordinate plane shows two intersecting lines and three distinct shaded regions, representing solutions to a system of inequalities.

Graph 1.40h+0.50c≤5. The boundary line is 1.40h+0.50c=5. We test (0, 0) and it makes the inequality true. We shade the side of the line that includes (0, 0).

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the x-axis from (5, 0) to (10, 0).

ⓒ To determine if 3 hamburgers and 1 cookie would meet Omar’s criteria, we see if the point (3, 2) is in the solution region. It is, so Omar might choose to eat 3 hamburgers and 1 cookie.

ⓓ To determine if 2 hamburgers and 4 cookies would meet Omar’s criteria, we see if the point (2, 4) is in the solution region. It is, Omar might choose to eat 2 hamburgers and 4 cookies.

We could also test the possible solutions by substituting the values into each inequality.

Tenison needs to eat at least an extra 1,000 calories a day to prepare for running a marathon. He has only $25 to spend on the extra food he needs and will spend it on $0.75 donuts which have 360 calories each and $2 energy drinks which have 110 calories.

ⓐ Write a system of inequalities that models this situation.
ⓑ Graph the system.
ⓒ Can he buy 8 donuts and 4 energy drinks and satisfy his caloric needs?
ⓓ Can he buy 1 donut and 3 energy drinks and satisfy his caloric needs?

Solution

ⓐ {0.75d+2e≤25360d+110e≥1000
ⓑ
The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.
ⓒ yes
ⓓ no

Philip’s doctor tells him he should add at least 1,000 more calories per day to his usual diet. Philip wants to buy protein bars that cost $1.80 each and have 140 calories and juice that costs $1.25 per bottle and have 125 calories. He doesn’t want to spend more than $12.

ⓐ Write a system of inequalities that models this situation.
ⓑ Graph the system.
ⓒ Can he buy 3 protein bars and 5 bottles of juice?
ⓓ Can he buy 5 protein bars and 3 bottles of juice?

Solution

ⓐ {140p+125j≥10001.80p+1.25j≤12
ⓑ
The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.
ⓒ yes
ⓓ no

Access these online resources for additional instruction and practice with solving systems of linear inequalities by graphing.

  • Solving Systems of Linear Inequalities by Graphing
  • Systems of Linear Inequalities

Key Concepts

  • Solutions of a System of Linear Inequalities: Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the x, y coordinate system that includes all the points whose ordered pairs make the inequalities true.
  • How to solve a system of linear inequalities by graphing.
    1. Graph the first inequality.
      Graph the boundary line.
      Shade in the side of the boundary line where the inequality is true.
    2. On the same grid, graph the second inequality.
      Graph the boundary line.
      Shade in the side of that boundary line where the inequality is true.
    3. The solution is the region where the shading overlaps.
    4. Check by choosing a test point.

Section Exercises

Practice Makes Perfect

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

In the following exercises, determine whether each ordered pair is a solution to the system.

{3x+y>52x−y≤10

ⓐ (3,−3) ⓑ (7,1)

{4x−y<10−2x+2y>−8

ⓐ (5,−2) ⓑ (−1,3)

Solution

ⓐ false ⓑ true

{y>23x−5x+12y≤4

ⓐ (6, −4) ⓑ (3, 0)

{y<32x+334x−2y<5

ⓐ (−4,−1) ⓑ (8, 3)

Solution

ⓐ false ⓑ true

{7x+2y>145x−y≤8

ⓐ (2, 3) ⓑ (7, −1)

{6x−5y<20−2x+7y>−8

ⓐ (1, −3) ⓑ (−4, 4)

Solution

ⓐ false ⓑ true

Solve a System of Linear Inequalities by Graphing

In the following exercises, solve each system by graphing.

{y≤3x+2y>x−1

{y<−2x+2y≥−x−1

Solution

The figure shows the graph of inequalities y less than minus two times x plus two and y greater than or equal to minus x minus one. Two intersecting lines are shown, one in red and the other in blue. The area bound by the two lines is shown in grey.
The solution is the grey region.

{y<2x−1y≤−12x+4

{y≥−23x+2y>2x−3

Solution

The figure shows the graph of the inequalities y greater than or equal to minus two by three x plus two and y greater than two times x minus three. Two intersecting lines, one in red and the other in blue, are shown. The region bound by them is shown in grey.
The solution is the grey region.

x−y>1y<−14x+3

{x+2y<4y<x−2

Solution

The figure shows the graph of the inequalities x minus two times y less than four and y less than x minus two. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey.
The solution is the grey region.

{3x−y≥6y≥−12x

{2x+4y≥8y≤34x

Solution

The figure shows the graph of the inequalities two times x plus four times y greater than or equal to eight and y less than or equal to minus three fourth of x. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{2x−5y<103x+4y≥12

{3x−2y≤6−4x−2y>8

Solution

The figure shows the graph of the inequalities three times x minus two times y less than or equal to six and minus four times x minus two times y greater than eight. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{2x+2y>−4−x+3y≥9

{2x+y>−6−x+2y≥−4

Solution

The figure shows the graph of the inequalities two times x plus y greater than minus six and minus x plus two times y greater than or equal to minus four. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{x−2y<3y≤1

{x−3y>4y≤−1

Solution

The figure shows the graph of the inequalities x minus three times y greater than four and y less than or equal to minus one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{y≥−12x−3x≤2

{y≤−23x+5x≥3

Solution

The figure shows the graph of the inequality y less than or equal to minus two by three times x plus five and x greater than or equal to three. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{y≥34x−2y<2

{y≤−12x+3y<1

Solution

The figure shows the graph of the inequalities y less than or equal to minus half x plus three and y less than one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{3x−4y<8x<1

{−3x+5y>10x>−1

Solution

The figure shows the graph of the inequalities minus three times x plus five times y greater than ten and x greater than minus one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{x≥3y≤2

{x≤−1y≥3

Solution

The figure shows the graph of the inequalities x less than or equal to minus one and y greater than or equal to three. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.
The solution is the grey region.

{2x+4y>4y≤−12x−2

{x−3y≥6y>13x+1

Solution

The figure shows the graph of the inequalities x minus three times y greater than or equal to six and y greater than one third of x plus one. Two non intersecting lines, one in blue and the other in red, are shown.
No solution.

{−2x+6y<06y>2x+4

{−3x+6y>124y≤2x−4

Solution

The figure shows the graph of the inequalities minus three times x plus six times y greater than twelve and four times y less than or equal to two times x minus four. Two non intersecting lines, one in blue and the other in red, are shown.
No solution.

{y≥−3x+23x+y>5

{y≥12x−1−2x+4y≥4

Solution

The figure shows the graph of the inequalities y greater than or equal to minus half x minus one and minus two times x plus four times y greater than or equal to four. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.
The solution is the grey region.

{y≤−14x−2x+4y<6

{y≥3x−1−3x+y>−4

Solution

The figure shows the graph of the inequalities y greater than or equal to three times x minus one and minus three times x plus y greater than minus four. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.
The solution is the grey region.

{3y>x+2−2x+6y>8

{y<34x−2−3x+4y<7

Solution

The figure shows the graph of the inequalities y less than three by fourth x minus two and minus three x plus four y less than seven. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.
The solution is the grey region.

Solve Applications of Systems of Inequalities

In the following exercises, translate to a system of inequalities and solve.

Caitlyn sells her drawings at the county fair. She wants to sell at least 60 drawings and has portraits and landscapes. She sells the portraits for $15 and the landscapes for $10. She needs to sell at least $800 worth of drawings in order to earn a profit.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Will she make a profit if she sells 20 portraits and 35 landscapes?
ⓓ Will she make a profit if she sells 50 portraits and 20 landscapes?

Jake does not want to spend more than $50 on bags of fertilizer and peat moss for his garden. Fertilizer costs $2 a bag and peat moss costs $5 a bag. Jake’s van can hold at most 20 bags.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can he buy 15 bags of fertilizer and 4 bags of peat moss?
ⓓ Can he buy 10 bags of fertilizer and 10 bags of peat moss?

Solution

ⓐ {f≥0p≥0f+p≤202f+5p≤50
ⓑ
The figure shows the graph of the inequalities f plus p less than or equal to twenty and two f and five p less than or equal to fifty. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.
ⓒ yes
ⓓ no

Reiko needs to mail her Christmas cards and packages and wants to keep her mailing costs to no more than $500. The number of cards is at least 4 more than twice the number of packages. The cost of mailing a card (with pictures enclosed) is $3 and for a package the cost is $7.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can she mail 60 cards and 26 packages?
ⓓ Can she mail 90 cards and 40 packages?

Juan is studying for his final exams in chemistry and algebra. he knows he only has 24 hours to study, and it will take him at least three times as long to study for algebra than chemistry.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can he spend 4 hours on chemistry and 20 hours on algebra?
ⓓ Can he spend 6 hours on chemistry and 18 hours on algebra?

Solution

ⓐ {c≥0a≥0c+a≤24a≥3c
ⓑ
The figure shows the graph of the inequalities c plus a less than or equal to twenty four and a greater than or equal to three times c. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.
ⓒ yes
ⓓ no

Jocelyn is pregnant and so she needs to eat at least 500 more calories a day than usual. When buying groceries one day with a budget of $15 for the extra food, she buys bananas that have 90 calories each and chocolate granola bars that have 150 calories each. The bananas cost $0.35 each and the granola bars cost $2.50 each.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could she buy 5 bananas and 6 granola bars?
ⓓ Could she buy 3 bananas and 4 granola bars?

Mark is attempting to build muscle mass and so he needs to eat at least an additional 80 grams of protein a day. A bottle of protein water costs $3.20 and a protein bar costs $1.75. The protein water supplies 27 grams of protein and the bar supplies 16 gram. If he has $10 dollars to spend

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he buy 3 bottles of protein water and 1 protein bar?
ⓓ Could he buy no bottles of protein water and 5 protein bars?

Solution

ⓐ {w≥0b≥027w+16b>803.20w+1.75b≤10
ⓑ
The figure shows the graph of the inequalities twenty seven times w plus sixteen times b greater than eighty and three point two times w plus one point seven five b less than or equal to ten. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.
ⓒ no
ⓓ yes

Jocelyn desires to increase both her protein consumption and caloric intake. She desires to have at least 35 more grams of protein each day and no more than an additional 200 calories daily. An ounce of cheddar cheese has 7 grams of protein and 110 calories. An ounce of parmesan cheese has 11 grams of protein and 22 calories.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could she eat 1 ounce of cheddar cheese and 3 ounces of parmesan cheese?
ⓓ Could she eat 2 ounces of cheddar cheese and 1 ounce of parmesan cheese?

Mark is increasing his exercise routine by running and walking at least 4 miles each day. His goal is to burn a minimum of 1500 calories from this exercise. Walking burns 270 calories/mile and running burns 650 calories.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he meet his goal by walking 3 miles and running 1 mile?
ⓓ Could he meet his goal by walking 2 miles and running 2 miles?

Solution

ⓐ {w≥0r≥0w+r≥4270w+650r≥1500
ⓑ
The figure shows the graph of the inequalities w plus r greater than or equals to four and two seventy w plus six fifty r greater than or equal to fifteen hundred. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.
ⓒ no
ⓓ yes

Writing Exercises

Graph the inequality x−y≥3. How do you know which side of the line x−y=3 should be shaded?

Graph the system {x+2y≤6y≥−12x−4. What does the solution mean?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The figure shows a table with four columns and four rows. The first row is the title row. The four titles are I can…, confidently, with some help and No – I don’t get it! In the second row of the first column, the text says ‘determine whether an ordered pair is a solution of a system of linear inequalities.’ In the third row of the first column, the text says ‘solve applications of systems of inequalities.’

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Chapter Review Exercises

Solve Systems of Linear Equations with Two Variables

Determine Whether an Ordered Pair is a Solution of a System of Equations.

In the following exercises, determine if the following points are solutions to the given system of equations.

{x+3y=−92x−4y=12
ⓐ (−3,−2)
ⓑ (0,−3)

{x+y=8y=x−4
ⓐ (6,2)
ⓑ (9,−1)

Solution

ⓐ yes ⓑ no

Solve a System of Linear Equations by Graphing

In the following exercises, solve the following systems of equations by graphing.

{3x+y=6x+3y=−6

{x+4y=−1x=3

Solution

The figure shows the graph of equations x plus four times y equal to minus one and x equal to three. Two intersecting lines are shown.
(3,−1)

{2x−y=54x−2y=10

{−x+2y=4y=12x−3

Solution

The figure shows the graph for the equations minus x plus two times y equal to four and y equal to half x minus three. Two parallel lines are shown.
no solution

In the following exercises, without graphing determine the number of solutions and then classify the system of equations.

{y=25x+2−2x+5y=10

{3x+2y=6y=−3x+4

Solution

one solution, consistent system, independent equations

{5x−4y=0y=54x−5

Solve a System of Equations by Substitution

In the following exercises, solve the systems of equations by substitution.

{3x−2y=2y=12x+3

Solution

(4,5)

{x−y=02x+5y=−14

{y=−2x+7y=23x−1

Solution

(3,1)

{y=−5x5x+y=6

{y=−13x+2x+3y=6

Solution

infinitely many solutions

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination

{x+y=12x−y=−10

{3x−8y=20x+3y=1

Solution

(4,−1)

{9x+4y=25x+3y=5

{13x−12y=134x−y=52

Solution

(6,2)

{−x+3y=82x−6y=−20

Choose the Most Convenient Method to Solve a System of Linear Equations

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

{6x−5y=273x+10y=−24

Solution

elimination

{y=3x−94x−5y=23

Solve Applications with Systems of Equations

Solve Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

Mollie wants to plant 200 bulbs in her garden, all irises and tulips. She wants to plant three times as many tulips as irises. How many irises and how many tulips should she plant?

Solution

50 irises and 150 tulips

Ashanti has been offered positions by two phone companies. The first company pays a salary of $22,000 plus a commission of $100 for each contract sold. The second pays a salary of $28,000 plus a commission of $25 for each contract sold. How many contract would need to be sold to make the total pay the same?

Leroy spent 20 minutes jogging and 40 minutes cycling and burned 600 calories. The next day, Leroy swapped times, doing 40 minutes of jogging and 20 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Solution

10 calories jogging and 10 calories cycling

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb drives for $68. Find the cost of each notebook and each thumb drive.

Solve Geometry Applications

In the following exercises, translate to a system of equations and solve.

The difference of two supplementary angles is 58 degrees. Find the measures of the angles.

Solution

119 degrees and 61 degrees

Two angles are complementary. The measure of the larger angle is five more than four times the measure of the smaller angle. Find the measures of both angles.

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

Solution

35 degrees and 55 degrees

Becca is hanging a 28 foot floral garland on the two sides and top of a pergola to prepare for a wedding. The height is four feet less than the width. Find the height and width of the pergola.

The perimeter of a city rectangular park is 1428 feet. The length is 78 feet more than twice the width. Find the length and width of the park.

Solution

Length = 502 feet, Width = 212 feet

Solve Uniform Motion Applications

In the following exercises, translate to a system of equations and solve.

Sheila and Lenore were driving to their grandmother’s house. Lenore left one hour after Sheila. Sheila drove at a rate of 45 mph, and Lenore drove at a rate of 60 mph. How long will it take for Lenore to catch up to Sheila?

Bob left home, riding his bike at a rate of 10 miles per hour to go to the lake. Cheryl, his wife, left 45 minutes (34 hour) later, driving her car at a rate of 25 miles per hour. How long will it take Cheryl to catch up to Bob?

Solution

12 an hour

Marcus can drive his boat 36 miles down the river in three hours but takes four hours to return upstream. Find the rate of the boat in still water and the rate of the current.

A passenger jet can fly 804 miles in 2 hours with a tailwind but only 776 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

the rate of the jet is 395 mph, the rate of the wind is 7 mph

Solve Mixture Applications with Systems of Equations

Solve Mixture Applications with Systems of Equations

For the following exercises, translate to a system of equations and solve.

Lynn paid a total of $2,780 for 261 tickets to the theater. Student tickets cost $10 and adult tickets cost $15. How many student tickets and how many adult tickets did Lynn buy?

Priam has dimes and pennies in a cup holder in his car. The total value of the coins is $4.21. The number of dimes is three less than four times the number of pennies. How many dimes and how many pennies are in the cup?

Solution

41 dimes and 11 pennies

Yumi wants to make 12 cups of party mix using candies and nuts. Her budget requires the party mix to cost her $1.29 per cup. The candies are $2.49 per cup and the nuts are $0.69 per cup. How many cups of candies and how many cups of nuts should she use?

A scientist needs 70 liters of a 40% solution of alcohol. He has a 30% and a 60% solution available. How many liters of the 30% and how many liters of the 60% solutions should he mix to make the 40% solution?

Solution

4623 liters of 30% solution, 2313 liters of 60% solution

Solve Interest Applications

For the following exercises, translate to a system of equations and solve.

Jack has $12,000 to invest and wants to earn 7.5% interest per year. He will put some of the money into a savings account that earns 4% per year and the rest into CD account that earns 9% per year. How much money should he put into each account?

When she graduates college, Linda will owe $43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was $1,585. What is the amount of each loan?

Solution

$29,000 for the federal loan, $14,000 for the private loan

Solve Systems of Equations with Three Variables

Solve Systems of Equations with Three Variables

In the following exercises, determine whether the ordered triple is a solution to the system.

{3x−4y−3z=22x−6y+z=32x+3y−2z=3
ⓐ (2,3,−1)
ⓑ (3,1,3)

{y=23x−2x+3y−z=15x−3y+z=−2
ⓐ (−6,5,12)
ⓑ (5,43,−3)

Solution

ⓐ no ⓑ no

Solve a System of Linear Equations with Three Variables

In the following exercises, solve the system of equations.

{3x−5y+4z=55x+2y+z=02x+3y−2z=3

{x+52y+z=−22x+2y+12z=−413x−y−z=1

Solution

(−3,2,−4)

{5x+3y=−62y+3z=−17x+z=1

{2x+3y+z=12x+y+z=93x+4y+2z=20

Solution

no solution

{−x−3y+2z=14−x+2y−3z=−43x+y−2z=6

Solve Applications using Systems of Linear Equations with Three Variables

After attending a major league baseball game, the patrons often purchase souvenirs. If a family purchases 4 t-shirts, a cap and 1 stuffed animal their total is $135. A couple buys 2 t-shirts, a cap and 3 stuffed animals for their nieces and spends $115. Another couple buys 2 t-shirts, a cap and 1 stuffed animal and their total is $85. What is the cost of each item?

Solution

25, 20, 15

Solve Systems of Equations Using Matrices

Write the Augmented Matrix for a System of Equations.

Write each system of linear equations as an augmented matrix.

{3x−y=−1−2x+2y=5

{4x+3y=−2x−2y−3z=72x−y+2z=−6

Solution

[430−21−2−372−12−6]

Write the system of equations that that corresponds to the augmented matrix.

[2−43−3|−2−1]

[10−31−200−12|−1−23]

Solution

{x−3z=−1x−2y=−2−y+2z=3

In the following exercises, perform the indicated operations on the augmented matrices.

[4−632|−31]

ⓐ Interchange rows 2 and 1.
ⓑ Multiply row 1 by 4.
ⓒ Multiply row 2 by 3 and add to row 1.

[1−3−222−14−2−3|4−3−1]

ⓐ Interchange rows 2 and 3.
ⓑ Multiply row 1 by 2.
ⓒ Multiply row 3 by −2 and add to row 2.

Solution


ⓐ [1−3−244−2−3−122−1−3]
ⓑ [2−6−484−2−3−122−1−3]
ⓒ [2−6−484−2−3−10−6−15]

Solve Systems of Equations Using Matrices

In the following exercises, solve each system of equations using a matrix.

{4x+y=6x−y=4

{2x−y+3z=−3−x+2y−z=10x+y+z=5

Solution

(−2,5,2)

{2y+3z=−15x+3y=−67x+z=1

{x+2y−3z=−1x−3y+z=12x−y−2z=2

Solution

no solution

{x+y−3z=−1y−z=0−x+2y=1

Solve Systems of Equations Using Determinants

Evaluate the Determinant of a 2 × 2 Matrix

In the following exercise, evaluate the determinate of the square matrix.

[8−45−3]

Solution

−4

Evaluate the Determinant of a 3 × 3 Matrix

In the following exercise, find and then evaluate the indicated minors.

|−1−324−2−1−20−3|; Find the minor ⓐ a1 ⓑ b1 ⓒ c2

In the following exercise, evaluate each determinant by expanding by minors along the first row.

|−2−3−45−67−120|

Solution

33

In the following exercise, evaluate each determinant by expanding by minors.

|354−130−261|

Use Cramer’s Rule to Solve Systems of Equations

In the following exercises, solve each system of equations using Cramer’s rule

{x−3y=−92x+5y=4

Solution

(−3,2)

{4x−3y+z=72x−5y−4z=33x−2y−2z=−7

{2x+5y=43y−z=34x+3z=−3

Solution

(−3,2,3)

{x+y−3z=−1y−z=0−x+2y=1

{3x+4y−3z=−22x+3y−z=−12x+y−2z=6

Solution

inconsistent

Solve Applications Using Determinants

In the following exercises, determine whether the given points are collinear.

(0,2), (−1,−1), and (−2,4)

Graphing Systems of Linear Inequalities

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

In the following exercises, determine whether each ordered pair is a solution to the system.

{4x+y>63x−y≤12

ⓐ (2,−1)
ⓑ (3,−2)

Solution

ⓐ yes ⓑ yes

{y>13x+2x−14y≤10

ⓐ (6,5)
ⓑ (15,8)

Solve a System of Linear Inequalities by Graphing

In the following exercises, solve each system by graphing.

{y≤3x+2y>−x-1

Solution

The figure shows the graph of inequalities y less than three times x plus one and y greater than or equal to minus x minus two. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.
The solution is the grey region.

{x−y>−1y<13x−2

{2x−3y<63x+4y≥12

Solution

The figure shows the graph of inequalities two times x minus three times y less six and three times x plus four times y greater than or equal to twelve. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.
The solution is the grey region.

{y≤−34x+1x≥−5

{x+3y<5y≥−13x+6

Solution

The figure shows the graph of inequalities x plus three times y less than five and y greater than or equal to minus one third x plus six. Two parallel lines, one in red and the other in blue, are shown. An area is shown in grey.
No solution.

{y≥2x−5−6x+3y>−4

Solve Applications of Systems of Inequalities

In the following exercises, translate to a system of inequalities and solve.

Roxana makes bracelets and necklaces and sells them at the farmers’ market. She sells the bracelets for $12 each and the necklaces for $18 each. At the market next weekend she will have room to display no more than 40 pieces, and she needs to sell at least $500 worth in order to earn a profit.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Should she display 26 bracelets and 14 necklaces?
ⓓ Should she display 39 bracelets and 1 necklace?

Solution

ⓐ {b≥0n≥0b+n≤4012b+18n≥500
ⓑ
The figure shows the graph of b plus n equal to forty and twelve b plus eighteen n equal to five hundred. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.
ⓒ yes
ⓓ no

Annie has a budget of $600 to purchase paperback books and hardcover books for her classroom. She wants the number of hardcover to be at least 5 more than three times the number of paperback books. Paperback books cost $4 each and hardcover books cost $15 each.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can she buy 8 paperback books and 40 hardcover books?
ⓓ Can she buy 10 paperback books and 37 hardcover books?

Chapter Practice Test

In the following exercises, solve the following systems by graphing.

{x−y=5x+2y=−4

Solution

The figure shows the graph of inequalities h equal to three p plus five and four times p plus fifteen times h equal to six hundred. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.
(2,−3)

{x−y>−2y≤3x+1

In the following exercises, solve each system of equations. Use either substitution or elimination.

{x+4y=6−2x+y=−3

Solution

(2,1)

{−3x+4y=25x−5y=−23

{x+y−z=−12x−y+2z=8−3x+2y+z=−9

Solution

(2,−2,1)

Solve the system of equations using a matrix.

{2x+y=7x−2y=6

{−3x+y+z=−4−x+2y−2z=12x−y−z=−1

Solution

(5,7,4)

Solve using Cramer’s rule.

{3x+y=−32x+3y=6

Evaluate the determinant by expanding by minors:
|3−2−22−14−10−3|.

Solution

7

In the following exercises, translate to a system of equations and solve.

Greg is paddling his canoe upstream, against the current, to a fishing spot 10 miles away. If he paddles upstream for 2.5 hours and his return trip takes 1.25 hours, find the speed of the current and his paddling speed in still water.

A pharmacist needs 20 liters of a 2% saline solution. He has a 1% and a 5% solution available. How many liters of the 1% and how many liters of the 5% solutions should she mix to make the 2% solution?

Solution

15 liters of 1% solution, 5 liters of 5% solution

Arnold invested $64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received $4,500 in interest in one year?

The church youth group is selling snacks to raise money to attend their convention. Amy sold 2 pounds of candy, 3 boxes of cookies and 1 can of popcorn for a total sales of $65. Brian sold 4 pounds of candy, 6 boxes of cookies and 3 cans of popcorn for a total sales of $140. Paulina sold 8 pounds of candy, 8 boxes of cookies and 5 can of popcorn for a total sales of $250. What is the cost of each item?

Solution

The candy cost $20; the cookies cost $5; and the popcorn cost $10.

The manufacturer of a granola bar spends $1.20 to make each bar and sells them for $2. The manufacturer also has fixed costs each month of $8,000.

ⓐ Find the cost function C when x granola bars are manufactured
ⓑ Find the revenue function R when x granola bars are sold.
ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid.
ⓓ Find the break-even point. Interpret what the break-even point means.

Translate to a system of inequalities and solve.

Andi wants to spend no more than $50 on Halloween treats. She wants to buy candy bars that cost $1 each and lollipops that cost $0.50 each, and she wants the number of lollipops to be at least three times the number of candy bars.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Can she buy 20 candy bars and 40 lollipops?

Solution

ⓐ {C≥0L≥0C+0.5L≤50L≥3C
ⓑ
The figure shows the graph of two equations. Two intersecting lines, one in red and the other in blue, are shown. The red line passes through origin. An area is shown in grey.
ⓒ no
ⓓ yes

system of linear inequalities
Two or more linear inequalities grouped together form a system of linear inequalities.

Introduction

A shimmering bounty of gold coins, divided into two distinct piles, one with intricate Austrian designs and the other with classic treasure chest motifs.
There are many different kinds of coins in circulation, but a new type of coin exists only in the virtual world. It is the bitcoin.

You may have coins and paper money in your wallet, but you may soon want to acquire a type of currency called bitcoins. They exist only in a digital wallet on your computer. You can use bitcoins to pay for goods at some companies, or save them as an investment. Although the future of bitcoins is uncertain, investment brokers are beginning to investigate ways to make business predictions using this digital currency. Understanding how bitcoins are created and obtained requires an understanding of a type of function known as a polynomial function. In this chapter you will investigate polynomials and polynomial functions and learn how to perform mathematical operations on them.

Add and Subtract Polynomials

Learning Objectives

By the end of this section, you will be able to:

  • Determine the degree of polynomials
  • Add and subtract polynomials
  • Evaluate a polynomial function for a given value
  • Add and subtract polynomial functions

Before you get started, take this readiness quiz.

Simplify: 3x2+3x+1+8x2+5x+5.
If you missed this problem, review Example 7 in Use the Language of Algebra.

Solution

11x2+8x+6

Subtract: (5n+8)−(2n−1).
If you missed this problem, review Example 5 in Use the Language of Algebra.

Solution

3n+9

Evaluate: 4xy2 when x=−2 and y=5.
If you missed this problem, review Example 10 in Integers.

Solution

−200

Determine the Degree of Polynomials

We have learned that a term is a constant or the product of a constant and one or more variables. A monomial is an algebraic expression with one term. When it is of the form axm, where a is a constant and m is a whole number, it is called a monomial in one variable. Some examples of monomials in one variable are 2x, 5y, 17z, and 4y2 . Monomials can also have more than one variable such as 5abc and −4a2b3c2.

Monomial

A monomial is an algebraic expression with one term.

A monomial in one variable is a term of the form axm, where a is a constant and m is a whole number.

A monomial, or two or more monomials combined by addition or subtraction, is a polynomial. Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term. A binomial has exactly two terms, and a trinomial has exactly three terms. There are no special names for polynomials with more than three terms.

Polynomials

polynomial—A monomial, or two or more algebraic terms combined by addition or subtraction is a polynomial.

monomial—A polynomial with exactly one term is called a monomial.

binomial—A polynomial with exactly two terms is called a binomial.

trinomial—A polynomial with exactly three terms is called a trinomial.

Here are some examples of polynomials.

Polynomial y+1 4a2−7ab+2b2 4x4+x3+8x2−9x+1
Monomial 14 8y2 −9x3y5 −13a3b2c
Binomial a+7b 4x2−y2 y2−16 3p3q−9p2q
Trinomial x2−7x+12 9m2+2mn−8n2 6k4−k3+8k z4+3z2−1

Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials.

The degree of a polynomial and the degree of its terms are determined by the exponents of the variable.

A monomial that has no variable, just a constant, is a special case. The degree of a constant is 0.

Degree of a Polynomial

The degree of a term is the sum of the exponents of its variables.

The degree of a constant is 0.

The degree of a polynomial is the highest degree of all its terms.

Let’s see how this works by looking at several polynomials. We’ll take it step by step, starting with monomials, and then progressing to polynomials with more terms.

Let's start by looking at a monomial. The monomial 8ab2 has two variables a and b. To find the degree we need to find the sum of the exponents. The variable a doesn't have an exponent written, but remember that means the exponent is 1. The exponent of b is 2. The sum of the exponents, 1+2, is 3 so the degree is 3.

The polynomial is 8 a b squared. The exponents of the variables are 1 and 2 so the degree of the monomial is 1 plus 2 which equals 3.

Here are some additional examples.

Monomial examples: 14 has degree 0, 8 a b squared has degree 3, negative 9 x cubed y to the fifth power has degree 8, negative 13 a has degree 1. Binomial examples: The terms in h plus 7 have degree 1 and 0 so the degree of the whole polynomial is 1. The terms in 7 b squared minus 3 b have degree 2 and 1 so the degree of the whole polynomial is 2. The terms in z squared y squared minus 25 have degree 4 and 0 so the degree of the whole polynomial is 4. The terms in 4 n cubed minus 8 n squared have degree 3 and 2 so the degree of the whole polynomial is 3. Trinomial examples: The terms in x squared minus 12 x plus 27 have degree 2, 1 and 0 so the degree of the whole polynomial is 2. The terms in 9 a squared plus 6 a b plus b squared have degree 2, 2, and 2 so the degree of the whole polynomial is 2. The terms in 6 m to the fourth power minus m cubed n squared plus 8 m n to the fifth power have degree 4, 5, and 6 so the degree of the whole polynomial is 6. The terms in z to the fourth power plus 3 z squared minus 1 have degree 4, 2, and 0 so the degree of the whole polynomial is 4. Polynomial examples: The terms in y minus 1 have degree 1 and 0 so the degree of the whole polynomial is 1. The terms in 3 y squared minus 2 y minus 5 have degree 2, 1, 0 so the degree of the whole polynomial is 2. The terms in 4 x to the fourth power plus x cubed plus eight x squared minus 9 x plus 1 have degree 4, 3, 2, 1, and 0 so the degree of the whole polynomial is 4.

Working with polynomials is easier when you list the terms in descending order of degrees. When a polynomial is written this way, it is said to be in standard form of a polynomial. Get in the habit of writing the term with the highest degree first.

Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. Then, find the degree of each polynomial.

ⓐ 7y2−5y+3 ⓑ −2a4b2 ⓒ 3x5−4x3−6x2+x−8 ⓓ 2y−8xy3 ⓔ 15

Solution
Polynomial Number of terms Type Degree of terms Degree of polynomial
ⓐ 7y2−5y+3 3 Trinomial 2, 1, 0 2
ⓑ −2a4b2 1 Monomial 6 6
ⓒ 3x5−4x3−6x2+x−8 5 Polynomial 5, 3, 2, 1, 0 5
ⓓ 2y−8xy3 2 Binomial 1, 4 4
ⓔ 15 1 Monomial 0 0

Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. Then, find the degree of each polynomial.

ⓐ −5 ⓑ 8y3−7y2−y−3 ⓒ −3x2y−5xy+9xy3 ⓓ 81m2−4n2 ⓔ −3x6y3z

Solution

ⓐ monomial, 0
ⓑ polynomial, 3 ⓒ trinomial, 4
ⓓ binomial, 2 ⓔ monomial, 10

Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. Then, find the degree of each polynomial.

ⓐ 64k3−8 ⓑ 9m3+4m2−2 ⓒ 56 ⓓ 8a4−7a3b−6a2b2−4ab3+7b4 ⓔ −p4q3

Solution

ⓐ binomial, 3 ⓑ trinomial, 3 ⓒ monomial, 0 ⓓ polynomial, 4 ⓔ monomial, 7

Add and Subtract Polynomials

We have learned how to simplify expressions by combining like terms. Remember, like terms must have the same variables with the same exponent. Since monomials are terms, adding and subtracting monomials is the same as combining like terms. If the monomials are like terms, we just combine them by adding or subtracting the coefficients.

Add or subtract: ⓐ 25y2+15y2 ⓑ 16pq3−(−7pq3).

Solution

ⓐ
25y2+15y2 Combine like terms.40y2

ⓑ
16pq3−(−7pq3) Combine like terms.23pq3

Add or subtract: ⓐ 12q2+9q2 ⓑ 8mn3−(−5mn3).

Solution

ⓐ 21q2 ⓑ 13mn3

Add or subtract: ⓐ −15c2+8c2 ⓑ −15y2z3−(−5y2z3).

Solution

ⓐ −7c2 ⓑ −10y2z3

Remember that like terms must have the same variables with the same exponents.

Simplify: ⓐ a2+7b2−6a2 ⓑ u2v+5u2−3v2.

Solution
ⓐ
Example demonstrating the simplification of an algebraic expression by combining like terms.
a2+7b2−6a2
Combine like terms. −5a2+7b2
ⓑ
Illustrates a polynomial with no like terms to combine, demonstrating an expression that remains unchanged after an attempt at simplification.
u2v+5u2−3v2
There are no like terms to combine.
In this case, the polynomial is unchanged.
u2v+5u2−3v2

Add: ⓐ 8y2+3z2−3y2 ⓑ m2n2−8m2+4n2.

Solution

ⓐ 5y2+3z2
ⓑ m2n2−8m2+4n2

Add: ⓐ 3m2+n2−7m2 ⓑ pq2−6p−5q2.

Solution

ⓐ −4m2+n2
ⓑ pq2−6p−5q2

We can think of adding and subtracting polynomials as just adding and subtracting a series of monomials. Look for the like terms—those with the same variables and the same exponent. The Commutative Property allows us to rearrange the terms to put like terms together.

Find the sum:(7y2−2y+9)+(4y2−8y−7).

Solution
Steps to simplify an algebraic expression by identifying and combining like terms.
Identify like terms. (7y2________−2y___+9)+(4y2________−8y___−7)
Rewrite without the parentheses,
rearranging to get the like terms together.
7y2+4y2__________________−2y−8y_______+9−7
Combine like terms. 11y2−10y+2

Find the sum: (7x2−4x+5)+(x2−7x+3).

Solution

8x2−11x+8

Find the sum: (14y2+6y−4)+(3y2+8y+5).

Solution

17y2+14y+1

Be careful with the signs as you distribute while subtracting the polynomials in the next example.

Find the difference: (9w2−7w+5)−(2w2−4).

Solution
Step-by-step subtraction of polynomials, illustrating the process of simplifying an algebraic expression.
(9w2−7w+5)−(2w2−4)
Distribute and identify like terms. 9w2________−7w___+5−2w2________+4
Rearrange the terms. 9w2−2w2____________________−7w___+5+4
Combine like terms. 7w2−7w+9

Find the difference: (8x2+3x−19)−(7x2−14).

Solution

x2+3x−5

Find the difference: (9b2−5b−4)−(3b2−5b−7).

Solution

6b2+3

To subtract a from b, we write it as b−a, placing the b first.

Subtract (p2+10pq−2q2) from (p2+q2).

Solution
Step-by-step simplification of an algebraic expression by distributing, rearranging, and combining like terms.
(p2+q2)−(p2+10pq−2q2)
Distribute. p2+q2−p2−10pq+2q2
Rearrange the terms, to put like terms together. p2−p2−10pq+q2+2q2
Combine like terms. −10pq+3q2

Subtract (a2+5ab−6b2) from (a2+b2).

Solution

−5ab+7b2

Subtract (m2−7mn−3n2) from (m2+n2).

Solution

7mn+4n2

Find the sum: (u2−6uv+5v2)+(3u2+2uv).

Solution
Steps to simplify an algebraic expression by combining like terms.
(u2−6uv+5v2)+(3u2+2uv)
Distribute. u2−6uv+5v2+3u2+2uv
Rearrange the terms to put like terms together. u2+3u2−6uv+2uv+5v2
Combine like terms. 4u2−4uv+5v2

Find the sum: (3x2−4xy+5y2)+(2x2−xy).

Solution

5x2−5xy+5y2

Find the sum: (2x2−3xy−2y2)+(5x2−3xy).

Solution

7x2−6xy−2y2

When we add and subtract more than two polynomials, the process is the same.

Simplify: (a3−a2b)−(ab2+b3)+(a2b+ab2).

Solution
Step-by-step simplification of a polynomial expression.
(a3−a2b)−(ab2+b3)+(a2b+ab2)
Distribute. a3−a2b−ab2−b3+a2b+ab2
Rewrite without the parentheses,
rearranging to get the like terms together.
a3−a2b+a2b−ab2+ab2−b3
Combine like terms. a3−b3

Simplify: (x3−x2y)−(xy2+y3)+(x2y+xy2).

Solution

x3-y3

Simplify: (p3−p2q)+(pq2+q3)−(p2q+pq2).

Solution

p3−2p2q+q3

Evaluate a Polynomial Function for a Given Value

A polynomial function is a function defined by a polynomial. For example, f(x)=x2+5x+6 and g(x)=3x−4 are polynomial functions, because x2+5x+6 and 3x−4 are polynomials.

Polynomial Function

A polynomial function is a function whose range values are defined by a polynomial.

In Graphs and Functions, where we first introduced functions, we learned that evaluating a function means to find the value of f(x) for a given value of x. To evaluate a polynomial function, we will substitute the given value for the variable and then simplify using the order of operations.

For the function f(x)=5x2−8x+4 find: ⓐ f(4) ⓑ f(−2) ⓒ f(0).

Solution
ⓐ
The image shows the function f(x) = 5x^2 - 8x + 4.
The image shows the text 'To find f(4), substitute 4 for x.' This is a common instruction in algebra for evaluating a function at a specific point.   The mathematical equation shows the evaluation of a function f(x) at x=4, represented as f(4) = 5(4)^2 - 8(4) + 4. The number 4 is highlighted in red throughout the expression.
Simplify the exponents. A mathematical equation is displayed, showing f(4) equals 5 multiplied by 16, minus 8 multiplied by 4, plus 4. It represents a function evaluation with numerical operations.
Multiply. A mathematical equation is displayed, showing 'f(4) = 80 - 32 + 4' in black text against a white background.
Simplify. A mathematical expression reads 'f(4) = 52' in black text against a white background.
ⓑ
A mathematical equation is displayed on a white background: f(x) = 5x^2 - 8x + 4.
To find f(-2), substitute -2 for x.  An algebraic equation showing f(-2) = 5(-2)^2 - 8(-2) + 4, with the number -2 highlighted in red in each instance it appears.
Simplify the exponents. Evaluating the function f(x) at x = -2: f(-2) = 5 * 4 - 8(-2) + 4. The calculation shows substituting -2 for x in the expression.
Multiply. A mathematical equation is shown, displaying f(-2) = 20 + 16 + 4, which simplifies to f(-2) = 40. This depicts a function evaluation with a numeric result.
Simplify. The image displays the mathematical expression 'f(-2) = 40' in a black serif font against a plain white background, indicating that the function f evaluated at -2 equals 40.
ⓒ
A mathematical equation displays the function f(x) = 5x^2 - 8x + 4, a quadratic equation in standard form.
The text reads: To find f(0), substitute 0 for x.   A mathematical equation is displayed, showing f(0) equals 5 multiplied by 0 squared, minus 8 multiplied by 0, plus 4. The zeros inside the parentheses are highlighted in red.
Simplify the exponents. The mathematical equation f(0) = 5 * 0 - 8(0) + 4 is shown, representing the evaluation of a function at x = 0.
Multiply. A mathematical equation is displayed on a white background, reading 'f(0) = 0 + 0 + 4' in black text.
Simplify. A mathematical expression on a white background states f(0) = 4, indicating that the function f evaluated at 0 is equal to 4.

For the function f(x)=3x2+2x−15, find ⓐ f(3) ⓑ f(−5) ⓒ f(0).

Solution

ⓐ 18 ⓑ 50 ⓒ −15

For the function g(x)=5x2−x−4, find ⓐ g(−2) ⓑ g(−1) ⓒ g(0).

Solution

ⓐ 18 ⓑ 2 ⓒ −4

The polynomial functions similar to the one in the next example are used in many fields to determine the height of an object at some time after it is projected into the air. The polynomial in the next function is used specifically for dropping something from 250 ft.

The polynomial function h(t)=−16t2+250 gives the height of a ball t seconds after it is dropped from a 250-foot tall building. Find the height after t=2 seconds.

Solution
This table illustrates the step-by-step calculation of the height of an object at t=2 seconds using the function h(t) = -16t^2 + 250, resulting in a height of 186 feet.
h(t)=−16t2+250
To find h(2), substitute t=2. h(2)=−16(2)2+250
Simplify. h(2)=−16·4+250
Simplify. h(2)=−64+250
Simplify. h(2)=186
After 2 seconds the height of the ball is 186 feet.

The polynomial function h(t)=−16t2+150 gives the height of a stone t seconds after it is dropped from a 150-foot tall cliff. Find the height after t=0 seconds (the initial height of the object).

Solution

The height is 150 feet.

The polynomial function h(t)=−16t2+175 gives the height of a ball t seconds after it is dropped from a 175-foot tall bridge. Find the height after t=3 seconds.

Solution

The height is 31 feet.

Add and Subtract Polynomial Functions

Just as polynomials can be added and subtracted, polynomial functions can also be added and subtracted.

Addition and Subtraction of Polynomial Functions

For functions f(x) and g(x),

(f+g)(x)=f(x)+g(x)(f−g)(x)=f(x)−g(x)

For functions f(x)=3x2−5x+7 and g(x)=x2−4x−3, find:

ⓐ (f+g)(x) ⓑ (f+g)(3) ⓒ (f−g)(x) ⓓ (f−g)(−2).

Solution
ⓐ
The image displays the sum of two functions, f and g, as (f + g)(x) = f(x) + g(x), illustrating that the sum of functions is defined by adding their individual outputs at each point x.
The image shows the instruction 'Substitute f(x) = 3x^2 - 5x + 7 and g(x) = x^2 - 4x - 3.' The expression for f(x) is in red, and g(x) is in blue. The image shows the sum of two polynomial functions, (f+g)(x), represented as the addition of (3x² - 5x + 7) and (x² - 4x - 3).
Rewrite without the parentheses. A mathematical expression shows the sum of two functions, (f+g)(x), as 3x^2 - 5x + 7 + x^2 - 4x - 3, written in a clear, digital font on a white background.
Put like terms together. A mathematical equation shows the sum of two functions, (f + g)(x), expanded as 3x^2 + x^2 - 5x - 4x + 7 - 3, likely as an intermediate step to simplification.
Combine like terms. The image displays the sum of two functions, (f + g)(x), expressed as the quadratic equation 4x^2 - 9x + 4.
ⓑ In part (a) we found (f+g)(x) and now are asked to find (f+g)(3).
Step-by-step evaluation of the function (f+g)(x) = 4x^2 - 9x + 4 at x=3.
(f+g)(x)=4x2−9x+4
To find (f+g)(3), substitute x=3. (f+g)(3)=4(3)2−9·3+4
(f+g)(3)=4·9−9·3+4
(f+g)(3)=36−27+4

Notice that we could have found (f+g)(3) by first finding the values of f(3) and g(3) separately and then adding the results.

Find f(3). The image displays the quadratic function f(x) = 3x^2 - 5x + 7 in black text against a white background.
The image shows a mathematical equation: f(3) = 3(3)^2 - 5(3) + 7. The number 3 is highlighted in red, indicating its substitution into the function.
A mathematical expression 'f(3) = 19' is displayed in black text on a white background, representing a function f evaluated at 3 equals 19.
Find g(3). The image displays the quadratic function g(x) = x^2 - 4x - 3 in black text against a white background, representing a mathematical equation.
A mathematical equation is displayed, showing a function g evaluated at 3. The equation reads: g(3) = 3^2 - 4(3) - 3, with the number '3' highlighted in red wherever it appears in the expression.
The image shows the mathematical equation g(3) = -6 on a white background. The function g evaluated at 3 equals negative 6.
Find (f+g)(3). A mathematical equation shows the sum of two functions, f and g, applied to x, is equal to the sum of each function applied individually to x: (f + g)(x) = f(x) + g(x).
The image shows the sum of two functions, f and g, evaluated at 3, which equals the sum of each function evaluated at 3: (f + g)(3) = f(3) + g(3).
The image shows the text 'Substitute f(3) = 19 and g(3) = -6.' The number 19 is highlighted in red, and the number -6 is highlighted in teal. A mathematical equation shows (f + g)(3) = 19 + (-6), representing the sum of two functions f and g evaluated at 3, equaling the addition of 19 and -6.
A mathematical equation showing the sum of two functions f and g evaluated at 3 equals 13, written as (f + g)(3) = 13.
ⓒ
The image displays the definition of the difference between two functions, stating that (f-g)(x) is equal to f(x) - g(x).
Two functions, f(x) = 3x^2 - 5x + 7 (red) and g(x) = x^2 - 4x - 3 (blue), are presented for substitution in a mathematical problem. Mathematical expression for (f-g)(x) where f(x) = 3x^2 - 5x + 7 (red) and g(x) = x^2 - 4x - 3 (blue), illustrating polynomial subtraction.
Rewrite without the parentheses. The equation shows the subtraction of two functions, (f-g)(x), which equals 3x^2 - 5x + 7 - x^2 + 4x + 3. This equation represents a polynomial expression with multiple terms.
Put like terms together. The image shows the mathematical expression (f - g)(x) = 3x^2 - x^2 - 5x + 4x + 7 + 3, representing the subtraction of two functions, f(x) and g(x).
Combine like terms. The image displays the subtraction of two functions, represented as (f - g)(x), which equals the quadratic expression 2x^2 - x + 10.
ⓓ
The image demonstrates the process of evaluating the function (f-g)(x) = 2x^2 - x + 10 at x = -2, showing step-by-step substitution and calculation to arrive at the result (f-g)(-2) = 20.

For functions f(x)=2x2−4x+3 and g(x)=x2−2x−6, find: ⓐ (f+g)(x) ⓑ (f+g)(3) ⓒ (f−g)(x) ⓓ (f−g)(−2).

Solution

ⓐ (f+g)(x)=3x2−6x−3 ⓑ (f+g)(3)=6
ⓒ (f−g)(x)=x2−2x+9
ⓓ (f−g)(−2)=17

For functions f(x)=5x2−4x−1 and g(x)=x2+3x+8, find ⓐ (f+g)(x) ⓑ (f+g)(3) ⓒ (f−g)(x) ⓓ (f−g)(−2).

Solution

ⓐ (f+g)(x)=6x2−x+7 ⓑ (f+g)(3)=58
ⓒ (f−g)(x)=4x2−7x−9
ⓓ (f−g)(−2)=21

Access this online resource for additional instruction and practice with adding and subtracting polynomials.

  • Adding and Subtracting Polynomials

Key Concepts

  • Monomial
    • A monomial is an algebraic expression with one term.
    • A monomial in one variable is a term of the form axm, where a is a constant and m is a whole number.
  • Polynomials
    • Polynomial—A monomial, or two or more algebraic terms combined by addition or subtraction is a polynomial.
    • monomial —A polynomial with exactly one term is called a monomial.
    • binomial — A polynomial with exactly two terms is called a binomial.
    • trinomial —A polynomial with exactly three terms is called a trinomial.
  • Degree of a Polynomial
    • The degree of a term is the sum of the exponents of its variables.
    • The degree of a constant is 0.
    • The degree of a polynomial is the highest degree of all its terms.

Practice Makes Perfect

Determine the Type of Polynomials

In the following exercises, determine if the polynomial is a monomial, binomial, trinomial, or other polynomial. Then, indicate the degree of the polynomial.


ⓐ 47x5−17x2y3+y2
ⓑ 5c3+11c2−c−8
ⓒ 59ab+13b
ⓓ 4
ⓔ 4pq+17

Solution

ⓐ trinomial, 5 ⓑ polynomial, 3 ⓒ binomial, 2 ⓓ monomial, 0
ⓔ binomial, 2


ⓐ x2−y2
ⓑ −13c4
ⓒ a2+2ab−7b2
ⓓ 4x2y2−3xy+8
ⓔ 19


ⓐ 8y−5x
ⓑ y2−5yz−6z2
ⓒ y3−8y2+2y−16
ⓓ 81ab4−24a2b2+3b
ⓔ −18

Solution

ⓐ binomial, 1ⓑ trinomial, 2
ⓒ polynomial, 3ⓓ trinomial, 5
ⓔ monomial, 0


ⓐ 11y2
ⓑ −73
ⓒ 6x2−3xy+4x−2y+y2
ⓓ 4y2+17z2
ⓔ 5c3+11c2−c−8


ⓐ 5a2+12ab−7b2
ⓑ 18xy2z
ⓒ 5x+2
ⓓ y3−8y2+2y−16
ⓔ −24

Solution

ⓐ trinomial, 2 ⓑ monomial, 4 ⓒ binomial, 1 ⓓ polynomial, 3
ⓔ monomial, 0


ⓐ 9y3−10y2+2y−6
ⓑ −12p3q
ⓒ a2+9ab+18b2
ⓓ 20x2y2−10a2b2+30
ⓔ 17


ⓐ 14s−29t
ⓑ z2−5z−6
ⓒ y3−8y2z+2yz2−16z3
ⓓ 23ab2−14
ⓔ −3

Solution

ⓐ binomial, 1 ⓑ trinomial, 2 ⓒ polynomial, 3 ⓓ binomial, 3
ⓔ monomial, 0


ⓐ 15xy
ⓑ 15
ⓒ 6x2−3xy+4x−2y+y2
ⓓ 10p−9q
ⓔ m4+4m3+6m2+4m+1

Add and Subtract Polynomials

In the following exercises, add or subtract the monomials.


ⓐ 7x2+5x2
ⓑ 4a−9a

Solution

ⓐ 12x2 ⓑ −5a


ⓐ 4y3+6y3
ⓑ −y−5y


ⓐ −12w+18w
ⓑ 7x2y−(−12x2y)

Solution

ⓐ 6w ⓑ 19x2y


ⓐ −3m+9m
ⓑ 15yz2−(−8yz2)

7x2+5x2+4a−9a

Solution

12x2−5a

4y3+6y3−y−5y

−12w+18w+7x2y−(−12x2y)

Solution

6w+19x2y

−3m+9m+15yz2−(−8yz2)


ⓐ −5b−17b
ⓑ 3xy−(−8xy)+5xy

Solution

ⓐ −22b ⓑ 16xy


ⓐ −10x−35x
ⓑ 17mn2−(−9mn2)+3mn2


ⓐ 12a+5b−22a
ⓑ pq2−4p−3q2

Solution

ⓐ −10a+5b
ⓑ pq2−4p−3q2


ⓐ 14x−3y−13x
ⓑ a2b−4a−5ab2


ⓐ 2a2+b2−6a2
ⓑ x2y−3x+7xy2

Solution

ⓐ −4a2+b2
ⓑ x2y−3x+7xy2


ⓐ 5u2+4v2−6u2
ⓑ 12a+8b


ⓐ xy2−5x−5y2
ⓑ 19y+5z

Solution

ⓐ xy2−5x−5y2
ⓑ 19y+5z

12a+5b−22a+pq2−4p−3q2

14x−3y−13x+a2b−4a−5ab2

Solution

x−3y+a2b−4a−5ab2

2a2+b2−6a2+x2y−3x+7xy2

5u2+4v2−6u2+12a+8b

Solution

−u2+4v2+12a+8b

xy2−5x−5y2+19y+5z

Add: 4a,−3b,−8a

Solution

−4a−3b

Add:4x,3y,−3x

Subtract 5x6 from −12x6

Solution

−17x6

Subtract 2p4 from −7p4

In the following exercises, add the polynomials.

(5y2+12y+4)+(6y2−8y+7)

Solution

11y2+4y+11

(4y2+10y+3)+(8y2−6y+5)

(x2+6x+8)+(−4x2+11x−9)

Solution

−3x2+17x−1

(y2+9y+4)+(−2y2−5y−1)

(8x2−5x+2)+(3x2+3)

Solution

11x2−5x+5

(7x2−9x+2)+(6x2−4)

(5a2+8)+(a2−4a−9)

Solution

6a2−4a−1

(p2−6p−18)+(2p2+11)

In the following exercises, subtract the polynomials.

(4m2−6m−3)−(2m2+m−7)

Solution

2m2−7m+4

(3b2−4b+1)−(5b2−b−2)

(a2+8a+5)−(a2−3a+2)

Solution

11a+3

(b2−7b+5)−(b2−2b+9)

(12s2−15s)−(s−9)

Solution

12s2−16s+9

(10r2−20r)−(r−8)

In the following exercises, subtract the polynomials.

Subtract (9x2+2) from (12x2−x+6)

Solution

3x2−x+4

Subtract (5y2−y+12) from (10y2−8y−20)

Subtract (7w2−4w+2) from (8w2−w+6)

Solution

w2+3w+4

Subtract (5x2−x+12) from (9x2−6x−20)

In the following exercises, find the difference of the polynomials.

Find the difference of (w2+w−42) and (w2−10w+24)

Solution

11w−66

Find the difference of (z2−3z−18) and (z2+5z−20)

In the following exercises, add the polynomials.

(7x2−2xy+6y2)+(3x2−5xy)

Solution

10x2−7xy+6y2

(−5x2−4xy−3y2)+(2x2−7xy)

(7m2+mn−8n2)+(3m2+2mn)

Solution

10m2+3mn−8n2

(2r2−3rs−2s2)+(5r2−3rs)

In the following exercises, add or subtract the polynomials.

(a2−b2)−(a2+3ab−4b2)

Solution

−3ab+3b2

(m2+2n2)−(m2−8mn−n2)

(p3−3p2q)+(2pq2+4q3)−(3p2q+pq2)

Solution

p3−6p2q+pq2+4q3

(a3−2a2b)+(ab2+b3)−(3a2b+4ab2)

(x3−x2y)−(4xy2−y3)+(3x2y−xy2)

Solution

x3+2x2y−5xy2+y3

(x3−2x2y)−(xy2−3y3)−(x2y−4xy2)

Evaluate a Polynomial Function for a Given Value

In the following exercises, find the function values for each polynomial function.

For the function f(x)=8x2−3x+2, find:
ⓐ f(5) ⓑ f(−2) ⓒ f(0)

Solution

ⓐ 187 ⓑ 40 ⓒ 2

For the function f(x)=5x2−x−7, find:
ⓐ f(−4) ⓑ f(1) ⓒ f(0)

For the function g(x)=4−36x, find:
ⓐ g(3) ⓑ g(0) ⓒ g(−1)

Solution

ⓐ −104 ⓑ 4 ⓒ 40

For the function g(x)=16−36x2, find:
ⓐ g(−1) ⓑ g(0) ⓒ g(2)

In the following exercises, find the height for each polynomial function.

A painter drops a brush from a platform 75 feet high. The polynomial function h(t)=−16t2+75 gives the height of the brush t seconds after it was dropped. Find the height after t=2 seconds.

Solution

The height is 11 feet.

A girl drops a ball off a 200-foot cliff into the ocean. The polynomial h(t)=−16t2+200 gives the height of the ball, in feet, t seconds after it is dropped. Find the height after t=3 seconds.

A manufacturer of stereo sound speakers has found that the revenue received from selling the speakers at a cost of p dollars each is given by the polynomial function R(p)=−4p2+420p. Find the revenue received when p=60 dollars.

Solution

The revenue is $10,800.

A manufacturer of the latest basketball shoes has found that the revenue received from selling the shoes at a cost of p dollars each is given by the polynomial R(p)=−4p2+420p. Find the revenue received when p=90 dollars.

The polynomial C(x)=6x2+90x gives the cost, in dollars, of producing a rectangular container whose top and bottom are squares with side x feet and height 6 feet. Find the cost of producing a box with x=4 feet.

Solution

The cost is $456.

The polynomial C(x)=6x2+90x gives the cost, in dollars, of producing a rectangular container whose top and bottom are squares with side x feet and height 4 feet. Find the cost of producing a box with x=6 feet.

Add and Subtract Polynomial Functions

In each example, find ⓐ (f + g)(x) ⓑ (f + g)(2) ⓒ (f − g)(x) ⓓ (f − g)(−3).

f(x)=2x2−4x+1 and g(x)=5x2+8x+3

Solution

ⓐ (f+g)(x)=7x2+4x+4 ⓑ (f+g)(2)=40
ⓒ (f−g)(x)=−3x2−12x−2
ⓓ (f−g)(−3)=7

f(x)=4x2−7x+3 and g(x)=4x2+2x−1

f(x)=3x3−x2−2x+3 and g(x)=3x3−7x

Solution


ⓐ  (f+g)(x)=6x3−x2−9x+3
ⓑ (f+g)(2)=29
ⓒ (f−g)(x)=−x2+5x+3
ⓓ (f−g)(−3)=−21

f(x)=5x3−x2+3x+4 and g(x)=8x3−1

Writing Exercises

Using your own words, explain the difference between a monomial, a binomial, and a trinomial.

Solution

Answers will vary.

Using your own words, explain the difference between a polynomial with five terms and a polynomial with a degree of 5.

Ariana thinks the sum 6y2+5y4 is 11y6. What is wrong with her reasoning?

Solution

Answers will vary.

Is every trinomial a second degree polynomial? If not, give an example.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The figure shows a table with six rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is "confidently", the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “identify polynomials, monomials, binomials, and trinomials”, “determine the degree of polynomials”, “add and subtract monomials”, “add and subtract polynomials”, and “evaluate a polynomial for a given value”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

binomial
A binomial is a polynomial with exactly two terms.
degree of a constant
The degree of any constant is 0.
degree of a polynomial
The degree of a polynomial is the highest degree of all its terms.
degree of a term
The degree of a term is the sum of the exponents of its variables.
monomial
A monomial is an algebraic expression with one term. A monomial in one variable is a term of the form axm, where a is a constant and m is a whole number.
polynomial
A monomial or two or more monomials combined by addition or subtraction is a polynomial.
standard form of a polynomial
A polynomial is in standard form when the terms of a polynomial are written in descending order of degrees.
trinomial
A trinomial is a polynomial with exactly three terms.
polynomial function
A polynomial function is a function whose range values are defined by a polynomial.

Properties of Exponents and Scientific Notation

Learning Objectives

By the end of this section, you will be able to:

  • Simplify expressions using the properties for exponents
  • Use the definition of a negative exponent
  • Use scientific notation

Before you get started, take this readiness quiz.

Simplify: (−2)(−2)(−2).
If you missed this problem, review Example 8 in Integers.

Solution

−8

Simplify: 8x24y.
If you missed this problem, review Example 1 in Fractions.

Solution

x3y

Name the decimal (−2.6)(4.21).
If you missed this problem, review Example 3 in Decimals.

Solution

−10.946

Simplify Expressions Using the Properties for Exponents

Remember that an exponent indicates repeated multiplication of the same quantity. For example, in the expression am, the exponent m tells us how many times we use the base a as a factor.

First example: a raised to the power of m equals a times a times a times a and so on until you have multiplied m different factors of a together. Second example: the quantity negative 9 raised to the power of 5 equals negative 9 times negative 9 times negative 9 times negative 9 times negative 9, a total of 5 factors of negative 9.

Let’s review the vocabulary for expressions with exponents.

Exponential Notation

The figure shows the letter a in a normal font with the label base and the letter m in a superscript font with the label exponent. This means we multiply the number a with itself, m times.

This is read a to the mth power.

In the expression am, the exponent m tells us how many times we use the base a as a factor.

When we combine like terms by adding and subtracting, we need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too.

First, we will look at an example that leads to the Product Property.

The mathematical expression x squared multiplied by x cubed, shown as x^2 ×•× x^3.
What does this mean? Visual demonstration of the addition of factors: 2 factors of 'x' + 3 factors of 'x' = 5 factors of 'x'.
The image displays x raised to the fifth power.

Notice that 5 is the sum of the exponents, 2 and 3. We see x2·x3 is x2+3 or x5.

The base stayed the same and we added the exponents. This leads to the Product Property for Exponents.

Product Property for Exponents

If a is a real number and m and n are integers, then

am·an=am+n

To multiply with like bases, add the exponents.

Simplify each expression: ⓐ y5·y6 ⓑ 2x·23x ⓒ 2a7·3a. ⓓ d4⋅ d5⋅ d2

Solution
ⓐ
   A mathematical expression showing 'y' raised to the power of 5, multiplied by 'y' raised to the power of 6, represented as y^5 * y^6.
Use the Product Property, am·an=am+n.    A mathematical expression showing the variable 'y' raised to the power of the sum '5+6'.
Simplify.    The image displays the mathematical expression y raised to the power of 11, written as y^11, on a white background.
ⓑ
   A mathematical expression displaying 2 to the power of x multiplied by 2 to the power of 3x, represented as 2^x  2^(3x).
Use the Product Property, am·an=am+n.    A mathematical expression showing the addition of two terms, '2x + 3x', with the 'x' variable and plus sign in a reddish hue, isolated against a white background.
Simplify.    A mathematical expression displays the number 2 with 'dx' as a superscript, written as 2^dx.
ⓒ
   A mathematical expression displays '2a^2 * 3a' in black text against a white background.
Rewrite, a=a1.    The image displays the mathematical expression '2a' ' multiplied by '3a' ', indicating the product of two terms, each containing a coefficient and a variable 'a' with a prime symbol.
Use the Commutative Property and
use the Product Property, am·an=am+n.
   A mathematical expression shows the product of 2, 3, and 'a' raised to the power of 'x+1', written as 2 * 3 * a^(x+1).
Simplify.    The image displays the mathematical expression '6a⁸' in black text against a plain white background.
ⓓ
  The mathematical expression d^4 multiplied by d^5 multiplied by d^2, which simplifies to d^(4+5+2) or d^11, illustrating the product rule for exponents.
Add the exponents, since bases are the same.   The image displays a mathematical expression with the letter 'd' as the base, and an exponent that is the sum of three numbers: 4, 5, and 2. The expression appears as d^(4+5+2).
Simplify.   A close-up of the mathematical expression q^11 on a white background.

Simplify each expression:

ⓐ b9·b8 ⓑ 42x·4x ⓒ 3p5·4p ⓓ x6·x4·x8.

Solution

ⓐ b17 ⓑ 43x ⓒ 12p6
ⓓ x18

Simplify each expression:

ⓐ x12·x4 ⓑ 10·10x ⓒ 2z·6z7 ⓓ b5·b9·b5.

Solution

ⓐ x16 ⓑ 10x+1 ⓐ 12z8
ⓓ b19

Now we will look at an exponent property for division. As before, we’ll try to discover a property by looking at some examples.

Consider x5x2 and x2x3
What do they mean? x·x·x·x·xx·x x·xx·x·x
Use the Equivalent Fractions Property. x·x·x·x·xx·x x·x·1x·x·x
Simplify. x3 1x

Notice, in each case the bases were the same and we subtracted exponents. We see x5x2 is x5−2 or x3. We see x2x3 is or 1x. When the larger exponent was in the numerator, we were left with factors in the numerator. When the larger exponent was in the denominator, we were left with factors in the denominator--notice the numerator of 1. When all the factors in the numerator have been removed, remember this is really dividing the factors to one, and so we need a 1 in the numerator. xx=1. This leads to the Quotient Property for Exponents.

Quotient Property for Exponents

If a is a real number, a≠0, and m and n are integers, then

aman=am−n,m>nandaman=1an−m,n>m

Simplify each expression: ⓐ x9x7 ⓑ 31032 ⓒ b8b12 ⓓ 7375.

Solution

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

ⓐ
Since 9>7, there are more factors of x in the numerator. A mathematical expression showing x to the power of 9 divided by x to the power of 7, which simplifies to x squared.
Use Quotient Property, aman=am−n. A mathematical expression featuring a large 'X' followed by a superscript '9', then a minus sign, and finally the number '7'. The 'X' is black, and the '9-7' portion is in a slightly faded red hue.
Simplify. The mathematical expression x squared, or x^2, is displayed in black text on a white background.
ⓑ
Since 10>2, there are more factors of 3 in the numerator. A mathematical fraction showing 3 to the power of 10 divided by 3 to the power of 2, an example of exponent rules where bases are the same.
Use Quotient Property, aman=am−n. A close-up view of the numbers '3 10-' displayed on a screen or sign. The '3' is in black, and '10-' is in a reddish-brown hue against a light background.
Simplify. A close-up of a white background with the number '3' in black, followed by a smaller, superscript 'g', resembling '3g'.

Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.

ⓒ
Since 12>8, there are more factors of b in the denominator. A mathematical expression showing b to the power of 8 divided by b to the power of 12.
Use Quotient Property, aman=1an−m. A mathematical expression showing the fraction 1 over b to the power of (12 minus 8).
Simplify. A mathematical expression showing the fraction one over b to the power of four, or 1/b^4.
ⓓ
Since 5>3, there are more factors of 3 in the denominator. A mathematical expression showing 7 raised to the power of 3, divided by 7 raised to the power of 5. This simplifies to 7 to the power of -2, or 1/49.
Use Quotient Property, aman=1an−m. A mathematical expression displaying the fraction 1 over 7 raised to the power of (s minus 3). The numerator '1' is in red, and the exponent 's-3' is also in a reddish-brown color.
Simplify. A fraction with 1 in the numerator and 7 squared (7^2) in the denominator, representing 1/49.
Simplify. The fraction one over forty-nine, represented numerically as 1/49, is displayed against a white background.

Notice that when the larger exponent is in the denominator, we are left with factors in the denominator.

Simplify each expression: ⓐ x15x10 ⓑ 61465 ⓒ x18x22 ⓓ 12151230.

Solution

ⓐ x5 ⓑ 69 ⓒ 1x4
ⓓ 11215

Simplify each expression: ⓐ y43y37 ⓑ 1015107 ⓒ m7m15 ⓓ 98919.

Solution

ⓐ y6 ⓑ 108 ⓒ 1m8
ⓓ 1911

A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like amam. We know,xx=1, for any x(x≠0) since any number divided by itself is 1.

The Quotient Property for Exponents shows us how to simplify amam. when m>n and when n<m by subtracting exponents. What if m=n? We will simplifyamam in two ways to lead us to the definition of the Zero Exponent Property. In general, for a≠0:

In the first way we write a to the power of m divided by a to the power of m as a to the power of the quantity m minus m. This is equal to a to the power of 0. In the second way we write a to the power of m divided by a to the power of m as a fraction with m factors of a in the numerator and a factors of m in the denominator. Simplifying this we can cross of all the factors and are left with the number 1. This shows that a to the power of 0 is equal to 1.

We see amam simplifies to a0 and to 1. So a0=1. Any non-zero base raised to the power of zero equals 1.

Zero Exponent Property

If a is a non-zero number, then a0=1.

If a is a non-zero number, then a to the power of zero equals 1.

Any non-zero number raised to the zero power is 1.

In this text, we assume any variable that we raise to the zero power is not zero.

Simplify each expression: ⓐ 90 ⓑ n0.

Solution

The definition says any non-zero number raised to the zero power is 1.

ⓐ
90 Use the definition of the zero exponent.1

ⓑ
n0 Use the definition of the zero exponent.1

To simplify the expression n raised to the zero power we just use the definition of the zero exponent. The result is 1.

Simplify each expression: ⓐ 110 ⓑ q0.

Solution

ⓐ 1 ⓑ 1

Simplify each expression: ⓐ 230 ⓑ r0.

Solution

ⓐ 1 ⓑ 1

Use the Definition of a Negative Exponent

We saw that the Quotient Property for Exponents has two forms depending on whether the exponent is larger in the numerator or the denominator. What if we just subtract exponents regardless of which is larger?

Let’s consider x2x5. We subtract the exponent in the denominator from the exponent in the numerator. We see x2x5 is x2−5 or x−3.

We can also simplify x2x5 by dividing out common factors:

In the figure the expression x raised to the power of 2 divided by x raised to the power of 5 is written as a fraction with 2 factors of x in the numerator divided by 5 factors of x in the denominator. Two factors are crossed off in both the numerator and denominator. This only leaves 3 factors of x in the denominator. The simplified fraction is 1 divided by x to the power of 3.

This implies that x−3=1x3 and it leads us to the definition of a negative exponent. If n is an integer and a≠0, then a−n=1an.

Let’s now look at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.

Steps demonstrating the simplification of the expression 1/(a^-n) using the definition of negative exponents.
1a−n
Use the definition of a negative exponent, a−n=1an. 11an
Simplify the complex fraction. 1·an1
Multiply. an

This implies 1a−n=an and is another form of the definition of Properties of Negative Exponents.

Properties of Negative Exponents

If n is an integer and a≠0, then a−n=1an or 1a−n=an.

The negative exponent tells us we can rewrite the expression by taking the reciprocal of the base and then changing the sign of the exponent.

Any expression that has negative exponents is not considered to be in simplest form. We will use the definition of a negative exponent and other properties of exponents to write the expression with only positive exponents.

For example, if after simplifying an expression we end up with the expression x−3, we will take one more step and write 1x3. The answer is considered to be in simplest form when it has only positive exponents.

Simplify each expression: ⓐ x−5 ⓑ 10−3 ⓒ 1y−4 ⓓ 13−2.

Solution
ⓐ
Demonstrates the negative exponent rule a^-n = 1/a^n with an example.
x−5
Use the definition of a negative exponent, a−n=1an. 1x5
ⓑ
Steps to evaluate 10^-3 by applying the negative exponent definition and simplifying to a fraction.
10−3
Use the definition of a negative exponent, a−n=1an. 1103
Simplify. 11000
ⓒ
This table demonstrates the application of the negative exponent property (1/a^-n = a^n) to simplify a mathematical expression.
1y−4
Use the property of a negative exponent, 1a−n=an. y4
ⓓ
Step-by-step simplification of 1/3^-2 using negative exponent properties.
13−2
Use the property of a negative exponent, 1a−n=an. 32
Simplify. 9

Simplify each expression: ⓐ z−3 ⓑ 10−7 ⓒ 1p−8 ⓓ 14−3.

Solution

ⓐ 1z3 ⓑ 1107 ⓒ p8 ⓓ 64

Simplify each expression: ⓐ n−2 ⓑ 10−4 ⓒ 1q−7 ⓓ 12−4.

Solution

ⓐ 1n2 ⓑ 110,000 ⓒ q7
ⓓ 16

Suppose now we have a fraction raised to a negative exponent. Let’s use our definition of negative exponents to lead us to a new property.

Step-by-step simplification of an expression with a negative exponent, illustrating how (3/4)^-2 equals (4/3)^2.
(34)−2
Use the definition of a negative exponent, a−n=1an. 1(34)2
Simplify the denominator. 1916
Simplify the complex fraction. 169
But we know that 169 is (43)2.
This tells us that (34)−2=(43)2

To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base—the fraction—and changed the sign of the exponent.

This leads us to the Quotient to a Negative Power Property.

Quotient to a Negative Power Property

If a and b are real numbers, a≠0,b≠0 and n is an integer, then

(ab)−n=(ba)n

Simplify each expression: ⓐ (57)−2 ⓑ (−xy)−3.

Solution
ⓐ
Steps to simplify a fractional expression with a negative exponent, applying the Quotient to a Negative Exponent Property.
(57)−2
Use the Quotient to a Negative Exponent Property, (ab)−n=(ba)n.
Take the reciprocal of the fraction and change the sign of the exponent.
(75)2
Simplify. 4925
ⓑ
This table illustrates the step-by-step simplification of an algebraic expression with a negative exponent, applying exponent properties.
(−xy)−3
Use the Quotient to a Negative Exponent Property, (ab)−n=(ba)n.
Take the reciprocal of the fraction and change the sign of the exponent.
(−yx)3
Simplify. −y3x3

Simplify each expression: ⓐ (23)−4 ⓑ (−mn)−2.

Solution

ⓐ 8116 ⓑ n2m2

Simplify each expression: ⓐ (35)−3 ⓑ (−ab)−4.

Solution

ⓐ 12527 ⓑ b4a4

Now that we have negative exponents, we will use the Product Property with expressions that have negative exponents.

Simplify each expression: ⓐ z−5·z−3 ⓑ (m4n−3)(m−5n−2) ⓒ (2x−6y8)(−5x5y−3).

Solution
ⓐ
Steps demonstrating the simplification of an expression involving multiplication of terms with negative exponents.
z−5·z−3
Add the exponents, since the bases are the same. z−5−3
Simplify. z−8
Use the definition of a negative exponent. 1z8
ⓑ
Step-by-step simplification of an algebraic expression using properties of exponents, including the commutative property and handling negative exponents.
(m4n−3)(m−5n−2)
Use the Commutative Property to get like bases together. m4m−5·n−2n−3
Add the exponents for each base. m−1·n−5
Take reciprocals and change the signs of the exponents. 1m1·1n5
Simplify. 1mn5
ⓒ
Step-by-step simplification of an algebraic expression using exponent rules, demonstrating each transformation.
(2x−6y8)(−5x5y−3)
Rewrite with the like bases together. 2(−5)·(x−6x5)·(y8y−3)
Multiply the coefficients and add the exponents of each variable. −10·x−1·y5
Use the definition of a negative exponent, a−n=1an. −10·1x·y5
Simplify. −10y5x

Simplify each expression:

ⓐ z−4·z−5 ⓑ (p6q−2)(p−9q−1) ⓒ (3u−5v7)(−4u4v−2).

Solution

ⓐ 1z9 ⓑ 1p3q3 ⓒ −12v5u

Simplify each expression:

ⓐ c−8·c−7 ⓑ (r5s−3)(r−7s−5) ⓒ (−6c−6d4)(−5c−2d−1).

Solution

ⓐ 1c15 ⓑ 1r2s8 ⓒ 30d3c8

Now let’s look at an exponential expression that contains a power raised to a power. See if you can discover a general property.

(x2)3 What does this mean?x2·x2·x2

How many factors altogether? An illustration showing x * x * x * x * x * x, grouped as three sets of 2 factors, totaling 6 factors. Each 'x' in the product is considered a factor.
So we have The image displays x raised to the sixth power.

Notice the 6 is the product of the exponents, 2 and 3. We see that (x2)3 is x2·3 or x6.

We multiplied the exponents. This leads to the Power Property for Exponents.

Power Property for Exponents

If a is a real number and m and n are integers, then

(am)n=am·n

To raise a power to a power, multiply the exponents.

Simplify each expression: ⓐ (y5)9 ⓑ (44)7 ⓒ (y3)6(y5)4.

Solution
ⓐ
The mathematical expression shows 'y to the power of 5, all raised to the power of 9', often simplified as y to the power of 45, demonstrating the power of a power rule in algebra.
Use the Power Property, (am)n=am·n. The image shows a mathematical expression with the letter 'y' in black, and a red exponent '5.9' directly above and to its right. The '5.9' appears slightly faded or lighter in color than the 'y'.
Simplify. A close-up shot of the mathematical expression y raised to the power of 45, or y^45, written in a dark font against a plain white background.
ⓑ
A mathematical expression featuring 4 raised to the power of 4, enclosed in parentheses, and then raised to the power of r.
Use the Power Property.          A numerical display showing '4.9' with the number '4' in black and larger, and '.9' in a smaller, reddish hue. It resembles a rating or a specific value.
Simplify. A numerical expression featuring the number 4 with a small, raised '2B' appearing as a superscript on a white background, suggesting a mathematical exponent.
ⓒ
Step-by-step simplification of an exponential expression using the Power Property and addition of exponents.
(y3)6(y5)4
Use the Power Property. y18·y20
Add the exponents. y38

Simplify each expression: ⓐ (b7)5 ⓑ (54)3 ⓒ (a4)5(a7)4.

Solution

ⓐ b35 ⓑ 512 ⓒ a48

Simplify each expression: ⓐ (z6)9 ⓑ (37)7 ⓒ (q4)5(q3)3.

Solution

ⓐ z54 ⓑ 349 ⓒ q29

We will now look at an expression containing a product that is raised to a power. Can you find this pattern?

This table illustrates the step-by-step expansion of the algebraic expression (2x)"3.
(2x)3
What does this mean? 2x·2x·2x
We group the like factors together. 2·2·2·x·x·x
How many factors of 2 and of x 23·x3

Notice that each factor was raised to the power and (2x)3 is 23·x3.

The exponent applies to each of the factors! This leads to the Product to a Power Property for Exponents.

Product to a Power Property for Exponents

If a and b are real numbers and m is a whole number, then

(ab)m=ambm

To raise a product to a power, raise each factor to that power.

Simplify each expression: ⓐ (−3mn)3 ⓑ (−4a2b)0 ⓒ (6k3)−2 ⓓ (5x−3)2.

Solution
ⓐ
 A mathematical expression reads as an open parenthesis, negative three, m, n, close parenthesis, raised to the power of three.
Use Power of a Product Property, (ab)m=ambm. A mathematical expression shows an open parenthesis, a minus sign, the number 3, a close parenthesis, a superscript 3 in red, the letter m with a superscript 3 in red, and the letter n with a superscript 3 in red.
Simplify.  A mathematical expression showing -27 multiplied by m to the power of 3, and n to the power of 3.
ⓑ
Step-by-step simplification of the expression (-4a^2b)^0 using the zero exponent rule.
(−4a2b)0
Use Power of a Product Property, (ab)m=ambm. (−4)0(a2)0(b)0
Simplify. 1·1·1
Multiply. 1
ⓒ
This table demonstrates the step-by-step simplification of the algebraic expression (6k^3)^-2 using exponent properties.
(6k3)−2
Use the Product to a Power Property, (ab)m=ambm. (6)−2(k3)−2
Use the Power Property, (am)n=am·n. 6−2k−6
Use the Definition of a negative exponent, a−n=1an. 162·1k6
Simplify. 136k6
ⓓ
Steps to simplify the algebraic expression (5x^-3)^2 using exponent properties.
(5x−3)2
Use the Product to a Power Property, (ab)m=ambm. 52(x−3)2
Simplify. 25·x−6
Rewrite x−6 using, a−n=1an. 25·1x6
Simplify. 25x6

Simplify each expression: ⓐ (2wx)5 ⓑ (−11pq3)0 ⓒ (2b3)−4 ⓓ (8a−4)2.

Solution

ⓐ 32w5x5 ⓑ 1 ⓒ 116b12
ⓓ 64a8

Simplify each expression: ⓐ (−3y)3 ⓑ (−8m2n3)0 ⓒ (−4x4)−2 ⓓ (2c−4)3.

Solution

ⓐ −27y3 ⓑ 1 ⓒ 116x8
ⓓ 8c12

Now we will look at an example that will lead us to the Quotient to a Power Property.

Steps demonstrating how to cube a fraction, showing its expansion and simplification to an exponential form.
(xy)3
This means xy·xy·xy
Multiply the fractions. x·x·xy·y·y
Write with exponents. x3y3

Notice that the exponent applies to both the numerator and the denominator.

We see that (xy)3 is x3y3.

This leads to the Quotient to a Power Property for Exponents.

Quotient to a Power Property for Exponents

If a and b are real numbers, b≠0, and m is an integer, then

(ab)m=ambm

To raise a fraction to a power, raise the numerator and denominator to that power.

Simplify each expression:

ⓐ (b3)4 ⓑ (kj)−3 ⓒ (2xy2z)3 ⓓ (4p−3q2)2.

Solution
ⓐ
  A mathematical expression showing the fraction b over 3, all raised to the power of 4.
Use Quotient to a Power Property, (ab)m=ambm.   A mathematical expression displaying b to the power of 4 divided by 3 to the power of 4, with both exponents in red. It can also be interpreted as the fraction b/3 raised to the power of 4.
Simplify.   A mathematical fraction showing 'b' raised to the power of 4, divided by 81.
ⓑ
The mathematical expression (k/j) raised to the power of -3.
Raise the numerator and denominator to the power. A mathematical fraction showing k to the power of negative 3 over j to the power of negative 3, with the negative exponents in red.
Use the definition of negative exponent. Math expression: (1/k^3) multiplied by an integral symbol with limits 1 to 3, all divided by 1.
Multiply. A mathematical expression displaying a fraction with 'j' cubed in the numerator and 'k' cubed in the denominator.
ⓒ
This table illustrates the step-by-step simplification of a rational expression using the Quotient and Product to a Power Properties.
(2xy2z)3
Use Quotient to a Power Property, (ab)m=ambm. (2xy2)3z3
Use the Product to a Power Property, (ab)m=ambm. 8x3y6z3
ⓓ
Step-by-step simplification of an exponential expression, illustrating the application of various exponent properties.
(4p−3q2)2
Use Quotient to a Power Property, (ab)m=ambm. (4p−3)2(q2)2
Use the Product to a Power Property, (ab)m=ambm. 42(p−3)2(q2)2
Simplify using the Power Property, (am)n=am·n. 16p−6q4
Use the definition of negative exponent. 16q4·1p6
Simplify. 16p6q4

Simplify each expression:

ⓐ (p10)4 ⓑ (mn)−7 ⓒ (3ab3c2)4 ⓓ (3x−2y3)3.

Solution

ⓐ p410000 ⓑ n7m7
ⓒ 81a4b12c8 ⓓ 27x6y9

Simplify each expression:

ⓐ (−2q)3 ⓑ (wx)−4 ⓒ (xy33z2)2 ⓓ (2m−2n−2)3.

Solution

ⓐ −8q3 ⓑ x4w4 ⓒ x2y69z4
ⓓ 8n6m6

We now have several properties for exponents. Let’s summarize them and then we’ll do some more examples that use more than one of the properties.

Summary of Exponent Properties

If a and b are real numbers, and m and n are integers, then

Property Description
Product Property am·an=am+n
Power Property (am)n=am·n
Product to a Power (ab)m=ambm
Quotient Property aman=am−n,a≠0
Zero Exponent Property a0=1,a≠0
Quotient to a Power Property (ab)m=ambm,b≠0
Properties of Negative Exponents a−n=1an and 1a−n=an
Quotient to a Negative Exponent (ab)−n=(ba)n

Simplify each expression by applying several properties:

ⓐ (3x2y)4(2xy2)3 ⓑ (x3)4(x−2)5(x6)5 ⓒ (2xy2x3y−2)2(12xy3x3y−1)−1.

Solution
ⓐ
Step-by-step simplification of an exponential expression using product to a power, commutative property, and exponent rules.
(3x2y)4(2xy2)3
Use the Product to a Power Property, (ab)m=ambm. (34x8y4)(23x3y6)
Simplify. (81x8y4)(8x3y6)
Use the Commutative Property. 81·8·x8·x3·y4·y6
Multiply the constants and add the exponents. 648x11y10
ⓑ
Step-by-step simplification of an algebraic expression involving exponents, demonstrating the application of exponent properties.
(x3)4(x−2)5(x6)5
Use the Power Property, (am)n=am·n. (x12)(x−10)(x30)
Add the exponents in the numerator. x2x30
Use the Quotient Property, aman=1an−m. 1x28
ⓒ
Step-by-step simplification of an algebraic expression using various exponent properties and algebraic rules.
(2xy2x3y−2)2(12xy3x3y−1)−1
Simplify inside the parentheses first. (2y4x2)2(12y4x2)−1
Use the Quotient to a Power Property, (ab)m=ambm. (2y4)2(x2)2(12y4)−1(x2)−1
Use the Product to a Power Property, (ab)m=ambm. 4y8x4·12−1y−4x−2
Simplify. 4y412x2
Simplify. y43x2

Simplify each expression:

ⓐ (c4d2)5(3cd5)4 ⓑ (a−2)3(a2)4(a4)5 ⓒ (3xy2x2y−3)2(9xy−3x3y2)−1.

Solution

ⓐ 81c24d30 ⓑ 1a18
ⓒ y15

Simplify each expression:

ⓐ (a3b2)6(4ab3)4 ⓑ (p−3)4(p5)3(p7)6 ⓒ (4x3y2x2y−1)2(8xy−3x2y)−1.

Solution

ⓐ 256a22b24 ⓑ 1p39
ⓒ 2x3y10

Use Scientific Notation

Working with very large or very small numbers can be awkward. Since our number system is base ten we can use powers of ten to rewrite very large or very small numbers to make them easier to work with. Consider the numbers 4,000 and 0.004.

Using place value, we can rewrite the numbers 4,000 and 0.004. We know that 4,000 means 4×1,000 and 0.004 means 4×11,000.

If we write the 1,000 as a power of ten in exponential form, we can rewrite these numbers in this way:

This table demonstrates different forms of numbers, including decimal, expanded, exponential, and scientific notations, emphasizing the use of powers of 10.
4,000 4×1,000 4×103
0.004 4×11,000 4×1103 4×10−3

When a number is written as a product of two numbers, where the first factor is a number greater than or equal to one but less than ten, and the second factor is a power of 10 written in exponential form, it is said to be in scientific notation.

Scientific Notation

A number is expressed in scientific notation when it is of the form

a×10nwhere1≤│a│<10andnis an integer.

It is customary in scientific notation to use as the × multiplication sign, even though we avoid using this sign elsewhere in algebra.

If we look at what happened to the decimal point, we can see a method to easily convert from decimal notation to scientific notation.

The figure shows two examples of converting from standard notation to scientific notation. In one example 4000 is converted to 4 times 10 to the power of 3. The decimal point in 4000 starts at the right and moves 3 places to the left to make the number 4. The 3 places moved make the exponent 3. In the other example, the number 0.004 is converted to 4 times 10 to the negative 3 power. The decimal point in 0.004 is moved 3 places to the right to make the number 4. The 3 places moved make the exponent negative 3.

In both cases, the decimal was moved 3 places to get the first factor between 1 and 10.

The power of 10 is positive when the number is larger than 1: 4,000=4×103

The power of 10 is negative when the number is between 0 and 1: 0.004=4×10−3

To convert a decimal to scientific notation.

  1. Move the decimal point so that the first factor is greater than or equal to 1 but less than 10.
  2. Count the number of decimal places, n, that the decimal point was moved.
  3. Write the number as a product with a power of 10. If the original number is.
    • greater than 1, the power of 10 will be 10n.
    • between 0 and 1, the power of 10 will be 10−n.
  4. Check.

Write in scientific notation: ⓐ 37,000 ⓑ 0.0052.

Solution
ⓐ
The original number, 37,000, is greater than 1
so we will have a positive power of 10.
37,000
Move the decimal point to get 3.7, a number
between 1 and 10.
The number 37,000 is shown with light blue wavy arrows underneath, indicating grouping or place value counting, with the last arrow pointing up to the digit 7.
Count the number of decimal places the point
was moved.
The image displays the text '4 places' in a clear, dark gray font on a white background.
Write as a product with a power of 10. The image displays the number 3.7 multiplied by 10 to the power of 4, written as 3.7 x 10^4, in a clear, dark gray font against a white background.
Check:    3.7×1043.7×10,00037,000
The number 37,000 is represented in scientific notation as 3.7 x 10^4.
ⓑ
The original number, 0.0052, is between 0
and 1 so we will have a negative power of 10.
0.0052
Move the decimal point to get 5.2, a number
between 1 and 10.
The number 0.0052 with three blue wavy arrows illustrating a decimal point shift three places to the right, indicating multiplication by 1000.
Count the number of decimal places the point
was moved.
The text '3 places' is displayed on a white background, indicating a quantity of locations.
Write as a product with a power of 10. A mathematical expression displaying 5.2 multiplied by 10 to the power of negative 3, written as 5.2 x 10^-3.
Check:5.2×10−35.2×11035.2×110005.2×0.0010.0052
The image displays the number 0.0052 expressed in scientific notation as 5.2 multiplied by 10 to the power of negative 3 (5.2 x 10^-3).

Write in scientific notation: ⓐ 96,000 ⓑ 0.0078.

Solution

ⓐ 9.6×104 ⓑ 7.8×10−3

Write in scientific notation: ⓐ 48,300 ⓑ 0.0129.

Solution

ⓐ 4.83×104
ⓑ 1.29×10−2

How can we convert from scientific notation to decimal form? Let’s look at two numbers written in scientific notation and see.

9.12×1049.12×10−4 9.12×10,0009.12×0.0001 91,2000.000912

If we look at the location of the decimal point, we can see an easy method to convert a number from scientific notation to decimal form.

The figure shows two examples of converting from scientific notation to standard notation. In one example 9.12 times 10 to the power of 4 is converted to 91200. The decimal point in 9.12 moves 4 places to the right to make the number 91200. In the other example, the number 9.12 times 10 to the power of -4 is converted to 0.000912. The decimal point in 9.12 is moved 4 places to the left to make the number 0.000912.

In both cases the decimal point moved 4 places. When the exponent was positive, the decimal moved to the right. When the exponent was negative, the decimal point moved to the left.

Convert scientific notation to decimal form.

  1. Determine the exponent, n, on the factor 10.
  2. Move the decimal n places, adding zeros if needed.
    • If the exponent is positive, move the decimal point n places to the right.
    • If the exponent is negative, move the decimal point |n| places to the left.
  3. Check.

Convert to decimal form: ⓐ 6.2×103 ⓑ −8.9×10−2.

Solution
ⓐ
A mathematical expression in scientific notation reads '6.2 multiplied by 10 to the power of 3' against a white background.
Determine the exponent, n, on the factor 10.
The exponent is 3.
Since the exponent is positive, move the
decimal point 3 places to the right.
The number 6.200 is displayed with a light blue wavy arrow pointing to the right, originating beneath the '200' digits, suggesting a focus on or action related to the trailing zeros.
Add zeros as needed for placeholders. The number 6,200 is prominently displayed in a clear, digital font against a plain white background.
The equation shows 6.2 multiplied by 10 to the power of 3 equals 6,200, illustrating the conversion from scientific notation to standard form.
ⓑ
A mathematical expression displays -8.9 x 10^-2 in a crisp, white background. The numbers and symbols are clear, presenting a standard scientific notation value.
Determine the exponent, n, on the factor 10. The exponent is −2.
Since the exponent is negative, move the
decimal point 2 places to the left.
A mathematical expression shows '- 8.9' with a blue wavy arrow pointing from the '8' to the minus sign, then curving downwards and to the right, passing beneath the entire number.
Add zeros as needed for placeholders. A close-up image shows the negative decimal number -0.089 against a white background.
The image displays the conversion of -8.9 x 10^-2 into its decimal form, -0.089, illustrating how multiplying by 10 to the power of a negative exponent shifts the decimal point to the left.

Convert to decimal form: ⓐ 1.3×103 ⓑ −1.2×10−4.

Solution

ⓐ 1,300 ⓑ −0.00012

Convert to decimal form: ⓐ −9.5×104 ⓑ 7.5×10−2.

Solution

ⓐ −950,000 ⓑ 0.075

When scientists perform calculations with very large or very small numbers, they use scientific notation. Scientific notation provides a way for the calculations to be done without writing a lot of zeros. We will see how the Properties of Exponents are used to multiply and divide numbers in scientific notation.

Multiply or divide as indicated. Write answers in decimal form: ⓐ (−4×105)(2×10−7) ⓑ 9×1033×10−2.

Solution
ⓐ
Step-by-step multiplication of numbers in scientific notation and conversion to decimal form.
(−4×105)(2×10−7)
Use the Commutative Property to rearrange the factors. −4·2·105·10−7
Multiply. −8×10−2
Change to decimal form by moving the decimal two places left. −0.08
ⓑ
Step-by-step simplification of a mathematical expression involving scientific notation, showing each stage of calculation.
9×1033×10−2
Separate the factors, rewriting as the product of two fractions. 93×10310−2
Divide. 3×105
Change to decimal form by moving the decimal five places right. 300,000

Multiply or divide as indicated. Write answers in decimal form:

ⓐ (−3×105)(2×10−8) ⓑ 8×1024×10−2.

Solution

ⓐ −0.006 ⓑ 20,000

Multiply or divide as indicated. Write answers in decimal form:

ⓐ (−3×10−2)(3×10−1) ⓑ 8×1042×10−1.

Solution

ⓐ −0.009 ⓑ 400,000

Access these online resources for additional instruction and practice with using multiplication properties of exponents.

  • Properties of Exponents
  • Negative exponents
  • Scientific Notation

Key Concepts

  • Exponential Notation
    The figure shows the letter a in a normal font with the label base and the letter m in a superscript font with the label exponent. This means we multiply the number a with itself, m times.
    This is read a to the mth power.
    In the expression am, the exponent m tells us how many times we use the base a as a factor.
  • Product Property for Exponents
    If a is a real number and m and n are integers, then
    am·an=am+n

    To multiply with like bases, add the exponents.
  • Quotient Property for Exponents
    If a is a real number, a≠0, and m and n are integers, then
    aman=am−n,m>nandaman=1an−m,n>m
  • Zero Exponent
    • If a is a non-zero number, then a0=1.
    • If a is a non-zero number, then a to the power of zero equals 1.
    • Any non-zero number raised to the zero power is 1.
  • Negative Exponent
    • If n is an integer and a≠0, then a−n=1an or 1a−n=an.
  • Quotient to a Negative Exponent Property
    If a,b are real numbers, a≠0,b≠0 and n is an integer, then
    (ab)−n=(ba)n
  • Power Property for Exponents
    If a is a real number and m,n are integers, then
    (am)n=am·n

    To raise a power to a power, multiply the exponents.
  • Product to a Power Property for Exponents
    If a and b are real numbers and m is a whole number, then
    (ab)m=ambm

    To raise a product to a power, raise each factor to that power.
  • Quotient to a Power Property for Exponents
    If a and are real numbers, b≠0, and m is an integer, then
    (ab)m=ambm

    To raise a fraction to a power, raise the numerator and denominator to that power.
  • Summary of Exponent Properties
    If a and b are real numbers, and m and n are integers, then

    Property Description
    Product Property am·an=am+n
    Power Property (am)n=am·n
    Product to a Power (ab)n=anbn
    Quotient Property aman=am−n,a≠0
    Zero Exponent Property a0=1,a≠0
    Quotient to a Power Property: (ab)m=ambm,b≠0
    Properties of Negative Exponents a−n=1an and 1a−n=an
    Quotient to a Negative Exponent (ab)−n=(ba)n
  • Scientific Notation
    A number is expressed in scientific notation when it is of the form
    a×10nwhere1≤a<10andnis an integer.
  • How to convert a decimal to scientific notation.
    1. Move the decimal point so that the first factor is greater than or equal to 1 but less than 10.
    2. Count the number of decimal places, n, that the decimal point was moved.
    3. Write the number as a product with a power of 10. If the original number is.
      • greater than 1, the power of 10 will be 10n.
      • between 0 and 1, the power of 10 will be 10−n.
    4. Check.
  • How to convert scientific notation to decimal form.
    1. Determine the exponent, n, on the factor 10.
    2. Move the decimal n places, adding zeros if needed.
      • If the exponent is positive, move the decimal point n places to the right.
      • If the exponent is negative, move the decimal point |n| places to the left.
    3. Check.

Practice Makes Perfect

Simplify Expressions Using the Properties for Exponents

In the following exercises, simplify each expression using the properties for exponents.

ⓐ d3·d6 ⓑ 45x·49x ⓒ 2y·4y3 ⓓ w·w2·w3

Solution

ⓐ d9 ⓑ 414x ⓒ 8y4 ⓓ w6

ⓐ x4·x2 ⓑ 89x·83 ⓒ 3z25·5z8 ⓓ y·y3·y5

ⓐ n19·n12 ⓑ 3x·36 ⓒ 7w5·8w ⓓ a4·a3·a9

Solution

ⓐ n31 ⓑ 3x+6 ⓒ 56w6
ⓓ a16

ⓐ q27·q15 ⓑ 5x·54x ⓒ 9u41·7u53
ⓓ c5·c11·c2

mx·m3

Solution

mx+3

ny·n2

ya·yb

Solution

ya+b

xp·xq

ⓐ x18x3 ⓑ 51253 ⓒ q18q36 ⓓ 102103

Solution

ⓐ x15 ⓑ 59 ⓒ 1q18 ⓓ 110

ⓐ y20y10 ⓑ 71672 ⓒ t10t40 ⓓ 8385

ⓐ p21p7 ⓑ 41644 ⓒ bb9 ⓓ 446

Solution

ⓐ p14 ⓑ 412 ⓒ 1b8 ⓓ 145

ⓐ u24u3 ⓑ 91595 ⓒ xx7 ⓓ 10103

ⓐ 200 ⓑ b0

Solution

ⓐ 1 ⓑ 1

ⓐ 130 ⓑ k0

ⓐ −270 ⓑ −(270)

Solution

ⓐ −1 ⓑ −1

ⓐ −150 ⓑ −(150)

Use the Definition of a Negative Exponent

In the following exercises, simplify each expression.

ⓐ a−2 ⓑ 10−3 ⓒ 1c−5 ⓓ 13−2

Solution

ⓐ 1a2 ⓑ 11000 ⓒ c5 ⓓ 9

ⓐ b−4 ⓑ 10−2 ⓒ 1b−3 ⓓ 15−2

ⓐ r−3 ⓑ 10−5 ⓒ 1q−10 ⓓ 110−3

Solution

ⓐ 1r3 ⓑ 1100,000 ⓒ q10
ⓓ 1,000

ⓐ s−8 ⓑ 10−2 ⓒ 1t−9 ⓓ 110−4

ⓐ (58)−2 ⓑ (−ba)−2

Solution

ⓐ 6425 ⓑ a2b2

ⓐ (310)−2 ⓑ (−2z)−3

ⓐ (49)−3 ⓑ (−uv)−5

Solution

ⓐ 72964 ⓑ −v5u5

ⓐ (72)−3 ⓑ (−3x)−3

ⓐ (−5)−2 ⓑ −5−2 ⓒ (−15)−2 ⓓ −(15)−2

Solution

ⓐ 125 ⓑ −125 ⓒ 25 ⓓ −25

ⓐ −5−3 ⓑ (−15)−3 ⓒ −(15)−3 ⓓ (−5)−3

ⓐ 3·5−1 ⓑ (3·5)−1

Solution

ⓐ 35 ⓑ 115

ⓐ 3·4−2 ⓑ (3·4)−2

In the following exercises, simplify each expression using the Product Property.

ⓐ b4b−8 ⓑ (w4x−5)(w−2x−4) ⓒ (−6c−3d9)(2c4d−5)

Solution

ⓐ 1b4 ⓑ w2x9 ⓒ −12cd4

ⓐ s3·s−7 ⓑ (m3n−3)(m−5n−1) ⓒ (−2j−5k8)(7j2k−3)

ⓐ a3·a−3 ⓑ (uv−2)(u−5v−3) ⓒ (−4r−2s−8)(9r4s3)

Solution

ⓐ 1 ⓑ 1u4v5 ⓒ −36r2s5

ⓐ y5·y−5 ⓑ (pq−4)(p−6q−3) ⓒ (−5m4n6)(8m−5n−3)

p5·p−2·p−4

Solution

1p

x4·x−2·x−3

In the following exercises, simplify each expression using the Power Property.

ⓐ (m4)2 ⓑ (103)6 ⓒ (x3)−4

Solution

ⓐ m8 ⓑ 1018 ⓒ 1x12

ⓐ (b2)7 ⓑ (38)2 ⓒ (k2)−5

ⓐ (y3)x ⓑ (5x)y ⓒ (q6)−8

Solution

ⓐ y3x ⓑ 5xy ⓒ 1q48

ⓐ (x2)y ⓑ (7a)b ⓒ (a9)−10

In the following exercises, simplify each expression using the Product to a Power Property.

ⓐ (−3xy)2 ⓑ (6a)0 ⓒ (5x2)−2 ⓓ (−4y−3)2

Solution

ⓐ 9x2y2 ⓑ 1 ⓒ 125x4
ⓓ 16y6

ⓐ (−4ab)2 ⓑ (5x)0 ⓒ (4y3)−3 ⓓ (−7y−3)2

ⓐ (−5ab)3 ⓑ (−4pq)0 ⓒ (−6x3)−2 ⓓ (3y−4)2

Solution

ⓐ −125a3b3 ⓑ 1 ⓒ 136x6 ⓓ 9y8

ⓐ (−3xyz)4 ⓑ (−7mn)0 ⓒ (−3x3)−2
ⓓ (2y−5)2

In the following exercises, simplify each expression using the Quotient to a Power Property.

ⓐ (p2)5 ⓑ (xy)−6 ⓒ (2xy2z)3 ⓓ (4p−3q2)2

Solution

ⓐ p532 ⓑ y6x6 ⓒ 8x3y6z3
ⓓ 16p6q4

ⓐ (x3)4 ⓑ (ab)−5 ⓒ (2x2y3z2)2 ⓓ (x3yz4)2

ⓐ (a3b)4 ⓑ (54m)−2 ⓒ (3a−2b3c2)−2 ⓓ (p−1q4r−4)2

Solution

ⓐ a481b4 ⓑ 16m225 ⓒ a4c49b6 ⓓ q8r8p2

ⓐ (x2y)3 ⓑ (103q)−4 ⓒ (2x3y43z2)5 ⓓ (5a3b−12c4)−3

In the following exercises, simplify each expression by applying several properties.

ⓐ (5t2)3(3t)2 ⓑ (t2)5(t−4)2(t3)7 ⓒ (2xy2x3y−2)2(12xy3x3y−1)−1

Solution

ⓐ 1125t8 ⓑ 1t19 ⓒ y43x2

ⓐ (10k4)3(5k6)2 ⓑ (q3)6(q−2)3(q4)8

ⓐ (m2n)2(2mn5)4 ⓑ (−2p−2)4(3p4)2(−6p3)2

Solution

ⓐ 16m8n22 ⓑ 4p6

ⓐ (3pq4)2(6p6q)2 ⓑ (−2k−3)2(6k2)4(9k4)2

Mixed Practice

In the following exercises, simplify each expression.

ⓐ 7n−1 ⓑ (7n)−1 ⓒ (−7n)−1

Solution

ⓐ 7n ⓑ 17n ⓒ −17n

ⓐ 6r−1 ⓑ (6r)−1 ⓒ (−6r)−1

ⓐ (3p)−2 ⓑ 3p−2 ⓒ −3p−2

Solution

ⓐ 19p2 ⓑ 3p2 ⓒ −3p2

ⓐ (2q)−4 ⓑ 2q−4 ⓒ −2q−4

(x2)4·(x3)2

Solution

x14

(y4)3·(y5)2

(a2)6·(a3)8

Solution

a36

(b7)5·(b2)6

(2m6)3

Solution

8m18

(3y2)4

(10x2y)3

Solution

1,000x6y3

(2mn4)5

(−2a3b2)4

Solution

16a12b8

(−10u2v4)3

(23x2y)3

Solution

827x6y3

(79pq4)2

(8a3)2(2a)4

Solution

1,024a10

(5r2)3(3r)2

(10p4)3(5p6)2

Solution

25,000p24

(4x3)3(2x5)4

(12x2y3)4(4x5y3)2

Solution

x18y18

(13m3n2)4(9m8n3)2

(3m2n)2(2mn5)4

Solution

144m8n22

(2pq4)3(5p6q)2

ⓐ (3x)2(5x) ⓑ (2y)3(6y)

Solution

ⓐ 45x3 ⓑ 48y4

ⓐ (12y2)3(23y)2 ⓑ (12j2)5(25j3)2

ⓐ (2r−2)3(4−1r)2 ⓑ (3x−3)3(3−1x5)4

Solution

ⓐ 12r4 ⓑ 13x11

(k−2k8k3)2

(j−2j5j4)3

Solution

1j3

(−4m−3)2(5m4)3(−10m6)3

(−10n−2)3(4n5)2(2n8)2

Solution

−4000n12

Use Scientific Notation

In the following exercises, write each number in scientific notation.

ⓐ 57,000 ⓑ 0.026

ⓐ 340,000 ⓑ 0.041

Solution

ⓐ 3.4×105 ⓑ 4.1×10−2

ⓐ 8,750,000 ⓑ 0.00000871

ⓐ 1,290,000 ⓑ 0.00000103

Solution

ⓐ 1.29×106
ⓑ 1.03×10−6

In the following exercises, convert each number to decimal form.

ⓐ 5.2×102 ⓑ 2.5×10−2

ⓐ −8.3×102 ⓑ 3.8×10−2

Solution

ⓐ −830 ⓑ 0.038

ⓐ 7.5×106 ⓑ −4.13×10−5

ⓐ 1.6×1010 ⓑ 8.43×10−6

Solution

ⓐ 16,000,000,000
ⓑ 0.00000843

In the following exercises, multiply or divide as indicated. Write your answer in decimal form.

ⓐ (3×10−5)(3×109) ⓑ 7×10−31×10−7

ⓐ (2×102)(1×10−4) ⓑ 5×10−21×10−10

Solution

ⓐ 0.02 ⓑ 500,000,000

ⓐ (7.1×10−2)(2.4×10−4) ⓑ 6×1043×10−2

ⓐ (3.5×10−4)(1.6×10−2) ⓑ 8×1064×10−1

Solution

ⓐ 0.0000056 ⓑ 20,000,000

Writing Exercises

Use the Product Property for Exponents to explain why x·x=x2.

Jennifer thinks the quotient a24a6 simplifies to a4. What is wrong with her reasoning?

Solution

Answers will vary.

Explain why −53=(−5)3 but −54≠(−5)4.

When you convert a number from decimal notation to scientific notation, how do you know if the exponent will be positive or negative?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “simplify expressions using the properties for exponents.”, “use the definition of a negative exponent”, and “use scientific notation”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ After reviewing this checklist, what will you do to become confident for all goals?

Product Property
According to the Product Property, a to the m times a to the n equals a to the m plus n.
Power Property
According to the Power Property, a to the m to the n equals a to the m times n.
Product to a Power
According to the Product to a Power Property, a times b in parentheses to the m equals a to the m times b to the m.
Quotient Property
According to the Quotient Property, a to the m divided by a to the n equals a to the m minus n as long as a is not zero.
Zero Exponent Property
According to the Zero Exponent Property, a to the zero is 1 as long as a is not zero.
Quotient to a Power Property
According to the Quotient to a Power Property, a divided by b in parentheses to the power of m is equal to a to the m divided by b to the m as long as b is not zero.
Properties of Negative Exponents
According to the Properties of Negative Exponents, a to the negative n equals 1 divided by a to the n and 1 divided by a to the negative n equals a to the n.
Quotient to a Negative Exponent
Raising a quotient to a negative exponent occurs when a divided by b in parentheses to the power of negative n equals b divided by a in parentheses to the power of n.

Multiply Polynomials

Learning Objectives

By the end of this section, you will be able to:

  • Multiply monomials
  • Multiply a polynomial by a monomial
  • Multiply a binomial by a binomial
  • Multiply a polynomial by a polynomial
  • Multiply special products
  • Multiply polynomial functions

Before you get started, take this readiness quiz.

Distribute: 2(x+3).
If you missed this problem, review Example 6 in Properties of Real Numbers.

Solution

2x+6

Simplify: ⓐ 92 ⓑ (−9)2 ⓒ −92.
If you missed this problem, review Example 8 in Integers.

Solution

ⓐ 81; ⓑ 81; ⓒ −81

Evaluate: 2x2−5x+3 for x=−2.
If you missed this problem, review Example 10 in Integers.

Solution

21

Multiply Monomials

We are ready to perform operations on polynomials. Since monomials are algebraic expressions, we can use the properties of exponents to multiply monomials.

Multiply: ⓐ (3x2)(−4x3) ⓑ (56x3y)(12xy2).

Solution
ⓐ
Steps demonstrating the multiplication of two monomials, illustrating the use of the commutative property to simplify the expression.
(3x2)(−4x3)
Use the Commutative Property to rearrange the terms. 3·(−4)·x2·x3
Multiply. −12x5
ⓑ
This table demonstrates the step-by-step multiplication of two monomial expressions, showing intermediate rearrangement and the final product.
(56x3y)(12xy2)
Use the Commutative Property to rearrange the terms. 56·12·x3·x·y·y2
Multiply. 10x4y3

Multiply: ⓐ (5y7)(−7y4) ⓑ (25a4b3)(15ab3).

Solution

ⓐ −35y11 ⓑ 6a5b6

Multiply: ⓐ (−6b4)(−9b5) ⓑ (23r5s)(12r6s7).

Solution

ⓐ 54b9 ⓑ 8r11s8

Multiply a Polynomial by a Monomial

Multiplying a polynomial by a monomial is really just applying the Distributive Property.

Multiply: ⓐ −2y(4y2+3y−5) ⓑ 3x3y(x2−8xy+y2).

Solution
ⓐ
The distributive property is illustrated by the expression -2y(4y^2 + 3y - 5), with red arrows showing -2y multiplying each term inside the parentheses.
Distribute. Expression demonstrating the distributive property: negative two times y, multiplied by four y squared, plus negative two times y, multiplied by three y, minus negative two times y, multiplied by five.
Multiply. -8y^3 - 6y^2 + 10y
ⓑ
This table illustrates the step-by-step process of simplifying a polynomial expression using distribution and multiplication.
3x3y(x2−8xy+y2)
Distribute. 3x3y·x2+(3x3y)·(−8xy)+(3x3y)·y2
Multiply. 3x5y−24x4y2+3x3y3

Multiply: ⓐ −3y(5y2+8y−7) ⓑ 4x2y2(3x2−5xy+3y2).

Solution

ⓐ −15y3−24y2+21y
ⓑ 12x4y2−20x3y3+12x2y4

Multiply: ⓐ 4x2(2x2−3x+5) ⓑ −6a3b(3a2−2ab+6b2).

Solution

ⓐ 8x4−12x3+20x2
ⓑ −18a5b+12a4b2−36a3b3

Multiply a Binomial by a Binomial

Just like there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a binomial times a binomial. We will start by using the Distributive Property.

Multiply: ⓐ (y+5)(y+8) ⓑ (4y+3)(2y−5).

Solution
ⓐ
An image illustrating the multiplication of two binomials, (y+5)(y+8), using the FOIL method, with red arrows showing the distribution of terms from the first binomial to the second.
Distribute (y+8). The mathematical expression shown is y(y + 8) + 5(y + 8), which can be factored as (y + 5)(y + 8).
Distribute again. The algebraic expression y^2 + 8y + 5y + 40.
Combine like terms. A quadratic equation is displayed, which reads y squared plus 13y plus 40.
ⓑ
An algebraic expression showing the product of two binomials: (4y + 3) multiplied by (2y - 5).
Distribute. A mathematical expression featuring two terms: the first is 4y multiplied by the quantity (2y - 5), and the second is 3 multiplied by the quantity (2y - 5), joined by an addition sign.
Distribute again. A mathematical expression featuring variables and constants: 8y^2 - 20y + 6y - 15.
Combine like terms. A mathematical expression displays a quadratic trinomial: 8y² - 14y - 15.

Multiply: ⓐ (x+8)(x+9) ⓑ (3c+4)(5c−2).

Solution

ⓐ x2+17x+72
ⓑ 15c2+14c−8

Multiply: ⓐ (5x+9)(4x+3) ⓑ (5y+2)(6y−3).

Solution

ⓐ 20x2+51x+27
ⓑ 30y2−3y−6

If you multiply binomials often enough you may notice a pattern. Notice that the first term in the result is the product of the first terms in each binomial. The second and third terms are the product of multiplying the two outer terms and then the two inner terms. And the last term results from multiplying the two last terms,

We abbreviate “First, Outer, Inner, Last” as FOIL. The letters stand for ‘First, Outer, Inner, Last’. We use this as another method of multiplying binomials. The word FOIL is easy to remember and ensures we find all four products.

Let’s multiply (x+3)(x+7) using both methods.

The figure shows how four terms in the product of two binomials can be remembered according to the mnemonic acronym FOIL. The example is the quantity x plus 3 in parentheses times the quantity x plus 7 in parentheses. The expression is expanded as in the previous examples by using the distributive property twice. After distributing the quantity x plus 7 in parentheses the result is x times the quantity x plus 7 in parentheses plus 3 times the quantity x plus 7 in parentheses. Then the x is distributed the x plus 7 and the 3 is distributed to the x plus 7 to get x squared plus 7 x plus 3 x plus 21. The letter F is written under the term x squared since it was the product of the first terms in the binomials. The letter O is written under the 7 x term sine it was the product of the outer terms in the binomials. The letter I is written under the 3 x term since it was the product of the inner terms in the binomials. The letter L is written under the 21 since it was the product of the last terms in the binomial. The original expression is shown again with four arrows connecting the first, outer, inner, and last terms in the binomials showing how the four terms can be determined directly from the factored form.

We summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!

Use the FOIL method to multiply two binomials.

The figure shows how to use the FOIL method to multiply two binomials. The example is the quantity a plus b in parentheses times the quantity c plus d in parentheses. The numbers a and c are labeled first and the numbers b and d are labeled last. The numbers b and c are labeled inner and the numbers a and d are labeled outer. A note on the side of the expression tells you to Say it as you multiply! FOIL First Outer Inner Last. The directions are then given in numbered steps. Step 1. Multiply the First terms. Step 2. Multiply the Outer terms. Step 3. Multiply the Inner terms. Step 4. Multiply the Last Terms. Step 5. Combine like terms when possible.

When you multiply by the FOIL method, drawing the lines will help your brain focus on the pattern and make it easier to apply.

Now we will do an example where we use the FOIL pattern to multiply two binomials.

Multiply: ⓐ (y−7)(y+4) ⓑ (4x+3)(2x−5).

Solution
  1. ⓐ
    The figure shows how to use the FOIL method to multiply two binomials. The example is the quantity y minus 7 in parentheses times the quantity y plus 4 in parentheses. Step 1. Multiply the First terms. The terms y and y are colored red with an arrow connecting them. The result is y squared and is shown above the letter F in the word FOIL. Step 2. Multiply the Outer terms. The terms y and 4 are colored red with an arrow connecting them. The result is 4 y and is shown above the letter O in the word FOIL. Step 3. Multiply the Inner terms. The terms negative 7 and y are colored red with an arrow connecting them. The result is negative 7 y squared and is shown above the letter I in the word FOIL. Step 4. Multiply the Last terms. The terms negative 7 and 4 are colored red with an arrow connecting them. The result is negative 28 and is shown above the letter L in the word FOIL. Step 5. Combine like terms. The simplified result is y squared minus 3 y minus 28.

  2. ⓑ
    The figure shows how to use the FOIL method to multiply two binomials. The example is the quantity 4 x plus 3 in parentheses times the quantity 2 x minus 5 in parentheses. The expression is show with four red arrows connecting the First. Outer, Inner, and Last terms. Step 1. Multiply the First terms 4 x and 2 x. The product of the first terms is 8 x squared and is shown above the letter F in the word FOIL. Step 2. Multiply the Outer terms 4 x and negative 5. The result is negative 20 x and is shown above the letter O in the word FOIL. Step 3. Multiply the Inner terms 3 and 2 x. The result is 6 x and is shown above the letter I in the word FOIL. Step 4. Multiply the Last terms 3 and negative 5. The result is negative 15 and is shown above the letter L in the word FOIL. Step 5. Combine like terms. The simplified result is 8 y squared minus 14 x minus 15.

Multiply: ⓐ (x−7)(x+5) ⓑ (3x+7)(5x−2).

Solution

ⓐ x2−2x−35
ⓑ 15x2+29x−14

Multiply: ⓐ (b−3)(b+6) ⓑ (4y+5)(4y−10).

Solution

ⓐ b2+3b−18
ⓑ 16y2−20y−50

The final products in the last example were trinomials because we could combine the two middle terms. This is not always the case.

Multiply: ⓐ (n2+4)(n−1) ⓑ (3pq+5)(6pq−11).

Solution
ⓐ
(n^2 + 4)(n - 1)
The algebraic expression (n^2 + 4)(n - 1) demonstrating the FOIL method for multiplying binomials, with red arrows indicating term distribution.
Step 1. Multiply the First terms. An algebraic expression featuring 'n^3' in red, followed by three blank terms, all connected by plus signs. The letters 'F', 'Q', 'I', 'L' are placed underneath each corresponding term.
Step 2. Multiply the Outer terms. A mathematical expression showing n cubed minus n squared (with n squared in red), followed by two blank terms, each preceded by a plus sign. The terms are underlined, with the letters F, Q, I, and L placed below them.
Step 3. Multiply the Inner terms. An amusing misuse of the FOIL method, applying its F, O, I, L labels to the terms of the polynomial n^3 - n^2 + 4n + __, typically used for binomial multiplication.
Step 4. Multiply the Last terms. A mathematical expression n^3 - n^2 + 4n - 4 is displayed, with the last term, -4, highlighted in red. Below it, the letters F O I L are arranged, corresponding to the terms of the polynomial.
Step 5. Combine like terms—there are none. The algebraic expression n^3 - n^2 + 4n - 4 is displayed.
ⓑ
A mathematical expression showing the product of two binomials: (3pq + 5)(6pq - 11).
The FOIL method demonstrated for (3pq + 5)(6pq - 11), with red arrows indicating the distribution of First, Outer, Inner, and Last terms for binomial multiplication.
Step 1. Multiply the First terms.      A mathematical expression illustrating the FOIL method, with '18p^2q^2' labeled as 'F' (First) and subsequent terms for 'O', 'I', and 'L' left as blanks for completion.
Step 2. Multiply the Outer terms. A math problem demonstrating the FOIL method with the first two terms given (18p^2q^2 for F, -33pq for O) and blanks for I and L, prompting completion of the binomial multiplication.
Step 3. Multiply the Inner terms. A mathematical expression 18p^2q^2 - 33pq + 30pq + ___ is shown, with labels F, O, I, L below the terms, illustrating the First, Outer, Inner, and a blank for the Last term of a FOIL expansion.
Step 4. Multiply the Last terms. A mathematical expression reads 18p^2q^2 - 33pq + 30pq - 55. The number 55 is highlighted in red. Below the terms are the letters F, Q, I, L, suggesting a mnemonic or a problem structure.
Step 5. Combine like terms. The image displays the algebraic expression 18p^2q^2 - 3pq - 55, featuring terms with variables p and q raised to powers, and a constant.

Multiply: ⓐ (x2+6)(x−8) ⓑ (2ab+5)(4ab−4).

Solution

ⓐ x3−8x2+6x−48
ⓑ 8a2b2+12ab−20

Multiply: ⓐ (y2+7)(y−9) ⓑ (2xy+3)(4xy−5).

Solution

ⓐ y3−9y2+7y−63
ⓑ 8x2y2+2xy−15

The FOIL method is usually the quickest method for multiplying two binomials, but it only works for binomials. You can use the Distributive Property to find the product of any two polynomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.

This figure shows the vertical multiplication of 23 and 46. The number 23 is above the number 46. Below this, there is the partial product 138 over the partial product 92. The final product is at the bottom and is 1058. Text on the right side of the image says “You start by multiplying 23 by 6 to get 138. Then you multiply 23 by 4, lining up the partial product in the correct columns. Last, you add the partial products.”

Now we’ll apply this same method to multiply two binomials.

Multiply using the Vertical Method: (3y−1)(2y−6).

Solution

It does not matter which binomial goes on the top.

This table illustrates the step-by-step vertical multiplication of two binomials, (3y-1) and (2y-6), showing partial products and the final algebraic expression.

Multiply 3y−1by−6.
Multiply 3y−1by2y.
Add like terms.
3y−1 ×2y−6___________ −18y+6 6y2−2y___________ 6y2−20y+6
partial product
partial product
product

Notice the partial products are the same as the terms in the FOIL method.

This figure has two columns. In the left column is the product of two binomials, 3y minus 1 and 2y minus 6. Below this is 6y squared minus 2y minus 18y plus 6. Below this is 6y squared minus 20y plus 6. In the right column is the vertical multiplication of 3y minus 1 and 2y minus 6. Below this is the partial product negative 18y plus 6. Below this is the partial product 6y squared minus 2y. Below this is 6y squared minus 20y plus 6.

Multiply using the Vertical Method: (5m−7)(3m−6).

Solution

15m2−51m+42

Multiply using the Vertical Method: (6b−5)(7b−3).

Solution

42b2−53b+15

We have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The methods are listed here all together, to help you remember them.

Multiplying Two Binomials

To multiply binomials, use the:

  • Distributive Property
  • FOIL Method
  • Vertical Method

Multiply a Polynomial by a Polynomial

We have multiplied monomials by monomials, monomials by polynomials, and binomials by binomials. Now we’re ready to multiply a polynomial by a polynomial. Remember, FOIL will not work in this case, but we can use either the Distributive Property or the Vertical Method.

Multiply (b+3)(2b2−5b+8) using ⓐ the Distributive Property and ⓑ the Vertical Method.

Solution
ⓐ
Multiplication of binomial by trinomial, showing the expression open parenthesis b plus 3 close parenthesis open parenthesis 2 b squared minus 5 b plus 8 close parenthesis, with 2 b squared minus 5 b plus 8 highlighted in red
Distribute. An algebraic expression showing the sum of two terms, b(2b^2 - 5b + 8) + 3(2b^2 - 5b + 8), which can be factored as (b+3)(2b^2 - 5b + 8).
Multiply. A mathematical expression 2b^3 - 5b^2 + 8b + 6b^2 - 15b + 24, displayed in a horizontal line, consisting of terms with variable 'b' raised to powers and a constant term.
Combine like terms.       A mathematical expression is displayed, reading '2b^3 + b^2 - 7b + 24' on a white background.
ⓑ It is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products this way.
Multiply (2b2−5b+8) by 3.
Multiply (2b2−5b+8) by b.
A vertical polynomial multiplication problem: (2b^2 - 5b + 8) times (b + 3). The first partial product, multiplying the top polynomial by 3, yields 6b^2 - 15b + 24, shown below the line.
Add like terms.    A mathematical expression reads '2b^3 - 5b^2 + 8b'.
   The image shows the algebraic expression 2b^3 + b^2 - 7b + 24, a polynomial in b, displayed in black text against a white background.

Multiply(y−3)(y2−5y+2) using ⓐ the Distributive Property and ⓑ the Vertical Method.

Solution

ⓐ y3−8y2+17y−6
ⓑ y3−8y2+17y−6

Multiply (x+4)(2x2−3x+5) using ⓐ the Distributive Property and ⓑ The Vertical Method.

Solution

ⓐ 2x3+5x2−7x+20
ⓑ y3−8y2+17y−6

We have now seen two methods you can use to multiply a polynomial by a polynomial. After you practice each method, you’ll probably find you prefer one way over the other. We list both methods are listed here, for easy reference.

Multiplying a Polynomial by a Polynomial

To multiply a trinomial by a binomial, use the:

  • Distributive Property
  • Vertical Method

Multiply Special Products

Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and multiplying them, there is less work to do if you learn to use a pattern. Let’s start by looking at three examples and look for a pattern.

Look at these results. Do you see any patterns?

The figure shows three examples of squaring a binomial. In the first example x plus 9 is squared to get x plus 9 times x plus 9 which is x squared plus 9 x plus 9 x plus 81 which simplifies to x squared plus 18 x plus 81. Colors show that x squared comes from the square of the x in the original binomial and 81 comes from the square of the 9 in the original binomial. In the second example y minus 7 is squared to get y minus y times y minus 7 which is y squared minus 7 y minus 7 y plus 49 which simplifies to y squared minus 14 y plus 49. Colors show that y squared comes from the square of the y in the original binomial and 49 comes from the square of the negative 7 in the original binomial. In the third example 2 x plus 3 is squared to get 2 x plus 3 times 2 x plus 3 which is 4 x squared plus 6 x plus 6 x plus 9 which simplifies to 4 x squared plus 12 x plus 9. Colors show that 4 x squared comes from the square of the 2 x in the original binomial and 9 comes from the square of the 3 in the original binomial.

What about the number of terms? In each example we squared a binomial and the result was a trinomial.

(a+b)2=___+___+___

Now look at the first term in each result. Where did it come from?

The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!

(a+b)2=a2+___+___

  To get the first term of the product, square the first term.

Where did the last term come from? Look at the examples and find the pattern.

The last term is the product of the last terms, which is the square of the last term.

(a+b)2=___+___+b2

  To get the last term of the product, square the last term.

Finally, look at the middle term. Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial.

(a+b)2=___+2ab+___ (a−b)2=___−2ab+___

  To get the middle term of the product, multiply the terms and double their product.

Putting it all together:

Binomial Squares Pattern

If a and b are real numbers,

The figure shows the result of squaring two binomials. The first example is a plus b squared equals a squared plus 2 a b plus b squared. The equation is written out again with each part labeled. The quantity a plus b squared is labeled binomial squared. The terms a squared is labeled first term squared. The term 2 a b is labeled 2 times product of terms. The term b squared is labeled last term squared. The second example is a minus b squared equals a squared minus 2 a b plus b squared. The equation is written out again with each part labeled. The quantity a minus b squared is labeled binomial squared. The terms a squared is labeled first term squared. The term negative 2 a b is labeled 2 times product of terms. The term b squared is labeled last term squared.

To square a binomial, square the first term, square the last term , double their product.

Multiply: ⓐ (x+5)2 ⓑ (2x−3y)2.

Solution
ⓐ
A mathematical expression featuring large parentheses enclosing two stacked lines: 'a + b' (with 'a' and 'b' in red) on the top and 'x + 5' on the bottom, with an exponent '2' outside.
Square the first term. The image shows two algebraic expressions: 'a² + 2ab + b²' in red, and 'x² + ___ + ___' in black with blank spaces, indicating a problem related to perfect square trinomials.
Square the last term. An image showing the algebraic identity a² + 2ab + b² and a fill-in-the-blank problem x² + ___ + 5² for completing the square, where the missing term corresponds to 2ab.
Double their product.       Algebraic expressions demonstrating the pattern for a perfect square trinomial: a^2 + 2ab + b^2 (top, general form) and x^2 + 2x5 + 5^2 (bottom, specific example).
Simplify. The image shows the mathematical expression x^2 + 10x + 25, which is a quadratic trinomial. This expression is a perfect square trinomial, as it can be factored into (x + 5)^2.
ⓑ
A mathematical expression featuring a column vector, represented by (a - b) over (2x - 3y) enclosed in large parentheses, all raised to the power of 2. The variables 'a' and 'b' are highlighted in red.
Use the pattern.         Algebraic identity for a perfect square trinomial: a^2 - 2ab + b^2, with (2x)^2 - 2(2x)(3y) + (3y)^2 as an application.
Simplify. The mathematical expression 4x^2 - 12xy + 9y^2 is shown.

Multiply: ⓐ (x+9)2 ⓑ (2c−d)2.

Solution

ⓐ x2+18x+81
ⓑ 4c2−4cd+d2

Multiply: ⓐ (y+11)2 ⓑ (4x−5y)2.

Solution

ⓐ y2+22y+121
ⓑ 16x2−40xy+25y2

We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference is called a conjugate pair and is of the form (a−b),(a+b).

Conjugate Pair

A conjugate pair is two binomials of the form

(a−b),(a+b).

The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.

There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier. Let’s look for the pattern by using FOIL to multiply some conjugate pairs.

The figure shows three examples of multiplying a binomial with its conjugate. In the first example x plus 9 is multiplied with x minus 9 to get x squared minus 9 x plus 9 x minus 81 which simplifies to x squared minus 81. Colors show that x squared comes from the square of the x in the original binomial and 81 comes from the square of the 9 in the original binomial. In the second example y minus 8 is multiplied with y plus 8 to get y squared plus 8 y minus 8 y minus 64 which simplifies to y squared minus 64. Colors show that y squared comes from the square of the y in the original binomial and 64 comes from the square of the 8 in the original binomial. In the third example 2 x minus 5 is multiplied with 2 x plus 5 to get 4 x squared plus 10 x minus 10 x minus 25 which simplifies to 4 x squared minus 25. Colors show that 4 x squared comes from the square of the 2 x in the original binomial and 25 comes from the square of the 5 in the original binomial.

What do you observe about the products?

The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.

Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term.

(a+b)(a−b)=a2−___

   To get the first term, square the first term.

The last term came from multiplying the last terms, the square of the last term.

(a+b)(a−b)=a2−b2

   To get the last term, square the last term.

Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.

The product of conjugates is always of the form a2−b2. This is called a difference of squares.

This leads to the pattern:

Product of Conjugates Pattern

If a and b are real numbers,

The figure shows the result of multiplying a binomial with its conjugate. The formula is a plus b times a minus b equals a squared minus b squared. The equation is written out again with labels. The product a plus b times a minus b is labeled conjugates. The result a squared minus b squared is labeled difference of squares.

The product is called a difference of squares.

To multiply conjugates, square the first term, square the last term, write it as a difference of squares.

Multiply using the product of conjugates pattern: ⓐ (2x+5)(2x−5) ⓑ (5m−9n)(5m+9n).

Solution
ⓐ
Are the binomials conjugates? A mathematical expression showing the product of two binomials, (2x + 5)(2x - 5), which is a classic example of the difference of squares factorization.
It is the product of conjugates. The mathematical expression (a+b)(a-b) paired with (2x+5)(2x-5), illustrating the difference of squares algebraic identity.
Square the first term, 2x. A mathematical image displaying the difference of squares formula, a² - b² in red, positioned above an incomplete expression (2x)² - _, indicating a problem for completion.
Square the last term, 5. Mathematical expression showing the difference of squares formula, a^2 - b^2, above its application: (2x)^2 - 5^2.
Simplify. The product is a difference of squares.   Two mathematical expressions are displayed against a white background. The top expression, in red, is 'a^2 - b^2'. Below it, in black, is '4x^2 - 25'. Both are examples of the difference of squares.
ⓑ
The mathematical expression (5m - 9n)(5m + 9n), representing the difference of squares, is displayed in bold black text on a white background.
This fits the pattern.               The algebraic identity (a-b)(a+b) and its application with (5m-9n)(5m+9n), showing the difference of squares.
Use the pattern. The image displays the algebraic identity a×2 - b×2 in red, followed by the specific example (5m)×2 - (9n)×2, demonstrating the difference of squares formula.
Simplify. The image displays the mathematical expression 25m^2 - 81n^2.

Multiply: ⓐ (6x+5)(6x−5) ⓑ (4p−7q)(4p+7q).

Solution

ⓐ 36x2−25
ⓑ 16p2−49q2

Multiply: ⓐ (2x+7)(2x−7) ⓑ (3x−y)(3x+y).

Solution

ⓐ 4x2−49 ⓑ 9x2−y2

We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.

Comparing the Special Product Patterns

Binomial Squares Product of Conjugates
(a+b)2=a2+2ab+b2 (a−b)(a+b)=a2−b2
(a−b)2=a2−2ab+b2
•  Squaring a binomial •  Multiplying conjugates
•  Product is a trinomial •  Product is a binomial.
•  Inner and outer terms with FOIL are the same. •  Inner and outer terms with FOIL are opposites.
•  Middle term is double the product of the terms •  There is no middle term.

Choose the appropriate pattern and use it to find the product:

ⓐ (2x−3)(2x+3) ⓑ (8x−5)2 ⓒ (6m+7)2 ⓓ (5x−6)(6x+5).

Solution

ⓐ (2x−3)(2x+3)

These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.

A mathematical expression displaying the difference of squares formula, (a - b)(a + b) and (2x - 3)(2x + 3), demonstrating a key algebraic identity.
Use the pattern.     The image displays the difference of squares formula, a² - b², followed by a specific example, (2x)² - 3².
Simplify. The image displays the mathematical expression 4x^2 - 9, presented in a clear, digital font on a white background.

ⓑ (8x−5)2

We are asked to square a binomial. It fits the binomial squares pattern.

A mathematical expression featuring two binomials within large parentheses, raised to the power of 2. The top line shows (a - b) with 'a' and 'b' in red, while the bottom line shows (8x - 5).
Use the pattern.     An algebraic expression demonstrating the perfect square trinomial identity: a² - 2ab + b² is equivalent to (8x)² - 2 * 8x * 5 + 5².
Simplify. A mathematical expression showing the quadratic trinomial 64x^2 - 80x + 25, which is a perfect square trinomial (8x - 5)^2.

ⓒ (6m+7)2

Again, we will square a binomial so we use the binomial squares pattern.

A mathematical expression displaying the fraction (a+b)/(6m+7) raised to the power of 2, with 'a' and 'b' highlighted in red.
Use the pattern.     An algebraic expression showing the perfect square trinomial formula, a^2 + 2ab + b^2, applied to specific values: (6m)^2 + 2 * 6m * 7 + 7^2, where 'a' is 6m and 'b' is 7.
Simplify. The mathematical expression 36m^2 + 84m + 49, which is a perfect square trinomial, is displayed on a white background.

ⓓ (5x−6)(6x+5)

This product does not fit the patterns, so we will use FOIL.

(5x−6)(6x+5) Use FOIL.30x2+25x−36x−30 Simplify.30x2−11x−30

Choose the appropriate pattern and use it to find the product:

ⓐ (9b−2)(2b+9) ⓑ (9p−4)2 ⓒ (7y+1)2 ⓓ (4r−3)(4r+3).

Solution

ⓐ FOIL; 18b2+77b−18
ⓑ Binomial Squares; 81p2−72p+16
ⓒ Binomial Squares; 49y2+14y+1
ⓓ Product of Conjugates; 16r2−9

Choose the appropriate pattern and use it to find the product:

ⓐ (6x+7)2 ⓑ (3x−4)(3x+4) ⓒ (2x−5)(5x−2) ⓓ (6n−1)2.

Solution

ⓐ Binomial Squares; 36x2+84x+49 ⓑ Product of Conjugates; 9x2−16 ⓒ FOIL; 10x2−29x+10 ⓓ Binomial Squares; 36n2−12n+1

Multiply Polynomial Functions

Just as polynomials can be multiplied, polynomial functions can also be multiplied.

Multiplication of Polynomial Functions

For functions f(x) and g(x),

(f·g)(x)=f(x)·g(x)

For functions f(x)=x+2 and g(x)=x2−3x−4, find: ⓐ (f·g)(x) ⓑ (f·g)(2).

Solution
ⓐ
This table illustrates the step-by-step process of multiplying two given polynomial functions, f(x) and g(x), to determine the combined function (f·g)(x).
(f·g)(x)=f(x)·g(x)
Substitute for f(x)andg(x). (f·g)(x)=(x+2)(x2−3x−4)
Multiply the polynomials. (f·g)(x)=x(x2−3x−4)+2(x2−3x−4)
Distribute. (f·g)(x)=x3−3x2−4x+2x2−6x−8
Combine like terms. (f·g)(x)=x3−x2−10x−8

ⓑ In part ⓐ we found (f·g)(x) and now are asked to find (f·g)(2).

Step-by-step evaluation of the function (f·g)(x) when x=2.
(f·g)(x)=x3−x2−10x−8
To find (f·g)(2), substitute x=2. (f·g)(2)=23−22−10·2−8
(f·g)(2)=8−4−20−8
(f·g)(2)=−24

For functions f(x)=x−5 and g(x)=x2−2x+3, find ⓐ (f·g)(x) ⓑ (f·g)(2).

Solution

ⓐ (f·g)(x)=x3−7x2+13x−15
ⓑ (f·g)(2)=−9

For functions f(x)=x−7 and g(x)=x2+8x+4, find ⓐ (f·g)(x) ⓑ (f·g)(2).

Solution

ⓐ (f·g)(x)=x3+x2−52x−28
ⓑ (f·g)(2)=−120

Access this online resource for additional instruction and practice with multiplying polynomials.

  • Introduction to special products of binomials

Key Concepts

  • How to use the FOIL method to multiply two binomials.
    The figure shows how to use the FOIL method to multiply two binomials. The example is the quantity a plus b in parentheses times the quantity c plus d in parentheses. The numbers a and c are labeled first and the numbers b and d are labeled last. The numbers b and c are labeled inner and the numbers a and d are labeled outer. A note on the side of the expression tells you to Say it as you multiply! FOIL First Outer Inner Last. The directions are then given in numbered steps. Step 1. Multiply the First terms. Step 2. Multiply the Outer terms. Step 3. Multiply the Inner terms. Step 4. Multiply the Last Terms. Step 5. Combine like terms when possible.
  • Multiplying Two Binomials: To multiply binomials, use the:
    • Distributive Property
    • FOIL Method
  • Multiplying a Polynomial by a Polynomial: To multiply a trinomial by a binomial, use the:
    • Distributive Property
    • Vertical Method
  • Binomial Squares Pattern
    If a and b are real numbers, The figure shows the result of squaring two binomials. The first example is a plus b squared equals a squared plus 2 a b plus b squared. The equation is written out again with each part labeled. The quantity a plus b squared is labeled binomial squared. The terms a squared is labeled first term squared. The term 2 a b is labeled 2 times product of terms. The term b squared is labeled last term squared. The second example is a minus b squared equals a squared minus 2 a b plus b squared. The equation is written out again with each part labeled. The quantity a minus b squared is labeled binomial squared. The terms a squared is labeled first term squared. The term negative 2 a b is labeled 2 times product of terms. The term b squared is labeled last term squared.
  • Product of Conjugates Pattern
    If a,b are real numbers
    The figure shows the result of multiplying a binomial with its conjugate. The formula is a plus b times a minus b equals a squared minus b squared. The equation is written out again with labels. The product a plus b times a minus b is labeled conjugates. The result a squared minus b squared is labeled difference of squares.
    The product is called a difference of squares.
    To multiply conjugates, square the first term, square the last term, write it as a difference of squares.
  • Comparing the Special Product Patterns
    Binomial Squares Product of Conjugates
    (a+b)2=a2+2ab+b2 (a−b)(a+b)=a2−b2
    (a−b)2=a2−2ab+b2
    •  Squaring a binomial •  Multiplying conjugates
    •  Product is a trinomial •  Product is a binomial.
    •  Inner and outer terms with FOIL are the same. •  Inner and outer terms with FOIL are opposites.
    •  Middle term is double the product of the terms •  There is no middle term.
  • Multiplication of Polynomial Functions:
    • For functions f(x) and g(x),
      (f·g)(x)=f(x)·g(x)

Practice Makes Perfect

Multiply Monomials

In the following exercises, multiply the monomials.


ⓐ (6y7)(−3y4)
ⓑ (47rs2)(14rs3)


ⓐ (−10x5)(−3x3)
ⓑ (58x3y)(24x5y)

Solution

ⓐ 30x8 ⓑ 15x8y2


ⓐ (−8u6)(−9u)
ⓑ (23x2y)(34xy2)


ⓐ (−6c4)(−12c)
ⓑ (35m3n2)(59m2n3)

Solution

ⓐ 72c5 ⓑ 13m5n5

Multiply a Polynomial by a Monomial

In the following exercises, multiply.


ⓐ −8x(x2+2x−15)
ⓑ 5pq3(p2−2pq+6q2)


ⓐ −5t(t2+3t−18);
ⓑ 9r3s(r2−3rs+5s2)

Solution

ⓐ −5t3−15t2+90t
ⓑ 9sr5−27s2r4+45s3r3


ⓐ −8y(y2+2y−15)
ⓑ −4y2z2(3y2+12yz−z2)


ⓐ −5m(m2+3m−18)
ⓑ −3x2y2(7x2+10xy−y2)

Solution

ⓐ −5m3−15m2+90m
ⓑ −21x4y2−30x3y3+3x2y4

Multiply a Binomial by a Binomial

In the following exercises, multiply the binomials using ⓐ the Distributive Property; ⓑ the FOIL method; ⓒ the Vertical Method.

(w+5)(w+7)

(y+9)(y+3)

Solution

y2+12y+27

(4p+11)(5p−4)

(7q+4)(3q−8)

Solution

21q2−44q−32

In the following exercises, multiply the binomials. Use any method.

(x+8)(x+3)

(y−6)(y−2)

Solution

y2−8y+12

(2t−9)(10t+1)

(6p+5)(p+1)

Solution

6p2+11p+5

(q−5)(q+8)

(m+11)(m−4)

Solution

m2+7m−44

(7m+1)(m−3)

(3r−8)(11r+1)

Solution

33r2−85r−8

(x2+3)(x+2)

(y2−4)(y+3)

Solution

y3+3y2−4y−12

(5ab−1)(2ab+3)

(2xy+3)(3xy+2)

Solution

6x2y2+13xy+6

(x2+8)(x2−5)

(y2−7)(y2−4)

Solution

y4−11y2+28

(6pq−3)(4pq−5)

(3rs−7)(3rs−4)

Solution

9r2s2−33rs+28

Multiply a Polynomial by a Polynomial

In the following exercises, multiply using ⓐ the Distributive Property; ⓑ the Vertical Method.

(x+5)(x2+4x+3)

(u+4)(u2+3u+2)

Solution

u3+7u2+14u+8

(y+8)(4y2+y−7)

(a+10)(3a2+a−5)

Solution

3a3+31a2+5a−50

(y2−3y+8)(4y2+y−7)

(2a2−5a+10)(3a2+a−5)

Solution

6a4−13a3+15a2+35a−50

Multiply Special Products

In the following exercises, multiply. Use either method.

(w−7)(w2−9w+10)

(p−4)(p2−6p+9)

Solution

p3−10p2+33p−36

(3q+1)(q2−4q−5)

(6r+1)(r2−7r−9)

Solution

6r3−41r2−61r−9

In the following exercises, square each binomial using the Binomial Squares Pattern.

(w+4)2

(q+12)2

Solution

q2+24q+144

(3x−y)2

(2y−3z)2

Solution

4y2−12yz+9z2

(y+14)2

(x+23)2

Solution

x2+43x+49

(15x−17y)2

(18x−19y)2

Solution

164x2−136xy+181y2

(3x2+2)2

(5u2+9)2

Solution

25u4+90u2+81

(4y3−2)2

(8p3−3)2

Solution

64p6−48p3+9

In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern.

(5k+6)(5k−6)

(8j+4)(8j−4)

Solution

64j2−16

(11k+4)(11k−4)

(9c+5)(9c−5)

Solution

81c2−25

(9c−2d)(9c+2d)

(7w+10x)(7w−10x)

Solution

49w2−100x2

(m+23n)(m−23n)

(p+45q)(p−45q)

Solution

p2−1625q2

(ab−4)(ab+4)

(xy−9)(xy+9)

Solution

x2y2−81

(12p3−11q2)(12p3+11q2)

(15m2−8n4)(15m2+8n4)

Solution

225m4−64n8

In the following exercises, find each product.

(p−3)(p+3)

(t−9)2

Solution

t2−18t+81

(m+n)2

(2x+y)(x−2y)

Solution

2x2−3xy−2y2

(2r+12)2

(3p+8)(3p−8)

Solution

9p2−64

(7a+b)(a−7b)

(k−6)2

Solution

k2−12k+36

(a5−7b)2

(x2+8y)(8x−y2)

Solution

8x3−x2y2+64xy−8y3

(r6+s6)(r6−s6)

(y4+2z)2

Solution

y8+4y4z+4z2

(x5+y5)(x5−y5)

(m3−8n)2

Solution

m6−16m3n+64n2

(9p+8q)2

(r2−s3)(r3+s2)

Solution

r5+r2s2−r3s3−s5

Mixed Practice

(10y−6)+(4y−7)

(15p−4)+(3p−5)

Solution

18p−9

(x2−4x−34)−(x2+7x−6)

(j2−8j−27)−(j2+2j−12)

Solution

−10j−15

(15f8)(20f3)

(14d5)(36d2)

Solution

9d7

(4a3b)(9a2b6)

(6m4n3)(7mn5)

Solution

42m5n8

−5m(m2+3m−18)

5q3(q2−2q+6)

Solution

5q5−10q4+30q3

(s−7)(s+9)

(y2−2y)(y+1)

Solution

y3−y2−2y

(5x−y)(x−4)

(6k−1)(k2+2k−4)

Solution

6k3+11k2−26k+4

(3x−11y)(3x−11y)

(11−b)(11+b)

Solution

121−b2

(rs−27)(rs+27)

(2x2−3y4)(2x2+3y4)

Solution

4x4−9y8

(m−15)2

(3d+1)2

Solution

9d2+6d+1

(4a+10)2

(3z+15)2

Solution

9z2+65z+125

Multiply Polynomial Functions

For functions f(x)=x+2 and g(x)=3x2−2x+4, find ⓐ (f·g)(x) ⓑ (f·g)(−1)

For functions f(x)=x−1 and g(x)=4x2+3x−5, find ⓐ (f·g)(x) ⓑ (f·g)(−2)

Solution


ⓐ (f·g)(x)=4x3−x2−8x+5
ⓑ (f·g)(−2)=−15

For functions f(x)=2x−7 and g(x)=2x+7, find ⓐ (f·g)(x) ⓑ (f·g)(−3)

For functions f(x)=7x−8 and g(x)=7x+8, find ⓐ (f·g)(x) ⓑ (f·g)(−2)

Solution

ⓐ (f·g)(x)=49x2−64
ⓑ (f·g)(−2)=132

For functions f(x)=x2−5x+2 and g(x)=x2−3x−1, find ⓐ (f·g)(x) ⓑ (f·g)(−1)

For functions f(x)=x2+4x−3 and g(x)=x2+2x+4, find ⓐ (f·g)(x) ⓑ (f·g)(1)

Solution


ⓐ  (f·g)(x)=x4+6x3+9x2+10x−12 ⓑ (f·g)(1)=14

Writing Exercises

Which method do you prefer to use when multiplying two binomials: the Distributive Property or the FOIL method? Why? Which method do you prefer to use when multiplying a polynomial by a polynomial: the Distributive Property or the Vertical Method? Why?

Multiply the following:

(x+2)(x−2)(y+7)(y−7)(w+5)(w−5)

Explain the pattern that you see in your answers.

Solution

Answers will vary.

Multiply the following:

(p+3)(p+3)(q+6)(q+6)(r+1)(r+1)

Explain the pattern that you see in your answers.

Why does (a+b)2 result in a trinomial, but (a−b)(a+b) result in a binomial?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A self-assessment chart for students to rate their ability to multiply various polynomial types (monomials, binomials, general polynomials, special products) using 'Confidently,' 'With some help,' or 'No-I don't get it!' categories.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

conjugate pair
A conjugate pair is two binomials of the form (a−b),(a+b). The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.

Dividing Polynomials

Learning Objectives

By the end of this section, you will be able to:

  • Dividing monomials
  • Dividing a polynomial by a monomial
  • Dividing polynomials using long division
  • Dividing polynomials using synthetic division
  • Dividing polynomial functions
  • Use the remainder and factor theorems

Before you get started, take this readiness quiz.

Add: 3d+xd.
If you missed this problem, review Example 5 in Fractions.

Solution

3+xd

Simplify: 30xy35xy.
If you missed this problem, review Example 2 in Fractions.

Solution

6y2

Combine like terms: 8a2+12a+1+3a2−5a+4.
If you missed this problem, review Example 7 in Use the Language of Algebra.

Solution

11a2+7a+5

Dividing Monomials

We are now familiar with all the properties of exponents and used them to multiply polynomials. Next, we’ll use these properties to divide monomials and polynomials.

Find the quotient: 54a2b3÷(−6ab5).

Solution
When we divide monomials with more than one variable, we write one fraction for each variable.
This table illustrates the step-by-step simplification of an algebraic expression involving division.
54a2b3÷(−6ab5)
Rewrite as a fraction. 54a2b3−6ab5
Use fraction multiplication. 54−6·a2a·b3b5
Simplify and use the Quotient Property. −9·a·1b2
Multiply. −9ab2

Find the quotient: −72a7b3÷(8a12b4).

Solution

−9a5b

Find the quotient: −63c8d3÷(7c12d2).

Solution

−9dc4

Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.

Find the quotient: 14x7y1221x11y6.

Solution
Be very careful to simplify 1421 by dividing out a common factor, and to simplify the variables by subtracting their exponents.
This table illustrates the simplification of a rational algebraic expression, showing the original expression, the instruction to simplify using the Quotient Property, and the resulting simplified expression.
14x7y1221x11y6
Simplify and use the Quotient Property. 2y63x4

Find the quotient: 28x5y1449x9y12.

Solution

4y27x4

Find the quotient: 30m5n1148m10n14.

Solution

58m5n3

Divide a Polynomial by a Monomial

Now that we know how to divide a monomial by a monomial, the next procedure is to divide a polynomial of two or more terms by a monomial.

The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition. The sum y5+25 simplifies to y+25.

Now we will do this in reverse to split a single fraction into separate fractions. For example, y+25 can be written y5+25.

This is the “reverse” of fraction addition and it states that if a, b, and c are numbers where c≠0, then a+bc=ac+bc. We will use this to divide polynomials by monomials.

Division of a Polynomial by a Monomial

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Find the quotient: (18x3y−36xy2)÷(−3xy).

Solution
This table demonstrates the step-by-step process of dividing a polynomial by a monomial, showing the transformation of the mathematical expression.
(18x3y−36xy2)÷(−3xy)
Rewrite as a fraction. 18x3y−36xy2−3xy
Divide each term by the divisor. Be careful with the signs! 18x3y−3xy−36xy2−3xy
Simplify. −6x2+12y

Find the quotient: (32a2b−16ab2)÷(−8ab).

Solution

−4a+2b

Find the quotient: (−48a8b4−36a6b5)÷(−6a3b3).

Solution

8a5b+6a3b2

Divide Polynomials Using Long Division

Divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.

This figure shows the long division of 875 divided by 25. 875 is labeled dividend and 25 is labeled divisor. The result of 35 is labeled quotient. The 3 in 35 is determined from the number of times we can divide 25 into 87. Multiplying 25 and 3 results in 75. 75 is subtracted from 87 to get 12. The 5 from 875 is dropped down to make 12 into 125. The 5 in 35 is determined from the number of times was can divide 25 into 125. Since 25 goes into 125 evenly there is no remainder. The result of subtracting 125 from 125 is 0 which is labeled remainder.

We check division by multiplying the quotient by the divisor.

If we did the division correctly, the product should equal the dividend.

35·25875✓

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

Find the quotient: (x2+9x+20)÷(x+5).

Solution
A mathematical expression showing polynomial division: (x^2 + 9x + 20) ÷ (x + 5).
Write it as a long division problem.
Be sure the dividend is in standard form.
A mathematical long division problem showing the polynomial x^2 + 9x + 20 being divided by the binomial x + 5.
Divide x2 by x. It may help to ask yourself, “What do I need
to multiply x by to get x2?”
Polynomial long division problem where x^2 + 9x + 20 is being divided by x + 5, with 'x' as the first term of the quotient.
Put the answer, x, in the quotient over the x term.
Multiply x times x+5. Line up the like terms under the dividend.
A step in polynomial long division, showing (x+5) dividing into x^2 + 9x + 20. The term 'x' is placed in the quotient, and x(x+5) = x^2 + 5x is written below, ready for subtraction.
Subtract x2+5x from x2+9x.
You may find it easier to change the signs and then add.
Then bring down the last term, 20.
Polynomial long division: dividing x^2 + 9x + 20 by x + 5. After placing 'x' in the quotient, -x^2 + (-5x) is subtracted from the dividend, resulting in 4x + 20, highlighted in red.

Divide 4x by x. It may help to ask yourself, “What do I
need to multiply x by to get 4x?”
Put the answer, 4, in the quotient over the constant term.
Polynomial long division of (x^2 + 9x + 20) by (x + 5) is demonstrated, showing the initial steps to arrive at the quotient x + 4. The calculation subtracts (-x^2 + (-5x)) to get 4x + 20.
Multiply 4 times x+5. Polynomial long division of x^2 + 9x + 20 by x + 5, yielding a quotient of x + 4. The steps illustrate the process of dividing polynomials, from the initial division to the final remainder.
Subtract 4x+20 from 4x+20. Polynomial long division problem for (x^2 + 9x + 20) divided by (x + 5), with the steps leading to a quotient of (x + 4) and a remainder of 0.
Check:
Multiply the quotient by the divisor. (x+4)(x+5)
You should get the dividend. x2+9x+20✓

Find the quotient: (y2+10y+21)÷(y+3).

Solution

y+7

Find the quotient: (m2+9m+20)÷(m+4).

Solution

m+5

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In the next example, we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Look back at the dividends in previous examples. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in this example will be x4−x2+5x−6. It is missing an x3 term. We will add in 0x3 as a placeholder.

Find the quotient: (x4−x2+5x−6)÷(x+2).

Solution

Notice that there is no x3 term in the dividend. We will add 0x3 as a placeholder.

A mathematical expression showing the division of a polynomial (x^4 - x^2 + 5x - 6) by a binomial (x + 2).
Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. A polynomial long division setup, showing x^4 + 0x^3 - x^2 + 5x - 6 being divided by x + 2, commonly used to find quotients and remainders of polynomial expressions.
Divide x4 by x.
Put the answer, x3, in the quotient over the x3 term.
Multiply x3 times x+2. Line up the like terms.
Subtract and then bring down the next term.
A step in polynomial long division where x^4 + 0x^3 - x^2 + 5x - 6 is divided by x + 2, showing the subtraction of (x^4 + 2x^3) from x^4 + 0x^3 to get -2x^3 - x^2. A note suggests changing signs and adding.
Divide −2x3 by x.
Put the answer, −2x2, in the quotient over the x2 term.
Multiply −2x2 times x+1. Line up the like terms
Subtract and bring down the next term.
A visual guide to polynomial long division, showing the steps to divide x^4 - x^2 + 5x - 6 by x + 2. It highlights the subtraction of terms, noting that changing signs and adding can be helpful.
Divide 3x2 by x.
Put the answer, 3x, in the quotient over the x term.
Multiply 3x times x+1. Line up the like terms.
Subtract and bring down the next term.
This image illustrates polynomial long division, providing a step-by-step example. A helpful tip suggests changing the signs and adding instead of directly subtracting terms during the process.
Divide −x by x.
Put the answer, −1, in the quotient over the constant term.
Multiply −1 times x+1. Line up the like terms.
Change the signs, add.

Write the remainder as a fraction with the divisor as the denominator.
This image illustrates the process of polynomial long division, dividing x^4 - x^2 + 5x - 6 by x + 2, with a helpful reminder to change signs when subtracting.
To check, multiply (x+2)(x3−2x2+3x−1−4x+2).
The result should be x4−x2+5x−6.

Find the quotient: (x4−7x2+7x+6)÷(x+3).

Solution

x3−3x2+2x+1+3x+3

Find the quotient: (x4−11x2−7x−6)÷(x+3).

Solution

x3−3x2−2x−1−3x+3

In the next example, we will divide by 2a+3. As we divide, we will have to consider the constants as well as the variables.

Find the quotient: (8a3+27)÷(2a+3).

Solution

This time we will show the division all in one step. We need to add two placeholders in order to divide.

A mathematical expression showing the division of a sum of cubes, (8a^3 + 27), by a binomial, (2a + 3), written horizontally with a division symbol.
A step-by-step example of polynomial long division showing (8a^3 + 27) divided by (2a + 3), resulting in 4a^2 - 6a + 9 with a remainder of 0. The intermediate multiplication steps are also indicated.

To check, multiply (2a+3)(4a2−6a+9).

The result should be 8a3+27.

Find the quotient: (x3−64)÷(x−4).

Solution

x2+4x+16

Find the quotient: (125x3−8)÷(5x−2).

Solution

25x2+10x+4

Divide Polynomials using Synthetic Division

As we have mentioned before, mathematicians like to find patterns to make their work easier. Since long division can be tedious, let’s look back at the long division we did in Example 4 and look for some patterns. We will use this as a basis for what is called synthetic division. The same problem in the synthetic division format is shown next.

The figure shows the long division of 1 x squared plus 9 x plus 20 divided by x plus 5 right next to the same problem done with synthetic division. In the long division problem, the coefficients of the dividend are 1 and 9 and 20 and the zero of the divisor is negative 5. In the synthetic division problem, we just write the numbers negative 5 1 9 20 with a line separating the negative 5. In the long division problem, the subtracted terms are 5 x and 20. In the synthetic division problem the second line is the numbers negative 5 and negative 20. The remainder of the problem is 0 and the quotient is x plus 4. The synthetic division puts these coefficients as the last line 1 4 0.

Synthetic division basically just removes unnecessary repeated variables and numbers. Here all the x and x2 are removed. as well as the −x2 and −4x as they are opposite the term above.

The first row of the synthetic division is the coefficients of the dividend. The −5 is the opposite of the 5 in the divisor.

The second row of the synthetic division are the numbers shown in red in the division problem.

The third row of the synthetic division are the numbers shown in blue in the division problem.

Notice the quotient and remainder are shown in the third row.

Synthetic division only works when the divisor is of the formx−c.

The following example will explain the process.

Use synthetic division to find the quotient and remainder when 2x3+3x2+x+8 is divided by x+2.

Solution
Write the dividend with decreasing powers of x. The polynomial expression 2x^3 + 3x^2 + x + 8.
Write the coefficients of the terms as the first
row of the synthetic division.
A sequence of numbers 2, 3, 1, 8 is displayed on a white background, partially enclosed by a dark gray bracket on the left.
Write the divisor as x−c and place c
in the synthetic division in the divisor box.
A synthetic division problem showing -2 as the divisor and 2, 3, 1, 8 as the coefficients of a polynomial.
Bring down the first coefficient to the third row. Illustrates the initial step of synthetic division: the first coefficient (2) of the polynomial is brought down below the line, with -2 as the divisor and 3, 1, 8 as subsequent coefficients.
Multiply that coefficient by the divisor and place the
result in the second row under the second coefficient.
An image showing the initial steps of synthetic division. The number -2 is being divided into a polynomial with coefficients 2, 3, 1, and 8. The leading coefficient, 2, is brought down, then multiplied by -2 to get -4.
Add the second column, putting the result in the third row. A mathematical division problem is depicted, likely synthetic division, showing the numbers -2, 2, 3, 1, 8. Below the 3, -4 is written, and below a horizontal line, -1 appears with a light blue arrow pointing down from 3 to -4 and then to -1, indicating subtraction. The number 2 is also shown below the first horizontal line.
Multiply that result by the divisor and place the
result in the second row under the third coefficient.
An image illustrating the initial steps of synthetic division, showing coefficients (2, 3, 1, 8) divided by -2, with multiplication and addition leading to the first few quotient terms.
Add the third column, putting the result in the third row. A step in synthetic division showing the coefficients 2, 3, 1, and 8 being divided by -2. The process shows bringing down 2, multiplying by -2 to get -4, then adding 3 and -4 to get -1. Next, -1 is multiplied by -2 to get 2, and 1 plus 2 yields 3.
Multiply that result by the divisor and place the
result in the third row under the third coefficient.
A mathematical example demonstrating the process of synthetic division, with a divisor of -2 and polynomial coefficients 2, 3, 1, and 8. Blue arrows indicate the steps of multiplication and addition.
Add the final column, putting the result in the third row. An example of synthetic division is shown, with -2 as the divisor and 2, 3, 1, 8 as the dividend coefficients. The result is a quotient with coefficients 2, -1, 3 and a remainder of 2.
The quotient is 2x2−1x+3 and the remainder is 2.

The division is complete. The numbers in the third row give us the result. The 2−13 are the coefficients of the quotient. The quotient is 2x2−1x+3. The 2 in the box in the third row is the remainder.

Check:

(quotient)(divisor)+remainder=dividend(2x2−1x+3)(x+2)+2=?2x3+3x2+x+8 2x3−x2+3x+4x2−2x+6+2=?2x3+3x2+x+8 2x3+3x2+x+8=2x3+3x2+x+8✓

Use synthetic division to find the quotient and remainder when 3x3+10x2+6x−2 is divided by x+2.

Solution

3x2+4x−2;2

Use synthetic division to find the quotient and remainder when 4x3+5x2−5x+3 is divided by x+2.

Solution

4x2−3x+1;1

In the next example, we will do all the steps together.

Use synthetic division to find the quotient and remainder when x4−16x2+3x+12 is divided by x+4.

Solution

The polynomial x4−16x2+3x+12 has its term in order with descending degree but we notice there is no x3 term. We will add a 0 as a placeholder for the x3 term. In x−c form, the divisor is x−(−4).

The figure shows the results of using synthetic division with the example of the polynomial x to the fourth power minus 16 x squared plus 3 x plus 12 divided by x plus 4. The divisor number if negative 4. The first row is 1 0 negative 16 3 12. The first column is 1 blank 1. The second column is negative 16 16 0. The third column is 3 0 3. The fourth column is 12 negative 12 0.

We divided a 4th degree polynomial by a 1st degree polynomial so the quotient will be a 3rd degree polynomial.

Reading from the third row, the quotient has the coefficients 1−403, which is x3−4x2+3. The remainder
is 0.

Use synthetic division to find the quotient and remainder when x4−16x2+5x+20 is divided by x+4.

Solution

x3−4x2+5;0

Use synthetic division to find the quotient and remainder when x4−9x2+2x+6 is divided by x+3.

Solution

x3−3x2+2;0

Divide Polynomial Functions

Just as polynomials can be divided, polynomial functions can also be divided.

Division of Polynomial Functions

For functions f(x) and g(x), where g(x)≠0,

(fg)(x)=f(x)g(x)

For functions f(x)=x2−5x−14 and g(x)=x+2, find: ⓐ (fg)(x) ⓑ (fg)(−4).

Solution
ⓐ
Equation shows f over g of x equals f of x divided by g of x. This is translated into a division problem showing x squared minus 5x minus 14 divided by x plus 2. The quotient is x minus 7.
This table illustrates the step-by-step polynomial division of two functions, f(x) and g(x), showing the substitution and the final simplified quotient.
Substitute for f(x) and g(x). (fg)(x)=x2−5x−14x+2
Divide the polynomials. (fg)(x)=x−7


ⓑ In part ⓐ we found (fg)(x) and now are asked to find (fg)(−4).
This table demonstrates the step-by-step evaluation of the function (f/g)(x) at x=-4.
(fg)(x)=x−7
To find (fg)(−4), substitute x=−4. (fg)(−4)=−4−7
(fg)(−4)=−11

For functions f(x)=x2−5x−24 and g(x)=x+3, find ⓐ (fg)(x) ⓑ (fg)(−3).

Solution

ⓐ (fg)(x)=x−8
ⓑ (fg)(−3)=−11

For functions f(x)=x2−5x−36 and g(x)=x+4, find ⓐ (fg)(x) ⓑ (fg)(−5).

Solution

ⓐ (fg)(x)=x−9
ⓑ undefined

Use the Remainder and Factor Theorem

Let’s look at the division problems we have just worked that ended up with a remainder. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as x−c, the value of the function at c,f(c), is the same as the remainder from the division problem.

Dividend Divisor x−c Remainder Function f(c)
x4−x2+5x−6 x−(−2) −4 f(x)=x4−x2+5x−6 −4
3x3−2x2−10x+8 x−2 4 f(x)=3x3−2x2−10x+8 4
x4−16x2+3x+15 x−(−4) 3 f(x)=x4−16x2+3x+15 3

To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and add the remainder. In function notation we could say, to get the dividend f(x), we multiply the quotient, q(x) times the divisor, x−c, and add the remainder, r.

A mathematical equation illustrating the Remainder Theorem: f(x) = q(x)(x - c) + r, where f(x) is a polynomial, q(x) is the quotient, (x-c) is the divisor, and r is the remainder.
If we evaluate this at c, we get: A mathematical equation is displayed: f(c) = q(c)(c - c) + r. The variable 'c' is highlighted in red within f(c), q(c), and (c - c).
A mathematical equation showing f(c) = q(c)(0) + r, where the term q(c)(0) simplifies to zero, resulting in f(c) = r.
The mathematical equation f(c) = r is displayed on a white background.

This leads us to the Remainder Theorem.

Remainder Theorem

If the polynomial function f(x) is divided by x−c, then the remainder is f(c).

Use the Remainder Theorem to find the remainder when f(x)=x3+3x+19 is divided by x+2.

Solution

To use the Remainder Theorem, we must use the divisor in the x−c form. We can write the divisor x+2 as x−(−2). So, our c is −2.

To find the remainder, we evaluate f(c) which is f(−2).

The image shows the mathematical function f(x) = x³ + 3x + 19.
To evaluate f(−2), substitute x=−2. A mathematical equation is displayed: f(-2) = (-2)^3 + 3(-2) + 19, showing the substitution of -2 into a function.
Simplify. A mathematical equation is displayed against a white background, reading 'f(-2) = -8 - 6 + 19'.
A mathematical equation displays 'f(-2) = 5' in a clear, bold font on a white background, representing the function f evaluated at -2 equals 5.
The remainder is 5 when f(x)=x3+3x+19 is divided by x+2.
Check:
Use synthetic division to check.
A synthetic division problem with a divisor of -2. The dividend's coefficients are 1, 0, 3, 19. The result shows quotient coefficients 1, -2, 7, and a remainder of 5.
The remainder is 5.

Use the Remainder Theorem to find the remainder when f(x)=x3+4x+15 is divided by x+2.

Solution

−1

Use the Remainder Theorem to find the remainder when f(x)=x3−7x+12 is divided by x+3.

Solution

6

When we divided 8a3+27 by 2a+3 in Example 6 the result was 4a2−6a+9. To check our work, we multiply 4a2−6a+9 by 2a+3 to get 8a3+27.

(4a2−6a+9)(2a+3)=8a3+27

Written this way, we can see that 4a2−6a+9 and 2a+3 are factors of 8a3+27. When we did the division, the remainder was zero.

Whenever a divisor, x−c, divides a polynomial function, f(x), and resulting in a remainder of zero, we say x−c is a factor of f(x).

The reverse is also true. If x−c is a factor of f(x) then x−c will divide the polynomial function resulting in a remainder of zero.

We will state this in the Factor Theorem.

Factor Theorem

For any polynomial function f(x),

  • if x−c is a factor of f(x), then f(c)=0
  • if f(c)=0, then x−c is a factor of f(x)

Use the Remainder Theorem to determine if x−4 is a factor of f(x)=x3−64.

Solution
The Factor Theorem tells us that x−4 is a factor of f(x)=x3−64 if f(4)=0.
Step-by-step evaluation of the function f(x) = x^3 - 64 at x=4.
f(x)=x3−64
To evaluate f(4) substitute x=4. f(4)=43−64
Simplify. f(4)=64−64
Subtract. f(4)=0

Since f(4)=0, x−4 is a factor of f(x)=x3−64.

Use the Factor Theorem to determine if x−5 is a factor of f(x)=x3−125.

Solution

yes

Use the Factor Theorem to determine if x−6 is a factor of f(x)=x3−216.

Solution

yes

Access these online resources for additional instruction and practice with dividing polynomials.

  • Dividing a Polynomial by a Binomial
  • Synthetic Division & Remainder Theorem

Key Concepts

  • Division of a Polynomial by a Monomial
    • To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
  • Division of Polynomial Functions
    • For functions f(x) and g(x), where g(x)≠0,
      (fg)(x)=f(x)g(x)
  • Remainder Theorem
    • If the polynomial function f(x) is divided by x−c, then the remainder is f(c).
  • Factor Theorem: For any polynomial function f(x),
    • if x−c is a factor of f(x), then f(c)=0
    • if f(c)=0, then x−c is a factor of f(x)

Section Exercises

Practice Makes Perfect

Divide Monomials

In the following exercises, divide the monomials.

15r4s9÷(15r4s9)

20m8n4÷(30m5n9)

Solution

2m33n5

18a4b8−27a9b5

45x5y9−60x8y6

Solution

−3y34x3

(10m5n4)(5m3n6)25m7n5

(−18p4q7)(−6p3q8)−36p12q10

Solution

−3q5p5

(6a4b3)(4ab5)(12a2b)(a3b)

(4u2v5)(15u3v)(12u3v)(u4v)

Solution

5v4u2

Divide a Polynomial by a Monomial

In the following exercises, divide each polynomial by the monomial.

(9n4+6n3)÷3n

(8x3+6x2)÷2x

Solution

4x2+3x

(63m4−42m3)÷(−7m2)

(48y4−24y3)÷(−8y2)

Solution

−6y2+3y

66x3y2−110x2y3−44x4y311x2y2

72r5s2+132r4s3−96r3s512r2s2

Solution

6r3+11r2s−8rs3

10x2+5x−4−5x

20y2+12y−1−4y

Solution

−5y−3+14y

Divide Polynomials using Long Division

In the following exercises, divide each polynomial by the binomial.

(y2+7y+12)÷(y+3)

(a2−2a−35)÷(a+5)

Solution

a−7

(6m2−19m−20)÷(m−4)

(4x2−17x−15)÷(x−5)

Solution

4x+3

(q2+2q+20)÷(q+6)

(p2+11p+16)÷(p+8)

Solution

p+3−8p+8

(3b3+b2+4)÷(b+1)

(2n3−10n+28)÷(n+3)

Solution

2n2−6n+8+4n+3

(z3+1)÷(z+1)

(m3+1000)÷(m+10)

Solution

m2−10m+100

(64x3−27)÷(4x−3)

(125y3−64)÷(5y−4)

Solution

25y2+20y+16

Divide Polynomials using Synthetic Division

In the following exercises, use synthetic Division to find the quotient and remainder.

x3−6x2+5x+14 is divided by x+1

x3−3x2−4x+12 is divided by x+2

Solution

x2−5x+6;0

2x3−11x2+11x+12 is divided by x−3

2x3−11x2+16x−12 is divided by x−4

Solution

2x2−3x+4;4

x4−5x2+13x+3 is divided by x+3

x4+x2+6x−10 is divided by x+2

Solution

x3−2x2+5x−4;−2

2x4−9x3+5x2−3x−6 is divided by x−4

3x4−11x3+2x2+10x+6 is divided by x−3

Solution

3x3−2x2−4x−2;0

Divide Polynomial Functions

In the following exercises, divide.

For functions f(x)=x2−13x+36 and g(x)=x−4, find ⓐ (fg)(x) ⓑ (fg)(−1)

For functions f(x)=x2−15x+54 and g(x)=x−9, find ⓐ (fg)(x) ⓑ (fg)(−5)

Solution

ⓐ (fg)(x)=x−6
ⓑ (fg)(−5)=−11

For functions f(x)=x3+x2−7x+2 and g(x)=x−2, find ⓐ (fg)(x) ⓑ (fg)(2)

For functions f(x)=x3+2x2−19x+12 and g(x)=x−3, find ⓐ (fg)(x) ⓑ (fg)(0)

Solution

ⓐ (fg)(x)=x2+5x−4
ⓑ (fg)(0)=−4

For functions f(x)=x2−3x+2 and g(x)=x+3, find ⓐ (fg)(x) ⓑ (fg)(3)

For functions f(x)=x2+2x−3 and g(x)=x+3 find ⓐ (fg)(x)ⓑ (fg)(3)

Solution


ⓐ fgx=x-1;x≠-3 ⓑ fg3=2

Use the Remainder and Factor Theorem

In the following exercises, use the Remainder Theorem to find the remainder.

f(x)=x3−8x+7 is divided by x+3

f(x)=x3−4x−9 is divided by x+2

Solution

−9

f(x)=2x3−6x−24 divided by x−3

f(x)=7x2−5x−8 divided by x−1

Solution

−6

In the following exercises, use the Factor Theorem to determine if x−c is a factor of the polynomial function.

Determine whether x+3 a factor of x3+8x2+21x+18

Determine whether x+4 a factor of x3+x2−14x+8

Solution

no

Determine whether x−2 a factor of x3−7x2+7x−6

Determine whether x−3 a factor of x3−7x2+11x+3

Solution

yes

Writing Exercises

James divides 48y+6 by 6 this way: 48y+66=48y. What is wrong with his reasoning?

Divide 10x2+x−122x and explain with words how you get each term of the quotient.

Solution

answer will vary

Explain when you can use synthetic division.

In your own words, write the steps for synthetic division for x2+5x+6 divided by x−2.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section

The figure shows a table with seven rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is "confidently", the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “divide monomials”, “divide a polynomial by using a monomial”, “divide polynomials using long division”, “divide polynomials using synthetic division”, “divide polynomial functions”, and “use the Remainder and Factor Theorem”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Chapter Review Exercises

Add and Subtract Polynomials

Types of Polynomials

In the following exercises, determine the type of polynomial.

16x2−40x−25

5m+9

Solution

binomial

−15

y2+6y3+9y4

Solution

trinomial

Add and Subtract Polynomials

In the following exercises, add or subtract the polynomials.

4p+11p

−8y3−5y3

Solution

−13y3

(4a2+9a−11)+(6a2−5a+10)

(8m2+12m−5)−(2m2−7m−1)

Solution

6m2+19m−4

(y2−3y+12)+(5y2−9)

(5u2+8u)−(4u−7)

Solution

5u2+4u+7

Find the sum of 8q3−27 and q2+6q−2.

Find the difference of x2+6x+8 and x2−8x+15.

Solution

14x−7

In the following exercises, simplify.

17mn2−(−9mn2)+3mn2

18a−7b−21a

Solution

−7b−3a

2pq2−5p−3q2

(6a2+7)+(2a2−5a−9)

Solution

8a2−5a−2

(3p2−4p−9)+(5p2+14)

(7m2−2m−5)−(4m2+m−8)

Solution

3m2−3m+3

(7b2−4b+3)−(8b2−5b−7)

Subtract (8y2−y+9) from (11y2−9y−5)

Solution

3y2−8y−14

Find the difference of (z2−4z−12) and (3z2+2z−11)

(x3−x2y)−(4xy2−y3)+(3x2y−xy2)

Solution

x3+2x2y−5xy2+y3

(x3−2x2y)−(xy2−3y3)−(x2y−4xy2)

Evaluate a Polynomial Function for a Given Value of the Variable

In the following exercises, find the function values for each polynomial function.

For the function f(x)=7x2−3x+5 find:
ⓐ f(5) ⓑ f(−2) ⓒ f(0)

Solution

ⓐ 165 ⓑ 39 ⓒ 5

For the function g(x)=15−16x2, find:
ⓐ g(−1) ⓑ g(0) ⓒ g(2)

A pair of glasses is dropped off a bridge 640 feet above a river. The polynomial function h(t)=−16t2+640 gives the height of the glasses t seconds after they were dropped. Find the height of the glasses when t=6.

Solution

The height is 64feet.

A manufacturer of the latest soccer shoes has found that the revenue received from selling the shoes at a cost of p dollars each is given by the polynomial R(p)=−5p2+360p. Find the revenue received when p=10 dollars.

Add and Subtract Polynomial Functions

In the following exercises, find ⓐ (f + g)(x) ⓑ (f + g)(3) ⓒ (f − g)(x) ⓓ (f − g)(−2)

f(x)=2x2−4x−7 and g(x)=2x2−x+5

Solution

ⓐ (f+g)(x)=4x2−5x−2 ⓑ (f+g)(3)=19
ⓒ (f−g)(x)=−3x−12
ⓓ (f−g)(−2)=−6

f(x)=4x3−3x2+x−1 and g(x)=8x3−1

Properties of Exponents and Scientific Notation

Simplify Expressions Using the Properties for Exponents

In the following exercises, simplify each expression using the properties for exponents.

p3·p10

Solution

p13

2·26

a·a2·a3

Solution

a6

x·x8

ya·yb

Solution

ya+b

2822

a6a

Solution

a5

n3n12

1x5

Solution

1x5

30

y0

Solution

1

(14t)0

12a0−15b0

Solution

−3

Use the Definition of a Negative Exponent

In the following exercises, simplify each expression.

6−2

(−10)−3

Solution

−11000

5·2−4

(8n)−1

Solution

18n

y−5

10−3

Solution

11000

1a−4

16−2

Solution

36

−5−3

(−15)−3

Solution

−125

−(12)−3

(−5)−3

Solution

−1125

(59)−2

(−3x)−3

Solution

−x327

In the following exercises, simplify each expression using the Product Property.

(y4)3

(32)5

Solution

310

(a10)y

x−3·x9

Solution

x6

r−5·r−4

(uv−3)(u−4v−2)

Solution

1u3v5

(m5)−1

p5·p−2·p−4

Solution

1p

In the following exercises, simplify each expression using the Power Property.

(k−2)−3

q4q20

Solution

1q16

b8b−2

n−3n−5

Solution

n2

In the following exercises, simplify each expression using the Product to a Power Property.

(−5ab)3

(−4pq)0

Solution

1

(−6x3)−2

(3y−4)2

Solution

9y8

In the following exercises, simplify each expression using the Quotient to a Power Property.

(35x)−2

(3xy2z)4

Solution

81x4y8z4

(4p−3q2)2

In the following exercises, simplify each expression by applying several properties.

(x2y)2(3xy5)3

Solution

27x7y17

(−3a−2)4(2a4)2(−6a2)3

(3xy34x4y−2)2(6xy48x3y−2)−1

Solution

3y44x4

In the following exercises, write each number in scientific notation.

2.568

5,300,000

Solution

5.3×106

0.00814

In the following exercises, convert each number to decimal form.

2.9×104

Solution

29,000

3.75×10−1

9.413×10−5

Solution

0.00009413

In the following exercises, multiply or divide as indicated. Write your answer in decimal form.

(3×107)(2×10−4)

(1.5×10−3)(4.8×10−1)

Solution

0.00072

6×1092×10−1

9×10−31×10−6

Solution

9,000

Multiply Polynomials

Multiply Monomials

In the following exercises, multiply the monomials.

(−6p4)(9p)

(13c2)(30c8)

Solution

10c10

(8x2y5)(7xy6)

(23m3n6)(16m4n4)

Solution

m7n109

Multiply a Polynomial by a Monomial

In the following exercises, multiply.

7(10−x)

a2(a2−9a−36)

Solution

a4−9a3−36a2

−5y(125y3−1)

(4n−5)(2n3)

Solution

8n4−10n3

Multiply a Binomial by a Binomial

In the following exercises, multiply the binomials using:

ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical Method.

(a+5)(a+2)

(y−4)(y+12)

Solution

y2+8y−48

(3x+1)(2x−7)

(6p−11)(3p−10)

Solution

18p2−93p+110

In the following exercises, multiply the binomials. Use any method.

(n+8)(n+1)

(k+6)(k−9)

Solution

k2−3k−54

(5u−3)(u+8)

(2y−9)(5y−7)

Solution

10y2−59y+63

(p+4)(p+7)

(x−8)(x+9)

Solution

x2+x−72

(3c+1)(9c−4)

(10a−1)(3a−3)

Solution

30a2−33a+3

Multiply a Polynomial by a Polynomial

In the following exercises, multiply using ⓐ the Distributive Property ⓑ the Vertical Method.

(x+1)(x2−3x−21)

(5b−2)(3b2+b−9)

Solution

15b3−b2−47b+18

In the following exercises, multiply. Use either method.

(m+6)(m2−7m−30)

(4y−1)(6y2−12y+5)

Solution

24y3−54y2+32y−5

Multiply Special Products

In the following exercises, square each binomial using the Binomial Squares Pattern.

(2x−y)2

(x+34)2

Solution

x2+32x+916

(8p3−3)2

(5p+7q)2

Solution

25p2+70pq+49q2

In the following exercises, multiply each pair of conjugates using the Product of Conjugates.

(3y+5)(3y−5)

(6x+y)(6x−y)

Solution

36x2−y2

(a+23b)(a−23b)

(12x3−7y2)(12x3+7y2)

Solution

144x6−49y4

(13a2−8b4)(13a2+8b4)

Divide Monomials

Divide Monomials

In the following exercises, divide the monomials.

72p12÷8p3

Solution

9p9

−26a8÷(2a2)

45y6−15y10

Solution

−3y4

−30x8−36x9

28a9b7a4b3

Solution

4a5b2

11u6v355u2v8

(5m9n3)(8m3n2)(10mn4)(m2n5)

Solution

4m9n4

(42r2s4)(54rs2)(6rs3)(9s)

Divide a Polynomial by a Monomial

In the following exercises, divide each polynomial by the monomial

(54y4−24y3)÷(−6y2)

Solution

−9y2+4y

63x3y2−99x2y3−45x4y39x2y2

12x2+4x−3−4x

Solution

−3x−1+34x

Divide Polynomials using Long Division

In the following exercises, divide each polynomial by the binomial.

(4x2−21x−18)÷(x−6)

(y2+2y+18)÷(y+5)

Solution

y−3+33y+5

(n3−2n2−6n+27)÷(n+3)

(a3−1)÷(a+1)

Solution

a2−a+1−2a+1

Divide Polynomials using Synthetic Division

In the following exercises, use synthetic Division to find the quotient and remainder.

x3−3x2−4x+12 is divided by x+2

2x3−11x2+11x+12 is divided by x−3

Solution

2x2−5x−4;0

x4+x2+6x−10 is divided by x+2

Divide Polynomial Functions

In the following exercises, divide.

For functions f(x)=x2−15x+54 and g(x)=x−9, find ⓐ (fg)(x)
ⓑ (fg)(−2)

Solution

ⓐ (fg)(x)=x−6
ⓑ (fg)(−2)=−8

For functions f(x)=x3+x2−7x+2 and g(x)=x−2, find ⓐ (fg)(x)
ⓑ (fg)(3)

Use the Remainder and Factor Theorem

In the following exercises, use the Remainder Theorem to find the remainder.

f(x)=x3−4x−9 is divided by x+2

Solution

−9

f(x)=2x3−6x−24 divided by x−3

In the following exercises, use the Factor Theorem to determine if x−c is a factor of the polynomial function.

Determine whether x−2 is a factor of x3−7x2+7x−6.

Solution

no

Determine whether x−3 is a factor of x3−7x2+11x+3.

Chapter Practice Test

For the polynomial 8y4−3y2+1

ⓐ Is it a monomial, binomial, or trinomial? ⓑ What is its degree?

Solution

ⓐ trinomial ⓑ 4

(5a2+2a−12)(9a2+8a−4)

(10x2−3x+5)−(4x2−6)

Solution

6x2−3x+11

(−34)3

x−3x4

Solution

x

5658

(47a18b23c5)0

Solution

1

4−1

(2y)−3

Solution

18y3

p−3·p−8

x4x−5

Solution

x9

(3x−3)2

24r3s6r2s7

Solution

4rs6

(x4y9x−3)2

(8xy3)(−6x4y6)

Solution

−48x5y9

4u(u2−9u+1)

(m+3)(7m−2)

Solution

7m2+19m−6

(n−8)(n2−4n+11)

(4x−3)2

Solution

16x2−24x+9

(5x+2y)(5x−2y)

(15xy3−35x2y)÷5xy

Solution

3y2−7x

(3x3−10x2+7x+10)÷(3x+2)

Use the Factor Theorem to determine if x+3 a factor of x3+8x2+21x+18.

Solution

yes

ⓐ Convert 112,000 to scientific notation. ⓑ Convert 5.25×10−4 to decimal form.

In the following exercises, simplify and write your answer in exponential notation.

(2.4×108)(2×10−5)

Solution

4.8×103

9×1043×10−1

For the function f(x)=6x2−3x−9 find:
ⓐ f(3) ⓑ f(−2) ⓒ f(0)

Solution

ⓐ 36 ⓑ 21 ⓒ −9

For f(x)=2x2−3x−5 and g(x)=3x2−4x+1, find
ⓐ (f+g)(x) ⓑ (f+g)(1)
ⓒ (f−g)(x) ⓓ (f−g)(−2)

For functions
f(x)=3x2−23x−36 and
g(x)=x−9, find
ⓐ (fg)(x) ⓑ (fg)(3)

Solution

ⓐ (fg)(x)=3x+4
ⓑ (fg)(3)=13

A hiker drops a pebble from a bridge 240 feet above a canyon. The function h(t)=−16t2+240 gives the height of the pebble t seconds after it was dropped. Find the height when t=3.

Introduction to Factoring

A photo of two scientists operating a microscope.
Scientists use factoring to calculate growth rates of infectious diseases such as viruses. (credit: “FotoshopTofs” / Pixabay)

An epidemic of a disease has broken out. Where did it start? How is it spreading? What can be done to control it? Answers to these and other questions can be found by scientists known as epidemiologists. They collect data and analyze it to study disease and consider possible control measures. Because diseases can spread at alarming rates, these scientists must use their knowledge of mathematics involving factoring. In this chapter, you will learn how to factor and apply factoring to real-life situations.

Greatest Common Factor and Factor by Grouping

Learning Objectives

By the end of this section, you will be able to:

  • Find the greatest common factor of two or more expressions
  • Factor the greatest common factor from a polynomial
  • Factor by grouping

Before you get started, take this readiness quiz.

Factor 56 into primes.
If you missed this problem, review Example 2 in Use the Language of Algebra.

Solution

2·2·2·7

Find the least common multiple (LCM) of 18 and 24.
If you missed this problem, review Example 3 in Use the Language of Algebra.

Solution

72

Multiply: −3a(7a+8b).
If you missed this problem, review Example 2 in Multiply Polynomials.

Solution

−21a2−24ab

Find the Greatest Common Factor of Two or More Expressions

Earlier we multiplied factors together to get a product. Now, we will reverse this process; we will start with a product and then break it down into its factors. Splitting a product into factors is called factoring.

8 times 7 is 56. Here 8 and 7 are factors and 56 is the product. An arrow pointing from 8 times 7 to 56 is labeled multiply. An arrow pointing from 56 to 8 times 7 is labeled factor. 2x open parentheses x plus 3 close parentheses equals 2x squared plus 6x. Here the left side of the equation is labeled factors and the right side is labeled products.

We have learned how to factor numbers to find the least common multiple (LCM) of two or more numbers. Now we will factor expressions and find the greatest common factor of two or more expressions. The method we use is similar to what we used to find the LCM.

Greatest Common Factor

The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.

We summarize the steps we use to find the greatest common factor.

Find the greatest common factor (GCF) of two expressions.

  1. Factor each coefficient into primes. Write all variables with exponents in expanded form.
  2. List all factors—matching common factors in a column. In each column, circle the common factors.
  3. Bring down the common factors that all expressions share.
  4. Multiply the factors.

The next example will show us the steps to find the greatest common factor of three expressions.

Find the greatest common factor of 21x3,9x2,15x.

Solution
Factor each coefficient into primes and write the
variables with exponents in expanded form.
Circle the common factors in each column.
Bring down the common factors.
A mathematical example demonstrating how to find the Greatest Common Factor (GCF) of 21x³, 9x², and 15x by breaking down each expression into its prime factors and identifying common terms.
Multiply the factors. The image displays the text 'GCF = 3x' in a bold, gray font against a plain white background, indicating a mathematical equation or a definition of the Greatest Common Factor.
The GCF of 21x3, 9x2 and 15x is 3x.

Find the greatest common factor: 25m4,35m3,20m2.

Solution

5m2

Find the greatest common factor: 14x3,70x2,105x.

Solution

7x

Factor the Greatest Common Factor from a Polynomial

It is sometimes useful to represent a number as a product of factors, for example, 12 as 2·6 or 3·4. In algebra, it can also be useful to represent a polynomial in factored form. We will start with a product, such as 3x2+15x, and end with its factors, 3x(x+5). To do this we apply the Distributive Property “in reverse.”

We state the Distributive Property here just as you saw it in earlier chapters and “in reverse.”

Distributive Property

If a, b, and c are real numbers, then

a(b+c)=ab+acandab+ac=a(b+c)

The form on the left is used to multiply. The form on the right is used to factor.

So how do you use the Distributive Property to factor a polynomial? You just find the GCF of all the terms and write the polynomial as a product!

How to Use the Distributive Property to factor a polynomial

Factor: 8m3−12m2n+20mn2.

Solution
Step 1 is find the GCF of all the terms in the polynomial. GCF of 8 m cubed, 12 m squared n and 20 mn squared is 4m. Step 1 is find the GCF of all the terms in the polynomial. GCF of 8 m cubed, 12 m squared n and 20 mn squared is 4m. In step 3, use the reverse Distributive Property to factor the expression as 4m open parentheses 2 m squared minus 3 mn plus 5 n squared close parentheses. Step 4 is to check by multiplying the factors. By multiplying the factors, we get the original polynomial.

Factor: 9xy2+6x2y2+21y3.

Solution

3y2(3x+2x2+7y)

Factor: 3p3−6p2q+9pq3.

Solution

3p(p2−2pq+3q2)

Factor the greatest common factor from a polynomial.

  1. Find the GCF of all the terms of the polynomial.
  2. Rewrite each term as a product using the GCF.
  3. Use the “reverse” Distributive Property to factor the expression.
  4. Check by multiplying the factors.

Factor as a Noun and a Verb

We use “factor” as both a noun and a verb:

Noun:7 is afactorof 14Verb:factor3 from3a+3

Factor: 5x3−25x2.

Solution
Find the GCF of 5x3 and 25x2. This image demonstrates how to find the Greatest Common Factor (GCF) of 5x³ and 25x². It shows the prime factorization of each expression and then identifies the common factors, which are multiplied together to get the GCF.
The image displays 'GCF = 5x^2' in a gray font on a white background, representing a mathematical equation for the greatest common factor.
The image shows the mathematical expression 5x^3 - 25x^2, which represents a polynomial with two terms, indicating a subtraction operation between 5 times x cubed and 25 times x squared.
Rewrite each term. A mathematical expression 5x^2 * x - 5x^2 * 5, which can be factored as 5x^2(x - 5). The numbers and variables are colored in red and black against a white background.
Factor the GCF. A mathematical expression shows 5x squared multiplied by the quantity x minus 5, written as 5x²(x-5).
Check:

5x2(x−5)5x2·x−5x2·55x3−25x2✓

Factor: 2x3+12x2.

Solution

2x2(x+6)

Factor: 6y3−15y2.

Solution

3y2(2y−5)

Factor: 8x3y−10x2y2+12xy3.

Solution
The GCF of 8x3y,−10x2y2,and12xy3
is 2xy.
This image demonstrates finding the GCF of 8x³y, 10x²y², and 12xy³ by listing their prime factors. Common factors (2, x, y) are circled vertically, and their product gives the GCF: 2xy.
The image displays a mathematical expression 'GCF = 2xy' in a clear, dark gray font against a plain white background.
     A mathematical expression featuring three terms: 8x^3y - 10x^2y^2 + 12xy^3.
Rewrite each term using the GCF, 2xy.       The image displays the mathematical expression 2xy * 4x^2 - 2xy * 5xy + 2xy * 6y^2, with the term '2xy' highlighted in red in each part, suggesting it's a common factor to be extracted.
Factor the GCF.      A mathematical expression showing the term 2xy multiplied by the trinomial (4x^2 - 5xy + 6y^2).
Check:

2xy(4x2−5xy+6y2)2xy·4x2−2xy·5xy+2xy·6y28x3y−10x2y2+12xy3✓

Factor: 15x3y−3x2y2+6xy3.

Solution

3xy(5x2−xy+2y2)

Factor: 8a3b+2a2b2−6ab3.

Solution

2ab(4a2+ab−3b2)

When the leading coefficient is negative, we factor the negative out as part of the GCF.

Factor: −4a3+36a2−8a.

Solution

The leading coefficient is negative, so the GCF will be negative.

A mathematical expression displays -4a cubed + 36a squared - 8a.
Rewrite each term using the GCF, −4a. The algebraic expression -4a * a^2 - (-4a) * 9a + (-4a) * 2, illustrating a common factor of -4a in a polynomial.
Factor the GCF. The image shows the algebraic expression -4a(a^2 - 9a + 2), which is a monomial multiplied by a trinomial.
Check:

−4a(a2−9a+2)−4a·a2−(−4a)·9a+(−4a)·2−4a3+36a2−8a✓

Factor: −4b3+16b2−8b.

Solution

−4b(b2−4b+2)

Factor: −7a3+21a2−14a.

Solution

−7a(a2−3a+2)

So far our greatest common factors have been monomials. In the next example, the greatest common factor is a binomial.

Factor: 3y(y+7)−4(y+7).

Solution

The GCF is the binomial y+7.

The mathematical expression is 3y(y + 7) - 4(y + 7). It shows a common factor (y+7) being multiplied by 3y and then subtracted from 4 times (y+7).
Factor the GCF, (y+7). An algebraic expression showing the product of two binomials: (y + 7)(3y - 4). The expression is displayed in black text against a white background.
Check on your own by multiplying.   

Factor: 4m(m+3)−7(m+3).

Solution

(m+3)(4m−7)

Factor: 8n(n−4)+5(n−4).

Solution

(n−4)(8n+5)

Factor by Grouping

Sometimes there is no common factor of all the terms of a polynomial. When there are four terms we separate the polynomial into two parts with two terms in each part. Then look for the GCF in each part. If the polynomial can be factored, you will find a common factor emerges from both parts. Not all polynomials can be factored. Just like some numbers are prime, some polynomials are prime.

How to Factor a Polynomial by Grouping

Factor by grouping: xy+3y+2x+6.

Solution
Step 1 is to group the terms with common factors. There is no greatest common factor in all the four terms of xy plus 3y plus 2x plus 6. So, separate the first two terms from the second two. Step 2 is to factor out the common factor in each group. By factoring the GCF from the first 2 terms, we get y open parentheses x plus 3 close parentheses plus 2x plus 6. Factoring the GCF from the second 2 terms, we get y open parentheses x plus 3 close parentheses plus 2 open parentheses x plus 3 close parentheses. Step 3 is to factor the common factor from the expression. Notice that each term has a common factor of x plus 3. By factoring this out, we get open parentheses x plus 3 close parentheses open parentheses y plus 2 close parentheses Step 4 is to check by multiplying the expressions to get the result xy plus 3y plus 2x plus 6.

Factor by grouping: xy+8y+3x+24.

Solution

(x+8)(y+3)

Factor by grouping: ab+7b+8a+56.

Solution

(a+7)(b+8)

Factor by grouping.

  1. Group terms with common factors.
  2. Factor out the common factor in each group.
  3. Factor the common factor from the expression.
  4. Check by multiplying the factors.

Factor by grouping: ⓐ x2+3x−2x−6 ⓑ 6x2−3x−4x+2.

Solution
ⓐ
Steps illustrating the factorization of a quadratic expression by grouping.
There is no GCF in all four terms. x2+3x−2x−6
Separate into two parts. x2+3x−2x−6
Factor the GCF from both parts. Be careful with the signs when factoring the GCF from the last two terms. x(x+3)−2(x+3)
Factor out the common factor. (x+3)(x−2)
Check on your own by multiplying.
ⓑ
Steps and corresponding mathematical expressions for factoring a quadratic equation by grouping, from initial expression to final factored form.
There is no GCF in all four terms. 6x2−3x−4x+2
Separate into two parts. 6x2−3x−4x+2
Factor the GCF from both parts. 3x(2x−1)−2(2x−1)
Factor out the common factor. (2x−1)(3x−2)
Check on your own by multiplying.

Factor by grouping: ⓐ x2+2x−5x−10 ⓑ 20x2−16x−15x+12.

Solution

ⓐ (x−5)(x+2)
ⓑ (5x−4)(4x−3)

Factor by grouping: ⓐ y2+4y−7y−28 ⓑ 42m2−18m−35m+15.

Solution

ⓐ (y+4)(y−7)
ⓑ (7m−3)(6m−5)

Key Concepts

  • How to find the greatest common factor (GCF) of two expressions.
    1. Factor each coefficient into primes. Write all variables with exponents in expanded form.
    2. List all factors—matching common factors in a column. In each column, circle the common factors.
    3. Bring down the common factors that all expressions share.
    4. Multiply the factors.
  • Distributive Property: If a, b, and c are real numbers, then
    a(b+c)=ab+acandab+ac=a(b+c)

    The form on the left is used to multiply. The form on the right is used to factor.
  • How to factor the greatest common factor from a polynomial.
    1. Find the GCF of all the terms of the polynomial.
    2. Rewrite each term as a product using the GCF.
    3. Use the “reverse” Distributive Property to factor the expression.
    4. Check by multiplying the factors.
  • Factor as a Noun and a Verb: We use “factor” as both a noun and a verb.
    Noun:7 is afactorof 14Verb:factor3 from3a+3
  • How to factor by grouping.
    1. Group terms with common factors.
    2. Factor out the common factor in each group.
    3. Factor the common factor from the expression.
    4. Check by multiplying the factors.

Practice Makes Perfect

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

10p3q,12pq2

Solution

2pq

8a2b3,10ab2

12m2n3,30m5n3

Solution

6m2n3

28x2y4,42x4y4

10a3,12a2,14a

Solution

2a

20y3,28y2,40y

35x3y2,10x4y,5x5y3

Solution

5x3y

27p2q3,45p3q4,9p4q3

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

6m+9

Solution

3(2m+3)

14p+35

9n−63

Solution

9(n−7)

45b−18

3x2+6x−9

Solution

3(x2+2x−3)

4y2+8y−4

8p2+4p+2

Solution

2(4p2+2p+1)

10q2+14q+20

8y3+16y2

Solution

8y2(y+2)

12x3−10x

5x3−15x2+20x

Solution

5x(x2−3x+4)

8m2−40m+16

24x3−12x2+15x

Solution

3x(8x2−4x+5)

24y3−18y2−30y

12xy2+18x2y2−30y3

Solution

6y2(2x+3x2−5y)

21pq2+35p2q2−28q3

20x3y−4x2y2+12xy3

Solution

4xy(5x2−xy+3y2)

24a3b+6a2b2−18ab3

−2x−4

Solution

−2(x+2)

−3b+12

−2x3+18x2−8x

Solution

−2x(x2−9x+4)

−5y3+35y2−15y

−4p3q−12p2q2+16pq2

Solution

−4pq(p2+3pq−4q)

−6a3b−12a2b2+18ab2

5x(x+1)+3(x+1)

Solution

(x+1)(5x+3)

2x(x−1)+9(x−1)

3b(b−2)−13(b−2)

Solution

(b−2)(3b−13)

6m(m−5)−7(m−5)

Factor by Grouping

In the following exercises, factor by grouping.

ab+5a+3b+15

Solution

(b+5)(a+3)

cd+6c+4d+24

8y2+y+40y+5

Solution

(y+5)(8y+1)

6y2+7y+24y+28

uv−9u+2v−18

Solution

(u+2)(v−9)

pq−10p+8q−80

u2−u+6u−6

Solution

(u−1)(u+6)

x2−x+4x−4

9p2+12p−15p−20

Solution

(3p−5)(3p+4)

16q2+20q−28q−35

mn−6m−4n+24

Solution

(n−6)(m−4)

r2−3r−r+3

2x2−14x−5x+35

Solution

(x−7)(2x−5)

4x2−36x−3x+27

Mixed Practice

In the following exercises, factor.

−18xy2−27x2y

Solution

−9xy(2y+3x)

−4x3y5−x2y3+12xy4

3x3−7x2+6x−14

Solution

(x2+2)(3x−7)

x3+x2+x+1

x2+xy+5x+5y

Solution

(x+y)(x+5)

5x3−3x2+5x−3

Writing Exercises

What does it mean to say a polynomial is in factored form?

Solution

Answers will vary.

How do you check result after factoring a polynomial?

The greatest common factor of 36 and 60 is 12. Explain what this means.

Solution

Answers will vary.

What is the GCF of y4,y5, and y10? Write a general rule that tells you how to find the GCF of ya,yb, and yc.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column I can, confidently, with some help and no I don’t get it. The first column has the following statements: find the greatest common factor of 2 or more expressions, factor the greatest common factor from a polynomial, factor by grouping. The remaining columns are blank.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved your goals in this section! Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific!

…with some help. This must be addressed quickly as topics you do not master become potholes in your road to success. Math is sequential - every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is critical and you must not ignore it. You need to get help immediately or you will quickly be overwhelmed. See your instructor as soon as possible to discuss your situation. Together you can come up with a plan to get you the help you need.

factoring
Splitting a product into factors is called factoring.
greatest common factor
The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.

Factor Trinomials

Learning Objectives

By the end of this section, you will be able to:

  • Factor trinomials of the form x2+bx+c
  • Factor trinomials of the form ax2+bx+c using trial and error
  • Factor trinomials of the form ax2+bx+c using the ‘ac’ method
  • Factor using substitution

Before you get started, take this readiness quiz.

Find all the factors of 72.
If you missed this problem, review Example 2 in Use the Language of Algebra.

Solution

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Find the product: (3y+4)(2y+5).
If you missed this problem, review Example 4 in Multiply Polynomials.

Solution

6y2+23y+20

Simplify: −9(6); −9(−6).
If you missed this problem, review Example 7 in Integers.

Solution

−54, 54

Factor Trinomials of the Form x2+bx+c

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.

Figure shows the equation open parentheses x plus 2 close parentheses open parentheses x plus 3 close parentheses equals x squared plus 5 x plus 6. The left side of the equation is labeled factors and the right is labeled product. An arrow pointing right is labeled multiply. An arrow pointing left is labeled factor.

To figure out how we would factor a trinomial of the form x2+bx+c, such as x2+5x+6 and factor it to (x+2)(x+3), let’s start with two general binomials of the form (x+m) and (x+n).

The image displays the algebraic expression (x + m)(x + n), which represents the product of two binomials. This is a common form in algebra, often encountered when expanding or factoring quadratic equations.
Foil to find the product. A mathematical expression showing x squared plus mx plus nx plus mn.
Factor the GCF from the middle terms. An algebraic expression is shown: x² + (m + n)x + mn. This is a quadratic expression with variables x, m, and n, demonstrating the expanded form of (x+m)(x+n).
Our trinomial is of the form x2+bx+c. This image illustrates the relationship between the standard quadratic form x^2 + bx + c and its factored form components x^2 + (m + n)x + mn, highlighting b = m + n and c = mn.

This tells us that to factor a trinomial of the form x2+bx+c, we need two factors (x+m) and (x+n) where the two numbers m and n multiply to c and add to b.

How to Factor a Trinomial of the form x2+bx+c

Factor: x2+11x+24.

Solution
Step 1 is to write the factors of x squared plus 11x plus 24 as two binomials with first terms x. Write two sets of parentheses and put x as the first term. Step 2 is to find two numbers m and n that multiply to c, m times n is c and add to b, m plus n is b. So, find two numbers that multiply to 24 and add to 11. Factors of 24 are 1 and 24, 2 and 12, 3 and 8, 4 and 6. Sum of factors: 1 plus 24 is 25, 2 plus 12 is 14, 3 plus 8 is 11 and 4 plus 6 is 10. Step 3 is to use m and n, in this case, 3 and 8, as the last terms of the binomials. So we get open parentheses x plus 3 close parentheses open parentheses x plus 8 close parentheses Step 4 is to check by multiplying the factors to get the original polynomial.

Factor: q2+10q+24.

Solution

(q+4)(q+6)

Factor: t2+14t+24.

Solution

(t+2)(t+12)

Let’s summarize the steps we used to find the factors.

Factor trinomials of the form x2+bx+c.

  1. Write the factors as two binomials with first terms x. x2+bx+c(x)(x)
  2. Find two numbers m and n that
    • multiply to c,m·n=c
    • add to b,m+n=b
  3. Use m and n as the last terms of the factors. (x+m)(x+n)
  4. Check by multiplying the factors.

In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

How do you get a positive product and a negative sum? We use two negative numbers.

Factor: y2−11y+28.

Solution
Again, with the positive last term, 28, and the negative middle term, −11y, we need two negative factors. Find two numbers that multiply 28 and add to −11.
Illustrates the initial steps in factoring the quadratic expression y^2 - 11y + 28, showing the setup for binomial factors and criteria for finding constants.
y2−11y+28
Write the factors as two binomials with first terms y. (y)(y)
Find two numbers that: multiply to 28 and add to −11.

Factors of 28 Sum of factors
−1,−28

−2,−14

−4,−7
−1+(−28)=−29

−2+(−14)=−16

−4+(−7)=−11*

Illustrates the formation and expansion of a binomial expression using specific constant terms.
Use −4,−7 as the last terms of the binomials. (y−4)(y−7)
Check:
(y−4)(y−7)y2−7y−4y+28y2−11y+28✓

Factor: u2−9u+18.

Solution

(u−3)(u−6)

Factor: y2−16y+63.

Solution

(y−7)(y−9)

Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

How do you get a negative product and a positive sum? We use one positive and one negative number.

When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.

Factor: 2x+x2−48.

Solution
Demonstrates initial steps for factoring a quadratic expression, presenting the procedure and corresponding algebraic forms.
2x+x2−48
First we put the terms in decreasing degree order. x2+2x−48
Factors will be two binomials with first terms x. (x)(x)

Factors of −48 Sum of factors
−1,48
−2,24
−3,16
−4,12
−6,8
−1+48=47
−2+24=22
−3+16=13
−4+12=8
−6+8=2*

This table illustrates the process of constructing a binomial expression from specified last terms and verifying its expansion.
Use−6,8as the last terms of the binomials. (x−6)(x+8)
Check:
(x−6)(x+8)x2−6q+8q−48x2+2x−48✓

Factor: 9m+m2+18.

Solution

(m+3)(m+6)

Factor: −7n+12+n2.

Solution

(n−3)(n−4)

Sometimes you’ll need to factor trinomials of the form x2+bxy+cy2 with two variables, such as x2+12xy+36y2. The first term, x2, is the product of the first terms of the binomial factors, x·x. The y2 in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in How To Factor trinomials.

Factor: r2−8rs−9s2.

Solution
We need r in the first term of each binomial and s in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
Initial steps for factoring the quadratic expression r^2 - 8rs - 9s^2, illustrating textual guidance and mathematical forms.
r2−8rs−9s2
Note that the first terms are r, last terms contain s. (rs)(rs)
Find the numbers that multiply to −9 and add to −8.

Factors of −9 Sum of factors
1,−9 −1+9=8
−1,9 1+(−9)=−8*
3,−3 3+(−3)=0

Steps for factoring a quadratic expression and checking the factored form.
Use1,−9as coefficients of the last terms. (r+s)(r−9s)
Check:
(r−9s)(r+s)r2+rs−9rs−9s2r2−8rs−9s2✓

Factor: a2−11ab+10b2.

Solution

(a−b)(a−10b)

Factor: m2−13mn+12n2.

Solution

(m−n)(m−12n)

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

Factor: u2−9uv−12v2.

Solution
We need u in the first term of each binomial and v in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
This table demonstrates the factoring process for the quadratic expression u^2 - 9uv - 12v^2, providing notes and structural forms.
u2−9uv−12v2
Note that the first terms are u, last terms contain v. (uv)(uv)
Find the numbers that multiply to −12 and add to −9.
Factors of −12 Sum of factors
1,−12
−1,12
2,−6
−2,6
3,−4
−3,4
1+(−12)=−11
−1+12=11
2+(−6)=−4
−2+6=4
3+(−4)=−1
−3+4=1

Note there are no factor pairs that give us −9 as a sum. The trinomial is prime.

Factor: x2−7xy−10y2.

Solution

prime

Factor: p2+15pq+20q2.

Solution

prime

Let’s summarize the method we just developed to factor trinomials of the form x2+bx+c.

Strategy for Factoring Trinomials of the Form x2+bx+c

When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.

x2+bx+c(x+m)(x+n)Whencis positive,mandnhave the same sign.bpositivebnegativem,npositivem,nnegativex2+5x+6x2−6x+8(x+2)(x+3)(x−4)(x−2)same signssame signsWhencis negative,mandnhave opposite signs.x2+x−12x2−2x−15(x+4)(x−3)(x−5)(x+3)opposite signsopposite signs

Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.

Factor Trinomials of the form ax2 + bx + c using Trial and Error

Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form ax2+bx+c.

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.

Factor completely: 4x3+16x2−20x.

Solution
Step-by-step example demonstrating the factoring process of a polynomial, including identifying the GCF and final trinomial factorization.
Is there a greatest common factor? 4x3+16x2−20x
Yes, GCF=4x. Factor it. 4x(x2+4x−5)
Binomial, trinomial, or more than three terms?
It is a trinomial. So “undo FOIL.” 4x(x)(x)
Use a table like the one shown to find two numbers that
multiply to −5 and add to 4.
4x(x−1)(x+5)

Factors of −5 Sum of factors
−1,5
1,−5
−1+5=4*
1+(−5)=−4

Details the step-by-step expansion and verification of the algebraic expression 4x(x-1)(x+5).
Check:
4x(x−1)(x+5)4x(x2+5x−x−5)4x(x2+4x−5)4x3+16x2−20x✓

Factor completely: 5x3+15x2−20x.

Solution

5x(x−1)(x+4)

Factor completely: 6y3+18y2−60y.

Solution

6y(y−2)(y+5)

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial 3x2+5x+2.

From our earlier work, we expect this will factor into two binomials.

3x2+5x+2()()

We know the first terms of the binomial factors will multiply to give us 3x2. The only factors of 3x2 are 1x,3x. We can place them in the binomials.

The polynomial is 3x squared plus 5x plus 2. There are two pairs of parentheses, with the first terms in them being x and 3x.

Check: Does 1x·3x=3x2?

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.

Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses.

Which factors are correct? To decide that, we multiply the inner and outer terms.

Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses. In each case, arrows are shown pairing the first term of the first factor with the last term of the second factor and the first term of the second factor with the last term of the first factor.

Since the middle term of the trinomial is 5x, the factors in the first case will work. Let’s use FOIL to check.

(x+1)(3x+2)3x2+2x+3x+23x2+5x+2✓

Our result of the factoring is:

3x2+5x+2(x+1)(3x+2)

How to Factor a Trinomial Using Trial and Error

Factor completely using trial and error: 3y2+22y+7.

Solution
Step 1 is to write the trinomial in descending order. The trinomial 3 y squared plus 22y plus 7 is already in descending order. Step 2 is to factor the GCF. Here, there is none. Step 3 is Find all the factor pairs of the first term. The only factors here are 1y and 3y. Since there is only one pair, we can put each as the first term in the parentheses. Step 4 is to find all the factor pairs of the third term. Here, the only pair is 1 and 7. Step 5 is to test all the possible combinations of the factors until the correct product is found. For possible factors open parentheses y plus 1 close parentheses open parentheses 37 plus 7 close parentheses, the product is 3 y squared plus 10y plus 7. For the possible factors open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses, the product is 3 y squared plus 22y plus 7, which is the correct product. Hence, the correct factors are open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses. Step 6 is to check by multiplying.

Factor completely using trial and error: 2a2+5a+3.

Solution

(a+1)(2a+3)

Factor completely using trial and error: 4b2+5b+1.

Solution

(b+1)(4b+1)

Factor trinomials of the form ax2+bx+c using trial and error.

  1. Write the trinomial in descending order of degrees as needed.
  2. Factor any GCF.
  3. Find all the factor pairs of the first term.
  4. Find all the factor pairs of the third term.
  5. Test all the possible combinations of the factors until the correct product is found.
  6. Check by multiplying.

Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

Factor completely using trial and error: 6b2−13b+5.

Solution
The trinomial is already in descending order. The image displays the quadratic expression 6b^2 - 13b + 5.
Find the factors of the first term. A quadratic expression, 6b^2 - 13b + 5, is shown with potential factors for the 6b^2 term listed below it: 1b * 6b and 2b * 3b, in red text.
Find the factors of the last term. Consider the signs.
Since the last term, 5, is positive its factors must both be
positive or both be negative. The coefficient of the
middle term is negative, so we use the negative factors.
The image shows the quadratic expression 6b^2 - 13b + 5, with potential factors for the first term (1b*6b, 2b*3b) and the constant term (-1, -5) listed below, indicating the process of factoring.

Consider all the combinations of factors.

6b2−13b+5
Possible factors Product
(b−1)(6b−5) 6b2−11b+5
(b−5)(6b−1) 6b2−31b+5
(2b−1)(3b−5) 6b2−13b+5*
(2b−5)(3b−1) 6b2−17b+5
This table demonstrates how to factor a trinomial and verify the factors by multiplication.
The correct factors are those whose product
is the original trinomial.
(2b−1)(3b−5)
Check by multiplying:
(2b−1)(3b−5)6b2−10b−3b+56b2−13b+5✓

Factor completely using trial and error: 8x2−14x+3.

Solution

(2x−3)(4x−1)

Factor completely using trial and error: 10y2−37y+7.

Solution

(2y−7)(5y−1)

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Factor completely using trial and error: 18x2−37xy+15y2.

Solution
The trinomial is already in descending order. The image shows the algebraic expression 18x^2 - 37xy + 15y^2.
Find the factors of the first term. Steps to factor the trinomial 18x^2 - 37xy + 15y^2 are shown, with initial factor pairs for the 18x^2 term listed as 1x*18x, 2x*9x, and 3x*6x.
Find the factors of the last term. Consider the signs.
Since 15 is positive and the coefficient of the middle
term is negative, we use the negative factors.
An image displaying the algebraic expression 18x^2 - 37xy + 15y^2, with potential factors for the first and last terms shown in red below it, indicating steps for factoring the trinomial.

Consider all the combinations of factors.

This table shows the possible factors and corresponding products of the trinomial 18 x squared minus 37xy plus 15 y squared. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: open parentheses x minus 1y close parentheses open parentheses 18x minus 15y close parentheses, highlighted. Factor, open parentheses x minus 15y close parentheses open parentheses 18x minus 1y close parentheses; product: 18 x squared minus 271xy plus 15 y squared. Factor open parentheses x minus 3y close parentheses open parentheses 18x minus 5 y close parentheses; product: 18 x squared minus 59xy plus 15 y squared. Factor: open parentheses x minus 5y close parentheses open parentheses 18x minus 3y close parentheses highlighted. Factor: open parentheses 2x minus 1y close parentheses open parentheses 9x minus 15y close parentheses highlighted. Factor: open parentheses 2x minus 15y close parentheses open parentheses 9x minus 1y close parentheses; product 18 x squared minus 137 xy plus 15y squared. Factor: open parentheses 2x minus 3y close parentheses open parentheses 9x minus 5y close parentheses; product: 18 x squared minus 37xy plus 15 y squared, which is the original trinomial. Factor: open parentheses 2x minus 57 close parentheses open parentheses 9x minus 3y close parentheses highlighted. Factor: open parentheses 3x minus 1y close parentheses open parentheses 6x minus 15y close parentheses highlighted. Factor: open parentheses 3x minus 15y close parentheses highlighted open parentheses 6x minus 1y close parentheses. Factor: open parentheses 3x minus 3y close parentheses highlighted open parentheses 6x minus 5y.
This table demonstrates verifying binomial factors by multiplying them to obtain the original trinomial, illustrating a check for correctness in factorization.
The correct factors are those whose product is the original trinomial. (2x−3y)(9x−5y)
Check by multiplying:
(2x−3y)(9x−5y)18x2−10xy−27xy+15y218x2−37xy+15y2✓

Factor completely using trial and error 18x2−3xy−10y2.

Solution

(3x+2y)(6x−5y)

Factor completely using trial and error: 30x2−53xy−21y2.

Solution

(3x+y)(10x−21y)

Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.

Factor completely using trial and error: −10y4−55y3−60y2.

Solution
A mathematical expression: negative ten y to the fourth power, minus fifty-five y cubed, minus sixty y squared.
Notice the greatest common factor, so factor it first. negative five y squared times open parenthesis two y squared plus eleven y plus twelve close parenthesis
Factor the trinomial. A mathematical expression -5y²(2y² + 11y + 12) illustrating steps to factor the quadratic trinomial. Red text shows factors for 2y² (y*2y) and factors for 12 (1*12, 2*6, 3*4).

Consider all the combinations.

This table shows the possible factors and product of the trinomial 2 y squared plus 11y plus 12. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: y plus 1, 2y plus 12 highlighted. Factor: y plus 12, 2y plus 1; product: 2 y squared plus 25y plus 12. Factor: y plus 2, 2y plus 6 highlighted. Factor: y plus 6, 2y plus 2 highlighted. Factor: y plus 3, 2y plus 4 highlighted. Factor: y plus 4, 2y plus 3; product: 2 y squared plus 11y plus 12. This is the original trinomial.
This table illustrates the factoring of an algebraic trinomial and provides the steps to verify the factored expression through multiplication.
The correct factors are those whose product
is the original trinomial. Remember to include
the factor −5y2.
−5y2(y+4)(2y+3)
Check by multiplying:
−5y2(y+4)(2y+3)−5y2(2y2+8y+3y+12)−10y4−55y3−60y2✓

Factor completely using trial and error: 15n3−85n2+100n.

Solution

5n(n−4)(3n−5)

Factor completely using trial and error: 56q3+320q2−96q.

Solution

8q(q+6)(7q−2)

Factor Trinomials of the Form ax2+bx+c using the “ac” Method

Another way to factor trinomials of the form ax2+bx+c is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!

How to Factor Trinomials using the “ac” Method

Factor using the ‘ac’ method: 6x2+7x+2.

Solution
Step 1 is to factor the GCF. There is none in 6 x squared plus 7x plus 2. Step 2 is to find the product of a and c. The product of 6 and 2 is 12. Step 3 is to find 2 numbers m and n such that mn is ac and m plus n is b. So we need to numbers that multiply to 12 and add to 7. Both factors must be positive. 3 times 4 is 12 and 3 plus 4 is 7. Step 4 is to split the middle term using m and n. So we rewrite 7 x as 3x plus 4x. It would give the same result if we used 4x plus 3x. Rewriting, we get 6 x squared plus 3x plus 4x plus 2. Notice that this is the same as the original polynomial. We just split the middle term to get a more useful form Step 5 is to factor by grouping. So, we get, 3x open parentheses 2x plus 1 close parentheses plus 2 open parentheses 2x plus 1 close parentheses. This is equal to 2x plus 1, 3x plus 2. Step 6 is to check by multiplying the factors.

Factor using the ‘ac’ method: 6x2+13x+2.

Solution

(x+2)(6x+1)

Factor using the ‘ac’ method: 4y2+8y+3.

Solution

(2y+1)(2y+3)

The “ac” method is summarized here.

Factor trinomials of the form ax2+bx+c using the “ac” method.

  1. Factor any GCF.
  2. Find the product ac.
  3. Find two numbers m and n that:
    Multiply toacm·n=a·cAdd tobm+n=bax2+bx+c
  4. Split the middle term using m and n. ax2+mx+nx+c
  5. Factor by grouping.
  6. Check by multiplying the factors.

Don’t forget to look for a common factor!

Factor using the ‘ac’ method: 10y2−55y+70.

Solution
Is there a greatest common factor?
Yes. The GCF is 5. The image shows the quadratic expression 10y^2 - 55y + 70.
Factor it. The image shows the mathematical expression 5(2y^2 - 11y + 14).
The trinomial inside the parentheses has a
leading coefficient that is not 1.
The image displays two algebraic expressions: 'ax^2 + bx + c' in red, representing a general quadratic equation, and '5(2y^2 - 11y + 14)' in black, a factored quadratic expression.
Find the product ac. ac=28
Find two numbers that multiply to ac (−4)(−7)=28
and add to b. −4+(−7)=−11
Split the middle term. Mathematical expression 5(2y^2 - 11y + 14) with arrows highlighting the -11y term, likely for breaking it down during factorization.
The algebraic expression 5(2y^2 - 7y - 4y + 14) is shown, with blue brackets indicating the grouping of terms (2y^2 - 7y) and (-4y + 14), typically a step in factorization.
Factor the trinomial by grouping. A mathematical expression featuring the difference of two products, 5(y(2y - 7)) - 2(2y - 7)), suitable for algebraic manipulation or solving.
A mathematical expression displays 5 multiplied by the quantity (y minus 2), which is then multiplied by the quantity (2y minus 7), representing a factored quadratic expression.
Check by multiplying all three factors.

5(y−2)(2y−7)5(2y2−7y−4y+14)5(2y2−11y+14)10y2−55y+70✓

Factor using the ‘ac’ method: 16x2−32x+12.

Solution

4(2x−3)(2x−1)

Factor using the ‘ac’ method: 18w2−39w+18.

Solution

3(3w−2)(2w−3)

Factor Using Substitution

Sometimes a trinomial does not appear to be in the ax2+bx+c form. However, we can often make a thoughtful substitution that will allow us to make it fit the ax2+bx+c form. This is called factoring by substitution. It is standard to use u for the substitution.

In the ax2+bx+c, the middle term has a variable, x, and its square, x2, is the variable part of the first term. Look for this relationship as you try to find a substitution.

Factor by substitution: x4−4x2−5.

Solution

The variable part of the middle term is x2 and its square, x4, is the variable part of the first term. (We know (x2)2=x4). If we let u=x2, we can put our trinomial in the ax2+bx+c form we need to factor it.

A mathematical expression reads 'x to the power of 4 minus 4x squared minus 5'.
Rewrite the trinomial to prepare for the substitution. A mathematical expression showing a quadratic in terms of x squared: (x^2)^2 - 4(x^2) - 5. The 'x^2' terms are highlighted in red.
Let u=x2 and substitute. The image shows the mathematical expression u squared minus 4u minus 5, with the 'u' characters in a reddish hue and the numbers and minus signs in grey/black.
Factor the trinomial. The mathematical expression (u+1)(u-5) is shown, representing the product of two binomials.
Replace u with x2. The image displays the mathematical expression (x^2 + 1)(x^2 - 5), which represents the product of two binomials involving x squared.
Check:

(x2+1)(x2−5)x4−5x2+x2−5x4−4x2−5✓

Factor by substitution: h4+4h2−12.

Solution

(h2−2)(h2+6)

Factor by substitution: y4−y2−20.

Solution

(y2+4)(y2−5)

Sometimes the expression to be substituted is not a monomial.

Factor by substitution: (x−2)2+7(x−2)+12

Solution

The binomial in the middle term, (x−2) is squared in the first term. If we let u=x−2 and substitute, our trinomial will be in ax2+bx+c form.

A mathematical expression is displayed, which reads as quantity x minus 2 squared, plus 7 times quantity x minus 2, plus 12. This is a quadratic expression in terms of (x-2).
Rewrite the trinomial to prepare for the substitution. A mathematical expression reads as 'open parenthesis x minus 2 close parenthesis squared plus 7 open parenthesis x minus 2 close parenthesis plus 12'.
Let u=x−2 and substitute. A quadratic expression is displayed against a white background, reading 'u^2 + 7u + 12' with the 'u' characters in red.
Factor the trinomial. An algebraic expression showing the product of two binomials, (u+3) and (u+4).
Replace u with x−2. A mathematical expression featuring (x-2) plus 3 times (x-2) plus 4. The 'x-2' terms are highlighted in red, indicating a common factor within the polynomial expression.
Simplify inside the parentheses. The mathematical expression (x+1)(x+2) is displayed in black text on a white background.

This could also be factored by first multiplying out the (x−2)2 and the 7(x−2) and then combining like terms and then factoring. Most students prefer the substitution method.

Factor by substitution: (x−5)2+6(x−5)+8.

Solution

(x−3)(x−1)

Factor by substitution: (y−4)2+8(y−4)+15.

Solution

(y−1)(y+1)

Access this online resource for additional instruction and practice with factoring.

  • Factor a trinomial using the AC method

Key Concepts

  • How to factor trinomials of the form x2+bx+c.
    1. Write the factors as two binomials with first terms x. x2+bx+c(x)(x)
    2. Find two numbers m and n that
      multiply toc,m·n=cadd tob,m+n=b
    3. Use m and n as the last terms of the factors. (x+m)(x+n)
    4. Check by multiplying the factors.
  • Strategy for Factoring Trinomials of the Form x2+bx+c: When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
    x2+bx+c(x+m)(x+n)Whencis positive,mandnhave the same sign.bpositivebnegativem,npositivem,nnegativex2+5x+6x2−6x+8(x+2)(x+3)(x−4)(x−2)same signssame signsWhencis negative,mandnhave opposite signs.x2+x−12x2−2x−15(x+4)(x−3)(x−5)(x+3)opposite signsopposite signs
    Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.
  • How to factor trinomials of the form ax2+bx+c using trial and error.
    1. Write the trinomial in descending order of degrees as needed.
    2. Factor any GCF.
    3. Find all the factor pairs of the first term.
    4. Find all the factor pairs of the third term.
    5. Test all the possible combinations of the factors until the correct product is found.
    6. Check by multiplying.
  • How to factor trinomials of the form ax2+bx+c using the “ac” method.
    1. Factor any GCF.
    2. Find the product ac.
    3. Find two numbers m and n that:
      Multiply toac.m·n=a·cAdd tob.m+n=bax2+bx+c
    4. Split the middle term using m and n. ax2+mx+nx+c
    5. Factor by grouping.
    6. Check by multiplying the factors.

Practice Makes Perfect

Factor Trinomials of the Form x2+bx+c

In the following exercises, factor each trinomial of the form x2+bx+c.

p2+11p+30

Solution

(p+5)(p+6)

w2+10w+21

n2+19n+48

Solution

(n+3)(n+16)

b2+14b+48

a2+25a+100

Solution

(a+5)(a+20)

u2+101u+100

x2−8x+12

Solution

(x−2)(x−6)

q2−13q+36

y2−18y+45

Solution

(y−3)(y−15)

m2−13m+30

x2−8x+7

Solution

(x−1)(x−7)

y2−5y+6

5p−6+p2

Solution

(p−1)(p+6)

6n−7+n2

8−6x+x2

Solution

(x−4)(x−2)

7x+x2+6

x2−12−11x

Solution

(x−12)(x+1)

−11−10x+x2

In the following exercises, factor each trinomial of the form x2+bxy+cy2. If the trinomial cannot be factored, answer “Prime.”

x2−2xy−80y2

Solution

(x+8y)(x−10y)

p2−8pq−65q2

m2−64mn−65n2

Solution

(m+n)(m−65n)

p2−2pq−35q2

a2+5ab−24b2

Solution

(a+8b)(a−3b)

r2+3rs−28s2

x2−3xy−14y2

Solution

Prime

u2−8uv−24v2

m2−5mn+30n2

Solution

Prime

c2−7cd+18d2

Factor Trinomials of the Form ax2+bx+c Using Trial and Error

In the following exercises, factor completely using trial and error.

p3−8p2−20p

Solution

p(p−10)(p+2)

q3−5q2−24q

3m3−21m2+30m

Solution

3m(m−5)(m−2)

11n3−55n2+44n

5x4+10x3−75x2

Solution

5x2(x−3)(x+5)

6y4+12y3−48y2

2t2+7t+5

Solution

(2t+5)(t+1)

5y2+16y+11

11x2+34x+3

Solution

(11x+1)(x+3)

7b2+50b+7

4w2−5w+1

Solution

(4w−1)(w−1)

5x2−17x+6

4q2−7q−2

Solution

(4q+1)(q−2)

10y2−53y−11

6p2−19pq+10q2

Solution

(2p−5q)(3p−2q)

21m2−29mn+10n2

4a2+17ab−15b2

Solution

(4a−3b)(a+5b)

6u2+5uv−14v2

−16x2−32x−16

Solution

−16(x+1)(x+1)

−81a2+153a+18

−30q3−140q2−80q

Solution

−10q(3q+2)(q+4)

−5y3−30y2+35y

Factor Trinomials of the Form ax2+bx+c using the ‘ac’ Method

In the following exercises, factor using the ‘ac’ method.

5n2+21n+4

Solution

(5n+1)(n+4)

8w2+25w+3

4k2−16k+15

Solution

(2k−3)(2k−5)

5s2−9s+4

6y2+y−15

Solution

(3y+5)(2y−3)

6p2+p−22

2n2−27n−45

Solution

(2n+3)(n−15)

12z2−41z−11

60y2+290y−50

Solution

10(6y−1)(y+5)

6u2−46u−16

48z3−102z2−45z

Solution

3z(8z+3)(2z−5)

90n3+42n2−216n

16s2+40s+24

Solution

8(2s+3)(s+1)

24p2+160p+96

48y2+12y−36

Solution

12(4y−3)(y+1)

30x2+105x−60

Factor Using Substitution

In the following exercises, factor using substitution.

x4−6x2−7

Solution

(x2+1)(x2−7)

x4+2x2−8

x4−3x2−28

Solution

(x2−7)(x2+4)

x4−13x2−30

(x−3)2−5(x−3)−36

Solution

(x−12)(x+1)

(x−2)2−3(x−2)−54

(3y−2)2−(3y−2)−2

Solution

(3y−4)(3y−1)

(5y−1)2−3(5y−1)−18

Mixed Practice

In the following exercises, factor each expression using any method.

u2−12u+36

Solution

(u−6)(u−6)

x2−14x−32

r2−20rs+64s2

Solution

(r−4s)(r−16s)

q2−29qr−96r2

12y2−29y+14

Solution

(4y−7)(3y−2)

12x2+36y−24z

6n2+5n−4

Solution

(2n−1)(3n+4)

3q2+6q+2

13z2+39z−26

Solution

13(z2+3z−2)

5r2+25r+30

3p2+21p

Solution

3p(p+7)

7x2−21x

6r2+30r+36

Solution

6(r+2)(r+3)

18m2+15m+3

24n2+20n+4

Solution

4(2n+1)(3n+1)

4a2+5a+2

x4−4x2−12

Solution

(x2+2)(x2−6)

x4−7x2−8

(x+3)2−9(x+3)−36

Solution

(x−9)(x+6)

(x+2)2−25(x+2)−54

Writing Exercises

Many trinomials of the form x2+bx+c factor into the product of two binomials (x+m)(x+n). Explain how you find the values of m and n.

Solution

Answers will vary.

Tommy factored x2−x−20 as (x+5)(x−4). Sara factored it as (x+4)(x−5). Ernesto factored it as (x−5)(x−4). Who is correct? Explain why the other two are wrong.

List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form ax2+bx+c.

Solution

Answers will vary.

How is the “ac” method similar to the “undo FOIL” method? How is it different?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 4 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: factor trinomials of the form x squared plus bx plus c, factor trinomials of the form a x squared plus b x plus c using trial and error, factor trinomials of the form a x squared plus bx plus c with using the “ac” method, factor using substitution.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Factor Special Products

Learning Objectives

By the end of this section, you will be able to:

  • Factor perfect square trinomials
  • Factor differences of squares
  • Factor sums and differences of cubes

Before you get started, take this readiness quiz.

Simplify: (3x2)3.
If you missed this problem, review Example 7 in Properties of Exponents and Scientific Notation.

Solution

27x6

Multiply: (m+4)2.
If you missed this problem, review Example 8 in Multiply Polynomials.

Solution

m2+8m+16

Multiply: (x−3)(x+3).
If you missed this problem, review Example 9 in Multiply Polynomials.

Solution

x2−9

We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.

Factor Perfect Square Trinomials

Some trinomials are perfect squares. They result from multiplying a binomial times itself. We squared a binomial using the Binomial Squares pattern in a previous chapter.

In open parentheses 3x plus 4 close parentheses squared, 3x is a and 4 is b. Writing it as a squared plus 2ab plus b squared, we get open parentheses 3x close parentheses squared plus 2 times 3x times 4 plus 4 squared. This is equal to 9 x squared plus 24x plus 16.

The trinomial 9x2+24x+16 is called a perfect square trinomial. It is the square of the binomial 3x+4.

In this chapter, you will start with a perfect square trinomial and factor it into its prime factors.

You could factor this trinomial using the methods described in the last section, since it is of the form ax2+bx+c. But if you recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, you will save yourself a lot of work.

Here is the pattern—the reverse of the binomial squares pattern.

Perfect Square Trinomials Pattern

If a and b are real numbers

a2+2ab+b2=(a+b)2a2−2ab+b2=(a−b)2

To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, a2. Next check that the last term is a perfect square, b2. Then check the middle term—is it the product, 2ab? If everything checks, you can easily write the factors.

How to Factor Perfect Square Trinomials

Factor: 9x2+12x+4.

Solution
Step 1 is to check if the trinomial fits the perfect square trinomials pattern, a squared plus 2ab plus b squared. For this we check if the first term is a perfect square. 9 x squared is the square of 3x. Next we check if the last term is a perfect square. 4 is the square of 2. Next we check if the middle term is 2ab. 12 x is twice 3x times 2. Hence we have a perfect square trinomial. Step 2 is to write this as the square of a binomial. We write it as open parentheses 3x plus 2 close parentheses squared. Step 3 is to check by multiplying.

Factor: 4x2+12x+9.

Solution

(2x+3)2

Factor: 9y2+24y+16.

Solution

(3y+4)2

The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern a2−2ab+b2, which factors to (a−b)2.

The steps are summarized here.

Factor perfect square trinomials.

Step 1.Does the trinomial fit the pattern?a2+2ab+b2a2−2ab+b2 Is the first term a perfect square?(a)2(a)2 Write it as a square. Is the last term a perfect square?(a)2(b)2(a)2(b)2 Write it as a square. Check the middle term. Is it2ab?(a)2↘2·a·b↙(b)2(a)2↘2·a·b↙(b)2 Step 2.Write the square of the binomial.(a+b)2(a−b)2 Step 3.Check by multiplying.

We’ll work one now where the middle term is negative.

Factor: 81y2−72y+16.

Solution

The first and last terms are squares. See if the middle term fits the pattern of a perfect square trinomial. The middle term is negative, so the binomial square would be (a−b)2.

A mathematical expression shows the quadratic polynomial 81y^2 - 72y + 16, which is a perfect square trinomial.
Are the first and last terms perfect squares?     Two mathematical expressions are shown: (9y)^2 on the left and (4)^2 on the right.
Check the middle term. A mathematical image illustrates the expansion of a binomial squared. The expressions (9y)^2 and (4)^2 point to 2(9y)(4), which simplifies to 72y. This shows the calculation of the 2ab term.
Does it match (a−b)2? Yes. An example of a perfect square trinomial (a-b)^2 = a^2 - 2ab + b^2, specifically (9y)^2 - 2 * 9y * 4 + 4^2.
Write as the square of a binomial. The mathematical expression (9y-4)^2 is shown in black text on a white background. It represents the quantity (9y minus 4) raised to the power of 2, indicating that the binomial should be squared.
Check by multiplying:

(9y−4)2(9y)2−2·9y·4+4281y2−72y+16✓

Factor: 64y2−80y+25.

Solution

(8y−5)2

Factor: 16z2−72z+81.

Solution

(4z−9)2

The next example will be a perfect square trinomial with two variables.

Factor: 36x2+84xy+49y2.

Solution
A mathematical expression displaying the quadratic trinomial 36x^2 + 84xy + 49y^2.
Test each term to verify the pattern.    The perfect square trinomial formula (a^2 + 2ab + b^2) demonstrated with an example: (6x)^2 + 2(6x)(7y) + (7y)^2.
Factor. The mathematical expression (6x + 7y)    ² is centered on a white background, representing the square of a binomial.
Check by multiplying.

(6x+7y)2(6x)2+2·6x·7y+(7y)236x2+84xy+49y2✓

Factor: 49x2+84xy+36y2.

Solution

(7x+6y)2

Factor: 64m2+112mn+49n2.

Solution

(8m+7n)2

Remember the first step in factoring is to look for a greatest common factor. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.

Factor: 100x2y−80xy+16y.

Solution
The image displays the algebraic expression 100x^2y - 80xy + 16y. This is a trinomial with three terms, each containing variables x and y, and constant coefficients.
Is there a GCF? Yes, 4y, so factor it out.     The mathematical expression 4y(25x^2 - 20x + 4) is displayed.
Is this a perfect square trinomial?
Verify the pattern. A mathematical expression showing the expansion of a perfect square trinomial inside brackets, specifically 4y[(5x)^2 - 2 * 5x * 2 + 2^2], with a^2 - 2ab + b^2 annotated in red above it.
Factor. A mathematical expression showing 4y multiplied by the quantity (5x minus 2) squared, written as 4y(5x - 2)×2.

Remember: Keep the factor 4y in the final product.

Check:

4y(5x−2)24y[(5x)2−2·5x·2+22]4y(25x2−20x+4)100x2y−80xy+16y✓

Factor: 8x2y−24xy+18y.

Solution

2y(2x−3)2

Factor: 27p2q+90pq+75q.

Solution

3q(3p+5)2

Factor Differences of Squares

The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

We have open parentheses 3x minus 4 close parentheses open parentheses 3x plus 4. This is of the form a minus b, a plus b. We rewrite as open parentheses 3x close parentheses squared minus 4 squared. Here, 3x is a and 4 is b. This is equal to 9 x squared minus 16.

A difference of squares factors to a product of conjugates.

Difference of Squares Pattern

If a and b are real numbers,

a squared minus b squared equals a minus b, a plus b. Here, a squared minus b squared is difference of squares and a minus b, a plus b are conjugates.

Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.

How to Factor a Binomial Using the Difference of Squares

Factor: 64y2−1.

Solution
Step 1 is to check if the binomial 64 y squared minus 1 fits the pattern. For that we check the following: Is this a difference? Yes. Are the first and last terms perfect squares? Yes. Step 2 is to write both terms as squares, So, we have open parentheses 8y close parentheses squared minus 1 squared. Step 3 is to write the product of conjugates 8y minus 1, 8y plus 1. Step 4 is to check. We multiply to get the original binomial

Factor: 121m2−1.

Solution

(11m−1)(11m+1)

Factor: 81y2−1.

Solution

(9y−1)(9y+1)

Factor differences of squares.

Step 1.Does the binomial fit the pattern?a2−b2Is this a difference?____−____Are the first and last terms perfect squares?Step 2.Write them as squares.(a)2−(b)2Step 3.Write the product of conjugates.(a−b)(a+b)Step 4.Check by multiplying.

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression a2+b2 is prime!

The next example shows variables in both terms.

Factor: 144x2−49y2.

Solution
Step-by-step factorization of a difference of squares polynomial, including identification, factoring, and verification.
144x2−49y2
Is this a difference of squares? Yes. (12x)2−(7y)2
Factor as the product of conjugates. (12x−7y)(12x+7y)
Check by multiplying.
(12x−7y)(12x+7y)144x2−49y2✓

Factor: 196m2−25n2.

Solution

(14m−5n)(14m+5n)

Factor: 121p2−9q2.

Solution

(11p−3q)(11p+3q)

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.

Also, to completely factor the binomial in the next example, we’ll factor a difference of squares twice!

Factor: 48x4y2−243y2.

Solution
Step-by-step factorization of the polynomial 48x^4y^2 - 243y^2, demonstrating GCF and difference of squares methods, with a final verification.
48x4y2−243y2
Is there a GCF? Yes, 3y2—factor it out! 3y2(16x4−81)
Is the binomial a difference of squares? Yes. 3y2((4x2)2−(9)2)
Factor as a product of conjugates. 3y2(4x2−9)(4x2+9)
Notice the first binomial is also a difference of squares! 3y2((2x)2−(3)2)(4x2+9)
Factor it as the product of conjugates. 3y2(2x−3)(2x+3)(4x2+9)
The last factor, the sum of squares, cannot be factored.
Check by multiplying:
3y2(2x−3)(2x+3)(4x2+9)
3y2(4x2−9)(4x2+9)
3y2(16x4−81)
48x4y2−243y2✓

Factor: 2x4y2−32y2.

Solution

2y2(x−2)(x+2)(x2+4)

Factor: 7a4c2−7b4c2.

Solution

7c2(a−b)(a+b)(a2+b2)

The next example has a polynomial with 4 terms. So far, when this occurred we grouped the terms in twos and factored from there. Here we will notice that the first three terms form a perfect square trinomial.

Factor: x2−6x+9−y2.

Solution

Notice that the first three terms form a perfect square trinomial.

A mathematical expression displays x squared minus 6x plus 9 minus y squared.
Factor by grouping the first three terms. A mathematical expression showing x^2 - 6x + 9 - y^2, with a brace underneath the first three terms (x^2 - 6x + 9) indicating they form a group.
Use the perfect square trinomial pattern.      The mathematical expression (x-3)^2 - y^2 is displayed, representing the difference of two squares with a binomial as the first squared term.
Is this a difference of squares? Yes.
Yes—write them as squares. Two mathematical expressions are displayed: a^2 - b^2 in red text, and (x-3)^2 - y^2 in black text, both representing the difference of squares.
Factor as the product of conjugates. The image shows an algebraic expression `((x-3)-y)((x-3)+y)`, which is an application of the difference of squares formula (A-B)(A+B).
A mathematical expression showing the product of two binomials: (x - 3 - y)(x - 3 + y). This is an example of the difference of squares formula.

You may want to rewrite the solution as (x−y−3)(x+y−3).

Factor: x2−10x+25−y2.

Solution

(x−5−y)(x−5+y)

Factor: x2+6x+9−4y2.

Solution

(x+3−2y)(x+3+2y)

Factor Sums and Differences of Cubes

There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.

a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)

We’ll check the first pattern and leave the second to you.

The algebraic identity for the sum of cubes: (a + b)(a^2 - ab + b^2).
Distribute. A mathematical expression showing the sum of two terms: a multiplied by (a squared minus ab plus b squared) plus b multiplied by (a squared minus ab plus b squared). This simplifies to a cubed plus b cubed.
Multiply. A mathematical expression: a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3. This expression simplifies to a^3 + b^3.
Combine like terms. The mathematical expression a^3 + b^3 is displayed, representing the sum of two cubes.

Sum and Difference of Cubes Pattern

a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)

The two patterns look very similar, don’t they? But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. If you recognize the pattern of the signs, it may help you memorize the patterns.

a cubed plus b cubed is open parentheses a plus b close parentheses open parentheses a squared minus ab plus b squared close parentheses. a cubed minus b cubed is open parentheses a minus close parentheses open parentheses a squared plus ab plus b squared close parentheses. In both cases, the sign of the first term on the right side of the equation is the same as the sign on the left side of the equation and the sign of the second term is the opposite of the sign on the left side.

The trinomial factor in the sum and difference of cubes pattern cannot be factored.

It will be very helpful if you learn to recognize the cubes of the integers from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers from 1 to 10 in Table 8.

n 1 2 3 4 5 6 7 8 9 10
n3 1 8 27 64 125 216 343 512 729 1000

How to Factor the Sum or Difference of Cubes

Factor: x3+64.

Solution
Step 1 is to check if the binomial fits the sum or difference of cubes pattern. For this, we check whether it is a sum or difference. x cubed plus 64 is a sum. Next we check if the first and last terms are perfect cubes. They are Step 2 is to rewrite as cubes. So we rewrite as x cubed plus 4 cubed. Step 3 is to use either the sum or difference of cubes pattern. Since this is a sum of cubes, we get open parentheses x plus 4 close parentheses open parentheses x squared minus 4x plus 4 squared. Step 4 is to simplify inside the parentheses. It is already simplified Step 5 is to check by multiplying the factors.

Factor: x3+27.

Solution

(x+3)(x2−3x+9)

Factor: y3+8.

Solution

(y+2)(y2−2y+4)

Factor the sum or difference of cubes.

  1. Does the binomial fit the sum or difference of cubes pattern?
    Is it a sum or difference?
    Are the first and last terms perfect cubes?
  2. Write them as cubes.
  3. Use either the sum or difference of cubes pattern.
  4. Simplify inside the parentheses.
  5. Check by multiplying the factors.

Factor: 27u3−125v3.

Solution
The mathematical expression 27u^3 - 125v^3 is displayed in black text on a white background, representing the difference of two cubes.
This binomial is a difference. The first and last
terms are perfect cubes.
Write the terms as cubes. A mathematical image showing the difference of cubes identity: a³ - b³ in red, followed by a specific application of the formula: (3u)³ - (5v)³.
Use the difference of cubes pattern. An image displaying the algebraic identity for the difference of cubes, (a - b)(a^2 + ab + b^2), with a specific example below it where a=3u and b=5v.
Simplify. Two lines of algebraic expressions, illustrating the difference of cubes identity: (a - b)(a^2 + ab + b^2) and (3u - 5v)(9u^2 + 15uv + 25v^2).
Check by multiplying. We’ll leave the check to you.

Factor: 8x3−27y3.

Solution

(2x−3y)(4x2+6xy+9y2)

Factor: 1000m3−125n3.

Solution

125(4m2 + 2mn + n2)(2m – n)

In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.

Factor: 6x3y+48y4.

Solution
A mathematical expression 6x^3y + 48y^4 is displayed in black text on a white background, featuring variables x and y raised to powers, coefficients, and an addition operator.
Factor the common factor. A mathematical expression is displayed, which reads as 6y multiplied by the sum of x cubed and 8y cubed, written as 6y(x^3 + 8y^3).
This binomial is a sum The first and last
terms are perfect cubes.
Write the terms as cubes. A mathematical expression 6y(x^3 + (2y)^3), with a hint above in red indicating the sum of cubes formula, a^3 + b^3.
Use the sum of cubes pattern. A mathematical expression featuring 6y multiplied by two parenthetical terms: (x + 2y) and (x^2 - x * 2y + (2y)^2
Simplify. The algebraic expression 6y(x + 2y)(x^2 - 2xy + 4y^2), which is a factored form related to the sum of cubes, simplifying to 6y(x^3 + 8y^3).

Check:

To check, you may find it easier to multiply the sum of cubes factors first, then multiply that product by 6y. We’ll leave the multiplication for you.

Factor: 500p3+4q3.

Solution

4(5p+q)(25p2−5pq+q2)

Factor: 432c3+686d3.

Solution

2(6c+7d)(36c2−42cd+49d2)

The first term in the next example is a binomial cubed.

Factor: (x+5)3−64x3.

Solution
The image displays the mathematical expression (x+5)² - 64x³.
This binomial is a difference. The first and
last terms are perfect cubes.
Write the terms as cubes. The image displays the difference of cubes formula a^3 - b^3 in red, above an example expression (x + 5)^3 - (4x)^3, demonstrating its algebraic application.
Use the difference of cubes pattern. A mathematical expression illustrating the difference of cubes factorization (a-b)(a^2+ab+b^2), with 'a' representing (x+5) and 'b' representing 4x.
Simplify. A mathematical expression featuring the product of two polynomials: (x + 5 - 4x) multiplied by (x^2 + 10x + 25 + 4x^2 + 20x + 16x^2).
A mathematical expression showing the product of two polynomials: (-3x + 5) and (21x^2 + 30x + 25).
Check by multiplying. We’ll leave the check to you.

Factor: (y+1)3−27y3.

Solution

(−2y+1)(13y2+5y+1)

Factor: (n+3)3−125n3.

Solution

(−4n+3)(31n2+21n+9)

Access this online resource for additional instruction and practice with factoring special products.

  • Factoring Binomials-Cubes #2

Key Concepts

  • Perfect Square Trinomials Pattern: If a and b are real numbers,
    a2+2ab+b2=(a+b)2a2−2ab+b2=(a−b)2
  • How to factor perfect square trinomials.
    Step 1.Does the trinomial fit the pattern?a2+2ab+b2a2−2ab+b2 Is the first term a perfect square?(a)2(a)2 Write it as a square. Is the last term a perfect square?(a)2(b)2(a)2(b)2 Write it as a square. Check the middle term. Is it2ab?(a)2↘2·a·b↙(b)2(a)2↘2·a·b↙(b)2 Step 2.Write the square of the binomial.(a+b)2(a−b)2 Step 3.Check by multiplying.
  • Difference of Squares Pattern: If a,b are real numbers,
    a squared minus b squared is a minus b, a plus b. Here, a squared minus b squared is the difference of squares and a minus b, a plus b are conjugates.
  • How to factor differences of squares.
    Step 1.Does the binomial fit the pattern?a2−b2Is this a difference?____−____Are the first and last terms perfect squares?Step 2.Write them as squares.(a)2−(b)2Step 3.Write the product of conjugates.(a−b)(a+b)Step 4.Check by multiplying.
  • Sum and Difference of Cubes Pattern
    a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)
  • How to factor the sum or difference of cubes.
    1. Does the binomial fit the sum or difference of cubes pattern?
      Is it a sum or difference?
      Are the first and last terms perfect cubes?
    2. Write them as cubes.
    3. Use either the sum or difference of cubes pattern.
    4. Simplify inside the parentheses
    5. Check by multiplying the factors.

Practice Makes Perfect

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

16y2+24y+9

Solution

(4y+3)2

25v2+20v+4

36s2+84s+49

Solution

(6s+7)2

49s2+154s+121

100x2−20x+1

Solution

(10x−1)2

64z2−16z+1

25n2−120n+144

Solution

(5n−12)2

4p2−52p+169

49x2+28xy+4y2

Solution

(7x+2y)2

25r2+60rs+36s2

100y2−20y+1

Solution

(10y−1)2

64m2−16m+1

10jk2+80jk+160j

Solution

10j(k+4)2

64x2y−96xy+36y

75u4−30u3v+3u2v2

Solution

3u2(5u−v)2

90p4+300p3q+250p2q2

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

25v2−1

Solution

(5v−1)(5v+1)

169q2−1

4−49x2

Solution

(2−7x)(2+7x)

121−25s2

6p2q2−54p2

Solution

6p2(q−3)(q+3)

98r3−72r

24p2+54

Solution

6(4p2+9)

20b2+140

121x2−144y2

Solution

(11x−12y)(11x+12y)

49x2−81y2

169c2−36d2

Solution

(13c−6d)(13c+6d)

36p2−49q2

16z4−1

Solution

(2z−1)(2z+1)(4z2+1)

m4−n4

162a4b2−32b2

Solution

2b2(3a−2)(3a+2)(9a2+4)

48m4n2−243n2

x2−16x+64−y2

Solution

(x−8−y)(x−8+y)

p2+14p+49−q2

a2+6a+9−9b2

Solution

(a+3−3b)(a+3+3b)

m2−6m+9−16n2

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

x3+125

Solution

(x+5)(x2−5x+25)

n6+512

z6−27

Solution

(z2−3)(z4+3z2+9)

v3−216

8−343t3

Solution

(2−7t)(4+14t+49t2)

125−27w3

8y3−125z3

Solution

(2y−5z)(4y2+10yz+25z2)

27x3−64y3

216a3+125b3

Solution

(6a+5b)(36a2−30ab+25b2)

27y3+8z3

7k3+56

Solution

7(k+2)(k2−2k+4)

6x3−48y3

2x2−16x2y3

Solution

2x2(1−2y)(1+2y+4y2)

−2x3y2−16y5

(x+3)3+8x3

Solution

9(x+1)(x2+3)

(x+4)3−27x3

(y−5)3−64y3

Solution

−(3y+5)(21y2−30y+25)

(y−5)3+125y3

Mixed Practice

In the following exercises, factor completely.

64a2−25

Solution

(8a−5)(8a+5)

121x2−144

27q2−3

Solution

3(3q−1)(3q+1)

4p2−100

16x2−72x+81

Solution

(4x−9)2

36y2+12y+1

8p2+2

Solution

2(4p2+1)

81x2+169

125−8y3

Solution

(5−2y)(25+10y+4y2)

27u3+1000

45n2+60n+20

Solution

5(3n+2)2

48q3−24q2+3q

x2−10x+25−y2

Solution

(x−5−y)(x−5+y)

x2+12x+36−y2

(x+1)3+8x3

Solution

(3x+1)(3x2+1)

(y−3)3−64y3

Writing Exercises

Why was it important to practice using the binomial squares pattern in the chapter on multiplying polynomials?

Solution

Answers will vary.

How do you recognize the binomial squares pattern?

Explain why n2+25≠(n+5)2. Use algebra, words, or pictures.

Solution

Answers will vary.

Maribel factored y2−30y+81 as (y−9)2. Was she right or wrong? How do you know?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: factor perfect square trinomials, factor differences of squares, factor sums and differences of cubes. The remaining columns are blank.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

General Strategy for Factoring Polynomials

Learning Objectives

By the end of this section, you will be able to:

  • Recognize and use the appropriate method to factor a polynomial completely

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

You have now become acquainted with all the methods of factoring that you will need in this course. The following chart summarizes all the factoring methods we have covered, and outlines a strategy you should use when factoring polynomials.

General Strategy for Factoring Polynomials

This chart shows the general strategies for factoring polynomials. It shows ways to find GCF of binomials, trinomials and polynomials with more than 3 terms. For binomials, we have difference of squares: a squared minus b squared equals a minus b, a plus b; sum of squares do not factor; sub of cubes: a cubed plus b cubed equals open parentheses a plus b close parentheses open parentheses a squared minus ab plus b squared close parentheses; difference of cubes: a cubed minus b cubed equals open parentheses a minus b close parentheses open parentheses a squared plus ab plus b squared close parentheses. For trinomials, we have x squared plus bx plus c where we put x as a term in each factor and we have a squared plus bx plus c. Here, if a and c are squares, we have a plus b whole squared equals a squared plus 2 ab plus b squared and a minus b whole squared equals a squared minus 2 ab plus b squared. If a and c are not squares, we use the ac method. For polynomials with more than 3 terms, we use grouping.

Use a general strategy for factoring polynomials.

  1. Is there a greatest common factor?
    Factor it out.
  2. Is the polynomial a binomial, trinomial, or are there more than three terms?
    If it is a binomial:
    • Is it a sum?
      Of squares? Sums of squares do not factor.
      Of cubes? Use the sum of cubes pattern.
    • Is it a difference?
      Of squares? Factor as the product of conjugates.
      Of cubes? Use the difference of cubes pattern.
    If it is a trinomial:
    • Is it of the form x2+bx+c? Undo FOIL.
    • Is it of the form ax2+bx+c?
      If a and c are squares, check if it fits the trinomial square pattern.
      Use the trial and error or “ac” method.
    If it has more than three terms:
    • Use the grouping method.
  3. Check.
    Is it factored completely?
    Do the factors multiply back to the original polynomial?

Remember, a polynomial is completely factored if, other than monomials, its factors are prime!

Factor completely: 7x3−21x2−70x.

Solution
This table demonstrates the step-by-step process of factoring a polynomial completely, including the verification of the factored form.
7x3−21x2−70x
Is there a GCF? Yes, 7x.
Factor out the GCF. 7x(x2−3x−10)
In the parentheses, is it a binomial, trinomial, or are there more terms?
Trinomial with leading coefficient 1.
“Undo” FOIL. 7x(x)(x)
7x(x+2)(x−5)
Is the expression factored completely? Yes.
Neither binomial can be factored.
Check your answer.
Multiply.
7x(x+2)(x−5)
7x(x2−5x+2x−10)
7x(x2−3x−10)
7x3−21x2−70x✓

Factor completely: 8y3+16y2−24y.

Solution

8y(y−1)(y+3)

Factor completely: 5y3−15y2−270y.

Solution

5y(y−9)(y+6)

Be careful when you are asked to factor a binomial as there are several options!

Factor completely: 24y2−150.

Solution
This table details the step-by-step factorization of the polynomial 24y^2 - 150, showcasing GCF extraction, difference of squares, and result verification.
24y2−150
Is there a GCF? Yes, 6.
Factor out the GCF. 6(4y2−25)
In the parentheses, is it a binomial, trinomial or are there more than three terms? Binomial.
Is it a sum? No.
Is it a difference? Of squares or cubes? Yes, squares. 6((2y)2−(5)2)
Write as a product of conjugates. 6(2y−5)(2y+5)
Is the expression factored completely?
Neither binomial can be factored.
Check:
Multiply.
6(2y−5)(2y+5)
6(4y2−25)
24y2−150✓

Factor completely: 16x3−36x.

Solution

4x(2x−3)(2x+3)

Factor completely: 27y2−48.

Solution

3(3y−4)(3y+4)

The next example can be factored using several methods. Recognizing the trinomial squares pattern will make your work easier.

Factor completely: 4a2−12ab+9b2.

Solution
Step-by-step guide to factoring the perfect square trinomial 4a^2 - 12ab + 9b^2, including identification, factorization, and verification.
4a2−12ab+9b2
Is there a GCF? No.
Is it a binomial, trinomial, or are there more terms?
Trinomial with a≠1. But the first term is a perfect square.
Is the last term a perfect square? Yes. (2a)2−12ab+(3b)2
Does it fit the pattern, a2−2ab+b2? Yes. (2a)2↘−12ab+−2(2a)(3b)↙(3b)2
Write it as a square. (2a−3b)2
Is the expression factored completely? Yes.
The binomial cannot be factored.
Check your answer.
Multiply.
(2a−3b)2
(2a)2−2·2a·3b+(3b)2
4a2−12ab+9b2✓

Factor completely: 4x2+20xy+25y2.

Solution

(2x+5y)2

Factor completely: 9x2−24xy+16y2.

Solution

(3x−4y)2

Remember, sums of squares do not factor, but sums of cubes do!

Factor completely 12x3y2+75xy2.

Solution
Step-by-step process for factoring the algebraic expression 12x^3y^2 + 75xy^2, including finding the GCF and verifying the result.
12x3y2+75xy2
Is there a GCF? Yes, 3xy2.
Factor out the GCF. 3xy2(4x2+25)
In the parentheses, is it a binomial, trinomial, or are there more than three terms? Binomial.
Is it a sum? Of squares? Yes. Sums of squares are prime.
Is the expression factored completely? Yes.
Check:
Multiply.
3xy2(4x2+25)
12x3y2+75xy2✓

Factor completely: 50x3y+72xy.

Solution

2xy(25x2+36)

Factor completely: 27xy3+48xy.

Solution

3xy(9y2+16)

When using the sum or difference of cubes pattern, being careful with the signs.

Factor completely: 24x3+81y3.

Solution
Is there a GCF? Yes, 3. The mathematical expression 24x³ + 81y³ is displayed.
Factor it out. A mathematical expression reads 3 multiplied by the sum of 8x cubed and 27y cubed, enclosed in parentheses. The expression is 3(8x^3 + 27y^3) on a white background.
In the parentheses, is it a binomial, trinomial,
of are there more than three terms? Binomial.
Is it a sum or difference? Sum.
Of squares or cubes? Sum of cubes. Mathematical expression: 3((2x) ^3 + (3y)^3). Red a^3 and b^3 above the terms suggest the sum of cubes formula, a^3 + b^3.
Write it using the sum of cubes pattern. A mathematical expression 3(2x + 3y)((2x)^2 - 2x * 3y + (3y)^3) is shown. Red annotations 'a', 'b', 'a^2', 'ab', 'b^2' suggest an algebraic identity, but the last term is cubed, not squared.
Is the expression factored completely? Yes. A mathematical expression showing the factorization 3(2x + 3y)(4x^2 - 6xy + 9y^2), which simplifies to 3(8x^3 + 27y^3).
Check by multiplying.

Factor completely: 250m3+432n3.

Solution

2(5m+6n)(25m2−30mn+36n2)

Factor completely: 2p3+54q3.

Solution

2(p+3q)(p2−3pq+9q2)

Factor completely: 3x5y−48xy.

Solution
This table demonstrates the step-by-step factorization of the polynomial 3x^5y - 48xy, concluding with a verification of the result.
3x5y−48xy
Is there a GCF? Factor out 3xy 3xy(x4−16)
Is the binomial a sum or difference? Of squares or cubes?
Write it as a difference of squares.
3xy((x2)2−(4)2)
Factor it as a product of conjugates 3xy(x2−4)(x2+4)
The first binomial is again a difference of squares. 3xy((x)2−(2)2)(x2+4)
Factor it as a product of conjugates. 3xy(x−2)(x+2)(x2+4)
Is the expression factored completely? Yes.
Check your answer.
Multiply.
3xy(x−2)(x+2)(x2+4)
3xy(x2−4)(x2+4)
3xy(x4−16)
3x5y−48xy✓

Factor completely: 4a5b−64ab.

Solution

4ab(a2+4)(a−2)(a+2)

Factor completely: 7xy5−7xy.

Solution

7xy(y2+1)(y−1)(y+1)

Factor completely: 4x2+8bx−4ax−8ab.

Solution
Step-by-step factorization of the polynomial 4x^2 + 8bx - 4ax - 8ab, demonstrating GCF extraction, grouping, and solution verification.
4x2+8bx−4ax−8ab
Is there a GCF? Factor out the GCF, 4. 4(x2+2bx−ax−2ab)
There are four terms. Use grouping. 4[x(x+2b)−a(x+2b)]4(x+2b)(x−a)
Is the expression factored completely? Yes.
Check your answer.
Multiply.
4(x+2b)(x−a)4(x2−ax+2bx−2ab)4x2+8bx−4ax−8ab✓

Factor completely: 6x2−12xc+6bx−12bc.

Solution

6(x+b)(x−2c)

Factor completely: 16x2+24xy−4x−6y.

Solution

2(4x−1)(2x+3y)

Taking out the complete GCF in the first step will always make your work easier.

Factor completely: 40x2y+44xy−24y.

Solution
Step-by-step guide to factoring the polynomial 40x^2y + 44xy - 24y, showing GCF and trinomial factorization, and verification.
40x2y+44xy−24y
Is there a GCF? Factor out the GCF, 4y. 4y(10x2+11x−6)
Factor the trinomial with a≠1. 4y(10x2+11x−6)
4y(5x−2)(2x+3)
Is the expression factored completely? Yes.
Check your answer.
Multiply.
4y(5x−2)(2x+3)
4y(10x2+11x−6)
40x2y+44xy−24y✓

Factor completely: 4p2q−16pq+12q.

Solution

4q(p−3)(p−1)

Factor completely: 6pq2−9pq−6p.

Solution

3p(2q+1)(q−2)

When we have factored a polynomial with four terms, most often we separated it into two groups of two terms. Remember that we can also separate it into a trinomial and then one term.

Factor completely: 9x2−12xy+4y2−49.

Solution
Detailed steps for factoring the algebraic expression 9x^2 - 12xy + 4y^2 - 49, identifying a perfect square trinomial and difference of squares, including verification.
9x2−12xy+4y2−49
Is there a GCF? No.
With more than 3 terms, use grouping. Last 2 terms have no GCF. Try grouping first 3 terms. 9x2−12xy+4y2−49
Factor the trinomial with a≠1. But the first term is a perfect square.
Is the last term of the trinomial a perfect square? Yes. (3x)2−12xy+(2y)2−49
Does the trinomial fit the pattern, a2−2ab+b2? Yes. (3x)2↘−12xy+−2(3x)(2y)↙(2y)2−49
Write the trinomial as a square. (3x−2y)2−49
Is this binomial a sum or difference? Of squares or cubes? Write it as a difference of squares. (3x−2y)2−72
Write it as a product of conjugates. ((3x−2y)−7)((3x−2y)+7)
(3x−2y−7)(3x−2y+7)
Is the expression factored completely? Yes.
Check your answer.
Multiply.
(3x−2y−7)(3x−2y+7)
9x2−6xy−21x−6xy+4y2+14y+21x−14y−49
9x2−12xy+4y2−49✓

Factor completely: 4x2−12xy+9y2−25.

Solution

(2x−3y−5)(2x−3y+5)

Factor completely: 16x2−24xy+9y2−64.

Solution

(4x−3y−8)(4x−3y+8)

Key Concepts

This chart shows the general strategies for factoring polynomials. It shows ways to find GCF of binomials, trinomials and polynomials with more than 3 terms. For binomials, we have difference of squares: a squared minus b squared equals a minus b, a plus b; sum of squares do not factor; sub of cubes: a cubed plus b cubed equals open parentheses a plus b close parentheses open parentheses a squared minus ab plus b squared close parentheses; difference of cubes: a cubed minus b cubed equals open parentheses a minus b close parentheses open parentheses a squared plus ab plus b squared close parentheses. For trinomials, we have x squared plus bx plus c where we put x as a term in each factor and we have a squared plus bx plus c. Here, if a and c are squares, we have a plus b whole squared equals a squared plus 2 ab plus b squared and a minus b whole squared equals a squared minus 2 ab plus b squared. If a and c are not squares, we use the ac method. For polynomials with more than 3 terms, we use grouping.
  • How to use a general strategy for factoring polynomials.
    1. Is there a greatest common factor?
      Factor it out.
    2. Is the polynomial a binomial, trinomial, or are there more than three terms?
      If it is a binomial:
      Is it a sum?
      Of squares? Sums of squares do not factor.
      Of cubes? Use the sum of cubes pattern.
      Is it a difference?
      Of squares? Factor as the product of conjugates.
      Of cubes? Use the difference of cubes pattern.
      If it is a trinomial:
      Is it of the form x2+bx+c? Undo FOIL.
      Is it of the form ax2+bx+c?
      If a and c are squares, check if it fits the trinomial square pattern.
      Use the trial and error or “ac” method.
      If it has more than three terms:
      Use the grouping method.
    3. Check.
      Is it factored completely?
      Do the factors multiply back to the original polynomial?

Practice Makes Perfect

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

2n2+13n−7

Solution

(2n−1)(n+7)

8x2−9x−3

a5+9a3

Solution

a3(a2+9)

75m3+12m

121r2−s2

Solution

(11r−s)(11r+s)

49b2−36a2

8m2−32

Solution

8(m−2)(m+2)

36q2−100

25w2−60w+36

Solution

(5w−6)2

49b2−112b+64

m2+14mn+49n2

Solution

(m+7n)2

64x2+16xy+y2

7b2+7b−42

Solution

7(b+3)(b−2)

30n2+30n+72

3x4y−81xy

Solution

3xy(x−3)(x2+3x+9)

4x5y−32x2y

k4−16

Solution

(k−2)(k+2)(k2+4)

m4−81

5x5y2−80xy2

Solution

5xy2(x2+4)(x+2)(x−2)

48x5y2−243xy2

15pq−15p+12q−12

Solution

3(5p+4)(q−1)

12ab−6a+10b−5

4x2+40x+84

Solution

4(x+3)(x+7)

5q2−15q−90

4u5+4u2v3

Solution

4u2(u+v)(u2−uv+v2)

5m4n+320mn4

4c2+20cd+81d2

Solution

prime

25x2+35xy+49y2

10m4−6250

Solution

10(m−5)(m+5)(m2+25)

3v4−768

36x2y+15xy−6y

Solution

3y(3x+2)(4x−1)

60x2y−75xy+30y

8x3−27y3

Solution

(2x−3y)(4x2+6xy+9y2)

64x3+125y3

y6−1

Solution

(y+1)(y−1)(y2−y+1)(y2+y+1)

y6+1

9x2−6xy+y2−49

Solution

(3x−y+7)(3x−y−7)

16x2−24xy+9y2−64

(3x+1)2−6(3x+1)+9

Solution

(3x−2)2

(4x−5)2−7(4x−5)+12

Writing Exercises

Explain what it mean to factor a polynomial completely.

Solution

Answers will vary.

The difference of squares y4−625 can be factored as (y2−25)(y2+25). But it is not completely factored. What more must be done to completely factor.

Of all the factoring methods covered in this chapter (GCF, grouping, undo FOIL, ‘ac’ method, special products) which is the easiest for you? Which is the hardest? Explain your answers.

Solution

Answers will vary.

Create three factoring problems that would be good test questions to measure your knowledge of factoring. Show the solutions.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 1 row and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statement: recognize and use the appropriate method to factor a polynomial completely. The remaining columns are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Polynomial Equations

Learning Objectives

By the end of this section, you will be able to:

  • Use the Zero Product Property
  • Solve quadratic equations by factoring
  • Solve equations with polynomial functions
  • Solve applications modeled by polynomial equations

Before you get started, take this readiness quiz.

Solve: 5y−3=0.
If you missed this problem, review Example 2 in Use a General Strategy to Solve Linear Equations.

Solution

y=35

Factor completely: n3−9n2−22n.
If you missed this problem, review Example 7 in Relations and Functions.

Solution

n(n−11)(n+2)

If f(x)=8x−16, find f(3) and solve f(x)=0.
If you missed this problem, review Example 9 in Graphs of Functions.

Solution

8;x=2

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one. Polynomial equations of degree one are linear equations are of the form ax+b=c.

We are now going to solve polynomial equations of degree two. A polynomial equation of degree two is called a quadratic equation. Listed below are some examples of quadratic equations:

x2+5x+6=03y2+4y=1064u2−81=0n(n+1)=42

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get n2+n.

The general form of a quadratic equation is ax2+bx+c=0, with a≠0. (If a=0, then 0·x2=0 and we are left with no quadratic term.)

Quadratic Equation

An equation of the form ax2+bx+c=0 is called a quadratic equation.

a,b,andcare real numbers anda≠0

To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

Use the Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Zero Product Property

If a·b=0, then either a=0 or b=0 or both.

We will now use the Zero Product Property, to solve a quadratic equation.

How to Solve a Quadratic Equation Using the Zero Product Property

Solve: (5n−2)(6n−1)=0.

Solution
The equation is open parentheses 5n minus 2 close parentheses open parentheses 6n minus 1 close parentheses equals 0. The product equals zero, so at least one factor must equal zero. Step 1 is set each factor equal to zero. So, 5n minus 2 equals 0 and 6n minus 1 equals 0. Step 2 is to solve the linear equations. So, we get n equal to 2 by 5 and n equal to 1 by 6. Step 3 is to check by substituting each solution separately into the original equation.

Solve: (3m−2)(2m+1)=0.

Solution

m=23,m=−12

Solve: (4p+3)(4p−3)=0.

Solution

p=−34,p=34

Use the Zero Product Property.

  1. Set each factor equal to zero.
  2. Solve the linear equations.
  3. Check.

Solve Quadratic Equations by Factoring

The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we must be sure to start with the quadratic equation in standard form, ax2+bx+c=0. Then we must factor the expression on the left.

How to Solve a Quadratic Equation by Factoring

Solve: 2y2=13y+45.

Solution
The equation is 2 y squared equals 13y plus 45. Step 1 is to write it in standard form a x squared plus bx plus c. So we have 2 y squared minus 13y minus 45 equals 0. Step 2 is to factor the quadratic expression. So we have 2y plus 5, y minus 9 equals 0. Step 3 is to use the zero product property. Setting each factor equal to zero, we have two linear equations: 2y plus 5 equals 0 and y minus 9 equals 0. Step 4 is to solve the linear equations. We get, y equals minus 5 by 2 and y equals 9. Step 5 is to check by substituting each solution separately into the original equation

Solve: 3c2=10c−8.

Solution

c=2,c=43

Solve: 2d2−5d=3.

Solution

d=3,d=−12

Solve a quadratic equation by factoring.

  1. Write the quadratic equation in standard form, ax2+bx+c=0.
  2. Factor the quadratic expression.
  3. Use the Zero Product Property.
  4. Solve the linear equations.
  5. Check. Substitute each solution separately into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form.

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Solve: 169x2=49.

Solution
Step-by-step solution for the quadratic equation 169x^2 = 49 using factoring and the Zero Product Property.
169x2=49
Write the quadratic equation in standard form. 169x2−49=0
Factor. It is a difference of squares. (13x−7)(13x+7)=0
Use the Zero Product Property to set each factor to 0.
Solve each equation.
13x−7=013x+7=013x=713x=−7x=713x=−713

Check:

We leave the check up to you.

Solve: 25p2=49.

Solution

p=75,p=−75

Solve: 36x2=121.

Solution

x=116,x=−116

In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Solve: (3x−8)(x−1)=3x.

Solution
This table demonstrates the step-by-step process of solving a quadratic equation by factoring, showing each mathematical transformation.
(3x−8)(x−1)=3x
Multiply the binomials. 3x2−11x+8=3x
Write the quadratic equation in standard form. 3x2−14x+8=0
Factor the trinomial. (3x−2)(x−4)=0
Use the Zero Product Property to set each factor to 0.
Solve each equation.
3x−2=0x−4=03x=2x=4
x=23
Check your answers. The check is left to you.

Solve: (2m+1)(m+3)=12m.

Solution

m=1,m=32

Solve: (k+1)(k−1)=8.

Solution

k=3,k=−3

In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Solve: 3x2=12x+63.

Solution
Step-by-step method for solving a quadratic equation by factoring, with corresponding mathematical expressions.
3x2=12x+63
Write the quadratic equation in standard form. 3x2−12x−63=0
Factor the greatest common factor first. 3(x2−4x−21)=0
Factor the trinomial. 3(x−7)(x+3)=0
Use the Zero Product Property to set each factor to 0.
Solve each equation.
3≠0x−7=0x+3=03≠0x=7x=−3
Check your answers. The check is left to you.

Solve: 18a2−30=−33a.

Solution

a=−52,a=23

Solve: 123b=−6−60b2.

Solution

b=−2,b=−120

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.

Solve: 9m3+100m=60m2.

Solution
Solution steps for the cubic equation 9m^3 + 100m = 60m^2, demonstrating factoring and the Zero Product Property.
9m3+100m=60m2
Bring all the terms to one side so that the other side is zero. 9m3−60m2+100m=0
Factor the greatest common factor first. m(9m2−60m+100)=0
Factor the trinomial. m(3m−10)(3m−10)=0
Use the Zero Product Property to set each factor to 0.
Solve each equation.
m=03m−10=03m−10=0m=0m=103m=103
Check your answers. The check is left to you.

Solve: 8x3=24x2−18x.

Solution

x=0,x=32

Solve: 16y2=32y3+2y.

Solution

y=0,y=14

Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

For the function f(x)=x2+2x−2,

ⓐ find x when f(x)=6 ⓑ find two points that lie on the graph of the function.

Solution
ⓐ
Demonstrates solving a quadratic equation (x^2 + 2x - 2 = 6) by factoring, with steps and solution verification.
f(x)=x2+2x−2
Substitute 6 for f(x). 6=x2+2x−2
Put the quadratic in standard form. x2+2x−8=0
Factor the trinomial. (x+4)(x−2)=0
Use the zero product property.
Solve.
x+4=0orx−2=0x=−4orx=2
Check:
f(x)=x2+2x−2f(x)=x2+2x−2f(−4)=(−4)2+2(−4)−2f(2)=22+2·2−2f(−4)=16−8−2f(2)=4+4−2f(−4)=6✓f(2)=6✓


ⓑ Since f(−4)=6 and f(2)=6, the points (−4,6) and (2,6) lie on the graph of the function.

For the function f(x)=x2−2x−8,

ⓐ find x when f(x)=7 ⓑ Find two points that lie on the graph of the function.

Solution

ⓐ x=−3 or x=5
ⓑ (−3,7) (5,7)

For the function f(x)=x2−8x+3,

ⓐ find x when f(x)=−4 ⓑ Find two points that lie on the graph of the function.

Solution

ⓐ x=1 or x=7
ⓑ (1,−4) (7,−4)

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function.

Zero of a Function

For any function f, if f(x)=0, then x is a zero of the function.

When f(x)=0, the point (x,0) is a point on the graph. This point is an x-intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.

For the function f(x)=3x2+10x−8, find

ⓐ the zeros of the function, ⓑ any x-intercepts of the graph of the function, ⓒ any y-intercepts of the graph of the function

Solution
ⓐ To find the zeros of the function, we need to find when the function value is 0.
Step-by-step solution to the quadratic equation 3x^2 + 10x - 8 = 0 by factoring, showing the process from substitution to finding the roots.
f(x)=3x2+10x−8
Substitute 0 for f(x). 0=3x2+10x−8
Factor the trinomial. (x+4)(3x−2)=0
Use the zero product property.
Solve.
x+4=0or3x−2=0x=−4orx=23


ⓑ An x-intercept occurs when y=0. Since f(−4)=0 and f(23)=0, the points (−4,0) and (23,0) lie on the graph. These points are x-intercepts of the function.


ⓒ A y-intercept occurs when x=0. To find the y-intercepts we need to find f(0).
Step-by-step evaluation of the function f(x) = 3x^2 + 10x - 8 for x=0, demonstrating substitution and simplification to find f(0).
f(x)=3x2+10x−8
Find f(0) by substituting 0 for x. f(0)=3·02+10·0−8
Simplify. f(0)=−8

Since f(0)=−8, the point (0,−8) lies on the graph. This point is the y-intercept of the function.

For the function f(x)=2x2−7x+5, find

ⓐ the zeros of the function, ⓑ any x-intercepts of the graph of the function, ⓒ any y-intercepts of the graph of the function.

Solution

ⓐ x=1 or x=52
ⓑ (1,0), (52,0) ⓒ (0,5)

For the function f(x)=6x2+13x−15, find

ⓐ the zeros of the function, ⓑ any x-intercepts of the graph of the function, ⓒ any y-intercepts of the graph of the function.

Solution

ⓐ x=−3 or x=56
ⓑ (−3,0), (56,0) ⓒ (0,−15)

Solve Applications Modeled by Polynomial Equations

The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.

Use a problem solving strategy to solve word problems.

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are looking for.
  3. Name what we are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
  5. Solve the equation using appropriate algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a polynomial equation.

The product of two consecutive odd integers is 323. Find the integers.

Solution
This table outlines the step-by-step process for solving a word problem to find consecutive odd integers whose product is 323, including variable definition, equation setup, and solution verification.
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for two consecutive integers.
Step 3. Name what we are looking for. Let n=the first integer.
n+2=next consecutive odd integer
Step 4. Translate into an equation. Restate the problem in a sentence. The product of the two consecutive odd integers is 323.
n(n+2)=323
Step 5. Solve the equation. n2+2n=323
Bring all the terms to one side. n2+2n−323=0
Factor the trinomial. (n−17)(n+19)=0
Use the Zero Product Property.
Solve the equations.
n−17=0n+19=0n=17n=−19
There are two values for n that are solutions to this problem. So there are two sets of consecutive odd integers that will work.
If the first integer is n=17 If the first integer is n=−19
then the next odd integer is then the next odd integer is
n+2 n+2
17+2 −19+2
19 −17
17,19 −17,−19
Step 6. Check the answer.
The results are consecutive odd integers
17,19and−19,−17.
17·19=323✓−19(−17)=323✓
Both pairs of consecutive integers are solutions.
Step 7. Answer the question The consecutive integers are 17, 19 and −19,−17.

The product of two consecutive odd integers is 255. Find the integers.

Solution

−15,−17 and 15, 17

The product of two consecutive odd integers is 483 Find the integers.

Solution

−23,−21 and 21, 23

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

Solution
Step 1. Read the problem. In problems involving
geometric figures, a sketch can help you visualize
the situation.
A rectangle is shown with its width labeled 'w' and its length labeled 'w + 4', indicating the dimensions are expressed in terms of variable 'w'.
Step 2. Identify what you are looking for. We are looking for the length and width.
Step 3. Name what you are looking for. Let w= the width of the bedroom.
The length is four feet more than the width. w+4= the length of the garden
Step 4. Translate into an equation.
Restate the important information in a sentence. The area of the bedroom is 117 square feet.
Use the formula for the area of a rectangle. A=l·w
Substitute in the variables. 117=(w+4)w
Step 5. Solve the equation Distribute first. 117=w2+4w
Get zero on one side. 117=w2+4w
Factor the trinomial. 0=w2+4w−117
Use the Zero Product Property. 0=(w+13)(w−9)
Solve each equation. 0=w+130=w−9
Since w is the width of the bedroom, it does not
make sense for it to be negative. We eliminate that value for w.
−13=w9=w
w=9 Width is 9 feet.     
Find the value of the length. w+4
9+4
13  Length is 13 feet.
Step 6. Check the answer.
Does the answer make sense?

An image illustrating the calculation of a rectangle's area. The width is 9 and the length is 13 (derived from w+4, where w=9). The area is calculated as 13 * 9 = 117.
Yes, this makes sense.
Step 7. Answer the question. The width of the bedroom is 9 feet and
the length is 13 feet.

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

Solution

The width is 5 feet and length is 6 feet.

A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

Solution

The width of the patio is 12 feet and the length is 15 feet.

In the next example, we will use the Pythagorean Theorem (a2+b2=c2). This formula gives the relation between the legs and the hypotenuse of a right triangle.

Figure shows a right triangle with the shortest side being a, the second side being b and the hypotenuse being c.

We will use this formula in the next example.

A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.

Figure shows a right triangle with the shortest side being x, the second side being x minus 7 and the hypotenuse being 17.
Solution
Step 1. Read the problem
Step 2. Identify what you are looking for. We are looking for the lengths of the
sides of the sail.
Step 3. Name what you are looking for.
One side is 7 less than the other.
Let x= length of a side of the sail.
x−7= length of other side
Step 4. Translate into an equation. Since this is a
right triangle we can use the Pythagorean Theorem.
a2+b2=c2
Substitute in the variables. x2+(x−7)2=172
Step 5. Solve the equation
Simplify.
x2+x2−14x+49=289
2x2−14x+49=289
It is a quadratic equation, so get zero on one side. 2x2−14x−240=0
Factor the greatest common factor. 2(x2−7x−120)=0
Factor the trinomial. 2(x−15)(x+8)=0
Use the Zero Product Property. 2≠0x−15=0x+8=0
Solve. 2≠0x=15x=−8
Since x is a side of the triangle, x=−8 does not
make sense.
2≠0x=15x=−8
Find the length of the other side.
   If the length of one side is
   then the length of the other side is
The image shows the mathematical equation 'x = 15' centered on a plain white background.
The mathematical expression 'X - 7' is displayed in grayscale against a white background.
The numbers '15' in red and '7' in black, separated by a black dash, on a white background. This appears to be a mathematical expression or part of a scoring display.
8 is the length of the other side.
Step 6. Check the answer in the problem
Do these numbers make sense?

This image illustrates the Pythagorean theorem in action with a right triangle. The sides 8, 15, and 17 form a Pythagorean triple, as 8^2 + 15^2 = 64 + 225 = 289, which equals 17^2.
Step 7. Answer the question The sides of the sail are 8, 15 and 17 feet.

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

Solution

5 feet and 12 feet

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.

Solution

The other leg is 24 feet and the hypotenuse is 25 feet.

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function h(t)=−16t2+64t+80 models the height, h, of the ball above the ground as a function of time, t. Find:

ⓐ the zeros of this function which tell us when the ball hits the ground, ⓑ when the ball will be 80 feet above the ground, ⓒ the height of the ball at t=2 seconds.

Solution
ⓐ The zeros of this function are found by solving h(t)=0. This will tell us when the ball will hit the ground.
Step-by-step solution of the quadratic equation -16t^2 + 64t + 80 = 0 by factoring.
h(t)=0
Substitute in the polynomial for h(t). −16t2+64t+80=0
Factor the GCF, −16. −16(t2−4t−5)=0
Factor the trinomial. −16(t−5)(t+1)=0
Use the Zero Product Property.
Solve.
t−5=0t+1=0t=5t=−1

The result t=5 tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result t=−1 is discarded.

ⓑ The ball will be 80 feet above the ground when h(t)=80.
This table details the step-by-step algebraic solution for determining when a projectile's height, h(t), reaches 80 feet, providing the equations and the interpretation of results.
h(t)=80
Substitute in the polynomial for h(t). −16t2+64t+80=80
Subtract 80 from both sides. −16t2+64t=0
Factor the GCF, −16t. −16t(t−4)=0
Use the Zero Product Property.
Solve.
−16t=0t−4=0t=0t=4
The ball will be at 80 feet the moment Dennis tosses the ball and then 4 seconds later, when the ball is falling.
ⓒ To find the height ball at t=2 seconds we find h(2).
This table demonstrates the step-by-step evaluation of the function h(t) = -16t^2 + 64t + 80 at t=2, determining the height of a ball.
h(t)=−16t2+64t+80
To find h(2) substitute 2 for t. h(2)=−16(2)2+64·2+80
Simplify. h(2)=144
After 2 seconds, the ball will be at 144 feet.

Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function h(t)=−16t2+48t+160 models the height, h, of the rock above the ocean as a function of time, t. Find:

ⓐ the zeros of this function which tell us when the rock will hit the ocean, ⓑ when the rock will be 160 feet above the ocean, ⓒ the height of the rock at t=1.5 seconds.

Solution

ⓐ 5 seconds; ⓑ 0 and 3 seconds; ⓒ 196 feet

Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function h(t)=−16t2+32t+128 models the height, h, of the penny above the ocean as a function of time, t. Find:

ⓐ the zeros of this function which is when the penny will hit the ocean, ⓑ when the penny will be 128 feet above the ocean, ⓒ the height the penny will be at t=1 seconds which is when the penny will be at its highest point.

Solution

ⓐ 4 seconds; ⓑ 0 and 2 seconds; ⓒ 144 feet

Access this online resource for additional instruction and practice with quadratic equations.

  • Beginning Algebra & Solving Quadratics with the Zero Property

Key Concepts

  • Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.
  • Quadratic Equation: An equation of the form ax2+bx+c=0 is called a quadratic equation.
    a,b,care real numbers anda≠0
  • Zero Product Property: If a·b=0, then either a=0 or b=0 or both.
  • How to use the Zero Product Property
    1. Set each factor equal to zero.
    2. Solve the linear equations.
    3. Check.
  • How to solve a quadratic equation by factoring.
    1. Write the quadratic equation in standard form, ax2+bx+c=0.
    2. Factor the quadratic expression.
    3. Use the Zero Product Property.
    4. Solve the linear equations.
    5. Check. Substitute each solution separately into the original equation.
  • Zero of a Function: For any function f, if f(x)=0, then x is a zero of the function.
  • How to use a problem solving strategy to solve word problems.
    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what we are looking for.
    3. Name what we are looking for. Choose a variable to represent that quantity.
    4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
    5. Solve the equation using appropriate algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.

Section Exercises

Practice Makes Perfect

Use the Zero Product Property

In the following exercises, solve.

(3a−10)(2a−7)=0

Solution

a=10/3,a=7/2

(5b+1)(6b+1)=0

6m(12m−5)=0

Solution

m=0,m=5/12

2x(6x−3)=0

(2x−1)2=0

Solution

x=1/2

(3y+5)2=0

Solve Quadratic Equations by Factoring

In the following exercises, solve.

5a2−26a=24

Solution

a=−45,a=6

4b2+7b=−3

4m2=17m−15

Solution

m=5/4,m=3

n2=5n−6

7a2+14a=7a

Solution

a=−1,a=0

12b2−15b=−9b

49m2=144

Solution

m=12/7,m=−12/7

625=x2

16y2=81

Solution

y=−9/4,y=9/4

64p2=225

121n2=36

Solution

n=−6/11,n=6/11

100y2=9

(x+6)(x−3)=−8

Solution

x=2,x=−5

(p−5)(p+3)=−7

(2x+1)(x−3)=−4x

Solution

x=3/2,x=−1

(y−3)(y+2)=4y

(3x−2)(x+4)=12x

Solution

x=2,x=−4/3

(2y−3)(3y−1)=8y

20x2−60x=−45

Solution

x=3/2

3y2−18y=−27

15x2−10x=40

Solution

x=2,x=−4/3

14y2−77y=−35

18x2−9=−21x

Solution

x=−3/2,x=1/3

16y2+12=−32y

16p3=24p2–9p

Solution

p=0,p=¾

m3−2m2=−m

2x3+72x=24x2

Solution

x=0,x=6

3y3+48y=24y2

36x3+24x2=−4x

Solution

x=0,x=–1/3

2y3+2y2=12y

Solve Equations with Polynomial Functions

In the following exercises, solve.

For the function, f(x)=x2−8x+8, ⓐ find when f(x)=−4 ⓑ Use this information to find two points that lie on the graph of the function.

Solution

ⓐ x=2 or x=6 ⓑ (2,−4) (6,−4)

For the function, f(x)=x2+11x+20, ⓐ find when f(x)=−8 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f(x)=8x2−18x+5, ⓐ find when f(x)=−4 ⓑ Use this information to find two points that lie on the graph of the function.

Solution

ⓐ x=32 or x=34
ⓑ (32,−4) (34,−4)

For the function, f(x)=18x2+15x−10, ⓐ find when f(x)=15 ⓑ Use this information to find two points that lie on the graph of the function.

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x-intercepts of the graph of the function ⓒ the y-intercept of the graph of the function.

f(x)=9x2−4

Solution

ⓐ x=23 or x=−23
ⓑ (23,0), (−23,0) ⓒ (0,−4)

f(x)=25x2−49

f(x)=6x2−7x−5

Solution

ⓐ x=53 or x=−12
ⓑ (53,0), (−12,0) ⓒ (0,−5)

f(x)=12x2−11x+2

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

The product of two consecutive odd integers is 143. Find the integers.

Solution

−13,−11 and 11, 13

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

Solution

−14,−12 and 12, 14

The product of two consecutive even integers is 288. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

Solution

Width: 4 feet; Length: 7 feet.

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

Solution

Width: 5 feet; Length: 11 feet.

A rectangular carport has area 150 square feet. The width of the carport is five feet less than twice its length. Find the width and the length of the carport.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

Solution

The sides are 6 feet and 8 feet.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

Solution

The building side is 8 feet, the hypotenuse is 17 feet, and the third side is 15 feet.

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure.

Juli is going to launch a model rocket in her back yard. When she launches the rocket, the function h(t)=−16t2+32t models the height, h, of the rocket above the ground as a function of time, t. Find:

ⓐ the zeros of this function, which tell us when the rocket will be on the ground. ⓑ the time the rocket will be 16 feet above the ground.

Solution

ⓐ 0 seconds and 2 seconds ⓑ 1 second

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48 feet above the ground, the function h(t)=−16t2+32t+48 models the height, h, of the ball above the ground as a function of time, t. Find:

ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 48 feet above the ground. ⓒ the height the ball will be at t=1 seconds which is when the ball will be at its highest point.

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Solution

Answers will vary.

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve quadratic equations by using the zero product property, solve quadratic equations by factoring and solve applications modeled by quadratic equations.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Greatest Common Factor and Factor by Grouping

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

12a2b3,15ab2

Solution

3ab2

12m2n3,42m5n3

15y3,21y2,30y

Solution

3y

45x3y2,15x4y,10x5y3

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

35y+84

Solution

7(5y+12)

6y2+12y−6

18x3−15x

Solution

3x(6x2−5)

15m4+6m2n

4x3−12x2+16x

Solution

4x(x2−3x+4)

−3x+24

−3x3+27x2−12x

Solution

−3x(x2−9x+4)

3x(x−1)+5(x−1)

Factor by Grouping

In the following exercises, factor by grouping.

ax−ay+bx−by

Solution

(a+b)(x−y)

x2y−xy2+2x−2y

x2+7x−3x−21

Solution

(x−3)(x+7)

4x2−16x+3x−12

m3+m2+m+1

Solution

(m2+1)(m+1)

5x−5y−y+x

Factor Trinomials

Factor Trinomials of the Form x2+bx+c

In the following exercises, factor each trinomial of the form x2+bx+c.

a2+14a+33

Solution

(a+3)(a+11)

k2−16k+60

m2+3m−54

Solution

(m+9)(m−6)

x2−3x−10

In the following examples, factor each trinomial of the form x2+bxy+cy2.

x2+12xy+35y2

Solution

(x+5y)(x+7y)

r2+3rs−28s2

a2+4ab−21b2

Solution

(a+7b)(a−3b)

p2−5pq−36q2

m2−5mn+30n2

Solution

Prime

Factor Trinomials of the Form ax2+bx+c Using Trial and Error

In the following exercises, factor completely using trial and error.

x3+5x2−24x

3y3−21y2+30y

Solution

3y(y−5)(y−2)

5x4+10x3−75x2

5y2+14y+9

Solution

(5y+9)(y+1)

8x2+25x+3

10y2−53y−11

Solution

(5y+1)(2y−11)

6p2−19pq+10q2

−81a2+153a+18

Solution

−9(9a+1)(a−2)

Factor Trinomials of the Form ax2+bx+c using the ‘ac’ Method

In the following exercises, factor.

2x2+9x+4

18a2−9a+1

Solution

(3a−1)(6a−1)

15p2+2p−8

15x2+6x−2

Solution

Prime

8a2+32a+24

3x2+3x−36

Solution

3(x+4)(x−3)

48y2+12y−36

18a2−57a−21

Solution

3(2a−7)(3a+1)

3n4−12n3−96n2

Factor using substitution

In the following exercises, factor using substitution.

x4−13x2−30

Solution

(x2−15)(x2+2)

(x−3)2−5(x−3)−36

Factor Special Products

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

25x2+30x+9

Solution

(5x+3)2

36a2−84ab+49b2

40x2+360x+810

Solution

10(2x+9)2

5k3−70k2+245k

75u4−30u3v+3u2v2

Solution

3u2(5u−v)2

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

81r2−25

169m2−n2

Solution

(13m+n)(13m−n)

25p2−1

9−121y2

Solution

(3+11y)(3−11y)

20x2−125

169n3−n

Solution

n(13n+1)(13n−1)

6p2q2−54p2

24p2+54

Solution

6(4p2+9)

49x2−81y2

16z4−1

Solution

(2z−1)(2z+1)(4z2+1)

48m4n2−243n2

a2+6a+9−9b2

Solution

(a+3−3b)(a+3+3b)

x2−16x+64−y2

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

a3−125

Solution

(a−5)(a2+5a+25)

b3−216

2m3+54

Solution

2(m+3)(m2−3m+9)

81m3+3

General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

24x3+44x2

Solution

4x2(6x+11)

24a4−9a3

16n2−56mn+49m2

Solution

(4n−7m)2

6a2−25a−9

5u4−45u2

Solution

5u2(u+3)(u−3)

n4−81

64j2+225

Solution

prime

5x2+5x−60

b3−64

Solution

(b−4)(b2+4b+16)

m3+125

2b2−2bc+5cb−5c2

Solution

(2b+5c)(b−c)

48x5y2−243xy2

5q2−15q−90

Solution

5(q+3)(q−6)

4u5v+4u2v3

10m4−6250

Solution

10(m−5)(m+5)(m2+25)

60x2y−75xy+30y

16x2−24xy+9y2−64

Solution

(4x−3y+8)(4x−3y−8)

Polynomial Equations

Use the Zero Product Property

In the following exercises, solve.

(a−3)(a+7)=0

(5b+1)(6b+1)=0

Solution

b=−1/5,b=−1/6

6m(12m−5)=0

(2x−1)2=0

Solution

x=1/2

3m(2m−5)(m+6)=0

Solve Quadratic Equations by Factoring

In the following exercises, solve.

x2+9x+20=0

Solution

x=−4,x=−5

y2−y−72=0

2p2−11p=40

Solution

p=−52,p=8

q3+3q2+2q=0

144m2−25=0

Solution

m=512,m=−512

4n2=36

(x+6)(x−3)=−8

Solution

x=2,x=−5

(3x−2)(x+4)=12x

16p3=24p2−9p

Solution

p=0,p=¾

2y3+2y2=12y

Solve Equations with Polynomial Functions

In the following exercises, solve.

For the function, f(x)=x2+11x+20, ⓐ find when f(x)=−8 ⓑ Use this information to find two points that lie on the graph of the function.

Solution

ⓐ x=−7 or x=−4
ⓑ (−7,−8) (−4,−8)

For the function, f(x)=9x2−18x+5, ⓐ find when f(x)=−3 ⓑ Use this information to find two points that lie on the graph of the function.

In each function, find: ⓐ the zeros of the function ⓑ the x-intercepts of the graph of the function ⓒ the y-intercept of the graph of the function.

f(x)=64x2−49

Solution

ⓐ x=78 or x=−78
ⓑ (78,0), (−78,0) ⓒ (0,−49)

f(x)=6x2−13x−5

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

The product of two consecutive odd numbers is 399. Find the numbers.

Solution

The numbers are −21 and −19 or 19 and 21.

The area of a rectangular shaped patio 432 square feet. The length of the patio is 6 feet more than its width. Find the length and width.

A ladder leans against the wall of a building. The length of the ladder is 9 feet longer than the distance of the bottom of the ladder from the building. The distance of the top of the ladder reaches up the side of the building is 7 feet longer than the distance of the bottom of the ladder from the building. Find the lengths of all three sides of the triangle formed by the ladder leaning against the building.

Solution

The lengths are 8, 15, and 17 ft.

Shruti is going to throw a ball from the top of a cliff. When she throws the ball from 80 feet above the ground, the function h(t)=−16t2+64t+80 models the height, h, of the ball above the ground as a function of time, t. Find: ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 80 feet above the ground. ⓒ the height the ball will be at t=2 seconds which is when the ball will be at its highest point.

Chapter Practice Test

In the following exercises, factor completely.

80a2+120a3

Solution

40a2(2+3a)

5m(m−1)+3(m−1)

x2+13x+36

Solution

(x+4)(x+9)

p2+pq−12q2

xy−8y+7x−56

Solution

(x−8)(y+7)

40r2+810

9s2−12s+4

Solution

(3s−2)2

6x2−11x−10

3x2−75y2

Solution

3(x+5y)(x−5y)

6u2+3u−18

x3+125

Solution

(x+5)(x2−5x+25)

32x5y2−162xy2

6x4−19x2+15

Solution

(3x2−5)(2x2−3)

3x3−36x2+108x

In the following exercises, solve

5a2+26a=24

Solution

a=4/5,a=−6

The product of two consecutive integers is 156. Find the integers.

The area of a rectangular place mat is 168 square inches. Its length is two inches longer than the width. Find the length and width of the placemat.

Solution

The width is 12 inches and the length is 14 inches.

Jing is going to throw a ball from the balcony of her condo. When she throws the ball from 80 feet above the ground, the function h(t)=−16t2+64t+80 models the height, h, of the ball above the ground as a function of time, t. Find: ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 128 feet above the ground. ⓒ the height the ball will be at t=4 seconds.

For the function, f(x)=x2−7x+5, ⓐ find when f(x)=−7 ⓑ Use this information to find two points that lie on the graph of the function.

Solution

ⓐ x=3 or x=4 ⓑ (3,−7) (4,−7)

For the function f(x)=25x2−81, find: ⓐ the zeros of the function ⓑ the x-intercepts of the graph of the function ⓒ the y-intercept of the graph of the function.

degree of the polynomial equation
The degree of the polynomial equation is the degree of the polynomial.
polynomial equation
A polynomial equation is an equation that contains a polynomial expression.
quadratic equation
Polynomial equations of degree two are called quadratic equations.
zero of the function
A value of x where the function is 0, is called a zero of the function.
Zero Product Property
The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero.

Introduction

A photo of American football players lining up on the line of scrimmage.
American football is the most watched spectator sport in the United States. People around the country are constantly tracking statistics for football and other sports. (credit: “keijj44” / Pixabay)

Twelve goals last season. Fifteen home runs. Nine touchdowns. Whatever the statistics, sports analysts know it. Their jobs depend on it. Compiling and analyzing sports data not only help fans appreciate their teams but also help owners and coaches decide which players to recruit, how to best use them in games, how much they should be paid, and which players to trade. Understanding this kind of data requires a knowledge of specific types of expressions and functions. In this chapter, you will work with rational expressions and perform operations on them. And you will use rational expressions and inequalities to solve real-world problems.

Multiply and Divide Rational Expressions

Learning Objectives

By the end of this section, you will be able to:

  • Determine the values for which a rational expression is undefined
  • Simplify rational expressions
  • Multiply rational expressions
  • Divide rational expressions
  • Multiply and divide rational functions

Before you get started, take this readiness quiz.

Simplify: 90y15y2.
If you missed this problem, review Example 2 in Properties of Exponents and Scientific Notation.

Solution

6y

Multiply: 1415·635.
If you missed this problem, review Example 2 in Fractions.

Solution

425

Divide: 1210÷825.
If you missed this problem, review Example 3 in Fractions.

Solution

154

We previously reviewed the properties of fractions and their operations. We introduced rational numbers, which are just fractions where the numerators and denominators are integers. In this chapter, we will work with fractions whose numerators and denominators are polynomials. We call this kind of expression a rational expression.

Rational Expression

A rational expression is an expression of the form pq, where p and q are polynomials and q≠0.

Here are some examples of rational expressions:

−24565x12y4x+1x2−94x2+3x−12x−8

Notice that the first rational expression listed above, −2456, is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero.

We will do the same operations with rational expressions that we did with fractions. We will simplify, add, subtract, multiply, divide and use them in applications.

Determine the Values for Which a Rational Expression is Undefined

If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator.

When we work with a numerical fraction, it is easy to avoid dividing by zero because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero.

So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero. That way, when we solve a rational equation for example, we will know whether the algebraic solutions we find are allowed or not.

Determine the values for which a rational expression is undefined.

  1. Set the denominator equal to zero.
  2. Solve the equation.

Determine the value for which each rational expression is undefined:

ⓐ 8a2b3c ⓑ 4b−32b+5 ⓒ x+4x2+5x+6.

Solution

The expression will be undefined when the denominator is zero.

ⓐ
Steps to find when a rational expression is undefined by setting its denominator to zero.
8a2b3c
Set the denominator equal to zero and solve
for the variable.
3c=0
c=0
8a2b3cis undefined forc=0.
ⓑ
Method for determining values that make a rational expression undefined by solving the denominator.
4b−32b+5
Set the denominator equal to zero and solve
for the variable.
2b+5=02b=−5b=−52
4b−32b+5is undefined forb=−52.
ⓒ
Steps to determine values for which a rational expression is undefined by solving its denominator.
x+4x2+5x+6
Set the denominator equal to zero and solve
for the variable.
x2+5x+6=0(x+2)(x+3)=0x+2=0orx+3=0x=−2orx=−3
x+4x2+5x+6is undefined forx=−2orx=−3.

Determine the value for which each rational expression is undefined.

ⓐ 3y28x ⓑ 8n−53n+1 ⓒ a+10a2+4a+3

Solution

ⓐ x=0 ⓑ n=−13
ⓒ a=−1,a=−3

Determine the value for which each rational expression is undefined.

ⓐ 4p5q ⓑ y−13y+2 ⓒ m−5m2+m−6

Solution

ⓐ q=0 ⓑ y=−23
ⓒ m=2,m=−3

Simplify Rational Expressions

A fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator. Similarly, a simplified rational expression has no common factors, other than 1, in its numerator and denominator.

Simplified Rational Expression

A rational expression is considered simplified if there are no common factors in its numerator and denominator.

For example,

x+2x+3is simplified because there are no common factors ofx+2andx+3. 2x3xis not simplified becausexis a common factor of2xand3x.

We use the Equivalent Fractions Property to simplify numerical fractions. We restate it here as we will also use it to simplify rational expressions.

Equivalent Fractions Property

If a, b, and c are numbers where b≠0,c≠0,

thenab=a·cb·canda·cb·c=ab.

Notice that in the Equivalent Fractions Property, the values that would make the denominators zero are specifically disallowed. We see b≠0,c≠0 clearly stated.

To simplify rational expressions, we first write the numerator and denominator in factored form. Then we remove the common factors using the Equivalent Fractions Property.

Be very careful as you remove common factors. Factors are multiplied to make a product. You can remove a factor from a product. You cannot remove a term from a sum.

The rational expression is the quantity 2 times 3 times 7 divided by the quantity 3 times 5 times 7 are 3 and 7. Its common factors are 3 and 7, which are factors of the product. When they are removed, the result is two-fifths. The rational expression is the product of 3 x and the quantity x minus 9 divided by the product of 5 and the quantity x minus 9. The common factor is x minus 9, which is a factor of the product. When it is removed, the result is 3 x divided by 5. The rational expression is the quantity x plus 5 divided by 5. There is an x both the numerator and denomiantor. However, it is a term of the sum in the numerator. The rational expression has no common factors.

Removing the x’s from x+5x would be like cancelling the 2’s in the fraction 2+52!

How to Simplify a Rational Expression

Simplify: x2+5x+6x2+8x+12.

Solution
Step 1 is to factor the numerator and denominator completely in the rational expression, the quantity x squared plus 5 x plus six divided by the quantity x squared 8 x plus 12. The numerator, x squared plus 5 x plus six, factors into the quantity x plus 2 times the quantity x plus 3. The denominator, x squared 8 x plus 12, factors into the quantity x plus 2 times the quantity x plus 6. Step 2 is to simplify the rational expression, the quantity x plus 2 times the quantity x plus 3 all divided by the quantity x plus 2 times the quantity x plus 6, by dividing out the common factor, x plus 6. The result of removing the common factor is the quantity x plus 3 divided by the quantity x plus 6, where x is not equal to 2 and x is not equal to -6.

Simplify: x2−x−2x2−3x+2.

Solution

x+1x−1,x≠2,x≠1

Simplify: x2−3x−10x2+x−2.

Solution

x−5x−1,x≠−2,x≠1

We now summarize the steps you should follow to simplify rational expressions.

Simplify a rational expression.

  1. Factor the numerator and denominator completely.
  2. Simplify by dividing out common factors.

Usually, we leave the simplified rational expression in factored form. This way, it is easy to check that we have removed all the common factors.

We’ll use the methods we have learned to factor the polynomials in the numerators and denominators in the following examples.

Every time we write a rational expression, we should make a statement disallowing values that would make a denominator zero. However, to let us focus on the work at hand, we will omit writing it in the examples.

Simplify: 3a2−12ab+12b26a2−24b2.

Solution
Step-by-step simplification of a rational algebraic expression, detailing factoring and cancellation.
3a2−12ab+12b26a2−24b2
Factor the numerator and denominator,
first factoring out the GCF.
3(a2−4ab+4b2)6(a2−4b2)
3(a−2b)(a−2b)6(a+2b)(a−2b)
Remove the common factors of a−2band3. 3(a−2b)(a−2b)3·2(a+2b)(a−2b)
a−2b2(a+2b)

Simplify: 2x2−12xy+18y23x2−27y2.

Solution

2(x−3y)3(x+3y)

Simplify: 5x2−30xy+25y22x2−50y2.

Solution

5(x−y)2(x+5y)

Now we will see how to simplify a rational expression whose numerator and denominator have opposite factors. We previously introduced opposite notation: the opposite of a is −a and −a=−1·a.

The numerical fraction, say 7−7 simplifies to −1. We also recognize that the numerator and denominator are opposites.

The fraction a−a, whose numerator and denominator are opposites also simplifies to −1.

Let’s look at the expressionb−a.b−a Rewrite.−a+b Factor out–1.−1(a−b)

This tells us that b−a is the opposite of a−b.

In general, we could write the opposite of a−b as b−a. So the rational expression a−bb−a simplifies to −1.

Opposites in a Rational Expression

The opposite of a−b is b−a.

a−bb−a=−1a≠b

An expression and its opposite divide to −1.

We will use this property to simplify rational expressions that contain opposites in their numerators and denominators. Be careful not to treat a+b and b+a as opposites. Recall that in addition, order doesn’t matter so a+b=b+a. So if a≠−b, then a+bb+a=1.

Simplify: x2−4x−3264−x2.

Solution
A mathematical expression displaying the fraction (x^2 - 4x - 32) over (64 - x^2).
Factor the numerator and the denominator. A mathematical fraction is shown. The numerator is (x-8)(x+4). The denominator is (8-x)(8+x).
Recognize the factors that are opposites. A mathematical expression showing the simplification of a rational function. The term (x-8) in the numerator and (8-x) in the denominator are canceled out, introducing a factor of -1, which combines with an existing -1.
Simplify. A mathematical expression showing a negative fraction: minus, then a fraction bar, with (x + 4) in the numerator and (x + 8) in the denominator.

Simplify: x2−4x−525−x2.

Solution

−x+1x+5

Simplify: x2+x−21−x2.

Solution

−x+2x+1

Multiply Rational Expressions

To multiply rational expressions, we do just what we did with numerical fractions. We multiply the numerators and multiply the denominators. Then, if there are any common factors, we remove them to simplify the result.

Multiplication of Rational Expressions

If p, q, r, and s are polynomials where q≠0,s≠0, then

pq·rs=prqs

To multiply rational expressions, multiply the numerators and multiply the denominators.

Remember, throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, x≠0,x≠3, and x≠4.

How to Multiply Rational Expressions

Simplify: 2xx2−7x+12·x2−96x2.

Solution
Step 1 is to factor each numerator and the denominator completely in 2 x divided by the quantity x squared minus 7 x plus 12 times the rational expression the quantity x squared minus 9 divided by 6 x squared. The denominator, x squared minus 7 x plus 12, factors into the quantity x minus 3 times the quantity x minus 4. The numerator x squared minus 9 factors into the quantity x minus 3 times the quantity x plus 3. Step 2 is to multiply the numerators 2 x and the quantity x minus 3 times the quantity x plus 3, and the denominators the quantity x minus 3 times the quantity x minus 4 and 6 x squared. It is helpful to write the monomials in the numerator and in the denominator. first. Step 3 is to simplify 2 x times the quantity x minus 3 times the quantity x plus 3 all divided by 2 times 3 times x times x times the quantity x minus 3 times the quantity x plus 4 by dividing out the common factor, x minus 3. Leaving the denominator in factored form, the result is the quantity x plus 3 divided by 3 x times the quantity x minus 4.

Simplify: 5xx2+5x+6·x2−410x.

Solution

x−22(x+3)

Simplify: 9x2x2+11x+30·x2−363x2.

Solution

3(x−6)x+5

Multiply rational expressions.

  1. Factor each numerator and denominator completely.
  2. Multiply the numerators and denominators.
  3. Simplify by dividing out common factors.

Multiply: 3a2−8a−3a2−25·a2+10a+253a2−14a−5.

Solution
This table demonstrates the step-by-step simplification of a rational expression by factoring and canceling common terms.
3a2−8a−3a2−25·a2+10a+253a2−14a−5
Factor the numerators and denominators
and then multiply.
(3a+1)(a−3)(a+5)(a+5)(a−5)(a+5)(3a+1)(a−5)
Simplify by dividing out
common factors.
(3a+1)(a−3)(a+5)(a+5)(a−5)(a+5)(3a+1)(a−5)
Simplify. (a−3)(a+5)(a−5)(a−5)
Rewrite (a−5)(a−5) using an exponent. (a−3)(a+5)(a−5)2

Simplify: 2x2+5x−12x2−16·x2−8x+162x2−13x+15.

Solution

x−4x−5

Simplify: 4b2+7b−21−b2·b2−2b+14b2+15b−4.

Solution

−(b+2)(b−1)(1+b)(b+4)

Divide Rational Expressions

Just like we did for numerical fractions, to divide rational expressions, we multiply the first fraction by the reciprocal of the second.

Division of Rational Expressions

If p, q, r, and s are polynomials where q≠0,r≠0,s≠0, then

pq÷rs=pq·sr

To divide rational expressions, multiply the first fraction by the reciprocal of the second.

Once we rewrite the division as multiplication of the first expression by the reciprocal of the second, we then factor everything and look for common factors.

How to Divide Rational Expressions

Divide: p3+q32p2+2pq+2q2÷p2−q26.

Solution
Step 1 is to rewrite the division of the rational expression, the quantity p cubed plus q cubes divided by the quantity 2 p squared plus 2 p q plus 2 q squared divided by the rational expression, the quantity p squared minus q squared all divided by 6. Do this by flipping the rational expression, the quantity p squared minus q squared all divided by 6, and changing division to multiplication. The result is the quantity p cubed plus q cubes divided by the quantity 2 p squared plus 2 p q plus 2 q squared times the quantity 6 divided by the quantity p squared minus q squared. Step 2 is to factor the numerators, the quantity p cubed plus q cubed and 6, and the denominators, the quantity 2 p squared plus 2 p q plus 2 squared and the quantity p squared minus q squared, completely. The result is the quantity p plus q times the quantity p squared minus p q plus q squared all times the quantity 2 times 3 divided by the quantity p minus q times the quantity p plus q. Step 3 is to multiply the numerators and denominators. The result is the quantity p plus q times the quantity p squared minus p q plus q squared times 2 times 3 all divided by the 2 times the quantity p squared plus p q plus q squared times the quantity p minus q times the quantity p plus q. Step 4 is to simplify the expression by dividing out the common factors, the quantity p plus q and 2. The result is 3 times the quantity p squared minus p q plus q squared all divided by the quantity p minus q times the quantity p squared plus p q plus q squared.

Simplify: x3+83x2−6x+12÷x2−46.

Solution

2x-2

Simplify: 2z2z2−1÷z3−z2+zz3+1.

Solution

2zz−1

Divide rational expressions.

  1. Rewrite the division as the product of the first rational expression and the reciprocal of the second.
  2. Factor the numerators and denominators completely.
  3. Multiply the numerators and denominators together.
  4. Simplify by dividing out common factors.

Recall from Use the Language of Algebra that a complex fraction is a fraction that contains a fraction in the numerator, the denominator or both. Also, remember a fraction bar means division. A complex fraction is another way of writing division of two fractions.

Divide: 6x2−7x+24x−82x2−7x+3x2−5x+6.

Solution
Step-by-step simplification of a complex rational algebraic expression, detailing each transformation from initial form to the final simplified result.
6x2−7x+24x−82x2−7x+3x2−5x+6
Rewrite with a division sign. 6x2−7x+24x−8÷2x2−7x+3x2−5x+6
Rewrite as product of first times reciprocal
of second.
6x2−7x+24x−8·x2−5x+62x2−7x+3
Factor the numerators and the
denominators, and then multiply.
(2x−1)(3x−2)(x−2)(x−3)4(x−2)(2x−1)(x−3)
Simplify by dividing out common factors. (2x−1)(3x−2)(x−2)(x−3)4(x−2)(2x−1)(x−3)
Simplify. 3x−24

Simplify: 3x2+7x+24x+243x2−14x−5x2+x−30.

Solution

x+24

Simplify: y2−362y2+11y−62y2−2y−608y−4.

Solution

2y+5

If we have more than two rational expressions to work with, we still follow the same procedure. The first step will be to rewrite any division as multiplication by the reciprocal. Then, we factor and multiply.

Perform the indicated operations: 3x−64x−4·x2+2x−3x2−3x−10÷2x+128x+16.

Solution
A mathematical problem showing the multiplication and division of three rational algebraic expressions: (3x-6)/(4x-4) * (x^2+2x-3)/(x^2-3x-10) / (2x+12)/(8x+16).
Rewrite the division as multiplication
by the reciprocal.
A multiplication problem involving three rational algebraic expressions: (3x-6)/(4x-4), (x^2+2x-3)/(x^2-3x-10), and (8x+16)/(2x+12). The last expression is in red.
Factor the numerators and the denominators. An algebraic expression showing the product of three rational functions: (3(x-2) / 4(x-1)) multiplied by ((x+3)(x-1) / (x+2)(x-5)) multiplied by (8(x+2) / 2(x+6)).
Multiply the fractions. Bringing the constants to
the front will help when removing common factors.
Simplify by dividing out common factors. Simplification of a rational algebraic expression, showing common factors (x-1) and (x+2) being cancelled from both the numerator and denominator, along with numerical terms 8, 4, and 2.
Simplify. An algebraic fraction with 3(x-2)(x+3) in the numerator and (x-5)(x+6) in the denominator.

Perform the indicated operations: 4m+43m−15·m2−3m−10m2−4m−32÷12m−366m−48.

Solution

2(m+1)(m+2)3(m+4)(m−3)

Perform the indicated operations: 2n2+10nn−1÷n2+10n+24n2+8n−9·n+48n2+12n.

Solution

(n+5)(n+9)2(n+6)(2n+3)

Multiply and Divide Rational Functions

We started this section stating that a rational expression is an expression of the form pq, where p and q are polynomials and q≠0. Similarly, we define a rational function as a function of the form R(x)=p(x)q(x) where p(x) and q(x) are polynomial functions and q(x) is not zero.

Rational Function

A rational function is a function of the form

R(x)=p(x)q(x)

where p(x) and q(x) are polynomial functions and q(x) is not zero.

The domain of a rational function is all real numbers except for those values that would cause division by zero. We must eliminate any values that make q(x)=0.

Determine the domain of a rational function.

  1. Set the denominator equal to zero.
  2. Solve the equation.
  3. The domain is all real numbers excluding the values found in Step 2.

Find the domain of R(x)=2x2−14x4x2−16x−48.

Solution

The domain will be all real numbers except those values that make the denominator zero. We will set the denominator equal to zero , solve that equation, and then exclude those values from the domain.

Procedure for finding values excluded from the domain of a rational function R(x).
Set the denominator to zero. 4x2−16x−48=0
Factor, first factor out the GCF. 4(x2−4x−12)=0
4(x−6)(x+2)=0
Use the Zero Product Property. 4≠0x−6=0x+2=0
Solve. x=6x=−2
The domain of R(x) is all real numbers
where x≠6 and x≠−2.

Find the domain of R(x)=2x2−10x4x2−16x−20.

Solution

The domain of R(x) is all real numbers where x≠5 and x≠−1.

Find the domain of R(x)=4x2−16x8x2−16x−64.

Solution

The domain of R(x) is all real numbers where x≠4 and x≠−2.

To multiply rational functions, we multiply the resulting rational expressions on the right side of the equation using the same techniques we used to multiply rational expressions.

Find R(x)=f(x)·g(x) where f(x)=2x−6x2−8x+15 and g(x)=x2−252x+10.

Solution
Step-by-step simplification of a rational algebraic expression R(x).
R(x)=f(x)·g(x)
R(x)=2x−6x2−8x+15·x2−252x+10
Factor each numerator and denominator. R(x)=2(x−3)(x−3)(x−5)·(x−5)(x+5)2(x+5)
Multiply the numerators and denominators. R(x)=2(x−3)(x−5)(x+5)2(x−3)(x−5)(x+5)
Remove common factors. R(x)=2(x−3)(x−5)(x+5)2(x−3)(x−5)(x+5)
Simplify. R(x)=1

Find R(x)=f(x)·g(x) where f(x)=3x−21x2−9x+14 and g(x)=2x2−83x+6.

Solution

R(x)=2

Find R(x)=f(x)·g(x) where f(x)=x2−x3x2+27x−30 and g(x)=x2−100x2−10x.

Solution

R(x)=13

To divide rational functions, we divide the resulting rational expressions on the right side of the equation using the same techniques we used to divide rational expressions.

Find R(x)=f(x)g(x) where f(x)=3x2x2−4x and g(x)=9x2−45xx2−7x+10.

Solution
This table illustrates the step-by-step process of dividing two rational functions and simplifying the resulting expression.
R(x)=f(x)g(x)
Substitute in the functions f(x),g(x). R(x)=3x2x2−4x9x2−45xx2−7x+10
Rewrite the division as the product of
f(x) and the reciprocal of g(x).
R(x)=3x2x2−4x·x2−7x+109x2−45x
Factor the numerators and denominators
and then multiply.
R(x)=3·x·x·(x−5)(x−2)x(x−4)·3·3·x·(x−5)
Simplify by dividing out common factors. R(x)=3·x·x(x−5)(x−2)x(x−4)·3·3·x(x−5)
R(x)=x−23(x−4)

Find R(x)=f(x)g(x) where f(x)=2x2x2−8x and g(x)=8x2+24xx2+x−6.

Solution

R(x)=x−24(x−8)

Find R(x)=f(x)g(x) where f(x)=15x23x2+33x and g(x)=5x−5x2+9x−22.

Solution

R(x)=x(x−2)x−1

Key Concepts

  • Determine the values for which a rational expression is undefined.
    1. Set the denominator equal to zero.
    2. Solve the equation.
  • Equivalent Fractions Property
    If a, b, and c are numbers where b≠0,c≠0, then ab=a·cb·c and a·cb·c=ab.
  • How to simplify a rational expression.
    1. Factor the numerator and denominator completely.
    2. Simplify by dividing out common factors.
  • Opposites in a Rational Expression
         The opposite of a−b is b−a.
        a−bb−a=−1a≠b
        An expression and its opposite divide to −1.
  • Multiplication of Rational Expressions
    If p, q, r, and s are polynomials where q≠0,s≠0, then
    pq·rs=prqs
  • How to multiply rational expressions.
    1. Factor each numerator and denominator completely.
    2. Multiply the numerators and denominators.
    3. Simplify by dividing out common factors.
  • Division of Rational Expressions
    If p, q, r, and s are polynomials where q≠0,r≠0,s≠0, then
    pq÷rs=pq·sr
  • How to divide rational expressions.
    1. Rewrite the division as the product of the first rational expression and the reciprocal of the second.
    2. Factor the numerators and denominators completely.
    3. Multiply the numerators and denominators together.
    4. Simplify by dividing out common factors.
  • How to determine the domain of a rational function.
    1. Set the denominator equal to zero.
    2. Solve the equation.
    3. The domain is all real numbers excluding the values found in Step 2.

Practice Makes Perfect

Determine the Values for Which a Rational Expression is Undefined

In the following exercises, determine the values for which the rational expression is undefined.

ⓐ 2x2z, ⓑ 4p−16p−5, ⓒ n−3n2+2n−8

Solution

ⓐ z=0 ⓑ p=56
ⓒ n=−4,n=2

ⓐ 10m11n, ⓑ 6y+134y−9, ⓒ b−8b2−36

ⓐ 4x2y3y, ⓑ 3x−22x+1, ⓒ u−1u2−3u−28

Solution

ⓐ y=0, ⓑ x=−12, ⓒ u=−4,u=7

ⓐ 5pq29q, ⓑ 7a−43a+5, ⓒ 1x2−4

Simplify Rational Expressions

In the following exercises, simplify each rational expression.

−4455

Solution

−45

5663

8m3n12mn2

Solution

2m23n

36v3w227vw3

8n−963n−36

Solution

83(n≠ 2)

12p−2405p−100

x2+4x−5x2−2x+1

Solution

x+5x−1

y2+3y−4y2−6y+5

a2−4a2+6a−16

Solution

a+2a+8

y2−2y−3y2−9

p3+3p2+4p+12p2+p−6

Solution

p2+4p−2

x3−2x2−25x+50x2−25

8b2−32b2b2−6b−80

Solution

4b(b−4)(b+5)(b−8)

−5c2−10c−10c2+30c+100

3m2+30mn+75n24m2−100n2

Solution

3(m+5n)4(m−5n)

5r2+30rs−35s2r2−49s2

a−55−a

Solution

−1

5−dd−5

20−5yy2−16

Solution

−5y+4

4v−3264−v2

w3+216w2−36

Solution

w2−6w+36w−6

v3+125v2−25

z2−9z+2016−z2

Solution

−z−54+z

a2−5a−3681−a2

Multiply Rational Expressions

In the following exercises, multiply the rational expressions.

1216·410

Solution

310

325·1624

5x2y412xy3·6x220y2

Solution

x38y

12a3bb2·2ab29b3

5p2p2−5p−36·p2−1610p

Solution

p(p−4)2(p−9)

3q2q2+q−6·q2−99q

2y2−10yy2+10y+25·y+56y

Solution

y−53(y+5)

z2+3zz2−3z−4·z−4z2

28−4b3b−3·b2+8b−9b2−49

Solution

−4(b+9)3(b+7)

72m−12m28m+32·m2+10m+24m2−36

3c2−16c+5c2−25·c2+10c+253c2−14c−5

Solution

(3c−1)(c+5)(3c+1)(c−5)

2d2+d−3d2−16·d2−8d+162d2−9d−18

6m2−13m+29−m2·m2−6m+96m2+23m−4

Solution

−(m−2)(m−3)(3+m)(m+4)

2n2−3n−1425−n2·n2−10n+252n2−13n+21

Divide Rational Expressions

In the following exercises, divide the rational expressions.

v−511−v÷v2−25v−11

Solution

−1v+5

10+ww−8÷100−w28−w

3s2s2−16÷s3+4s2+16ss3−64

Solution

3ss+4

r2−915÷r3−275r2+15r+45

p3+q33p2+3pq+3q2÷p2−q212

Solution

4(p2−pq+q2)(p−q)(p2+pq+q2)

v3−8w32v2+4vw+8w2÷v2−4w24

x2+3x−104x÷(2x2+20x+50)

Solution

x−28x(x+5)

2y2−10yz−48z22y−1÷(4y2−32yz)

2a2−a−215a+20a2+7a+12a2+8a+16

Solution

2a−75

3b2+2b−812b+183b2+2b−82b2−7b−15

12c2−122c2−3c+14c+46c2−13c+5

Solution

3(3c−5)

4d2+7d−235d+10d2−47d2−12d−4

For the following exercises, perform the indicated operations.

10m2+80m3m−9·m2+4m−21m2−9m+20÷5m2+10m2m−10

Solution

4(m+8)(m+7)3(m−4)(m+2)

4n2+32n3n+2·3n2−n−2n2+n−30÷108n2−24nn+6

12p2+3pp+3÷p2+2p−63p2−p−12·p−79p3−9p2

Solution

(4p+1)(p−4)3p(p+9)(p−1)

6q+39q2−9q÷q2+14q+33q2+4q−5·4q2+12q12q+6

Multiply and Divide Rational Functions

In the following exercises, find the domain of each function.

R(x)=x3−2x2−25x+50x2−25

Solution

x≠5 and x≠−5

R(x)=x3+3x2−4x−12x2−4

R(x)=3x2+15x6x2+6x−36

Solution

x≠2 and x≠−3

R(x)=8x2−32x2x2−6x−80

For the following exercises, find R(x)=f(x)·g(x) where f(x) and g(x) are given.

f(x)=6x2−12xx2+7x−18
g(x)=x2−813x2−27x

Solution

R(x)=2

f(x)=x2−2xx2+6x−16
g(x)=x2−64x2−8x

f(x)=4xx2−3x−10
g(x)=x2−258x2

Solution

R(x)=x+52x(x+2)

f(x)=2x2+8xx2−9x+20
g(x)=x−5x2

For the following exercises, find R(x)=f(x)g(x) where f(x) and g(x) are given.

f(x)=27x23x−21
g(x)=3x2+18xx2+13x+42

Solution

R(x)=3x(x+7)x−7

f(x)=24x22x−8
g(x)=4x3+28x2x2+11x+28

f(x)=16x24x+36
g(x)=4x2−24xx2+4x−45

Solution

R(x)=x(x−5)x−6

f(x)=24x22x−4
g(x)=12x2+36xx2−11x+18

Writing Exercises

Explain how you find the values of x for which the rational expression x2−x−20x2−4 is undefined.

Solution

Answers will vary.

Explain all the steps you take to simplify the rational expression p2+4p−219−p2.

ⓐ Multiply 74·910 and explain all your steps. ⓑ Multiply nn−3·9n+3 and explain all your steps. ⓒ Evaluate your answer to part ⓑ when n=7. Did you get the same answer you got in part ⓐ ? Why or why not?

Solution

Answers will vary.

ⓐ Divide 245÷6 and explain all your steps. ⓑ Divide x2−1x÷(x+1) and explain all your steps. ⓒ Evaluate your answer to part ⓑ when x=5. Did you get the same answer you got in part ⓐ ? Why or why not?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and six rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was determine the values for which a rational expression is undefined. In row 3, the I can was simplify rationale expressions. In row 4, the I can was multiply rational expressions. In row 5, the I can was divide rational expressions. In row 6, the I can was multiply and divide rational functions. There is the nothing in the other columns.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved your goals in this section! Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific!

…with some help. This must be addressed quickly as topics you do not master become potholes in your road to success. Math is sequential - every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is critical and you must not ignore it. You need to get help immediately or you will quickly be overwhelmed. See your instructor as soon as possible to discuss your situation. Together you can come up with a plan to get you the help you need.

rational expression
A rational expression is an expression of the form pq, where p and q are polynomials and q≠0.
simplified rational expression
A simplified rational expression has no common factors, other than 1, in its numerator and denominator.
rational function
A rational function is a function of the form R(x)=p(x)q(x) where p(x) and q(x) are polynomial functions and q(x) is not zero.

Add and Subtract Rational Expressions

Learning Objectives

By the end of this section, you will be able to:

  • Add and subtract rational expressions with a common denominator
  • Add and subtract rational expressions whose denominators are opposites
  • Find the least common denominator of rational expressions
  • Add and subtract rational expressions with unlike denominators
  • Add and subtract rational functions

Before you get started, take this readiness quiz.

Add: 710+815.
If you missed this problem, review Example 6 in Fractions.

Solution

3730

Subtract: 3x4−89.
If you missed this problem, review Example 5 in Fractions.

Solution

27x−3236

Subtract: 6(2x+1)−4(x−5).
If you missed this problem, review Example 12 in Properties of Real Numbers.

Solution

8x+26

Add and Subtract Rational Expressions with a Common Denominator

What is the first step you take when you add numerical fractions? You check if they have a common denominator. If they do, you add the numerators and place the sum over the common denominator. If they do not have a common denominator, you find one before you add.

It is the same with rational expressions. To add rational expressions, they must have a common denominator. When the denominators are the same, you add the numerators and place the sum over the common denominator.

Rational Expression Addition and Subtraction

If p, q, and r are polynomials where r≠0, then

pr+qr=p+qrandpr−qr=p−qr

To add or subtract rational expressions with a common denominator, add or subtract the numerators and place the result over the common denominator.

We always simplify rational expressions. Be sure to factor, if possible, after you subtract the numerators so you can identify any common factors.

Remember, too, we do not allow values that would make the denominator zero. What value of x should be excluded in the next example?

Add: 11x+28x+4+x2x+4.

Solution

Since the denominator is x+4, we must exclude the value x=−4.

This table illustrates the step-by-step process of adding two rational expressions with a common denominator and simplifying the resulting expression.
11x+28x+4+x2x+4,x≠−4
The fractions have a common denominator,
so add the numerators and place the sum
over the common denominator.
11x+28+x2x+4
Write the degrees in descending order. x2+11x+28x+4
Factor the numerator. (x+4)(x+7)x+4
Simplify by removing common factors. (x+4)(x+7)x+4
Simplify. x+7

The expression simplifies to x+7 but the original expression had a denominator of x+4 so x≠−4.

Simplify: 9x+14x+7+x2x+7.

Solution

x+2

Simplify: x2+8xx+5+15x+5.

Solution

x+3

To subtract rational expressions, they must also have a common denominator. When the denominators are the same, you subtract the numerators and place the difference over the common denominator. Be careful of the signs when you subtract a binomial or trinomial.

Subtract: 5x2−7x+3x2−3x-18−4x2+x−9x2−3x-18.

Solution
Steps to subtract and simplify rational algebraic expressions.
5x2−7x+3x2−3x-18−4x2+x−9x2−3x-18
Subtract the numerators and place the
difference over the common denominator.
5x2−7x+3−(4x2+x−9)x2−3x-18
Distribute the sign in the numerator. 5x2−7x+3−4x2−x+9x2−3x−18
Combine like terms. x2−8x+12x2−3x−18
Factor the numerator and the denominator. (x−2)(x−6)(x+3)(x−6)
Simplify by removing common factors. (x−2)(x−6)(x+3)(x−6)
(x−2)(x+3)

Subtract: 4x2−11x+8x2−3x+2−3x2+x−3x2−3x+2.

Solution

x−11x−2

Subtract: 6x2−x+20x2−81−5x2+11x−7x2−81.

Solution

x−3x+9

Add and Subtract Rational Expressions Whose Denominators are Opposites

When the denominators of two rational expressions are opposites, it is easy to get a common denominator. We just have to multiply one of the fractions by −1−1.

Let’s see how this works.

A mathematical expression showing the sum of two fractions: 7 over d plus 5 over negative d.
Multiply the second fraction by −1−1. A mathematical expression showing the sum of two fractions. The first fraction is 7 divided by d. The second fraction has a numerator of (-1) multiplied by 5, and a denominator of (-1) multiplied by (-d).
The denominators are the same. A mathematical expression showing the addition of two fractions with a common denominator 'd'. The expression is 7/d + (-5)/d.
Simplify. A mathematical fraction displaying the number 2 over the letter d, indicating 2 divided by d, centered on a white background.

Be careful with the signs as you work with the opposites when the fractions are being subtracted.

Subtract: m2−6mm2−1−3m+21−m2.

Solution
The image displays a mathematical expression showing the subtraction of two algebraic fractions: (m^2 - 6m) / (m^2 - 1) - (3m + 2) / (1 - m^2).
The denominators are opposites, so multiply the
second fraction by −1−1.
An algebraic expression showing the subtraction of two rational terms: (m^2 - 6m)/(m^2 - 1) minus [-1(3m+2)]/[-1(1-m^2)], illustrating fraction manipulation.
Simplify the second fraction. A mathematical expression displaying the subtraction of two algebraic fractions: (m^2 - 6m) / (m^2 - 1) - (-3m - 2) / (m^2 - 1). Both fractions share the same denominator, m^2 - 1.
The denominators are the same. Subtract the numerators. A mathematical expression showing a fraction with m^2 - 6m - (-3m - 2) in the numerator and m^2 - 1 in the denominator.
Distribute. A mathematical expression displaying the fraction (m^2 - 6m + 3m + 2) / (m^2 - 1), which simplifies to (m^2 - 3m + 2) / (m^2 - 1).
Combine like terms. A mathematical expression displays a fraction with m squared minus 3m plus 2 in the numerator, and m squared minus 1 in the denominator.
Factor the numerator and denominator. A mathematical fraction, where the numerator is (m-1)(m-2) and the denominator is (m-1)(m+1).
Simplify by removing common factors. Cancellation of the common factor (m-1) from the numerator and denominator of the algebraic fraction ((m-1)(m-2))/((m-1)(m+1)).
Simplify. A mathematical fraction is displayed, with 'm - 2' in the numerator and 'm + 1' in the denominator, set against a plain white background.

Subtract: y2−5yy2−4−6y−64−y2.

Solution

y+3y+2

Subtract: 2n2+8n−1n2−1−n2−7n−11−n2.

Solution

3n−2n−1

Find the Least Common Denominator of Rational Expressions

When we add or subtract rational expressions with unlike denominators, we will need to get common denominators. If we review the procedure we used with numerical fractions, we will know what to do with rational expressions.

Let’s look at this example: 712+518. Since the denominators are not the same, the first step was to find the least common denominator (LCD).

To find the LCD of the fractions, we factored 12 and 18 into primes, lining up any common primes in columns. Then we “brought down” one prime from each column. Finally, we multiplied the factors to find the LCD.

When we add numerical fractions, once we found the LCD, we rewrote each fraction as an equivalent fraction with the LCD by multiplying the numerator and denominator by the same number. We are now ready to add.

Seven-twelfths plus five-eighteenths. Write the prime factorizations of each denominator and line up the common factors. The denominator of the first fraction is 12. The prime factorization of 12 is 2 times 2 times 3. The denominator of the second fraction is 18. The prime factorization of 18 is 2 times 3 times 3. Bringing down a factor from each column, the lowest common denominator of 12 and 18 is 2 times 2 times 3 times 3, which is 36. Write both fractions using the lowest common denominator. To do this multiply the numerator and denominator of the first fraction by 3 and multiply the numerator and denominator of the second fraction by 2. The result is 7 times 3 all divided by 12 times 3 plus 5 times 2 all divided by 18 times 2. Simplify each fraction. 7 times 3 is 21 and 12 times 3 is 36. 5 times 2 is 10 and 18 times 2 is 36. The result is twenty-one thirty-sixths plus ten thirty-sixths.

We do the same thing for rational expressions. However, we leave the LCD in factored form.

Find the least common denominator of rational expressions.

  1. Factor each denominator completely.
  2. List the factors of each denominator. Match factors vertically when possible.
  3. Bring down the columns by including all factors, but do not include common factors twice.
  4. Write the LCD as the product of the factors.

Remember, we always exclude values that would make the denominator zero. What values of x should we exclude in this next example?

ⓐ Find the LCD for the expressions 8x2−2x−3,3xx2+4x+3 and ⓑ rewrite them as equivalent rational expressions with the lowest common denominator.

Solution

ⓐ

Find the LCD for 8x2−2x−3,3xx2+4x+3.
Factor each denominator completely, lining up common factors.

Bring down the columns.
The image displays the factorization of two quadratic polynomials, x^2 - 2x - 3 and x^2 + 4x + 3, into their binomial factors. Below these factorizations, the Least Common Denominator (LCD) is calculated using these factors.
Write the LCD as the product of the factors. The image displays text on a white background, stating, 'The LCD is (x + 1)(x - 3)(x + 3).'

ⓑ

Two algebraic fractions are displayed side-by-side. The first fraction is 8 over (x^2 - 2x - 3), followed by a prime symbol. The second fraction is 3x over (x^2 + 4x + 3).
Factor each denominator. The image shows two algebraic fractions separated by a comma. The first fraction is 8 over (x+1)(x-3). The second fraction is 3x over (x+1)(x+3).
Multiply each denominator by the ‘missing’
LCD factor and multiply each numerator by the same factor.
Two algebraic fractions showing common factors highlighted in red, indicating terms like (x+3) and (x-3) that can be cancelled for simplification.
Simplify the numerators. Two algebraic fractions are presented. The first is (8x + 24) / ((x + 1)(x - 3)(x + 3)) and the second is (3x^2 - 9x) / ((x + 1)(x + 3)(x - 3)).

ⓐ Find the LCD for the expressions 2x2−x−12,1x2−16 ⓑ rewrite them as equivalent rational expressions with the lowest common denominator.

Solution

ⓐ (x−4)(x+3)(x+4)
ⓑ 2x+8(x−4)(x+3)(x+4),
x+3(x−4)(x+3)(x+4)

ⓐ Find the LCD for the expressions 3xx2−3x–10,5x2+3x+2 ⓑ rewrite them as equivalent rational expressions with the lowest common denominator.

Solution

ⓐ (x+2)(x−5)(x+1)
ⓑ 3x2+3x(x+2)(x−5)(x+1),
5x−25(x+2)(x−5)(x+1)

Add and Subtract Rational Expressions with Unlike Denominators

Now we have all the steps we need to add or subtract rational expressions with unlike denominators.

How to Add Rational Expressions with Unlike Denominators

Add: 3x−3+2x−2.

Solution
Step 1 is to determine if the rational expressions 3 divided by the quantity x minus 3 and 2 divided by the quantity x minus 2 have a common factors. The denominators x minus 3 and x minus 2 do not have any common factors, which means the lowest common denominator of the rational expressions is the quantity x minus 3 times the quantity x minus 2. Rewrite each rational expression with the least common denominator. Multiply the numerator and denominator of 3 divided by the quantity x minus 3 by the quantity x minus 2. Multiply the numerator and denominator of 2 divided by the quantity x minus 2 by the quantity x minus 2. The result is the rational expression 3 times the quantity x minus 2 all divided by the quantity x minus 3 times the quantity x minus 2 plus the rational expression 2 times the quantity x minus 3 divided by the quantity x minus 2 times the quantity x minus 3. Simplify the numerators and keep the denominators factored. The numerator of the first rational expression, 3 times the quantity x minus 2, simplifies to 3 x minus 6. The numerator of the second rational expression, 2 times the quantity x minus 3, simplifies to 2 x minus 6. The result is the rational expression the quantity 3 x minus 6 all divided by the quantity x minus 3 times the quantity x minus 2 plus the rational expression, the quantity 2 x minus 6 all divided by the quantity x minus 3 times the quantity x minus 2. Step 2 is to add or subtract the rational expressions by adding the numerators, the quantity 3 x minus 6 and the quantity 2 x minus 6, and placing the sum over the denominator, the quantity x minus 3 times the quantity x minus 2. The result is the quantity 3 x minus 6 plus 2 x minus 6 all divided by the quantity x minus 3 times the quantity x minus 2. Simplify the numerator by combining like terms. The result is the quantity 5 x minus 12 all divided by the quantity x minus 3 times the quantity x minus 2. Step 3. Notice that 5 x minus 12 cannot be factored, so the answer is simplified.

Add: 2x−2+5x+3.

Solution

7x−4(x−2)(x+3)

Add:4m+3+3m+4.

Solution

7m+25(m+3)(m+4)

The steps used to add rational expressions are summarized here.

Add or subtract rational expressions.

  1. Determine if the expressions have a common denominator.
    • Yes – go to step 2.
    • No – Rewrite each rational expression with the LCD.
      • Find the LCD.
      • Rewrite each rational expression as an equivalent rational expression with the LCD.
  2. Add or subtract the rational expressions.
  3. Simplify, if possible.

Avoid the temptation to simplify too soon. In the example above, we must leave the first rational expression as 3x−6(x−3)(x−2) to be able to add it to 2x−6(x−2)(x−3). Simplify only after you have combined the numerators.

Add: 8x2−2x−3+3xx2+4x+3.

Solution
A mathematical expression showing the sum of two algebraic fractions: 8 over (x^2 - 2x - 3) plus 3x over (x^2 + 4x + 3).
Do the expressions have a common denominator? No.
Rewrite each expression with the LCD.
Find the LCD.x2−2x−3=(x+1)(x−3)x2+4x+3=(x+1)(x+3)______________________________LCD=(x+1)(x−3)(x+3)
Rewrite each rational expression as an
equivalent rational expression with the LCD.
A mathematical expression showing the addition of two rational terms with identical denominators. The numerators are 8(x+3) and 3x(x-3), while the common denominator is (x+1)(x-3)(x+3). Some factors are highlighted in red.
Simplify the numerators. An algebraic problem displaying the sum of two rational expressions, each with a common denominator of (x+1)(x-3)(x+3). The numerators are 8x+24 and 3x^2-9x.
Add the rational expressions. A fraction with the numerator 3x^2 - x + 24 and the denominator (x + 1)(x - 3)(x + 3).
Simplify the numerator. A fraction with the numerator 3x^2 - x + 24 and the denominator (x + 1)(x - 3)(x + 3).
The numerator is prime, so there are
no common factors.

Add: 1m2−m−2+5mm2+3m+2.

Solution

5m2−9m+2(m+1)(m−2)(m+2)

Add:2nn2−3n−10+6n2+5n+6.

Solution

2n2+12n−30(n+2)(n−5)(n+3)

The process we use to subtract rational expressions with different denominators is the same as for addition. We just have to be very careful of the signs when subtracting the numerators.

Subtract: 8yy2−16−4y−4.

Solution
A mathematical expression shows a subtraction of two fractions: 8y over the quantity y squared minus 16, minus 4 over the quantity y minus 4.
Do the expressions have a common denominator? No.
Rewrite each expression with the LCD.
Find the LCD.y2−16=(y−4)(y+4)​y−4=y−4____________LCD=(y−4)(y+4)
Rewrite each rational expression as an
equivalent rational expression with the LCD.
An algebraic expression showing the subtraction of two rational terms, each with a numerator of 8y and 4(y+4) respectively, and a common denominator of (y-4)(y+4).
Simplify the numerators. A mathematical expression showing the subtraction of two rational expressions with a common denominator (y-4)(y+4).
Subtract the rational expressions. An algebraic fraction featuring (8y - 4y - 16) in the numerator and (y - 4)(y + 4) in the denominator.
Simplify the numerator. A mathematical expression displaying a fraction. The numerator is '4y - 16', and the denominator is '(y - 4)(y + 4)'.
Factor the numerator to look for common factors. A mathematical fraction shown with the expression 4(y-4) in the numerator and (y-4)(y+4) in the denominator. This represents an algebraic expression that can be simplified.
Remove common factors A mathematical expression showing the cancellation of the term (y-4) in both the numerator 4(y-4) and the denominator (y-4)(y+4) of a fraction during simplification.
Simplify. A fraction with a numerator of 4 and a denominator of (y + 4).

Subtract: 2xx2−4−1x+2.

Solution

1x−2

Subtract: 3z+3−6zz2−9.

Solution

−3z−3

There are lots of negative signs in the next example. Be extra careful.

Subtract:−3n−9n2+n−6−n+32−n.

Solution
A mathematical expression showing the subtraction of two algebraic fractions. The first fraction is (-3n-9)/(n^2+n-6) and the second is (n+3)/(2-n).
Factor the denominator. A mathematical expression featuring two fractions separated by a minus sign. The first fraction is (-3n - 9) divided by (n - 2)(n + 3). The second fraction is (n + 3) divided by (2 - n).
Since n−2 and 2−n are opposites, we will
multiply the second rational expression by −1−1.
Mathematical expression involving the subtraction of two rational functions. The second term includes a -1 factor in both its numerator and denominator, shown in red.
The image displays a mathematical instruction: 'Write (-1)(2 - n) as n - 2.' The term (-1) is highlighted in red, while the rest of the text is black. A mathematical expression showing the subtraction of two algebraic fractions. The first fraction is (-3n - 9) divided by ((n - 2)(n + 3)), and the second is (-1)(n + 3) divided by (n - 2).
Simplify. Remember, a−(−b)=a+b. A mathematical expression showing the sum of two algebraic fractions: (-3n - 9) / ((n - 2)(n + 3)) + (n + 3) / (n - 2).
Do the rational expressions have a
common denominator? No.
Find the LCD.n2+n−6=(n−2)(n+3)n−2=(n−2)_________________LCD=(n−2)(n+3)
Rewrite each rational expression as an
equivalent rational expression with the LCD.
An algebraic expression representing the sum of two rational functions with a common denominator of (n-2)(n+3).
Simplify the numerators. An algebraic expression showing the addition of two rational functions with a common denominator of (n-2)(n+3).
Add the rational expressions. An algebraic fraction with a numerator of -3n - 9 + n^2 + 6n + 9 and a denominator of (n-2)(n+3).
Simplify the numerator. A mathematical expression showing the fraction (n^2 + 3n) / ((n - 2)(n + 3)).
Factor the numerator to look for common factors. A rational algebraic expression showing the term (n+3) being canceled out from both the numerator n(n+3) and the denominator (n-2)(n+3).
Simplify. A mathematical expression displaying the fraction n divided by the quantity (n-2).

Subtract :3x−1x2−5x−6−26−x.

Solution

5x+1(x−6)(x+1)

Subtract: −2y−2y2+2y−8−y−12−y.

Solution

y+3y+4

Things can get very messy when both fractions must be multiplied by a binomial to get the common denominator.

Subtract: 4a2+6a+5−3a2+7a+10.

Solution
A mathematical expression featuring two fractions subtracted from each other: 4 divided by (a squared plus 6a plus 5) minus 3 divided by (a squared plus 7a plus 10).
Factor the denominators. A mathematical expression showing the subtraction of two algebraic fractions. The first fraction is 4 over (a+1)(a+5), and the second is 3 over (a+2)(a+5).
Do the rational expressions have a
common denominator? No.
Find the LCD.a2+6a+5=(a+1)(a+5)a2+7a+10=(a+5)(a+2)____________________________LCD=(a+1)(a+5)(a+2)
Rewrite each rational expression as an
equivalent rational expression with the LCD.
A mathematical expression showing the subtraction of two algebraic fractions. The first fraction is 4(a+2) over (a+1)(a+5)(a+2), and the second is 3(a+1) over (a+2)(a+5)(a+1).
Simplify the numerators. A mathematical expression displaying the subtraction of two algebraic fractions: (4a + 8)/((a + 1)(a + 5)(a + 2)) - (3a + 3)/((a + 2)(a + 5)(a + 1)).
Subtract the rational expressions. A mathematical fraction with the numerator '4a + 8 - (3a - 3)' and the denominator '(a + 1)(a + 5)(a + 2)'.
Simplify the numerator. A mathematical fraction with the numerator '4a + 8 - 3a + 3' and the denominator '(a + 1)(a + 5)(a + 2)'.
A fraction with numerator (a+5) and denominator (a+1)(a+5)(a+2).
Look for common factors. An algebraic fraction with (a+5) in the numerator and (a+1)(a+5)(a+2) in the denominator, showing (a+5) being canceled out from both the numerator and the denominator.
Simplify. A mathematical fraction displaying 1 in the numerator and the product of (a+1) and (a+2) in the denominator.

Subtract: 3b2−4b−5−2b2−6b+5.

Solution

1(b+1)(b−1)

Subtract: 4x2−4−3x2−x−2.

Solution

1(x+2)(x+1)

We follow the same steps as before to find the LCD when we have more than two rational expressions. In the next example, we will start by factoring all three denominators to find their LCD.

Simplify: 2uu−1+1u−2u−1u2−u.

Solution
A mathematical expression displaying three rational terms: (2u/(u-1)) + (1/u) - ((2u-1)/(u^2-u)).
Do the expressions have a common denominator? No.
Rewrite each expression with the LCD.
Find the LCD. u−1=(u−1) u=u u2−u=u(u−1)_______________ LCD=u(u−1)
Rewrite each rational expression as an
equivalent rational expression with the LCD.
A multi-term algebraic expression featuring fractions with a common denominator, u(u-1). The expression is (2u * u) / ((u - 1)u) + (1 * (u - 1)) / (u * (u - 1)) - (2u - 1) / (u(u - 1)), with some 'u' and '(u-1)' terms highlighted in red.
An algebraic expression showing the sum and difference of three rational functions involving the variable 'u'.
Write as one rational expression. A mathematical fraction with numerator 2u^2 + u - 1 - 2u + 1 and denominator u(u - 1).
Simplify. A mathematical expression presented as a fraction: (2u^2 - u) divided by u(u - 1).
Factor the numerator, and remove
common factors.
A mathematical expression showing a fraction with 'u-dot(2u-1)' in the numerator and 'u-dot(u-1)' in the denominator, set against a plain white background.
Simplify. A mathematical expression displaying a fraction with (2u - 1) in the numerator and (u - 1) in the denominator.

Simplify: vv+1+3v−1−6v2−1.

Solution

v+3v+1

Simplify: 3ww+2+2w+7−17w+4w2+9w+14.

Solution

3ww+7

Add and subtract rational functions

To add or subtract rational functions, we use the same techniques we used to add or subtract polynomial functions.

Find R(x)=f(x)−g(x) where f(x)=x+5x−2 and g(x)=5x+18x2−4.

Solution
A mathematical equation showing R(x) as the difference between two functions, f(x) and g(x), written as R(x) = f(x) - g(x).
Substitute in the functions f(x),g(x). A mathematical equation displays R(x) = (x+5)/(x-2) - (5x+18)/(x^2-4), showing a subtraction of two rational expressions.
Factor the denominators. A mathematical equation showing R(x) as the difference between two rational expressions: (x+5)/(x-2) and (5x+18)/((x-2)(x+2)).
Do the expressions have a common denominator? No.
Rewrite each expression with the LCD.
Find the LCD. x−2=(x−2) x2−4=(x−2)(x+2)___________________ LCD=(x−2)(x+2)
Rewrite each rational expression as an
equivalent rational expression with the LCD.
An algebraic expression showing R(x) as the difference of two rational functions with a common denominator (x-2)(x+2). The numerator of the first fraction is (x+5)(x+2) and the second is 5x+18.
Write as one rational expression. A mathematical equation defining the rational function R(x) with a numerator (x+5)(x+2)-(5x+18) and a denominator (x-2)(x+2).
Simplify. A rational function is displayed as R(x) = (x^2 + 7x + 10 - 5x - 18) / ((x - 2)(x + 2)). The numerator is a quadratic expression and the denominator is a product of two linear terms.
A rational function is displayed as R(x) = (x^2 + 2x - 8) / ((x - 2)(x + 2)).
Factor the numerator, and remove
common factors.
A mathematical equation shows R(x) as a fraction with the numerator (x + 4) times (x - 2) and the denominator (x - 2) times (x + 2), where the (x - 2) terms are crossed out in both the numerator and the denominator.
Simplify. A mathematical expression for R(x) is shown, where R(x) is equal to the fraction (x + 4) divided by (x + 2).

Find R(x)=f(x)−g(x) where f(x)=x+1x+3 and g(x)=x+17x2−x−12.

Solution

x−7x−4

Find R(x)=f(x)+g(x) where f(x)=x−4x+3 and g(x)=4x+6x2−9.

Solution

x2−3x+18(x+3)(x−3)

Access this online resource for additional instruction and practice with adding and subtracting rational expressions.

  • Add and Subtract Rational Expressions- Unlike Denominators

Key Concepts

  • Rational Expression Addition and Subtraction
    If p, q, and r are polynomials where r≠0, then
    pr+qr=p+qr and pr−qr=p−qr
  • How to find the least common denominator of rational expressions.
    1. Factor each expression completely.
    2. List the factors of each expression. Match factors vertically when possible.
    3. Bring down the columns.
    4. Write the LCD as the product of the factors.
  • How to add or subtract rational expressions.
    1. Determine if the expressions have a common denominator.
      • Yes – go to step 2.
      • No – Rewrite each rational expression with the LCD.
        • Find the LCD.
        • Rewrite each rational expression as an equivalent rational expression with the LCD.
    2. Add or subtract the rational expressions.
    3. Simplify, if possible.

Practice Makes Perfect

Add and Subtract Rational Expressions with a Common Denominator

In the following exercises, add.

215+715

Solution

35

724+1124

3c4c−5+54c−5

Solution

3c+54c−5

7m2m+n+42m+n

2r22r−1+15r−82r−1

Solution

r+8

3s23s−2+13s−103s−2

2w2w2−16+8ww2−16

Solution

2ww−4

7x2x2−9+21xx2−9

In the following exercises, subtract.

9a23a−7−493a−7

Solution

3a+7

25b25b−6−365b−6

3m26m−30−21m−306m−30

Solution

m−22

2n24n−32−18n−164n−32

6p2+3p+4p2+4p−5−5p2+p+7p2+4p−5

Solution

p+3p+5

5q2+3q−9q2+6q+8−4q2+9q+7q2+6q+8

5r2+7r−33r2−49−4r2+5r+30r2−49

Solution

r+9r+7

7t2−t−4t2−25−6t2+12t−44t2−25

Add and Subtract Rational Expressions whose Denominators are Opposites

In the following exercises, add or subtract.

10v2v−1+2v+41−2v

Solution

4

20w5w−2+5w+62−5w

10x2+16x−78x−3+2x2+3x−13−8x

Solution

x+2

6y2+2y−113y−7+3y2−3y+177−3y

z2+6zz2−25−3z+2025−z2

Solution

z+4z−5

a2+3aa2−9−3a−279−a2

2b2+30b−13b2−49−2b2−5b−849−b2

Solution

4b−3b−7

c2+5c−10c2−16−c2−8c−1016−c2

Find the Least Common Denominator of Rational Expressions

In the following exercises, ⓐ find the LCD for the given rational expressions ⓑ rewrite them as equivalent rational expressions with the lowest common denominator.

5x2−2x−8,2xx2−x−12

Solution

ⓐ (x+2)(x−4)(x+3)
ⓑ 5x+15(x+2)(x−4)(x+3),
2x2+4x(x+2)(x−4)(x+3)

8y2+12y+35,3yy2+y−42

9z2+2z−8,4zz2−4

Solution

ⓐ (z−2)(z+4)(z+2)
ⓑ 9z+18(z−2)(z+4)(z+2),
4z2+16z(z−2)(z+4)(z+2)

6a2+14a+45,5aa2−81

4b2+6b+9,2bb2−2b−15

Solution

ⓐ (b+3)(b+3)(b−5)
ⓑ 4b−20(b+3)(b+3)(b−5),
2b2+6b(b+3)(b+3)(b−5)

5c2−4c+4,3cc2−7c+10

23d2+14d−5,5d3d2−19d+6

Solution

ⓐ (d+5)(3d−1)(d−6)
ⓑ 2d−12(d+5)(3d−1)(d−6),
5d2+25d(d+5)(3d−1)(d−6)

35m2−3m−2,6m5m2+17m+6

Add and Subtract Rational Expressions with Unlike Denominators

In the following exercises, perform the indicated operations.

710x2y+415xy2

Solution

21y+8x30x2y2

112a3b2+59a2b3

3r+4+2r−5

Solution

5r−7(r+4)(r−5)

4s−7+5s+3

53w−2+2w+1

Solution

11w+1(3w−2)(w+1)

42x+5+2x−1

2yy+3+3y−1

Solution

2y2+y+9(y+3)(y−1)

3zz−2+1z+5

5ba2b−2a2+2bb2−4

Solution

b(5b+10+2a2)a2(b−2)(b+2)

4cd+3c+1d2−9

−3m3m−3+5mm2+3m−4

Solution

−mm+4

84n+4+6n2−n−2

3rr2+7r+6+9r2+4r+3

Solution

3(r2+6r+18)(r+1)(r+6)(r+3)

2ss2+2s−8+4s2+3s−10

tt−6−t−2t+6

Solution

2(7t−6)(t−6)(t+6)

x−3x+6−xx+3

5aa+3−a+2a+6

Solution

4a2+25a−6(a+3)(a+6)

3bb−2−b−6b−8

6m+6−12mm2−36

Solution

−6m−6

4n+4−8nn2−16

−9p−17p2−4p−21−p+17−p

Solution

p+2p+3

−13q−8q2+2q−24−q+24−q

−2r−16r2+6r−16−52−r

Solution

3r−2

2t−30t2+6t−27−23−t

2x+710x−1+3

Solution

4(8x+1)10x−1

8y−45y+2−6

3x2−3x−4−2x2−5x+4

Solution

x−5(x−4)(x+1)(x−1)

4x2−6x+5−3x2−7x+10

5x2+8x−9−4x2+10x+9

Solution

1(x−1)(x+1)

32x2+5x+2−12x2+3x+1

5aa−2+9a−2a+18a2−2a

Solution

5a2+7a−36a(a−2)

2bb−5+32b−2b−152b2−10b

cc+2+5c−2−10cc2−4

Solution

c−5c+2

6dd−5+1d+4−7d−5d2−d−20

3dd+2+4d−d+8d2+2d

Solution

3(d+1)d+2

2qq+5+3q−3−13q+15q2+2q−15

Add and Subtract Rational Functions

In the following exercises, find ⓐ R(x)=f(x)+g(x) ⓑ R(x)=f(x)−g(x).

f(x)=−5x−5x2+x−6 and
g(x)=x+12−x

Solution

ⓐ R(x)=−(x+8)(x+1)(x−2)(x+3) ⓑ R(x)=x+1x+3

f(x)=−4x−24x2+x−30 and
g(x)=x+75−x

f(x)=6xx2−64 and
g(x)=3x−8

Solution

ⓐ 3(3x+8)(x−8)(x+8)
ⓑ R(x)=3x+8

f(x)=5x+7 and
g(x)=10xx2−49

Writing Exercises

Donald thinks that 3x+4x is 72x. Is Donald correct? Explain.

Solution

Answers will vary.

Explain how you find the Least Common Denominator of x2+5x+4 and x2−16.

Felipe thinks 1x+1y is 2x+y. ⓐ Choose numerical values for x and y and evaluate 1x+1y. ⓑ Evaluate 2x+y for the same values of x and y you used in part ⓐ. ⓒ Explain why Felipe is wrong. ⓓ Find the correct expression for 1x+1y.

Solution

ⓐ Answers will vary.
ⓑ Answers will vary.
ⓒ Answers will vary.
ⓓ x+yxy

Simplify the expression 4n2+6n+9−1n2−9 and explain all your steps.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and six rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was add and subtract rational expressions with a common denominator. In row 3, the I can was add and subtract rational expressions with denominators that are opposites. In row 4, the I can find the least common denominator of rational expressions. In row 5, the I can was add and subtract rational expressions with unlike denominators. In row 6, the I can was add or subtract rational functions. There is the nothing in the other columns.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Simplify Complex Rational Expressions

Learning Objectives

By the end of this section, you will be able to:

  • Simplify a complex rational expression by writing it as division
  • Simplify a complex rational expression by using the LCD

Before you get started, take this readiness quiz.

Simplify: 35910.
If you missed this problem, review Example 4 in Fractions.

Solution

23

Simplify: 1−1342+4·5.
If you missed this problem, review Example 8 in Fractions.

Solution

154

Solve: 12x+14=18.
If you missed this problem, review Example 9 in Use a General Strategy to Solve Linear Equations.

Solution

x=-4

Simplify a Complex Rational Expression by Writing it as Division

Complex fractions are fractions in which the numerator or denominator contains a fraction. We previously simplified complex fractions like these:

3458x2xy6

In this section, we will simplify complex rational expressions, which are rational expressions with rational expressions in the numerator or denominator.

Complex Rational Expression

A complex rational expression is a rational expression in which the numerator and/or the denominator contains a rational expression.

Here are a few complex rational expressions:

4y−38y2−91x+1yxy−yx2x+64x−6−4x2−36

Remember, we always exclude values that would make any denominator zero.

We will use two methods to simplify complex rational expressions.

We have already seen this complex rational expression earlier in this chapter.

6x2−7x+24x−82x2−8x+3x2−5x+6

We noted that fraction bars tell us to divide, so rewrote it as the division problem:

(6x2−7x+24x−8)÷(2x2−8x+3x2−5x+6).

Then, we multiplied the first rational expression by the reciprocal of the second, just like we do when we divide two fractions.

This is one method to simplify complex rational expressions. We make sure the complex rational expression is of the form where one fraction is over one fraction. We then write it as if we were dividing two fractions.

Simplify the complex rational expression by writing it as division: 6x−43x2−16.

Solution
Step-by-step simplification of a complex rational expression involving algebraic terms.
6x−43x2−16
Rewrite the complex fraction as division. 6x−4÷3x2−16
Rewrite as the product of first times the
reciprocal of the second.
6x−4·x2−163
Factor. 3·2x−4·(x−4)(x+4)3
Multiply. 3·2(x−4)(x+4)3(x−4)
Remove common factors. 3·2(x−4)(x+4)3(x−4)
Simplify. 2(x+4)

Are there any value(s) of x that should not be allowed? The original complex rational expression had denominators of x−4 and x2−16. This expression would be undefined if x=4 or x=−4.

Simplify the complex rational expression by writing it as division: 2x2−13x+1.

Solution

23(x−1)

Simplify the complex rational expression by writing it as division: 1x2−7x+122x−4.

Solution

12(x−3)

Fraction bars act as grouping symbols. So to follow the Order of Operations, we simplify the numerator and denominator as much as possible before we can do the division.

Simplify the complex rational expression by writing it as division: 13+1612−13.

Solution
A fraction with the sum of 1/3 and 1/6 in the numerator, and the difference of 1/2 and 1/3 in the denominator, representing a complex mathematical expression.
Simplify the numerator and denominator.
Find the LCD and add the fractions in the numerator.
Find the LCD and subtract the fractions in the
denominator.
A complex fraction illustrating the addition of 1/3 and 1/6 in the numerator, and the subtraction of 1/3 from 1/2 in the denominator, with common denominator steps shown.
Simplify the numerator and denominator. A fraction where the numerator is 2/6 + 1/6 and the denominator is 3/6 - 2/6, demonstrating addition and subtraction of fractions with a common denominator.
Rewrite the complex rational expression as a division
problem.
A mathematical expression showing the division of two fractions: 3/6 ÷ 1/6.
Multiply the first by the reciprocal of the second. A mathematical expression showing the multiplication of two fractions: 3/6 multiplied by 6/1. The 'dot' symbol is used to denote multiplication between the fractions.
Simplify. 3

Simplify the complex rational expression by writing it as division: 12+2356+112.

Solution

1411

Simplify the complex rational expression by writing it as division: 34−1318+56.

Solution

1023

We follow the same procedure when the complex rational expression contains variables.

How to Simplify a Complex Rational Expression using Division

Simplify the complex rational expression by writing it as division: 1x+1yxy−yx.

Solution

Step 1 is to simplify the sum in the numerator and the difference in the denominator of complex rational expression, the quantity 1 divided by x plus 1 divided by y all divided by the quantity x divided by y minus y divided by x. The common denominator of the fractions in the complex rational expression is x y. Multiply the numerator and denominator of 1 divided by x by y over y. Multiply the numerator and denominator of 1 divided by y by x over x. Multiply the numerator and denominator of x divided by y by x over x. Multiply the numerator and denominator of y over x by y over y. The result is the quantity y divided by x y plus x divided by x y all divided by the quantity x squared divided by x y minus y squared divided by x y. Add the fractions in the numerator and subtract the fractions in the denominator. The result is the sum of y and x divided by x y all divided by the difference between x squared and y squared divided by x y. We now have just one rational expression in the numerator and one in the denominator.

Step 2 is to rewrite the complex rational expression as a division problem. Write the numerator divided by the denominator. The result is the quantity of the sum y and x divided by x y all divided by the quantity of the difference between x squared and y squared divided by x y. Step 3 is to divided the expressions. Multiply the first expression by the reciprocal of the second expression. The result is the quantity of the sum y and x divided by x y times the quantity x y divided by the difference between x squared and y squared. Factor any expressions if possible. The result is the product of x y and the sum of y and x all divided by the product of x y, the difference between x and y, and the sum of x and y. Remove the common factors, x y and the sum of x and y. Simplify. The result is 1 divided by the quantity x minus y.

Simplify the complex rational expression by writing it as division: 1x+1y1x−1y.

Solution

y+xy−x

Simplify the complex rational expression by writing it as division: 1a+1b1a2−1b2.

Solution

abb−a

We summarize the steps here.

Simplify a complex rational expression by writing it as division.

  1. Simplify the numerator and denominator.
  2. Rewrite the complex rational expression as a division problem.
  3. Divide the expressions.

Simplify the complex rational expression by writing it as division: n−4nn+51n+5+1n−5.

Solution
A complex fraction with a numerator of n minus 4n over n plus 5, and a denominator of 1 over n plus 5 plus 1 over n minus 5.
Simplify the numerator and denominator.
Find common denominators for the numerator and
denominator.
An algebraic expression showing a complex fraction. The numerator involves subtraction of two fractions, and the denominator involves addition of two fractions. Variables 'n' are present.
Simplify the numerators. A large algebraic fraction. The numerator is the difference of (n^2+5n)/(n+5) and 4n/(n+5). The denominator is the sum of (n-5)/((n+5)(n-5)) and (n+5)/((n-5)(n+5)).
Subtract the rational expressions in the numerator and
add in the denominator.
A mathematical expression showing a complex algebraic fraction. The numerator is (n^2 + 5n - 4n) / (n + 5) and the denominator is (n - 5 + n + 5) / ((n + 5)(n - 5)).
Simplify. (We now have one rational expression over
one rational expression.)
A complex algebraic fraction is displayed, with the numerator being (n^2 + n)/(n + 5) and the denominator being 2n/((n + 5)(n - 5)).
Rewrite as fraction division. A mathematical expression showing the division of two algebraic fractions: (n^2 + n) / (n + 5) divided by (2n) / ((n + 5)(n - 5)).
Multiply the first times the reciprocal of the second. A mathematical expression showing the product of two fractions: (n^2 + n) / (n + 5) multiplied by (n + 5)(n - 5) / (2n).
Factor any expressions if possible. A mathematical expression showing a fraction with n(n+1)(n+5)(n-5) in the numerator and (n+5)2n in the denominator, set against a plain white background.
Remove common factors. A fraction with algebraic terms in the numerator and denominator, showing cancellation of common factors 'n', '(n+5)', and '2' (implicitly from '2n').
Simplify. A mathematical expression showing the fraction (n+1)(n-5) all divided by 2.

Simplify the complex rational expression by writing it as division: b−3bb+52b+5+1b−5.

Solution

b(b+2)(b−5)3b−5

Simplify the complex rational expression by writing it as division: 1−3c+41c+4+c3.

Solution

3c+3

Simplify a Complex Rational Expression by Using the LCD

We “cleared” the fractions by multiplying by the LCD when we solved equations with fractions. We can use that strategy here to simplify complex rational expressions. We will multiply the numerator and denominator by the LCD of all the rational expressions.

Let’s look at the complex rational expression we simplified one way in Example 2. We will simplify it here by multiplying the numerator and denominator by the LCD. When we multiply by LCDLCD we are multiplying by 1, so the value stays the same.

Simplify the complex rational expression by using the LCD: 13+1612−13.

Solution
A complex fraction with (1/3 + 1/6) in the numerator and (1/2 - 1/3) in the denominator, illustrating basic fraction arithmetic.
The LCD of all the fractions in the whole expression is 6.
Clear the fractions by multiplying the numerator and
denominator by that LCD.
A mathematical expression presented as a fraction. The numerator is 6 multiplied by the sum of 1/3 and 1/6. The denominator is 6 multiplied by the difference of 1/2 and 1/3. The '6's are red.
Distribute. A mathematical expression showing a fraction where the numerator is (6 * 1/3) + (6 * 1/6) and the denominator is (6 * 1/2) - (6 * 1/3). The number 6 is highlighted in red in each term.
Simplify. A mathematical expression showing a fraction with 2+1 in the numerator and 3-2 in the denominator.
The fraction 3 over 1 is displayed in black text on a white background.
The number '3' is prominently displayed in a bold, black font against a stark white background.

Simplify the complex rational expression by using the LCD: 12+15110+15.

Solution

73

Simplify the complex rational expression by using the LCD: 14+3812−516.

Solution

103

We will use the same example as in Example 3. Decide which method works better for you.

How to Simplify a Complex Rational Expressing using the LCD

Simplify the complex rational expression by using the LCD: 1x+1yxy−yx.

Solution
Step 1 is to find the least common denominator of the complex rational expression, the sum of the quantity 1 divided by x and the quantity 1 divided by y all divided by the difference between the quantity x divided by y and the quantity y divided by x. Step 2 is to multiply the numerator and the denominator by the least common denominator, x y. The result is x y times the sum of the quantity 1 divided by x and the quantity 1 divided by y all divided by x y times the difference between the quantity x divided by y and the quantity y divided by x. Step 3 is to simplify the expression. Distribute x y in the numerator and the denominator. The result is x y times 1 divided by x plus x y times 1 divided by y all divided by x y times x divided by y plus x y times y divided by x. It simplifies to the sum of y and x divided by the quantity x squared minus y squared. Write the denominator as the difference of squares, the quantity x minus y times the quantity x plus y. The result is the quantity y plus x all divided by the quantity x minus y times the quantity x plus y. Remove the common factor, y plus x, from the numerator and denominator. The result is 1 divided by the quantity x minus y.

Simplify the complex rational expression by using the LCD: 1a+1bab+ba.

Solution

b+aa2+b2

Simplify the complex rational expression by using the LCD: 1x2−1y21x+1y.

Solution

y−xxy

Simplify a complex rational expression by using the LCD.

  1. Find the LCD of all fractions in the complex rational expression.
  2. Multiply the numerator and denominator by the LCD.
  3. Simplify the expression.

Be sure to start by factoring all the denominators so you can find the LCD.

Simplify the complex rational expression by using the LCD: 2x+64x−6−4x2−36.

Solution
A complex fraction with 2/(x+6) in the numerator, and in the denominator, 4/(x-6) minus 4/(x^2-36).
Find the LCD of all fractions in the complex rational
expression. The LCD is x2−36=(x+6)(x−6).
Multiply the numerator and denominator by the LCD. An algebraic fraction where both numerator and denominator are multiplied by (x+6)(x-6) as a step to simplify the expression and eliminate nested fractions.
Simplify the expression.
Distribute in the denominator. A mathematical expression displaying a fraction. The numerator has (x+6)(x-6) times 2/(x+6). The denominator subtracts two terms, both starting with (x+6)(x-6).
Simplify. A complex algebraic fraction showing several terms being canceled out, including (x+6) and (x-6), across the numerator and the two subtracted terms in the denominator.
Simplify. A mathematical expression showing the fraction 2(x - 6) divided by 4(x + 6) - 4.
To simplify the denominator, distribute
and combine like terms.
A mathematical expression showing a fraction with 2(x - 6) in the numerator and 4x + 20 in the denominator.
Factor the denominator. A mathematical fraction with 2(x-6) in the numerator and 4(x+5) in the denominator.
Remove common factors. A mathematical fraction showing the simplification of 2(x-6) over 2*2(x+5), with one '2' in the numerator and one in the denominator crossed out.
Simplify. A mathematical expression displaying a fraction with (x - 6) in the numerator and 2(x + 5) in the denominator.
Notice that there are no more factors
common to the numerator and denominator.

Simplify the complex rational expression by using the LCD: 3x+25x−2−3x2−4.

Solution

3(x−2)5x+7

Simplify the complex rational expression by using the LCD: 2x−7−1x+76x+7−1x2−49.

Solution

x+216x−43

Be sure to factor the denominators first. Proceed carefully as the math can get messy!

Simplify the complex rational expression by using the LCD: 4m2−7m+123m−3−2m−4.

Solution
A complex algebraic fraction with 4 over the quadratic m^2 - 7m + 12 in the numerator, and the difference of two rational expressions, 3/(m-3) - 2/(m-4), in the denominator.
Find the LCD of all fractions in the
complex rational expression.
The LCD is (m−3)(m−4).
Multiply the numerator and
denominator by the LCD.
Algebraic expression demonstrating the simplification of a complex fraction by multiplying both the numerator and denominator by (m-3)(m-4).
Simplify. A complex algebraic fraction showing cancellation of terms (m-3) and (m-4) in both the numerator and denominator, simplifying to 4 over (m-4)3 - (m-3)2.
Simplify. A mathematical expression showing the fraction 4 divided by the quantity 3 times (m minus 4) minus 2 times (m minus 3).
Distribute. A mathematical expression showing the fraction 4 over the quantity 3m minus 12 minus 2m plus 6, which simplifies to 4 over m minus 6.
Combine like terms. A fraction is displayed with a numerator of 4 and a denominator of m minus 6.

Simplify the complex rational expression by using the LCD: 3x2+7x+104x+2+1x+5.

Solution

35x+22

Simplify the complex rational expression by using the LCD: 4yy+5+2y+63yy2+11y+30.

Solution

2(2y2+13y+5)3y

Simplify the complex rational expression by using the LCD: yy+11+1y−1.

Solution
A complex fraction with y/(y+1) in the numerator and 1 + 1/(y-1) in the denominator, representing a rational algebraic expression.
Find the LCD of all fractions in the complex rational expression.
The LCD is (y+1)(y−1).
Multiply the numerator and denominator by the LCD. An algebraic complex fraction. The numerator is (y+1)(y-1)y/(y+1) and the denominator is (y+1)(y-1)(1 + 1/(y-1)). The factor (y+1)(y-1) is shown in red.
Distribute in the denominator and simplify. Simplifying a rational algebraic expression by cancelling common factors (y+1) and (y-1) from the numerator and denominator of the fraction.
Simplify. A mathematical fraction with (y-1)y in the numerator and (y+1)(y-1) + (y+1) in the denominator.
Simplify the denominator and leave the
numerator factored.
A mathematical expression displaying a fraction. The numerator is y(y-1) and the denominator is y^2 - 1 + y + 1.
A mathematical expression showing the fraction y(y-1) over y^2 + y.
Factor the denominator and remove factors
common with the numerator.
A mathematical fraction is shown, with y(y-1) as the numerator and y(y+1) as the denominator.
Simplify. A mathematical expression showing the fraction (y - 1) / (y + 1). The numerator is 'y minus 1' and the denominator is 'y plus 1', separated by a horizontal fraction bar.

Simplify the complex rational expression by using the LCD: xx+31+1x+3.

Solution

xx+4

Simplify the complex rational expression by using the LCD: 1+1x−13x+1.

Solution

x(x+1)3(x−1)

Access this online resource for additional instruction and practice with complex fractions.

  • Complex Fractions

Key Concepts

  • How to simplify a complex rational expression by writing it as division.
    1. Simplify the numerator and denominator.
    2. Rewrite the complex rational expression as a division problem.
    3. Divide the expressions.
  • How to simplify a complex rational expression by using the LCD.
    1. Find the LCD of all fractions in the complex rational expression.
    2. Multiply the numerator and denominator by the LCD.
    3. Simplify the expression.

Practice Makes Perfect

Simplify a Complex Rational Expression by Writing it as Division

In the following exercises, simplify each complex rational expression by writing it as division.

2aa+44a2a2−16

Solution

a−42a

3bb−5b2b2−25

5c2+5c−1410c+7

Solution

12(c−2)

8d2+9d+1812d+6

12+5623+79

Solution

1213

12+3435+710

23−1934+56

Solution

2057

12−1623+34

nm+1n1n−nm

Solution

n2+mm−n2

1p+pqqp−1q

1r+1t1r2−1t2

Solution

rtt−r

2v+2w1v2−1w2

x−2xx+31x+3+1x−3

Solution

(x+1)(x−3)2

y−2yy−42y−4+2y+4

2−2a+31a+3+a2

Solution

4a+1

4+4b−51b−5+b4

Simplify a Complex Rational Expression by Using the LCD

In the following exercises, simplify each complex rational expression by using the LCD.

13+1814+112

Solution

118

14+1916+112

56+29718−13

Solution

19

16+41535−12

cd+1d1d−dc

Solution

c2+cc−d2

1m+mnnm−1n

1p+1q1p2−1q2

Solution

pqq−p

2r+2t1r2−1t2

2x+53x−5+1x2−25

Solution

2x−103x+16

5y−43y+4+2y2−16

5z2−64+3z+81z+8+2z−8

Solution

3z−193z+8

3s+6+5s−61s2−36+4s+6

4a2−2a−151a−5+2a+3

Solution

43a−7

5b2−6b−273b−9+1b+3

5c+2−3c+75cc2+9c+14

Solution

2c+295c

6d−4−2d+72dd2+3d−28

2+1p−35p−3

Solution

2p−55

nn−23+5n−2

mm+54+1m−5

Solution

m(m−5)(4m−19)(m+5)

7+2q−21q+2

In the following exercises, simplify each complex rational expression using either method.

34−2712+514

Solution

1324

vw+1v1v−vw

2a+41a2−16

Solution

2(a−4)

3b2−3b−405b+5−2b−8

3m+3n1m2−1n2

Solution

3mnn−m

2r−91r+9+3r2−81

x−3xx+23x+2+3x−2

Solution

(x−1)(x−2)6

yy+32+1y−3

Writing Exercises

In this section, you learned to simplify the complex fraction 3x+2xx2−4 two ways: rewriting it as a division problem or multiplying the numerator and denominator by the LCD. Which method do you prefer? Why?

Solution

Answers will vary.

Efraim wants to start simplifying the complex fraction 1a+1b1a−1b by cancelling the variables from the numerator and denominator, 1a+1b1a−1b. Explain what is wrong with Efraim’s plan.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was simplify a complex rational expression by writing it as division. In row 3, the I can was simplify a complex rational expression by using the least common denominator.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

complex rational expression
A complex rational expression is a rational expression in which the numerator and/or denominator contains a rational expression.

Solve Rational Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve rational equations
  • Use rational functions
  • Solve a rational equation for a specific variable

Before you get started, take this readiness quiz.

Solve: 16x+12=13.
If you missed this problem, review Example 9 in Use a General Strategy to Solve Linear Equations.

Solution

x=−1

Solve: n2−5n−36=0.
If you missed this problem, review Example 2 in Polynomial Equations.

Solution

n=9,n=−4

Solve the formula 5x+2y=10 for y.
If you missed this problem, review Example 4 in Solve a Formula for a Specific Variable.

Solution

y=10−5x2

After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations. We have simplified many rational expressions so far in this chapter. Now we will solve a rational equation.

Rational Equation

A rational equation is an equation that contains a rational expression.

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

Rational ExpressionRational Equation 18x+12 y+6y2−36 1n−3+1n+4 18x+12=14 y+6y2−36=y+1 1n−3+1n+4=15n2+n−12

Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution to a rational equation.

Extraneous Solution to a Rational Equation

An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

We note any possible extraneous solutions, c, by writing x≠c next to the equation.

How to Solve a Rational Equation

Solve: 1x+13=56.

Solution
Step 1 is to find any value of the variable that makes the denominator of the zero. Remember that if x is equal to 0, then 1 divided by x is undefined. So the equation becomes the sum of 1 divided by x and one-third is equal to five-sixths, where x is not equal to 0. Step 2 is to find the least common denominator of all the fractions in the problem, 1 divided by x, one-third, and five-sixths. The least common denominator is 6 x. Step 3 is to clear the fractions in the equation by multiplying each side by the least common denominator. The result is 6 x times the sum of 1 divided by x and one-third is equal to 6 x times five-sixths. Simplify using the distributive property. The result is 6 x times the quantity1 divided by x plus 6 x times one-third is equal to 6 x times five-sixths, which simplifies to 6 plus 2 x is equal to 5 x. This simplifies to 6 is equal to 3 x. Step 4 is to solve the equation that results. The result is 2 is equal to x Step 5 is to check the solution. Remember that any solutions that makes the original expression undefined must be discarded. The solution is not 0. Substitute x is equal to 2 into the original equation, 1 divided by x plus one-third is equal to five-sixths. Is one-half plus one-third is equal to five-sixths a true equation? Is three-sixths plus two-sixth is equal to five-sixths a true equation? Three-sixths plus two-sixth is equal to five-sixths. Five-sixth is equal to five-sixth is a true equation. So, the solution is x is equal to 2

Solve: 1y+23=15.

Solution

y=−157

Solve: 23+15=1x.

Solution

x=1513

The steps of this method are shown.

Solve equations with rational expressions.

  1. Note any value of the variable that would make any denominator zero.
  2. Find the least common denominator of all denominators in the equation.
  3. Clear the fractions by multiplying both sides of the equation by the LCD.
  4. Solve the resulting equation.
  5. Check:
    • If any values found in Step 1 are algebraic solutions, discard them.
    • Check any remaining solutions in the original equation.

We always start by noting the values that would cause any denominators to be zero.

How to Solve a Rational Equation using the Zero Product Property

Solve: 1−5y=−6y2.

Solution
A mathematical equation is displayed: 1 - 5/y = -6/y^2. This is a rational equation that can be transformed into a quadratic equation by multiplying all terms by y^2 and rearranging them.
Note any value of the variable that would make
any denominator zero.
A mathematical equation is displayed: 1 - 5/y = -6/y^2, with the condition that y is not equal to 0. This is a rational equation that can be solved for y.
Find the least common denominator of all denominators in
the equation. The LCD is y2.
Clear the fractions by multiplying both sides of
the equation by the LCD.
A mathematical equation is displayed, showing y^2 multiplied by (1 - 5/y) on the left side, and y^2 multiplied by (-6/y^2) on the right side, both separated by an equals sign.
Distribute. A mathematical equation is displayed on a white background, reading 'y^2 * 1 - y^2(5/y) = y^2(-6/y^2)'. The term 'y^2' is highlighted in red in the first two instances.
Multiply. The image shows the quadratic equation y^2 - 5y = -6.
Solve the resulting equation. First
write the quadratic equation in standard form.
A mathematical equation is displayed, reading y squared minus 5y plus 6 equals 0.
Factor. A mathematical equation is displayed on a white background: (y-2)(y-3)=0. This quadratic equation is in factored form, indicating its roots are y=2 and y=3.
Use the Zero Product Property. A mathematical expression states 'y - 2 = 0 or y - 3 = 0' in black text against a white background.
Solve. The image shows the mathematical expression y = 2 or y = 3, indicating two possible values for the variable y.
Check.
We did not get 0 as an algebraic solution.


Two columns demonstrate checking solutions y=2 and y=3 for the equation 1 - 5/y = 6/y^2, showing step-by-step calculations that confirm both values are correct.
The solution is y=2, y=3.

Solve: 1−2x=15x2.

Solution

x=−3,x=5

Solve: 1−4y=12y2.

Solution

y=−2,y=6

In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.

Solve: 2x+2+4x−2=x−1x2−4.

Solution
A rational equation displayed on a white background, reading '2/(x+2) + 4/(x-2) = (x-1)/(x^2-4)'.
Note any value of the variable
that would make any denominator
zero.
A rational algebraic equation is shown: 2/(x+2) + 4/(x-2) = (x-1)/((x+2)(x-2)), with the conditions x not equal to -2 and x not equal to 2, ensuring denominators are non-zero.
Find the least common
denominator of all denominators
in the equation.
The LCD is (x+2)(x−2).
Clear the fractions by multiplying
both sides of the equation by the
LCD.
An algebraic equation is shown, with the term (x+2)(x-2) highlighted in red and appearing on both sides of the equality, multiplying other rational expressions.
Distribute. An algebraic equation with rational expressions. Each term is multiplied by the common factor (x+2)(x-2), aiming to clear the denominators before solving for x.
Remove common factors. A mathematical equation demonstrating the process of clearing denominators by multiplying each term by the least common denominator, (x+2)(x-2), leading to cancellation of terms and simplification.
Simplify. A mathematical equation is displayed against a white background: 2(x-2) + 4(x+2) = x-1. The equation involves a single variable 'x' with parentheses, multiplication, addition, and subtraction.
Distribute. A mathematical equation is displayed on a white background, which reads '2x - 4 + 4x + 8 = x - 1'.
Solve. A mathematical equation is displayed, showing '6x + 4 = x - 1' in black text against a plain white background, ready for solving.
A mathematical equation on a white background shows '5X = 5'.
An image displays the mathematical expression X = -1.
Check:
We did not get 2 or −2 as algebraic solutions.

Verifying the solution x = -1 for the rational equation. By substituting x = -1 into the original equation, both sides simplify to 2/3, confirming the validity of the solution.
The solution is x=−1.

Solve: 2x+1+1x−1=1x2−1.

Solution

x=23

Solve: 5y+3+2y−3=5y2−9.

Solution

y=2

In the next example, the first denominator is a trinomial. Remember to factor it first to find the LCD.

Solve: m+11m2−5m+4=5m−4−3m−1.

Solution
An algebraic equation showing (m+11) divided by (m^2-5m+4) equals 5 divided by (m-4) minus 3 divided by (m-1).
Note any value of the variable that
would make any denominator zero.
Use the factored form of the quadratic
denominator.
A mathematical equation is displayed, showing a rational expression equal to the difference of two other rational expressions, with specified restrictions for the variable 'm'.
Find the least common denominator
of all denominators in the equation.
The LCD is (m−4)(m−1).
Clear the fractions by
multiplying both sides of the
equation by the LCD.
An algebraic equation showing a step where both sides are multiplied by the common denominator (m-4)(m-1) to clear fractions, with the multiplying factor highlighted in red.
Distribute. The algebraic simplification of an equation by multiplying each term by the common denominator (m-4)(m-1) to eliminate fractions and solve for m.
Remove common factors. An algebraic equation is shown undergoing simplification, with terms (m-4) and (m-1) visibly crossed out. This step demonstrates the cancellation of common factors to clear denominators and simplify the rational expression.
Simplify. A mathematical equation is displayed on a white background: m + 11 = 5(m - 1) - 3(m - 4).
Solve the resulting equation. A mathematical equation is displayed, showing m + 11 = 5m - 5 - 3m + 12 on a white background.
4 = m
Check.
The only algebraic solution
was 4, but we said that 4 would make
a denominator equal to zero. The algebraic solution is an
extraneous solution.
There is no solution to this equation.

Solve: x+13x2−7x+10=6x−5−4x−2.

Solution

There is no solution.

Solve: y−6y2+3y−4=2y+4+7y−1.

Solution

There is no solution.

The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.

Solve: yy+6=72y2−36+4.

Solution
A mathematical equation is displayed on a white background, featuring the variable 'y' in a rational expression: y/(y+6) = 72/(y^2-36) + 4.
Factor all the denominators,
so we can note any value of
the variable that would make
any denominator zero.
A mathematical equation is displayed: y/(y+6) = 72/((y-6)(y+6)) + 4, with conditions y ≠ 6 and y ≠ -6. The equation involves algebraic fractions and the variable 'y'.
Find the least common denominator.
The LCD is (y−6)(y+6).
Clear the fractions. A mathematical equation where both sides are multiplied by (y-6)(y+6) to simplify a rational expression, likely a step in solving for y.
Simplify. A mathematical equation is shown: (y-6) * y = 72 + (y-6)(y+6) * 4.
Simplify. A quadratic equation is displayed, showing y(y-6) = 72 + 4(y^2-36).
Solve the resulting equation. A quadratic equation is displayed: y^2 - 6y = 72 + 4y^2 - 144. This equation involves the variable 'y' raised to the power of two, along with linear terms and constants, making it suitable for algebraic manipulation to find the values of 'y' that satisfy the equation.
A mathematical equation is displayed, showing '0 = 3y² + 6y - 72' in a black font against a white background.
A mathematical equation is displayed on a white background: 0 = 3(y^2 + 2y - 24).
A mathematical equation is displayed, reading 0 = 3(y+6)(y-4). This equation represents a quadratic function in factored form, where the roots of y are -6 and 4.
The image displays the equations y=-6 and y=4 written vertically on a white background, suggesting two distinct values for the variable y.
Check.

Verification of a mathematical solution. The image demonstrates checking if y=4 is a valid solution to the equation y/(y+6) = 72/(y^2-36) + 4, after y=-6 was identified as extraneous. The steps confirm that y=4 is a correct solution.
The solution is y=4.

Solve: xx+4=32x2−16+5.

Solution

x=3

Solve: yy+8=128y2−64+9.

Solution

y=7

In some cases, all the algebraic solutions are extraneous.

Solve: x2x−2−23x+3=5x2−2x+912x2−12.

Solution
An algebraic equation showing the subtraction of two rational expressions equal to a third rational expression. The equation is x/(2x-2) - 2/(3x+3) = (5x^2-2x+9)/(12x^2-12).
We will start by factoring all
denominators, to make it easier
to identify extraneous solutions and the LCD.
An algebraic equation showing the difference of two rational expressions equal to a third rational expression. The equation is x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)).
Note any value of the variable
that would make any denominator zero.
A complex algebraic equation is displayed, involving rational expressions with variable x in both numerators and denominators. It states that x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)), with restrictions x ≠ 1 and x ≠ -1.
Find the least common
denominator.
The LCD is 12(x−1)(x+1).
Clear the fractions. An algebraic equation illustrating the step of multiplying both sides by the common factor 12(x-1)(x+1). The left side consists of a difference of two fractions, and the right side is a single fractional term.
Simplify. A mathematical equation is displayed on a white background: 6(x+1)x - 4(x-1)2 = 5x^2 - 2x + 9. Some parts of the equation, specifically the variables 'x' within parentheses and the '1's, are highlighted in red.
Simplify. A mathematical equation is displayed on a white background: 6x(x+1) - 4 * 2(x-1) = 5x^2 - 2x + 9. The numbers and variables are in a dark gray font.
Solve the resulting equation. A mathematical equation shows 6x squared plus 6x minus 8x plus 8 equals 5x squared minus 2x plus 9, appearing on a white background.
The mathematical equation x^2 - 1 = 0 is displayed on a white background, suggesting a problem or concept from algebra or pre-calculus.
The image displays the algebraic equation (x-1)(x+1) = 0, representing a quadratic equation in factored form.
The mathematical expression 'x = 1 or x = -1' is displayed vertically on a white background.
Check.

 x=1 and x=−1 are extraneous solutions.
The equation has no solution.

Solve: y5y−10−53y+6=2y2−19y+5415y2−60.

Solution

There is no solution.

Solve: z2z+8−34z−8=3z2−16z−168z2+16z−64.

Solution

There is no solution.

Solve: 43x2−10x+3+33x2+2x−1=2x2−2x−3.

Solution
A complex algebraic equation is displayed, showing the sum of two fractions equaling a third. The equation is 4/(3x^2 - 10x + 3) + 3/(3x^2 + 2x - 1) = 2/(x^2 - 2x - 3).
Factor all the denominators, so we can note any value of the variable that would make any denominator
zero.
An algebraic equation showing the sum of two fractions equaling a third fraction: 4 over (3x-1)(x-3) plus 3 over (3x-1)(x+1) equals 2 over (x-3)(x+1).
x≠−1,x≠13,x≠3
Find the least common denominator. The LCD is (3x−1)(x+1)(x−3).
Clear the fractions.
An algebraic equation showing the product of three binomials, (3x-1)(x+1)(x-3), multiplied by a sum of two fractions on the left, equating to the same product multiplied by a single fraction on the right.
Simplify. A mathematical equation is displayed, showing '4(x+1) + 3(x-3) = 2(3x-1)'.
Distribute. A mathematical equation is displayed on a white background, which reads '4x + 4 + 3x - 9 = 6x - 2'.
Simplify. A mathematical equation is displayed on a white background, reading '7x - 5 = 6x - 2' in black text.
The equation 'x = 3' is displayed in the center of a plain white background.
The only algebraic solution was x=3, but we said that x=3 would make a denominator equal to zero. The algebraic solution is an extraneous solution.
There is no solution to this equation.

Solve: 15x2+x−6−3x−2=2x+3.

Solution

There is no solution.

Solve: 5x2+2x−3−3x2+x−2=1x2+5x+6.

Solution

There is no solution.

Use Rational Functions

Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.

For rational function, f(x)=2x−6x2−8x+15, ⓐ find the domain of the function, ⓑ solve f(x)=1, and ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain of a rational function is all real numbers except those that make the rational expression undefined. So to find them, we will set the denominator equal to zero and solve.

This table illustrates the step-by-step solution of a quadratic equation by factoring, concluding with its domain.
x2−8x+15=0
Factor the trinomial. (x−3)(x−5)=0
Use the Zero Product Property. x−3=0x−5=0
Solve. x=3x=5
The domain is all real numbers except x≠3,x≠5.
ⓑ
The mathematical equation f(x) = 1 is displayed in a clean, sans-serif font on a white background.
Substitute in the rational expression. A rational equation is displayed: (2x - 6) / (x^2 - 8x + 15) = 1. The equation involves a linear expression in the numerator and a quadratic expression in the denominator, set equal to 1.
Factor the denominator. The image displays a mathematical equation: a fraction with (2x - 6) in the numerator and the product of (x - 3) and (x - 5) in the denominator, set equal to 1. This is a rational equation to be solved for x.
Multiply both sides by the LCD,
(x−3)(x−5).
An algebraic step demonstrating the multiplication of both sides of an equation by the common denominator (x-3)(x-5) to eliminate the rational expression.
Simplify. A mathematical equation is displayed, showing '2x - 6 = x^2 - 8x + 15', a quadratic equation that can be rearranged and solved for x.
Solve. A mathematical equation is displayed against a white background. The equation reads '0 = x^2 - 10x + 21'.
Factor. A mathematical equation is shown with a blurred background. The equation reads '0 = (x - 7)(x - 3)', representing a quadratic equation in factored form.
Use the Zero Product Property. Two mathematical expressions are displayed horizontally. The expression on the left reads 'x - 7 = 0' and the expression on the right reads 'x - 3 = 0'.
Solve. Two mathematical expressions are displayed against a white background: 'x = 7' on the left and 'x = 3' on the right, both in dark grey font.
However, x=3 is outside the domain of this function, so we discard that root as extraneous.

ⓒ The value of the function is 1 when x=7. So the points on the graph of this function when f(x)=1 is (7,1))

For rational function, f(x)=8−xx2−7x+12, ⓐ find the domain of the function ⓑ solve f(x)=3 ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain is all real numbers except x≠3 and x≠4. ⓑ x=2,x=143
ⓒ (2,3),(143,3)

For rational function, f(x)=x−1x2−6x+5, ⓐ find the domain of the function ⓑ solve f(x)=4 ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain is all real numbers except x≠1 and x≠5. ⓑ x=214 ⓒ (214,4)

Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.

m=y−y1x−x1 Multiply both sides of the equation byx−x1.m(x−x1)=(y−y1x−x1)(x−x1) Simplify.m(x−x1)=y−y1 Rewrite the equation with theyterms on the left.y−y1=m(x−x1)

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through a point in Chapter 3. We will add one more step to solve for y.

Solve:m=y−2x−3 for y.

Solution
A mathematical equation shows the slope 'm' defined as a fraction: (y - 2) over (x - 3). The variable 'y' in the numerator is highlighted in red.
Note any value of the variable that would
make any denominator zero.
A mathematical equation for slope is shown as m = (y - 2) / (x - 3), with the condition x ≠ 3 to prevent division by zero. The variable 'y' is highlighted in red.
Clear the fractions by multiplying both sides of
the equation by the LCD, x−3.
A mathematical equation is displayed, showing (x - 3)m = (x - 3) multiplied by the fraction (y - 2) over (x - 3). The variable 'y' in the numerator is highlighted in red.
Simplify. A mathematical equation is displayed with the text 'xm - 3m = y - 2' in black and red lettering on a white background. The variable 'y' is highlighted in red.
Isolate the term with y. A mathematical equation is displayed, reading 'xm - 3m + 2 = y' with the 'y' highlighted in red. The equation appears in a dark gray font against a plain white background.

Solve: m=y−5x−4for y.

Solution

y=mx−4m+5

Solve: m=y−1x+5 for y.

Solution

y=mx+5m+1

Remember to multiply both sides by the LCD in the next example.

Solve: 1c+1m=1 for c.

Solution
A mathematical equation is displayed, showing '1/c + 1/m = 1 for c' in black text on a white background.
Note any value of the variable that would make
any denominator zero.
A mathematical equation displays '1/c + 1/m = 1, c not equal 0, m not equal 0'.
Clear the fractions by multiplying both sides of
the equations by the LCD, cm.
Mathematical equation: cm(1/c + 1/m) = cm(1). The expression shows a step in algebraic simplification, where the terms inside the parentheses simplify to 1.
Distribute. A mathematical equation is displayed against a white background, reading 'cm(1/c) + cm(1/m) = cm(1)', with 'cm' in red and the rest in black.
Simplify. A mathematical equation on a white background reads 'm + c = cm'.
Collect the terms with c to the right. A mathematical equation shows 'm = 500 - (', suggesting an incomplete expression or problem where 'm' is being defined in relation to '500' minus some value within parentheses.
Factor the expression on the right. The image displays a mathematical equation, m = c(m - 1), on a white background. This equation relates the variables 'm' and 'c', suggesting a problem or concept in algebra or mathematics.
To isolate c, divide both sides by m−1. The equation m/(m-1) = c(m-1)/(m-1) is displayed, with the common denominator 'm-1' highlighted in red.
Simplify by removing common factors. A mathematical equation is displayed, showing a fraction m divided by (m - 1), which is set equal to c. The equation reads: m/(m-1) = c.
Notice that even though we excluded c=0,m=0 from the original equation, we must also now state that m≠1.

Solve: 1a+1b=c for a.

Solution

a=bcb−1

Solve: 2x+13=1y for y.

Solution

y=3xx+6

Access this online resource for additional instruction and practice with equations with rational expressions.

  • Equations with Rational Expressions

Key Concepts

  • How to solve equations with rational expressions.
    1. Note any value of the variable that would make any denominator zero.
    2. Find the least common denominator of all denominators in the equation.
    3. Clear the fractions by multiplying both sides of the equation by the LCD.
    4. Solve the resulting equation.
    5. Check:
      • If any values found in Step 1 are algebraic solutions, discard them.
      • Check any remaining solutions in the original equation.

Practice Makes Perfect

Solve Rational Equations

In the following exercises, solve each rational equation.

1a+25=12

Solution

a=10

63−2d=49

45+14=2v

Solution

v=4021

38+2y=14

1−2m=8m2

Solution

m=−2,m=4

1+4n=21n2

1+9p=−20p2

Solution

p=−5,p=−4

1−7q=−6q2

53v−2=74v

Solution

v=14

82w+1=3w

3x+4+7x−4=8x2−16

Solution

x=−45

5y−9+1y+9=18y2−81

8z−10−7z+10=5z2−100

Solution

z=−145

9a+11−6a−11=6a2−121

−10q−2−7q+4=1

Solution

q=−18,q=−1

2s+7−3s−3=1

v−10v2−5v+4=3v−1−6v−4

Solution

no solution

w+8w2−11w+28=5w−7+2w−4

x−10x2+8x+12=3x+2+4x+6

Solution

no solution

y−5y2−4y−5=1y+1+1y−5

b+33b+b24=1b

Solution

b=−8

c+312c+c36=14c

dd+3=18d2−9+4

Solution

d=2

mm+5=50m2−25+6

nn+2−3=8n2−4

Solution

n=1

pp+7−8=98p2−49

q3q−9−34q+12=7q2+6q+6324q2−216

Solution

no solution

r3r−15−14r+20=3r2+17r+4012r2−300

s2s+6−25s+5=5s2−3s−710s2+40s+30

Solution

s=54

t6t−12−52t+10=t2−23t+7012t2+36t−120

2x2+2x−8−1x2+9x+20=4x2+3x−10

Solution

x=−43

5x2+4x+3+2x2+x−6=3x2−x−2

3x2−5x−6+3x2−7x+6=6x2−1

Solution

no solution

2x2+2x−3+3x2+4x+3=6x2−1

Solve Rational Equations that Involve Functions

For rational function, f(x)=x−2x2+6x+8, ⓐ find the domain of the function ⓑ solve f(x)=5 ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain is all real numbers except x≠−2 and x≠−4. ⓑ x=−3,x=−145 ⓒ (−3,5),(−145,5)

For rational function, f(x)=x+1x2−2x−3, ⓐ find the domain of the function ⓑ solve f(x)=1 ⓒ find the points on the graph at this function value.

For rational function, f(x)=2−xx2−7x+10, ⓐ find the domain of the function ⓑ solve f(x)=2 ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain is all real numbers except x≠2 and x≠5. ⓑ x=92, ⓒ (92,2)

For rational function, f(x)=5−xx2+5x+6,
ⓐ find the domain of the function
ⓑ solve f(x)=3
ⓒ the points on the graph at this function value.

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

Cr=2π for r.

Solution

r=C2π

Ir=P for r.

v+3w−1=12 for w.

Solution

w=2v+7

x+52−y=43 for y.

a=b+3c−2 for c.

Solution

c=b+3+2aa

m=n2−n for n.

1p+2q=4 for p.

Solution

p=q4q−2

3s+1t=2 for s.

2v+15=3w for w.

Solution

w=15v10+v

6x+23=1y for y.

m+3n−2=45 for n.

Solution

n=5m+234

r=s3−t for t.

Ec=m2 for c.

Solution

c=Em2

RT=W for T.

3x−5y=14 for y.

Solution

y=20x12−x

c=2a+b5 for a.

Writing Exercises

Your class mate is having trouble in this section. Write down the steps you would use to explain how to solve a rational equation.

Solution

Answers will vary.

Alek thinks the equation yy+6=72y2−36+4 has two solutions, y=−6 and y=4. Explain why Alek is wrong.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve rational equations. In row 3, the I can was solve rational equations involving functions. In row 4, the I can was solve rational equations for a specific variable.

ⓑ On a scale of 1−10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

extraneous solution to a rational equation
An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.
rational equation
A rational equation is an equation that contains a rational expression.

Solve Applications with Rational Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve proportions
  • Solve similar figure applications
  • Solve uniform motion applications
  • Solve work applications
  • Solve direct variation problems
  • Solve inverse variation problems

Before you get started, take this readiness quiz.

Solve: 2(n−1)−4=−10.
If you missed this problem, review Example 13 in Use a General Strategy to Solve Linear Equations.

Solution

n=-2

An express train and a charter bus leave Chicago to travel to Champaign. The express train can make the trip in two hours and the bus takes five hours for the trip. The speed of the express train is 42 miles per hour faster than the speed of the bus. Find the speed of the bus.
If you missed this problem, review Example 5 in Solve Mixture and Uniform Motion Applications.

Solution

The speed of the bus is 28 mph.

Solve 13x+14x=56.
If you missed this problem, review Example 9 in Use a General Strategy to Solve Linear Equations.

Solution

x=107

Solve Proportions

When two rational expressions are equal, the equation relating them is called a proportion.

Proportion

A proportion is an equation of the form ab=cd, where b≠0,d≠0.

The proportion is read “a is to b as c is to d.”

The equation 12=48 is a proportion because the two fractions are equal. The proportion 12=48 is read “1 is to 2 as 4 is to 8.”

Since a proportion is an equation with rational expressions, we will solve proportions the same way we solved rational equations. We’ll multiply both sides of the equation by the LCD to clear the fractions and then solve the resulting equation.

Solve: nn+14=57.

Solution
A mathematical equation displays 'n over n plus 14 equals 5 over 7' with the condition 'n is not equal to negative 14' to prevent division by zero, presented in a clean, sans-serif font.
Multiply both sides by LCD. A mathematical equation is displayed, showing 7(n + 14) multiplied by the fraction n/(n + 14) on the left side, equaling 7(n + 14) multiplied by 5/7 on the right side. The numbers 7 and 14 are in red.
Remove common factors on each side. A mathematical equation is displayed, reading '7n = 5(n + 14)'.
Simplify. A mathematical equation is displayed, showing '7n = 5n + 70' on a white background.
Solve for n. The equation 2n = 70 is displayed on a white background.
The image displays the text 'n = 35' in gray font on a white background.
Check.
A mathematical equation displays the fraction n over n plus 14, which is set equal to the fraction 5 over 7. The equation is n/(n+14) = 5/7, presented in a clean, sans-serif font.
The image displays a mathematical instruction, A math problem showing the fraction 35 over (35 + 14) on the left side, an equals sign with a question mark above it, and the fraction 5/7 on the right side, implying the need to solve for the missing numerator.
Simplify. A mathematical equation showing the fraction 35/49 being compared to 5/7, with a question mark over the equals sign indicating an inquiry into their equivalence. Both fractions simplify to 5/7.
Show common factors. A mathematical expression showing the fraction (5 multiplied by 7) over (7 multiplied by 7) being compared to the fraction 5 over 7, with a question mark above the equals sign.
Simplify. The image displays a mathematical equation: 5/7 = 5/7, followed by a checkmark, indicating that the statement is correct.

Solve the proportion: yy+55=38.

Solution

y=33

Solve the proportion: zz−84=−15.

Solution

z=14

Notice in the last example that when we cleared the fractions by multiplying by the LCD, the result is the same as if we had cross-multiplied.

The equation n divided by the quantity n plus 14 is equal to 5 divided by 7 can be solved by multiplying each side by the least common denominator, 7 times the quantity n plus 14. Multiplying by the least common denominator is a way to clear the fractions. The result is 7 n is equal to 5 times the quantity n plus 14. The equation n divided by the quantity n plus 14 is equal to 5 divided by 7 can also be solved using cross multiplication. Multiply n and 7. Multiply the quantity n plus 14 and 5. The result is also 7 n is equal to 5 times the quantity n plus 14. Cross multiplication also clears fractions.

For any proportion, ab=cd, we get the same result when we clear the fractions by multiplying by the LCD as when we cross-multiply.

Multiply each side of a proportion a divided by b is equal to c divided by d by the least common denominator, b d, to clear the fractions. The result is a d is equal to b c. Cross multiply to clear the fractions in the proportion a divided by b is equal to c divided by d. The cross products are a times d and b times c. The result is also a d is equal to b c.

To solve applications with proportions, we will follow our usual strategy for solving applications But when we set up the proportion, we must make sure to have the units correct—the units in the numerators must match each other and the units in the denominators must also match each other.

When pediatricians prescribe acetaminophen to children, they prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of the child’s weight. If Zoe weighs 80 pounds, how many milliliters of acetaminophen will her doctor prescribe?

Solution
Identify what we are asked to find,
and choose a variable to represent it.
How many ml of acetaminophen will the
doctor prescribe?
Let a = ml of acetaminophen.
Write a sentence that gives the
information to find it.
If 5 ml is prescribed for every
25 pounds, how much will be
prescribed for 80 pounds?
Translate into a proportion—be
careful of the units.
This image shows the mathematical identity 'ml/pounds = ml/pounds', a statement that is always true but might provoke questions about its context or implication, perhaps in a dimensional analysis problem. A mathematical equation is presented, showing the fraction 5/25 equal to the fraction a/80, set against a plain white background.
Multiply both sides by the LCD, 400. A mathematical equation is displayed on a white background: 400(5/25) = 400(a/80). The numbers '400' on both sides of the equation are highlighted in red.
Remove common factors on each side. Equation showing simplification steps. Twenty-five multiplied by sixteen times five twenty-fifths equals eighty multiplied by five times a over eighty. The twenty-fives cancel, and the eighties cancel.
Simplify, but don’t multiply on the left. Notice
what the next step will be.
A mathematical equation shows '16 ⋅ 5 = 5a'.
Solve for a. An algebraic equation is shown, with '16 dot 5' over 5 on the left side and '5a' over 5 on the right side. The number '5' in the denominator of both fractions is highlighted in red.
A clear image displaying the mathematical equation '16 = q' on a white background.
Check.
Is the answer reasonable?

Verifying a proportion solution: 'a=16' is substituted into 5/25 = a/80, simplifying to 1/5 = 1/5, confirming the estimate of 16ml.
Write a complete sentence. The pediatrician would prescribe 16 ml of
acetaminophen to Zoe.

Pediatricians prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of a child’s weight. How many milliliters of acetaminophen will the doctor prescribe for Emilia, who weighs 60 pounds?

Solution

The pediatrician will prescribe 12 ml of acetaminophen to Emilia.

For every 1 kilogram (kg) of a child’s weight, pediatricians prescribe 15 milligrams (mg) of a fever reducer. If Isabella weighs 12 kg, how many milligrams of the fever reducer will the pediatrician prescribe?

Solution

The pediatrician will prescribe 180 mg of fever reducer to Isabella.

Solve similar figure applications

When you shrink or enlarge a photo on a phone or tablet, figure out a distance on a map, or use a pattern to build a bookcase or sew a dress, you are working with similar figures. If two figures have exactly the same shape, but different sizes, they are said to be similar. One is a scale model of the other. All their corresponding angles have the same measures and their corresponding sides have the same ratio.

Similar Figures

Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides have the same ratio.

For example, the two triangles in Figure 1 are similar. Each side of ΔABC is four times the length of the corresponding side of ΔXYZ.

The first figure is triangle A B C with side A B 12 units long, side B C 16 units long, and side A C 20 units long. The second figure is triangle X Y Z with side X Y 3 units long, side Y X 4 units long, and side X Z is 5 units long. The measure of angle A is equal to the measure of angle X. The measure of angle B is equal to the measure of angle Y. The measure of angle C is equal to the measure of angle Z. 16 divided by 4 is equal to 20 divided 5 is equal to 12 divided by 3.

This is summed up in the Property of Similar Triangles.

Property of Similar Triangles

If ΔABC is similar to ΔXYZ, then their corresponding angle measure are equal and their corresponding sides have the same ratio.

The first figure is triangle A B C with side A B c units long, side B C a units long, and side A C b units long. The second figure is triangle X Y Z with side X Y x units long, side Y Z x units long, and side X Z y units long. The measure of angle A is equal to the measure of angle X. The measure of angle B is equal to the measure of angle Y. The measure of angle C is equal to the measure of angle Z. a divided by x is equal to b divided by y is equal to c divided by z.

To solve applications with similar figures we will follow the Problem-Solving Strategy for Geometry Applications we used earlier.

On a map, San Francisco, Las Vegas, and Los Angeles form a triangle. The distance between the cities is measured in inches. The figure on the left below represents the triangle formed by the cities on the map. If the actual distance from Los Angeles to Las Vegas is 270 miles, find the distance from Los Angeles to San Francisco.

The first figure is a triangle labeled “Distances on map.” The triangle is formed by San Francisco, Las Vegas, and Los Angeles. The distance between San Francisco and Las Vegas is 2.1 inches. The distance between Las Vegas and Los Angeles is 1 inch. The distance between Los Angeles and San Francisco is 1.3 inches. The second figure is a triangle labeled “Actual Distances.” The triangle is formed by San Francisco, Las Vegas, and Los Angeles. The distance between Las Vegas and Los Angeles is 270 miles. The distance between Los Angeles and San Francisco is labeled x.
Solution

Since the triangles are similar, the corresponding sides are proportional.

Read the problem. Draw the figures and label
it with the given information.
The figures are shown above.
Identify what we are looking for. the actual distance from Los Angeles
to San Francisco
Name the variables. Let x = distance from Los Angeles
to San Francisco.
Translate into an equation.
Since the triangles are similar, the
corresponding sides are proportional. We’ll
make the numerators “miles” and
the denominators “inches”.
A mathematical equation showing a proportion: x miles / 1.3 inches = 270 miles / 1 inch, used to calculate an unknown distance based on a given scale.
Solve the equation. A mathematical equation reads: 1.3(x/1.3) = 1.3(270/1). The number 1.3 is highlighted in red, indicating a multiplication step to solve for x.
The equation 'x = 351' is prominently displayed in the center of a plain white background, rendered in a dark gray, sans-serif font.
Check.
On the map, the distance from Los Angeles
to San Francisco is more than
the distance from Los Angeles to
Las Vegas. Since 351 is more than 270
the answer makes sense.

A mathematical problem showing the verification of x=351 in a proportion. It demonstrates that 351 miles / 1.3 inches is equal to 270 miles / 1 inch, confirming the solution.
Answer the question. The distance from Los Angeles to
San Francisco is 351 miles.

On the map, Seattle, Portland, and Boise form a triangle. The distance between the cities is measured in inches. The figure on the left below represents the triangle formed by the cities on the map. The actual distance from Seattle to Boise is 400 miles.

The figure is a triangle formed by Portland, Seattle, and Boise. The distance between Portland and Seattle is 1.5 inches. The distance between Seattle and Boise is 4 inches. The distance between Boise and Portland is 3.5 inches.

Find the actual distance from Seattle to Portland.

Solution

The distance is 150 miles.

Find the actual distance from Portland to Boise.

Solution

The distance is 350 miles.

We can use similar figures to find heights that we cannot directly measure.

Tyler is 6 feet tall. Late one afternoon, his shadow was 8 feet long. At the same time, the shadow of a tree was 24 feet long. Find the height of the tree.

Solution
Read the problem and draw a figure.
We are looking for h, the height of the tree. Math problem illustrating similar triangles to determine the height 'h' of a tree. A 6-unit tall person stands 8 units from the end of a 24-unit long shadow cast by the tree under the sun.
We will use similar triangles to write an equation.
The small triangle is similar to the large triangle. A mathematical equation is displayed on a white background, showing 'h/24 = 6/8' in black text.
Solve the proportion. A mathematical equation is shown with the expression 24 multiplied by 6/8 on the left side of the equals sign, and 24 multiplied by h/24 on the right side.
The image displays the equation '18 = h' in a bold, sans-serif font on a white background, representing a simple algebraic expression.
Simplify.
Check.

Verifying h=18 in the proportion 6/8 = h/24, simplifying to 3/4 = 3/4. This confirms the relationship between Tyler's height and shadow length, applying it to a tree's height.

A telephone pole casts a shadow that is 50 feet long. Nearby, an 8 foot tall traffic sign casts a shadow that is 10 feet long. How tall is the telephone pole?

Solution

The telephone pole is 40 feet tall.

A pine tree casts a shadow of 80 feet next to a 30 foot tall building which casts a 40 feet shadow. How tall is the pine tree?

Solution

The pine tree is 60 feet tall.

Solve Uniform Motion Applications

We have solved uniform motion problems using the formula D=rt in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

This chart has two columns and three rows. The first row is a header row and the second column is labeled “Rate times Time is equal to Distance.” There is nothing in the rest of the chart.

The formula D=rt assumes we know r and t and use them to find D. If we know D and r and need to find t, we would solve the equation for t and get the formula t=Dr.

We have also explained how flying with or against the wind affects the speed of a plane. We will revisit that idea in the next example.

An airplane can fly 200 miles into a 30 mph headwind in the same amount of time it takes to fly 300 miles with a 30 mph tailwind. What is the speed of the airplane?

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

Math problem setup showing 30 mph wind: 300 miles with wind (r+30) and 200 miles against wind (r-30).
We fill in the chart to organize the information.
We are looking for the speed of the airplane. Let r = the speed of the airplane.
When the plane flies with the wind,
the wind increases its speed and so the rate is r + 30.
When the plane flies against the wind,
the wind decreases its speed and the rate is r − 30.
Write in the rates.
Write in the distances.


Since D=r·t, we solve for t and get t=Dr.

We divide the distance by the rate in each row, and place the expression in the time column.
A table showing rate (r +/- 30), time, and distance calculations for headwind and tailwind scenarios, with distances of 200 and 300 units respectively.
We know the times are equal and so we write
our equation.
200r−30=300r+30
We multiply both sides by the LCD. (r+30)(r−30)(200r−30)=(r+30)(r−30)(300r+30)
Simplify. (r+30)(200)=(r−30)300
200r+6000=300r−9000
Solve. 15000=100r
Check.
Is 150 mph a reasonable speed for an airplane? Yes.
If the plane is traveling 150 mph and the wind is 30 mph,
Tailwind150+30=180mph300180=53hoursHeadwind150−30=120mph200120=53hours
The times are equal, so it checks. The plane was traveling 150 mph.

Link has an electric bike which runs at a constant speed. That speed will be reduced by the amount of any headwind and increased by the amount of any tailwind. Link can ride his bike 20 miles into a 3 mph headwind in the same amount of time he can ride 30 miles with a 3 mph tailwind. What is Link’s biking speed?

Solution

Link’s biking speed is 15 mph.

On a river that flows at 7 miles per hour, Danica can take a motorboat 5 miles upstream, in the same amount of time she can take her motorboat 12 miles downstream. How fast is Danica's boat in a lake?

Solution

The speed of Danica’s boat is 17 mph.

In the next example, we will know the total time resulting from travelling different distances at different speeds.

Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

A diagram depicting an 8-mile run and a 24-mile bike ride over a total duration of 3 hours.
We fill in the chart to organize the information.
We are looking for Jazmine’s running speed. Let r = Jazmine’s running speed.
Her biking speed is 4 miles faster than her
running speed.
r + 4 = her biking speed
The distances are given, enter them into the chart.

Since D=r·t, we solve for t and get t=Dr.

We divide the distance by the rate in each row, and place the expression in the time column.
A table illustrates a rate-time-distance problem, showing variables and expressions for running and biking activities. 'r' represents the running rate, and the total time for both activities is 3.
Write a word sentence. Her time plus the time biking is 3 hours.
Translate the sentence to get the equation. 8r+24r+4=3
Solve. r(r+4)(8r+24r+4)=3·r(r+4)8(r+4)+24r=3r(r+4)8r+32+24r=3r2+12r32+32r=3r2+12r0=3r2−20r−320=(3r+4)(r−8)
(3r+4)=0(r−8)=0
r=−43r=8
Check.
A negative speed does not make sense in this problem,
so r=8 is the solution.
Is 8 mph a reasonable running speed? Yes.
If Jazmine’s running rate is 4, then her biking rate,
r+4, which is 8+4=12.
Run8mph8miles8mph=1hour Bike12mph24miles12mph=2hours
Total 3 hours. Jazmine’s running speed is 8 mph.

Dennis went cross-country skiing for 6 hours on Saturday. He skied 20 mile uphill and then 20 miles back downhill, returning to his starting point. His uphill speed was 5 mph slower than his downhill speed. What was Dennis’ speed going uphill and his speed going downhill?

Solution

Dennis’s uphill speed was 5 mph and his downhill speed was 10 mph.

Joon drove 4 hours to his home, driving 208 miles on the interstate and 40 miles on country roads. If he drove 15 mph faster on the interstate than on the country roads, what was his rate on the country roads?

Solution

Joon’s rate on the country roads was 50 mph.

Once again, we will use the uniform motion formula solved for the variable t.

Hamilton rode his bike downhill 12 miles on the river trail from his house to the ocean and then rode uphill to return home. His uphill speed was 8 miles per hour slower than his downhill speed. It took him 2 hours longer to get home than it took him to get to the ocean. Find Hamilton’s downhill speed.

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

A diagram for a distance-rate-time problem. It shows a 12-mile distance. One scenario is 8 mph slower and takes 2 hours longer than another, implying a calculation for speed or time.
We fill in the chart to organize the information.
We are looking for Hamilton’s downhill speed. Let h = Hamilton’s downhill speed.
His uphill speed is 8 miles per hour slower.
Enter the rates into the chart.
h − 8 = Hamilton’s uphill speed
The distance is the same in both directions.
12 miles.

Since D=r·t, we solve for t and get t=Dr.

We divide the distance by the rate in each row, and place the expression in the time column.
A table showing rate, time, and distance for downhill and uphill travel. Downhill: rate 'h', time '12/h', distance '12'. Uphill: rate 'h-8', time '12/(h-8)', distance '12'.
Write a word sentence about the line. He took 2 hours longer uphill than downhill.
The uphill time is 2 more than the downhill time.
Translate the sentence to get the equation. 12h−8=12h+2
Solve. h(h−8)(12h−8)=h(h−8)(12h+2)12h=12(h−8)+2h(h−8)12h=12h−96+2h2−16h0=2h2−16h−960=2(h2−8h−48)0=2(h−12)(h+4)h−12=0h+4=0h=12h=−4
Check.
Is 12 mph a reasonable speed for biking downhill? Yes.
Downhill12 mph12 miles12 mph=1 hourUphill12−8=4 mph12 miles4 mph=3 hours.
The uphill time is 2 hours more that the downhill time.
Hamilton’s downhill speed is 12 mph.

Kayla rode her bike 75 miles home from college one weekend and then rode the bus back to college. It took her 2 hours less to ride back to college on the bus than it took her to ride home on her bike, and the average speed of the bus was 10 miles per hour faster than Kayla’s biking speed. Find Kayla’s biking speed.

Solution

Kayla’s biking speed was 15 mph.

Victoria jogs 12 miles to the park along a flat trail and then returns by jogging on an 20 mile hilly trail. She jogs 1 mile per hour slower on the hilly trail than on the flat trail, and her return trip takes her two hours longer. Find her rate of jogging on the flat trail.

Solution

Victoria jogged 6 mph on the flat trail.

Solve Work Applications

The weekly gossip magazine has a big story about the Princess’ baby and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours to do the job and Press #2 takes 12 hours to do the job. How long will it take the printer to get the magazine printed with both presses running together?

This is a typical ‘work’ application. There are three quantities involved here—the time it would take each of the two presses to do the job alone and the time it would take for them to do the job together.

If Press #1 can complete the job in 6 hours, in one hour it would complete 16 of the job.

If Press #2 can complete the job in 12 hours, in one hour it would complete 112 of the job.

We will let t be the number of hours it would take the presses to print the magazines with both presses running together. So in 1 hour working together they have completed 1t of the job.

We can model this with the word equation and then translate to a rational equation. To find the time it would take the presses to complete the job if they worked together, we solve for t.

A chart will help us organize the information. We are looking for how many hours it would take to complete the job with both presses running together.

Let t = the number of hours needed to
complete the job together.
Enter the hours per job for Press #1,
Press #2, and when they work together.

If a job on Press #1 takes 6 hours, then in 1 hour 16 of the job is completed.

Similarly find the part of the job completed/hours for Press #2 and when thet both together.

Write a word sentence.
A table displays work rates for two presses. Press #1 takes 6 hours, completing 1/6 of the job per hour. Press #2 takes 12 hours, completing 1/12 per hour. Together, they take 't' hours, completing 1/t per hour.
The part completed by Press #1 plus the part completed by Press #2 equals the amount completed together.
Translate into an equation. A work-rate problem showing how Press #1 (1/6) and Press #2 (1/12) combine their efforts. The equation 1/6 + 1/12 = 1/t determines their total time 't' to complete the work together.
Solve. A mathematical equation displays the sum of two fractions, 1/6 and 1/12, equaling 1/t, demonstrating a problem involving inverse proportions or rates.
Mutiply by the LCD, 12t A mathematical equation is shown: 12t (1/6 + 1/12) = 12t (1/t).
Simplify. A step-by-step solution to the linear equation 2t + t = 12, showing the simplification to 3t = 12 and the final solution t = 4.
When both presses are running it
takes 4 hours to do the job.

Keep in mind, it should take less time for two presses to complete a job working together than for either press to do it alone.

Suppose Pete can paint a room in 10 hours. If he works at a steady pace, in 1 hour he would paint 110 of the room. If Alicia would take 8 hours to paint the same room, then in 1 hour she would paint 18 of the room. How long would it take Pete and Alicia to paint the room if they worked together (and didn’t interfere with each other’s progress)?

Solution

This is a ‘work’ application. A chart will help us organize the information. We are looking for the numbers of hours it will take them to paint the room together.

In one hour Pete did 110 of the job. Alicia did 18 of the job. And together they did 1t of the job.

Let t be the number of hours needed
to paint the room together.
Enter the hours per job for Pete, Alicia, and when they work together.

In 1 hour working together, they have completed 1t of the job.
Similarly, find the part of the job
completed/hour by Pete and then by Alicia.
Table showing the hours Pete (10), Alicia (8), and both together (t) take to complete a job, and their respective hourly work rates (1/10, 1/8, 1/t).
Write a word sentence. The work completed by Pete plus the work
completed by Alicia equals the total
work completed.
Work completed by:

A math equation showing how to combine individual work rates (1/10 for Pete, 1/8 for Alicia) to find the combined work rate (1/t) for both working together. A mathematical equation is displayed on a white background, showing the sum of two fractions equaling another fraction: 1/10 + 1/8 = 1/t.
Multiply by the LCD, 40t. A mathematical equation shows '40t (1/10 + 1/8) = 40t (1/t)' on a white background. The term '40t' is highlighted in red on both sides of the equation.
Distribute. A mathematical equation is displayed on a white background, showing 40t multiplied by 1/10, added to 40t multiplied by 1/8, which equals 40t multiplied by 1/t.
Simplify and solve. A multi-step algebraic equation is solved, showing 4t + 5t = 40 simplifying to 9t = 40, and finally resulting in t = 40/9.
We’ll write as a mixed number
so that we can convert it to hours
and minutes.
A mathematical expression states t = 4 and 4/9 hours, indicating a time duration.
Remember, 1 hour = 60 minutes. A mathematical equation shows 't = 4 hours + 4/9 (60 minutes)', calculating a total time by adding 4 hours to four-ninths of 60 minutes.
Multiply, and then round to the
nearest minute.
The equation displays 't = 4 hours + 27 minutes' in a simple black font against a white background.
It would take Pete and Alica about
4 hours and 27 minutes to paint the room.

One gardener can mow a golf course in 4 hours, while another gardener can mow the same golf course in 6 hours. How long would it take if the two gardeners worked together to mow the golf course?

Solution

When the two gardeners work together it takes 2 hours and 24 minutes.

Daria can weed the garden in 7 hours, while her mother can do it in 3. How long will it take the two of them working together?

Solution

When Daria and her mother work together it takes 2 hours and 6 minutes.

Ra’shon can clean the house in 7 hours. When his sister helps him it takes 3 hours. How long does it take his sister when she cleans the house alone?

Solution

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take Ra’shon’s sister to complete the job by herself.

Let s be the number of hours Ra’shon’s
sister takes to clean the house alone.
Enter the hours per job for Ra’shon, his
sister, and when they work together.
If Ra’shon takes 7 hours, then in 1 hour 17
of the job is completed.
If Ra’shon’s sister takes s hours, then in
1 hour 1s of the job is completed.
Table shows hours to clean a house for Ra'shon (7), his sister (s), and together (3), alongside their hourly work rates (1/7, 1/s, 1/3 respectively), suggesting a work-rate problem setup.
Write a word sentence. The part completed by Ra’shon plus the part
by his sister equals the amount completed together.
Translate to an equation. A math problem illustrates the combined work rate of Ra'shon and his sister with the equation: 1/7 + 1/s = 1/3, where 's' represents the sister's time to complete the work.
Solve. A mathematical equation showing the sum of two fractions: 1/7 + 1/5 = 1/3.
Multiply by the LCD, 21s. The image displays the first two steps of solving an algebraic equation. The equation starts as 21s(1/7 + 1/s) = (1/3)21s and simplifies to 3s + 21 = 7s.
Simplify. A mathematical equation showing the steps to solve for 's'. The equation -4s = -21 is solved by dividing both sides by -4, resulting in s = -21/-4, which simplifies to s = 21/4.
Write as a mixed number to
convert it to hours and minutes.
The image displays the equation s = 5 1/4 hours, indicating a duration of five and a quarter hours.
There are 60 minutes in 1 hour. A mathematical equation showing the conversion of time: s = 5 hours + 1/4 (60 minutes) simplifies to s = 5 hours + 15 minutes, demonstrating how a quarter hour equals 15 minutes.
It would take Ra’shon’s sister 5 hours and
15 minutes to clean the house alone.

Alice can paint a room in 6 hours. If Kristina helps her it takes them 4 hours to paint the room. How long would it take Kristina to paint the room by herself?

Solution

Kristina can paint the room in 12 hours.

Tracy can lay a slab of concrete in 3 hours, with Jordan’s help they can do it in 2 hours. If Jordan works alone, how long will it take?

Solution

It will take Jordan 6 hours.

Solve Direct Variation Problems

When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities. We will discuss direct variation and inverse variation in this section.

Lindsay gets paid $15 per hour at her job. If we let s be her salary and h be the number of hours she has worked, we could model this situation with the equation

s=15h

Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

Direct Variation

For any two variables x and y, y varies directly with x if

y=kx,wherek≠0

The constant k is called the constant of variation.

In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates x and y. Then we can use that equation to find values of y for other values of x.

We’ll list the steps here.

Solve direct variation problems.

  1. Write the formula for direct variation.
  2. Substitute the given values for the variables.
  3. Solve for the constant of variation.
  4. Write the equation that relates x and y using the constant of variation.

Now we’ll solve an application of direct variation.

When Raoul runs on the treadmill at the gym, the number of calories, c, he burns varies directly with the number of minutes, m, he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

ⓐ Write the equation that relates c and m. ⓑ How many calories would he burn if he ran on the treadmill for 25 minutes?

Solution
ⓐ
The number of calories, c, varies directly with
the number of minutes, m, on the treadmill,
and c=315 when m=18.
Write the formula for direct variation. The equation y = kx is shown, representing direct proportionality where y varies directly with x, and k is the constant of proportionality.
We will use c in place of y and m in place of x. A mathematical equation 'c = km' is displayed in black text on a white background.
Substitute the given values for the variables. The image displays the mathematical equation 315 = k   • 18, presented in a red and black color scheme on a white background. This equation asks for the value of 'k' that makes the statement true.
Solve for the constant of variation. A mathematical equation displays 315/18 = (k * 18)/18, with the second 18 on the right side of the equation highlighted in red.
A mathematical equation displays '17.5 = k' in a clear, digital font against a white background, representing a simple algebraic statement or a numerical value assigned to a variable.
Write the equation that relates c and m. A white background displays the mathematical equation 'c = km' in black text, where 'c' equals 'k' multiplied by 'm'.
Substitute in the constant of variation. The image displays text that reads 'c = 17.5m' on a plain white background, likely representing a variable 'c' with a value of 17.5 meters, possibly from a physics or engineering problem.
ⓑ
Find c when m = 25.
Write the equation that relates c and m. The image displays the text 'c = 17.5m' in a dark grey, sans-serif font on a plain white background. The text is centrally aligned and clearly visible.
Substitute the given value for m. The image displays the equation c = 17.5(25). The variable 'c', the equals sign, '17.5', and the opening parenthesis are in black. The number '25' within the parentheses is highlighted in red, followed by a black closing parenthesis.
Simplify. The image displays the equation 'c = 437,5' in dark gray text against a plain white background.
Raoul would burn 437.5 calories if
he used the treadmill for 25 minutes.

The number of calories, c, burned varies directly with the amount of time, t, spent exercising. Arnold burned 312 calories in 65 minutes exercising.

ⓐ Write the equation that relates c and t. ⓑ How many calories would he burn if he exercises for 90 minutes?

Solution

ⓐ c=4.8t ⓑ He would burn 432 calories.

The distance a moving body travels, d, varies directly with time, t, it moves. A train travels 100 miles in 2 hours

ⓐ Write the equation that relates d and t. ⓑ How many miles would it travel in 5 hours?

Solution

ⓐ d=50t ⓑ It would travel 250 miles.

Solve Inverse Variation Problems

Many applications involve two variable that vary inversely. As one variable increases, the other decreases. The equation that relates them is y=kx.

Inverse Variation

For any two variables x and y, y varies inversely with x if

y=kx,wherek≠0

The constant k is called the constant of variation.

The word ‘inverse’ in inverse variation refers to the multiplicative inverse. The multiplicative inverse of x is 1x.

We solve inverse variation problems in the same way we solved direct variation problems. Only the general form of the equation has changed. We will copy the procedure box here and just change ‘direct’ to ‘inverse’.

Solve inverse variation problems.

  1. Write the formula for inverse variation.
  2. Substitute the given values for the variables.
  3. Solve for the constant of variation.
  4. Write the equation that relates x and y using the constant of variation.

The frequency of a guitar string varies inversely with its length. A 26 in.-long string has a frequency of 440 vibrations per second.

ⓐ Write the equation of variation. ⓑ How many vibrations per second will there be if the string’s length is reduced to 20 inches by putting a finger on a fret?

Solution
ⓐ
The frequency varies
inversely with the length.
Name the variables. Let f = frequency.
L = length
Write the formula for inverse variation. The image displays the mathematical equation for inverse proportionality, represented as y = k/x, where 'y' is inversely proportional to 'x' and 'k' is the constant of proportionality.
We will use f in place of y and L in place of x. A mathematical formula displaying the relationship f = k/L, where 'f' equals 'k' divided by 'L'. This equation could represent concepts such as frequency, force, or flux in relation to a constant and a length.
Substitute the given values for the variables. A mathematical equation displays f equals 440 when L equals 26, with the numbers 440 in light blue and 26 in red for emphasis.
A mathematical equation is displayed on a white background, showing 440 equals k divided by 26.
Solve for the constant of variation A mathematical equation is displayed: 26 multiplied by 440 equals 26 multiplied by the fraction k over 26.
The equation 11,440 = k is displayed in a clear, white background, showing a numerical value assigned to the variable 'k'.
Write the equation that relates f and L. A mathematical formula shows f equals k divided by L.
Substitute the constant of variation A mathematical formula is shown with f equals to a fraction, where the numerator is 11,440 and the denominator is L.
ⓑ
Steps demonstrating how to calculate the frequency (f) of a guitar string for a given length (L=20 inches) using a direct proportion formula.
Find f when L=20.
Write the equation that relates fandL. f=11,440L
Substitute the given value for L. f=11,44020
Simplify. f=572
A 20″-guitar string has frequency 572
vibrations per second.

The number of hours it takes for ice to melt varies inversely with the air temperature. Suppose a block of ice melts in 2 hours when the temperature is 65 degrees Celsius.

ⓐ Write the equation of variation. ⓑ How many hours would it take for the same block of ice to melt if the temperature was 78 degrees?

Solution

ⓐ h=130t ⓑ 123 hours

Xander’s new business found that the daily demand for its product was inversely proportional to the price, p. When the price is $5, the demand is 700 units.

ⓐ Write the equation of variation. ⓑ What is the demand if the price is raised to $7?

Solution

ⓐ x=3500p ⓑ 500 units

Access this online resource for additional instruction and practice with applications of rational expressions

  • Applications of Rational Expressions

Key Concepts

  • A proportion is an equation of the form ab=cd, where b≠0,d≠0. The proportion is read “a is to b as c is to d.”
  • Property of Similar Triangles
    If ΔABC is similar to ΔXYZ, then their corresponding angle measure are equal and their corresponding sides have the same ratio.
    The first figure is triangle A B C with side A B c units long, side B C a units long, and side A C b units long. The second figure is triangle X Y Z with side X Y x units long, side Y Z x units long, and side X Z y units long. The measure of angle A is equal to the measure of angle X. The measure of angle B is equal to the measure of angle Y. The measure of angle C is equal to the measure of angle Z. a divided by x is equal to b divided by y is equal to c divided by z.
  • Direct Variation
    • For any two variables x and y, y varies directly with x if y=kx, where k≠0. The constant k is called the constant of variation.
    • How to solve direct variation problems.
      1. Write the formula for direct variation.
      2. Substitute the given values for the variables.
      3. Solve for the constant of variation.
      4. Write the equation that relates x and y.
  • Inverse Variation
    • For any two variables x and y, y varies inversely with x if y=kx, where k≠0. The constant k is called the constant of variation.
    • How to solve inverse variation problems.
      1. Write the formula for inverse variation.
      2. Substitute the given values for the variables.
      3. Solve for the constant of variation.
      4. Write the equation that relates x and y.

Practice Makes Perfect

Solve Proportions

In the following exercises, solve each proportion.

x56=78

Solution

x=49

5672=y9

98154=−7p

Solution

p=−11

72156=−6q

aa+12=47

Solution

a=16

bb−16=119

m+9025=m+3015

Solution

m=60

n+104=40−n6

2p+48=p+186

Solution

p=30

q−22=2q−718

In the following exercises, solve.

Kevin wants to keep his heart rate at 160 beats per minute while training. During his workout he counts 27 beats in 10 seconds.

ⓐ How many beats per minute is this? ⓑ Has Kevin met his target heart rate?

Solution

ⓐ 162 beats per minute ⓑ yes

Jesse’s car gets 30 miles per gallon of gas.

ⓐ If Las Vegas is 285 miles away, how many gallons of gas are needed to get there and then home? ⓑ If gas is $3.09 per gallon, what is the total cost of the gas for the trip?

Pediatricians prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of a child’s weight. How many milliliters of acetaminophen will the doctor prescribe for Jocelyn, who weighs 45 pounds?

Solution

9 ml

A veterinarian prescribed Sunny, a 65-pound dog, an antibacterial medicine in case an infection emerges after her teeth were cleaned. If the dosage is 5 mg for every pound, how much medicine was Sunny given?

A new energy drink advertises 106 calories for 8 ounces. How many calories are in 12 ounces of the drink?

Solution

159 calories

One 12-ounce can of soda has 150 calories. If Josiah drinks the big 32-ounce size from the local mini-mart, how many calories does he get?

Kyra is traveling to Canada and will change $250 US dollars into Canadian dollars. At the current exchange rate, $1 US is equal to $1.3 Canadian. How many Canadian dollars will she get for her trip?

Solution

325 Canadian dollars

Maurice is traveling to Mexico and needs to exchange $450 into Mexican pesos. If each dollar is worth 12.29 pesos, how many pesos will he get for his trip?

Ronald needs a morning breakfast drink that will give him at least 390 calories. Orange juice has 130 calories in one cup. How many cups does he need to drink to reach his calorie goal?

Solution

3 cups

Sonya drinks a 32-ounce energy drink containing 80 calories per 12 ounce. How many calories did she drink?

Phil wants to fertilize his lawn. Each bag of fertilizer covers about 4,000 square feet of lawn. Phil’s lawn is approximately 13,500 square feet. How many bags of fertilizer will he have to buy?

Solution

4 bags

An oatmeal cookie recipe calls for 12 cup of butter to make 4 dozen cookies. Hilda needs to make 10 dozen cookies for the bake sale. How many cups of butter will she need?

Solve Similar Figure Applications

In the following exercises, the triangles are similar. Find the length of the indicated side.


The first figure is triangle A B C with side A B 15 units long, side B C 9 units long, and side A C b units long. The second figure is triangle X Y Z with side X Y 10 units long, side Y Z x units long, and side X Z 8 units long.

ⓐ side x ⓑ side b

Solution

ⓐ 6 ⓑ 12


The first figure is triangle DEF with side D E 5 halves units long, side E F d units long, and side D F 1 unit long. The second figure is triangle N P Q with side N P q units long, side P Q 11 halves units long, and side N Q 9 units long.

ⓐ side d ⓑ side q

In the following exercises, use the map shown. On the map, New York City, Chicago, and Memphis form a triangle. The actual distance from New York to Chicago is 800 miles.

The figure is a triangle formed by Memphis, Chicago, and New York. The distance between Memphis and Chicago is 5.4 inches. The distance between Chicago and New York is 8 inches. The distance between New York and Memphis is 9.5 inches.

Find the actual distance from New York to Memphis.

Solution

950 miles

Find the actual distance from Chicago to Memphis.

In the following exercises, use the map shown. On the map, Atlanta, Miami, and New Orleans form a triangle. The actual distance from Atlanta to New Orleans is 420 miles.

The figure is a triangle formed by New Orleans, Atlanta, and Miami. The distance between New Orleans and Atlanta is 2.1 inches. The distance between Atlanta and Miami is 3 inches. The distance between Miami and New Orleans is 3.4 inches.

Find the actual distance from New Orleans to Miami.

Solution

680 miles

Find the actual distance from Atlanta to Miami.

In the following exercises, answer each question.

A 2-foot-tall dog casts a 3-foot shadow at the same time a cat casts a one foot shadow. How tall is the cat ?

Solution

23 foot (8 in.)

Larry and Tom were standing next to each other in the backyard when Tom challenged Larry to guess how tall he was. Larry knew his own height is 6.5 feet and when they measured their shadows, Larry’s shadow was 8 feet and Tom’s was 7.75 feet long. What is Tom’s height?

The tower portion of a windmill is 212 feet tall. A six foot tall person standing next to the tower casts a seven-foot shadow. How long is the windmill’s shadow?

Solution

247.3 feet

The height of the Statue of Liberty is 305 feet. Nikia, who is standing next to the statue, casts a 6-foot shadow and she is 5 feet tall. How long should the shadow of the statue be?

Solve Uniform Motion Applications

In the following exercises, solve the application problem provided.

Mary takes a sightseeing tour on a helicopter that can fly 450 miles against a 35-mph headwind in the same amount of time it can travel 702 miles with a 35-mph tailwind. Find the speed of the helicopter.

Solution

160 mph

A private jet can fly 1,210 miles against a 25-mph headwind in the same amount of time it can fly 1694 miles with a 25-mph tailwind. Find the speed of the jet.

A boat travels 140 miles downstream in the same time as it travels 92 miles upstream. The speed of the current is 6mph. What is the speed of the boat?

Solution

29 mph

Darrin can skateboard 2 miles against a 4-mph wind in the same amount of time he skateboards 6 miles with a 4-mph wind. Find the speed Darrin skateboards with no wind.

Jane spent 2 hours exploring a mountain with a dirt bike. First, she rode 40 miles uphill. After she reached the peak she rode for 12 miles along the summit. While going uphill, she went 5 mph slower than when she was on the summit. What was her rate along the summit?

Solution

30 mph

Laney wanted to lose some weight so she planned a day of exercising. She spent a total of 2 hours riding her bike and jogging. She biked for 12 miles and jogged for 6 miles. Her rate for jogging was 10 mph less than biking rate. What was her rate when jogging?

Byron wanted to try out different water craft. He went 62 miles downstream in a motor boat and 27 miles downstream on a jet ski. His speed on the jet ski was 10 mph faster than in the motor boat. Bill spent a total of 4 hours on the water. What was his rate of speed in the motor boat?

Solution

20 mph

Nancy took a 3-hour drive. She went 50 miles before she got caught in a storm. Then she drove 68 miles at 9 mph less than she had driven when the weather was good. What was her speed driving in the storm?

Chester rode his bike uphill 24 miles and then back downhill at 2 mph faster than his uphill. If it took him 2 hours longer to ride uphill than downhill, what was his uphill rate?

Solution

4 mph

Matthew jogged to his friend’s house 12 miles away and then got a ride back home. It took him 2 hours longer to jog there than ride back. His jogging rate was 25 mph slower than the rate when he was riding. What was his jogging rate?

Hudson travels 1080 miles in a jet and then 240 miles by car to get to a business meeting. The jet goes 300 mph faster than the rate of the car, and the car ride takes 1 hour longer than the jet. What is the speed of the car?

Solution

60 mph

Nathan walked on an asphalt pathway for 12 miles. He walked the 12 miles back to his car on a gravel road through the forest. On the asphalt he walked 2 miles per hour faster than on the gravel. The walk on the gravel took one hour longer than the walk on the asphalt. How fast did he walk on the gravel.

John can fly his airplane 2800 miles with a wind speed of 50 mph in the same time he can travel 2400 miles against the wind. If the speed of the wind is 50 mph, find the speed of his airplane.

Solution

650 mph

Jim’s speedboat can travel 20 miles upstream against a 3-mph current in the same amount of time it travels 22 miles downstream with a 3-mph current speed . Find the speed of the Jim’s boat.

Hazel needs to get to her granddaughter’s house by taking an airplane and a rental car. She travels 900 miles by plane and 250 miles by car. The plane travels 250 mph faster than the car. If she drives the rental car for 2 hours more than she rode the plane, find the speed of the car.

Solution

50 mph

Stu trained for 3 hours yesterday. He ran 14 miles and then biked 40 miles. His biking speed is 6 mph faster than his running speed. What is his running speed?

When driving the 9-hour trip home, Sharon drove 390 miles on the interstate and 150 miles on country roads. Her speed on the interstate was 15 more than on country roads. What was her speed on country roads?

Solution

50 mph

Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?

Dana enjoys taking her dog for a walk, but sometimes her dog gets away, and she has to run after him. Dana walked her dog for 7 miles but then had to run for 1 mile, spending a total time of 2.5 hours with her dog. Her running speed was 3 mph faster than her walking speed. Find her walking speed.

Solution

3 mph

Ken and Joe leave their apartment to go to a football game 45 miles away. Ken drives his car 30 mph faster Joe can ride his bike. If it takes Joe 2 hours longer than Ken to get to the game, what is Joe’s speed?

Solve Work Applications

Mike, an experienced bricklayer, can build a wall in 3 hours, while his son, who is learning, can do the job in 6 hours. How long does it take for them to build a wall together?

Solution

2 hours

It takes Sam 4 hours to rake the front lawn while his brother, Dave, can rake the lawn in 2 hours. How long will it take them to rake the lawn working together?

Mia can clean her apartment in 6 hours while her roommate can clean the apartment in 5 hours. If they work together, how long would it take them to clean the apartment?

Solution

2 hours and 44 minutes

Brian can lay a slab of concrete in 6 hours, while Greg can do it in 4 hours. If Brian and Greg work together, how long will it take?

Josephine can correct her students test papers in 5 hours, but if her teacher’s assistant helps, it would take them 3 hours. How long would it take the assistant to do it alone?

Solution

7 hours and 30 minutes

Washing his dad’s car alone, eight year old Levi takes 2.5 hours. If his dad helps him, then it takes 1 hour. How long does it take Levi’s dad to wash the car by himself?

At the end of the day Dodie can clean her hair salon in 15 minutes. Ann, who works with her, can clean the salon in 30 minutes. How long would it take them to clean the shop if they work together?

Solution

10 min

Ronald can shovel the driveway in 4 hours, but if his brother Donald helps it would take 2 hours. How long would it take Donald to shovel the driveway alone?

Solve Direct Variation Problems

In the following exercises, solve.

If y varies directly as x and y=14,​ when x=3. find the equation that relates x and y.

Solution

y=143x

If a varies directly as b and a=16,​​ when b=4. find the equation that relates a and b.

If p varies directly as q and p=9.6,​​ when q=3. find the equation that relates p and q.

Solution

p=3.2q

If v varies directly as w and v=8,​ when ​w=12. find the equation that relates v and w.

The price, P, that Eric pays for gas varies directly with the number of gallons, g, he buys. It costs him $50 to buy 20 gallons of gas.

ⓐ Write the equation that relates P and g. ⓑ How much would 33 gallons cost Eric?

Solution

ⓐ P=2.5g ⓑ $82.50

Joseph is traveling on a road trip. The distance, d, he travels before stopping for lunch varies directly with the speed, v, he travels. He can travel 120 miles at a speed of 60 mph.

ⓐ Write the equation that relates d and v. ⓑ How far would he travel before stopping for lunch at a rate of 65 mph?

The mass of a liquid varies directly with its volume. A liquid with mass 16 kilograms has a volume of 2 liters.

ⓐ Write the equation that relates the mass to the volume. ⓑ What is the volume of this liquid if its mass is 128 kilograms?

Solution

ⓐ m=8v ⓑ 16 liters

The length that a spring stretches varies directly with a weight placed at the end of the spring. When Sarah placed a 10-pound watermelon on a hanging scale, the spring stretched 5 inches.

ⓐ Write the equation that relates the length of the spring to the weight. ⓑ What weight of watermelon would stretch the spring 6 inches?

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 6 inch will support a maximum load of 108 pounds.

ⓐ Write the equation that relates the load to the diagonal of the cross-section. ⓑ What load will a beam with a 10-inch diagonal support?

Solution

ⓐ L=3d2 ⓑ 300 pounds

The area of a circle varies directly as the square of the radius. A circular pizza with a radius of 6 inches has an area of 113.04 square inches.

ⓐ Write the equation that relates the area to the radius. ⓑ What is the area of a personal pizza with a radius 4 inches?

Solve Inverse Variation Problems

In the following exercises, solve.

If y varies inversely with x and y=5 when x=4, find the equation that relates x and y.

Solution

y=20x

If p varies inversely with q and p=2 when q=1, find the equation that relates p and q.

If v varies inversely with w and v=6 when w=12, find the equation that relates v and w.

Solution

v=3w

If a varies inversely with b and a=12 when b=13, find the equation that relates a and b.

In the following exercises, write an inverse variation equation to solve the following problems.

The fuel consumption (mpg) of a car varies inversely with its weight. A Toyota Corolla weighs 2800 pounds getting 33 mpg on the highway.

ⓐ Write the equation that relates the mpg to the car’s weight. ⓑ What would the fuel consumption be for a Toyota Sequoia that weighs 5500 pounds?

Solution

ⓐ g=92,400w ⓑ 16.8 mpg

A car’s value varies inversely with its age. Jackie bought a 10-year-old car for $2,400.

ⓐ Write the equation that relates the car’s value to its age. ⓑ What will be the value of Jackie’s car when it is 15 years old?

The time required to empty a tank varies inversely as the rate of pumping. It took Ada 5 hours to pump her flooded basement using a pump that was rated at 200 gpm (gallons per minute).

ⓐ Write the equation that relates the number of hours to the pump rate. ⓑ How long would it take Ada to pump her basement if she used a pump rated at 400 gpm?

Solution

ⓐ t=1000r ⓑ 2.5 hours

On a string instrument, the length of a string varies inversely as the frequency of its vibrations. An 11-inch string on a violin has a frequency of 400 cycles per second.

ⓐ Write the equation that relates the string length to its frequency. ⓑ What is the frequency of a 10 inch string?

Paul, a dentist, determined that the number of cavities that develops in his patient’s mouth each year varies inversely to the number of minutes spent brushing each night. His patient, Lori, had four cavities when brushing her teeth 30 seconds (0.5 minutes) each night.

ⓐ Write the equation that relates the number of cavities to the time spent brushing. ⓑ How many cavities would Paul expect Lori to have if she had brushed her teeth for 2 minutes each night?

Solution

ⓐ c=2t ⓑ 1 cavity

Boyle’s law states that if the temperature of a gas stays constant, then the pressure varies inversely to the volume of the gas. Braydon, a scuba diver, has a tank that holds air under a pressure of 220 psi.

ⓐ Write the equation that relates pressure to volume. ⓑ If the pressure increases to 330 psi, how much air can Braydon’s tank hold?

The cost of a ride service varies directly with the distance traveled. It costs $35 for a ride from the city center to the airport, 14 miles away.

ⓐ Write the equation that relates the cost, c, with the number of miles, m. ⓑ What would it cost to travel 22 miles with this service?

Solution

ⓐ c=2.5m ⓑ $55

The number of hours it takes Jack to drive from Boston to Bangor is inversely proportional to his average driving speed. When he drives at an average speed of 40 miles per hour, it takes him 6 hours for the trip.

ⓐ Write the equation that relates the number of hours, h, with the speed, s. ⓑ How long would the trip take if his average speed was 75 miles per hour?

Writing Exercises

Marisol solves the proportion 144a=94 by ‘cross multiplying,’ so her first step looks like 4·144=9·a. Explain how this differs from the method of solution shown in Example 2.

Solution

Answers will vary.

Paula and Yuki are roommates. It takes Paula 3 hours to clean their apartment. It takes Yuki 4 hours to clean the apartment. The equation 13+14=1t can be used to find t, the number of hours it would take both of them, working together, to clean their apartment. Explain how this equation models the situation.

In your own words, explain the difference between direct variation and inverse variation.

Solution

Answers will vary.

Make up an example from your life experience of inverse variation.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and seven rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve proportions. In row 3, the I can was solve similar figure applications. In row 4, the I can was solve uniform motion applications. In row 5, the I can was solve work applications. In row 6, the I can was solve direct variation problems. In row 7, the I can was solve inverse variation problems.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

proportion
When two rational expressions are equal, the equation relating them is called a proportion.
similar figures
Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides have the same ratio.

Solve Rational Inequalities

Learning Objectives

By the end of this section, you will be able to:

  • Solve rational inequalities
  • Solve an inequality with rational functions

Before you get started, take this readiness quiz.

Find the value of x−5 when ⓐ x=6 ⓑ x=−3 ⓒ x=5.
If you missed this problem, review Example 6 in Use the Language of Algebra.

Solution

ⓐ 1; ⓑ −8; ⓒ 0

Solve: 8−2x<12.
If you missed this problem, review Example 5 in Solve Linear Inequalities.

Solution

x>−2

Write in interval notation: −3≤x<5.
If you missed this problem, review Example 2 in Solve Linear Inequalities.

Solution

[−3,5)

Solve Rational Inequalities

We learned to solve linear inequalities after learning to solve linear equations. The techniques were very much the same with one major exception. When we multiplied or divided by a negative number, the inequality sign reversed.

Having just learned to solve rational equations we are now ready to solve rational inequalities. A rational inequality is an inequality that contains a rational expression.

Rational Inequality

A rational inequality is an inequality that contains a rational expression.

Inequalities such as 32x>1,2xx−3<4,2x−3x−6≥x, and 14−2x2≤3x are rational inequalities as they each contain a rational expression.

When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities. We especially must remember that when we multiply or divide by a negative number, the inequality sign must reverse.

Another difference is that we must carefully consider what value might make the rational expression undefined and so must be excluded.

When we solve an equation and the result is x=3, we know there is one solution, which is 3.

When we solve an inequality and the result is x>3, we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a zero partition number and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.

This figure shows the solution, the interval 3 to infinity, of the inequality x is greater than 3 on a number line. The values range from negative 5 to 5 on the number line. The inequality is modeled by an open parenthesis at the zero partition number 3 and shading the right.

To solve a rational inequality, we first must write the inequality with only one quotient on the left and 0 on the right.

Next we determine the zero partition numbers to use to divide the number line into intervals. A zero partition number is a number which make the rational expression zero or undefined.

We then will evaluate the factors of the numerator and denominator, and find the quotient in each interval. This will identify the interval, or intervals, that contains all the solutions of the rational inequality.

We write the solution in interval notation being careful to determine whether the endpoints are included.

Solve and write the solution in interval notation: x−1x+3≥0.

Solution

Step 1. Write the inequality as one quotient on the left and zero on the right.

Our inequality is in this form. x−1x+3≥0

Step 2. Determine the zero partition numbers—the points where the rational expression will be zero or undefined.

The rational expression will be zero when the numerator is zero. Since x−1=0 when x=1, then 1 is a zero partition number.

The rational expression will be undefined when the denominator is zero. Since x+3=0 when x=−3, then −3 is a zero partition number.

The zero partition numbers are 1 and −3.

Step 3. Use the zero partition numbers to divide the number line into intervals.

This figure shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0.

The number line is divided into three intervals:

(−∞,−3)(−3,1)(1,∞)

Step 4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

Interval(−∞,−3)

The number −4 is in the interval (−∞,−3). Test x=−4 in the expression in the numerator and the denominator.

This figure labels the expression, x minus 1, as the “numerator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 5. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 1. It labels the result “negative”.

Above the number line, mark the factor x−1 negative and mark the factor x+3 negative.

Since a negative divided by a negative is positive, mark the quotient positive in the interval (−∞,−3).

This figure shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is negative, which is positive. It shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3.
Interval(−3,1)

The number 0 is in the interval (−3,1). Test x=0.

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 0 is substituted into the expression for x, the result is negative 1. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 0 is substituted into the expression for x, the result is 3. It labels the result “positive”.

Above the number line, mark the factor x−1 negative and mark x+3 positive.

Since a negative divided by a positive is negative, the quotient is marked negative in the interval (−3,1).

This figure shows a shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line divided into three intervals by its zero partition numbers marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1.
Interval(1,∞)

The number 2 is in the interval (1,∞). Test x=2.

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 2 is substituted into the expression for x, the result is 1. It labels the result as “positive”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 2 is substituted into the expression for x, the result is 5. It labels the result “positive”.

Above the number line, mark the factor x−1 positive and mark x+3 positive.

Since a positive divided by a positive is positive, mark the quotient positive in the interval (1,∞).

The figure shows that in the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line is divided into intervals by zero partition numbers at negative 3 and 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.

Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals (−∞,−3) and (1,∞) are solutions.

But what about the zero partition numbers?

The zero partition number x=−3 makes the denominator 0, so it must be excluded from the solution and we mark it with a parenthesis.

The zero partition number x=1 makes the whole rational expression 0. The inequality requires that the rational expression be greater than or equal to 0. So, 1 is part of the solution and we will mark it with a bracket.

The number line is divided into intervals by zero partition numbers at negative 3 and 1. A closed parenthesis is used at 3 and an open bracket is used at 1. The number is shaded to the left of 3 and to the right of 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.

Recall that when we have a solution made up of more than one interval we use the union symbol, ∪, to connect the two intervals. The solution in interval notation is (−∞,−3)∪[1,∞).

Solve and write the solution in interval notation: x−2x+4≥0.

Solution

(−∞,−4)∪[2,∞)

Solve and write the solution in interval notation: x+2x−4≥0.

Solution

(−∞,−2]∪(4,∞)

We summarize the steps for easy reference.

Solve a rational inequality.

  1. Write the inequality as one quotient on the left and zero on the right.
  2. Determine the zero partition numbers–the points where the rational expression will be zero or undefined.
  3. Use the zero partition numbers to divide the number line into intervals.
  4. Test a value in each interval. Above the number line show the sign of each factor of the numerator and denominator in each interval. Below the number line show the sign of the quotient.
  5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

The next example requires that we first get the rational inequality into the correct form.

Solve and write the solution in interval notation: 4xx−6<1.

Solution
4xx−6<1
Subtract 1 to get zero on the right. 4xx−6−1<0
Rewrite 1 as a fraction using the LCD. 4xx−6−x−6x−6<0
Subtract the numerators and place the
difference over the common denominator.
4x−(x−6)x−6<0
Simplify. 3x+6x−6<0
Factor the numerator to show all factors. 3(x+2)x−6<0
Find the zero partition numbers.
The quotient will be zero when the numerator is zero.
The quotient is undefined when the denominator is zero.
x+2=0x−6=0x=−2x=6
Use the zero partition numbers to divide the number line into intervals.
A number line with vertical dashed lines at x=-2 and x=6, indicating boundaries or vertical asymptotes.
Test a value in each interval.
A sign analysis table for algebraic expressions x+2 and x-6. It evaluates each expression within intervals (-∞, -2), (-2, 6), and (6, ∞), showing the computed value and its sign.
Above the number line show the sign of each factor of the rational expression in each interval.
Below the number line show the sign of the quotient.
A sign analysis chart shows the behavior of the rational expression (x+2)/(x-6) on a number line. It illustrates that the expression is positive for x < -2 or x > 6, and negative for -2 < x < 6.
Determine the intervals where the inequality is correct. We want the quotient to be negative, so the solution includes the points between −2 and 6. Since the inequality is strictly less than, the endpoints are not included.
We write the solution in interval notation as (−2, 6).

Solve and write the solution in interval notation: 3xx−3<1.

Solution

(−32,3)

Solve and write the solution in interval notation: 3xx−4<2.

Solution

(−8,4)

In the next example, the numerator is always positive, so the sign of the rational expression depends on the sign of the denominator.

Solve and write the solution in interval notation: 5x2−2x−15>0.

Solution
The inequality is in the correct form. 5x2−2x−15>0
Factor the denominator. 5(x+3)(x−5)>0
Find the zero partition numbers.
The quotient is 0 when the numerator is 0.
Since the numerator is always 5, the quotient cannot be 0.
The quotient will be undefined when the
denominator is zero.
(x+3)(x−5)=0x=−3, x=5
Use the zero partition numbers to divide the number line into intervals.
Sign chart showing the intervals where 5/((x+3)(x-5)) is positive or negative. It details the signs of (x+3) and (x-5) across a number line with critical points at -3 and 5.
Test values in each interval.
Above the number line show the sign of each
factor of the denominator in each interval.
Below the number line, show the sign of the quotient.
Write the solution in interval notation. (−∞,−3)∪(5,∞)

Solve and write the solution in interval notation: 1x2+2x−8>0.

Solution

(−∞,−4)∪(2,∞)

Solve and write the solution in interval notation: 3x2+x−12>0.

Solution

(−∞,−4)∪(3,∞)

The next example requires some work to get it into the needed form.

Solve and write the solution in interval notation: 13−2x2<53x.

Solution
13−2x2<53x
Subtract 53x to get zero on the right. 13−2x2−53x<0
Rewrite to get each fraction with the LCD 3x2. 1⋅x23⋅x2−2⋅3x2⋅3−5⋅x3x⋅x<0
Simplify. x23x2−63x2−5x3x2<0
Subtract the numerators and place the
difference over the common denominator.
x2−5x−63x2<0
Factor the numerator. (x−6)(x+1)3x2<0
Find the zero partition numbers. 3x2=0x−6=0x+1=0x=0x=6x=−1
Use the zero partition numbers to divide the number
line into intervals.
This image displays a sign chart for the expression (x-6)(x+1) / (3x^2), showing the signs of its factors and the overall expression across intervals defined by critical points -1, 0, and 6.
Above the number line show the sign of each
factor in each interval. Below the number line, show the sign of the quotient.
Since, 0 is excluded, the solution is the two
intervals, (−1,0) and (0,6).
(−1,0)∪(0,6)

Solve and write the solution in interval notation: 12+4x2<3x.

Solution

(2,4)

Solve and write the solution in interval notation: 13+6x2<3x.

Solution

(3,6)

Solve an Inequality with Rational Functions

When working with rational functions, it is sometimes useful to know when the function is greater than or less than a particular value. This leads to a rational inequality.

Given the function R(x)=x+3x−5, find the values of x that make the function less than or equal to 0.

Solution

We want the function to be less than or equal to 0.

R(x)≤0
Substitute the rational expression for R(x). x+3x−5≤0x≠5
Find the zero partition numbers. x+3=0x−5=0 x=−3x=5
Use the zero partition numbers to divide the number line into intervals.
A sign chart illustrating the intervals where the rational expression (x+3)/(x-5) is positive or negative, based on the signs of its numerator and denominator.
Test values in each interval. Above the
number line, show the sign of each factor
in each interval. Below the number line,
show the sign of the quotient
Write the solution in interval notation. Since
5 is excluded we, do not include it in the interval.
[−3,5)

Given the function R(x)=x−2x+4, find the values of x that make the function less than or equal to 0.

Solution

(−4,2]

Given the function R(x)=x+1x−4, find the values of x that make the function less than or equal to 0.

Solution

[−1,4)

In economics, the function C(x) is used to represent the cost of producing x units of a commodity. The average cost per unit can be found by dividing C(x) by the number of items x. Then, the average cost per unit is c(x)=C(x)x.

The function C(x)=10x+3000 represents the cost to produce x, number of items. Find ⓐ the average cost function, c(x) ⓑ how many items should be produced so that the average cost is less than $40.

Solution
ⓐ
This table illustrates the derivation of the average cost function c(x) from the total cost function C(x).
C(x)=10x+3000
The average cost function is c(x)=C(x)x.
To find the average cost function, divide the
cost function by x.
c(x)=C(x)xc(x)=10x+3000x
The average cost function is c(x)=10x+3000x.
ⓑ
Step-by-step solution of a rational inequality, demonstrating algebraic manipulation and identification of partition numbers.
We want the function c(x) to be less than 40. c(x)<40
Substitute the rational expression for c(x). 10x+3000x<40x≠0
Subtract 40 to get 0 on the right. 10x+3000x−40<0
Rewrite the left side as one quotient by finding
the LCD and performing the subtraction.
10x+3000x−40(xx)<0
10x+3000x−40xx<0
10x+3000−40xx<0
−30x+3000x<0
Factor the numerator to show all factors. −30(x−100)x<0
Find the zero partition numbers. −30(x−100)=0x=0−30≠0x−100=0x=100

More than 100 items must be produced to keep the average cost below $40 per item.

The function C(x)=20x+6000 represents the cost to produce x, number of items. Find ⓐ the average cost function, c(x) ⓑ how many items should be produced so that the average cost is less than $60?

Solution

ⓐ c(x)=20x+6000x
ⓑ More than 150 items must be produced to keep the average cost below $60 per item.

The function C(x)=5x+900 represents the cost to produce x, number of items. Find ⓐ the average cost function, c(x) ⓑ how many items should be produced so that the average cost is less than $20?

Solution

ⓐ c(x)=5x+900x ⓑ More than 60 items must be produced to keep the average cost below $20 per item.

Key Concepts

  • Solve a rational inequality.
    1. Write the inequality as one quotient on the left and zero on the right.
    2. Determine the zero partition numbers–the points where the rational expression will be zero or undefined.
    3. Use the zero partition numbers to divide the number line into intervals.
    4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.
    5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Section Exercises

Practice Makes Perfect

Solve Rational Inequalities

In the following exercises, solve each rational inequality and write the solution in interval notation.

x−3x+4≥0

Solution

(−∞,−4)∪[3,∞)

x+6x−5≥0

x+1x−3≤0

Solution

[−1,3)

x−4x+2≤0

x−7x−1>0

Solution

(−∞,1)∪(7,∞)

x+8x+3>0

x−6x+5<0

Solution

(−5,6)

x+5x−2<0

3xx−5<1

Solution

(−52,5)

5xx−2<1

6xx−6>2

Solution

(−∞,−3)∪(6,∞)

3xx−4>2

2x+3x−6≤1

Solution

[−9,6)

4x−1x−4≤1

3x−2x−4≥2

Solution

(−∞,−6]∪(4,∞)

4x−3x−3≥2

1x2+7x+12>0

Solution

(−∞,−4)∪(−3,∞)

1x2−4x−12>0

3x2−5x+4<0

Solution

(1,4)

4x2+7x+12<0

22x2+x−15≥0

Solution

(−∞,−3)∪(52,∞)

63x2−2x−5≥0

−26x2−13x+6≤0

Solution

(−∞,23)∪(32,∞)

−110x2+11x−6≤0

12+12x2>5x

Solution

(−∞,0)∪(0,4)∪(6,∞)

13+1x2>43x

12−4x2≤1x

Solution

[−2,0)∪(0,4]

12−32x2≥1x

1x2−16<0

Solution

(−4,4)

4x2−25>0

4x−2≥3x+1

Solution

[−10,−1)∪(2,∞)

5x−1≤4x+2

Solve an Inequality with Rational Functions

In the following exercises, solve each rational function inequality and write the solution in interval notation.

Given the function R(x)=x−5x−2, find the values of x that make the function less than or equal to 0.

Solution

(2,5]

Given the function R(x)=x+1x+3, find the values of x that make the function greater than or equal to 0.

Given the function R(x)=x−6x+2, find the values of x that make the function less than or equal to 0.

Solution

(−2,6]

Given the function R(x)=x+1x−4, find the values of x that make the function less than or equal to 0.

Writing Exercises

Write the steps you would use to explain solving rational inequalities to your little brother.

Solution

Answers will vary.

Create a rational inequality whose solution is (−∞,−2]∪[4,∞).

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve rational inequalities. In row 3, the I can was solve an inequality with rational functions.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Chapter Review Exercises

Simplify, Multiply, and Divide Rational Expressions

Determine the Values for Which a Rational Expression is Undefined

In the following exercises, determine the values for which the rational expression is undefined.

5a+33a−2

Solution

a≠23

b−7b2−25

5x2y28y

Solution

y≠0

x−3x2−x−30

Simplify Rational Expressions

In the following exercises, simplify.

1824

Solution

34

9m418mn3

x2+7x+12x2+8x+16

Solution

x+3x+4

7v−3525−v2

Multiply Rational Expressions

In the following exercises, multiply.

58·415

Solution

16

3xy28y3·16y224x

72x−12x28x+32·x2+10x+24x2−36

Solution

−3x2

2y2+y−34−y2·y2−4y+42y2+11y+12

Divide Rational Expressions

In the following exercises, divide.

x2−4x−12x2+8x+12÷x2−363x

Solution

3x(x+6)(x+6)

y2−164÷y3−642y2+8y+32

11+ww−9÷121−w29−w

Solution

−111−w

3y2−12y−634y+3÷(6y2−42y)

c2−643c2+26c+16c2−4c−3215c+10

Solution

5c+4

8a2+16aa−4·a2+2a−24a2+7a+10÷2a2−6aa+5

Multiply and Divide Rational Functions

Find R(x)=f(x)·g(x) where f(x)=9x2+9xx2−3x−4 and g(x)=x2−163x2+12x.

Solution

R(x)=3

Find R(x)=f(x)g(x) where f(x)=27x23x−21 and
g(x)=9x2+54xx2−x−42.

Add and Subtract Rational Expressions

Add and Subtract Rational Expressions with a Common Denominator

In the following exercises, perform the indicated operations.

715+815

Solution

1

4a22a−1−12a−1

y2+10yy+5+25y+5

Solution

y+5

7x2x2−9+21xx2−9

x2x−7−3x+28x−7

Solution

x+4

y2y+11−121y+11

4q2−q+3q2+6q+5−3q2+q+6q2+6q+5

Solution

q-3q+5

5t2+4t2+3t2−25−4t2−8t−32t2−25

Add and Subtract Rational Expressions Whose Denominators Are Opposites

In the following exercises, add and subtract.

18w6w−1+3w−21−6w

Solution

15w+26w−1

a2+3aa2−4−3a+84−a2

2b2+3b−15b2−49−b2+16b−149−b2

Solution

3b2+19b−16b2−49

8y2−10y+72y−5+2y2+7y+25−2y

Find the Least Common Denominator of Rational Expressions

In the following exercises, find the LCD.

7a2−3a−10,3aa2−a−20

Solution

(a+2)(a−5)(a+4)

6n2−4,2nn2−4n+4

53p2+17p−6,2m3p2+23p−8

Solution

(3p−1)(p+6)(p+8)

Add and Subtract Rational Expressions with Unlike Denominators

In the following exercises, perform the indicated operations.

75a+32b

2c−2+9c+3

Solution

11c−12(c−2)(c+3)

3xx2−9+5x2+6x+9

2xx2+10x+24+3xx2+8x+16

Solution

5x2+26x(x+4)(x+4)(x+6)

5qp2q−p2+4qq2−1

3yy+2−y+2y+8

Solution

2(y2+10y−2)(y+2)(y+8)

−3w−15w2+w−20−w+24−w

7m+3m+2−5

Solution

2m−7m+2

nn+3+2n−3−n+9n2−9

8aa2−64−4a+8

Solution

4a−8

512x2y+720xy3

Add and Subtract Rational Functions

In the following exercises, find R(x)=f(x)+g(x) where f(x) and g(x) are given.

f(x)=2x2+12x−11x2+3x−10,g(x)=x+12−x

Solution

R(x)=x+8x+5

f(x)=−4x+31x2+x−30,g(x)=5x+6

In the following exercises, find R(x)=f(x)−g(x) where f(x) and g(x) are given.

f(x)=4xx2−121,g(x)=2x−11

Solution

R(x)=2x+11

f(x)=7x+6,g(x)=14xx2−36

Simplify Complex Rational Expressions

Simplify a Complex Rational Expression by Writing It as Division

In the following exercises, simplify.

7xx+214x2x2−4

Solution

x−22x

25+5613+14

x−3xx+51x+5+1x−5

Solution

(x+2)(x−5)2

2m+mnnm−1n

Simplify a Complex Rational Expression by Using the LCD

In the following exercises, simplify.

13+1814+112

Solution

118

3a2−1b1a+1b2

2z2−49+1z+79z+7+12z−7

Solution

z−521z+21

3y2−4y−322y−8+1y+4

7.4 Solve Rational Equations

Solve Rational Equations

In the following exercises, solve.

12+23=1x

Solution

x=67

1−2m=8m2

1b−2+1b+2=3b2−4

Solution

b=32

3q+8−2q−2=1

v−15v2−9v+18=4v−3+2v−6

Solution

no solution

z12+z+33z=1z

Solve Rational Equations that Involve Functions

For rational function, f(x)=x+2x2−6x+8, ⓐ find the domain of the function ⓑ solve f(x)=1 ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain is all real numbers except x≠2 and x≠4. ⓑ x=1,x=6
ⓒ (1,1),(6,1)

For rational function, f(x)=2−xx2+7x+10, ⓐ find the domain of the function ⓑ solve f(x)=2 ⓒ find the points on the graph at this function value.

Solve a Rational Equation for a Specific Variable

In the following exercises, solve for the indicated variable.

Vl=hw for l.

Solution

l=Vhw

1x−2y=5 for y.

x=y+5z−7 for z.

Solution

z=y+5+7xx

P=kV for V.

Solve Applications with Rational Equations

Solve Proportions

In the following exercises, solve.

x4=35

Solution

x=125

3y=95

ss+20=37

Solution

s=15

t−35=t+29

Solve Using Proportions

In the following exercises, solve.

Rachael had a 21-ounce strawberry shake that has 739 calories. How many calories are there in a 32-ounce shake?

Solution

1126 calories

Leo went to Mexico over Christmas break and changed $525 dollars into Mexican pesos. At that time, the exchange rate had $1 US is equal to 16.25 Mexican pesos. How many Mexican pesos did he get for his trip?

Solve Similar Figure Applications

In the following exercises, solve.

ΔABC is similar to ΔXYZ. The lengths of two sides of each triangle are given in the figure. Find the lengths of the third sides.
The first figure is triangle A B C with side A B 8 units long, side B C 7 units long, and side A C b units long. The second figure is triangle X Y Z with side X Y 2 and two-thirds units long, side Y Z x units long, and side X Z 3 units long.

Solution

b=9;x=213

On a map of Europe, Paris, Rome, and Vienna form a triangle whose sides are shown in the figure below. If the actual distance from Rome to Vienna is 700 miles, find the distance from
ⓐ Paris to Rome
ⓑ Paris to Vienna
The figure is a triangle formed by Paris, Vienna, and Rome. The distance between Paris and Vienna is 7.7 centimeters. The distance between Vienna and Rome is 7 centimeters. The distance between Rome and Paris is 8.9 centimeters.

Francesca is 5.75 feet tall. Late one afternoon, her shadow was 8 feet long. At the same time, the shadow of a nearby tree was 32 feet long. Find the height of the tree.

Solution

23 feet

The height of a lighthouse in Pensacola, Florida is 150 feet. Standing next to the statue, 5.5-foot-tall Natasha cast a 1.1-foot shadow. How long would the shadow of the lighthouse be?

Solve Uniform Motion Applications

In the following exercises, solve.

When making the 5-hour drive home from visiting her parents, Lolo ran into bad weather. She was able to drive 176 miles while the weather was good, but then driving 10 mph slower, went 81 miles when it turned bad. How fast did she drive when the weather was bad?

Solution

45 mph

Mark is riding on a plane that can fly 490 miles with a headwind of 20 mph in the same time that it can fly 350 miles against a tailwind of 20 mph. What is the speed of the plane?

Josue can ride his bicycle 8 mph faster than Arjun can ride his bike. It takes Arjun 3 hours longer than Josue to ride 48 miles. How fast can Josue ride his bike?

Solution

16 mph

Curtis was training for a triathlon. He ran 8 kilometers and biked 32 kilometers in a total of 3 hours. His running speed was 8 kilometers per hour less than his biking speed. What was his running speed?

Solve Work Applications

In the following exercises, solve.

Brandy can frame a room in 1 hour, while Jake takes 4 hours. How long could they frame a room working together?

Solution

48 minutes

Prem takes 3 hours to mow the lawn while her cousin, Barb, takes 2 hours. How long will it take them working together?

Jeffrey can paint a house in 6 days, but if he gets a helper he can do it in 4 days. How long would it take the helper to paint the house alone?

Solution

12 days

Marta and Deb work together writing a book that takes them 90 days. If Marta worked alone it would take her 120 days. How long would it take Deb to write the book alone?

Solve Direct Variation Problems

In the following exercises, solve.

If y varies directly as x when y=9 and x=3, find x when y=21.

Solution

x=7

If y varies inversely as x when y=20 and x=2, find y when x=4.

Vanessa is traveling to see her fiancé. The distance, d, varies directly with the speed, v, she drives. If she travels 258 miles driving 60 mph, how far would she travel going 70 mph?

Solution

301 mph

If the cost of a pizza varies directly with its diameter, and if an 8” diameter pizza costs $12, how much would a 6” diameter pizza cost?

The distance to stop a car varies directly with the square of its speed. It takes 200 feet to stop a car going 50 mph. How many feet would it take to stop a car going 60 mph?

Solution

288 feet

Solve Inverse Variation Problems

In the following exercises, solve.

If m varies inversely with the square of n, when m=4 and n=6 find m when n=2.

The number of tickets for a music fundraiser varies inversely with the price of the tickets. If Madelyn has just enough money to purchase 12 tickets for $6 each, how many tickets can Madelyn afford to buy if the price increased to $8?

Solution

9 tickets

On a string instrument, the length of a string varies inversely with the frequency of its vibrations. If an 11-inch string on a violin has a frequency of 360 cycles per second, what frequency does a 12-inch string have?

Solve Rational Inequalities

Solve Rational Inequalities

In the following exercises, solve each rational inequality and write the solution in interval notation.

x−3x+4≤0

Solution

(−4,3]

5xx−2>1

3x−2x−4≤2

Solution

[−6,4)

1x2−4x−12<0

12−4x2≥1x

Solution

(−∞,−2]∪[4,∞)

4x−2<3x+1

Solve an Inequality with Rational Functions

In the following exercises, solve each rational function inequality and write the solution in interval notation

Given the function, R(x)=x−5x−2, find the values of x that make the function greater than or equal to 0.

Solution

(−∞,2)∪[5,∞)

Given the function, R(x)=x+1x+3, find the values of x that make the function less than or equal to 0.

The function
C(x)=150x+100,000 represents the cost to produce x, number of items. Find ⓐ the average cost function, c(x) ⓑ how many items should be produced so that the average cost is less than $160.

Solution


ⓐ c(x)=150x+100000x
ⓑ More than 10,000 items must be produced to keep the average cost below $160 per item.

Tillman is starting his own business by selling tacos at the beach. Accounting for the cost of his food truck and ingredients for the tacos, the function C(x)=2x+6,000 represents the cost for Tillman to produce x, tacos. Find ⓐ the average cost function, c(x) for Tillman’s Tacos ⓑ how many tacos should Tillman produce so that the average cost is less than $4.

Practice Test

In the following exercises, simplify.

4a2b12ab2

Solution

a3b

6x−18x2−9

In the following exercises, perform the indicated operation and simplify.

4xx+2·x2+5x+612x2

Solution

x+33x

2y2y2−1÷y3−y2+yy3+1

6x2−x+20x2−81−5x2+11x−7x2−81

Solution

x−3x+9

−3a3a−3+5aa2+3a−4

2n2+8n−1n2−1−n2−7n−11−n2

Solution

3n−2n−1

10x2+16x−78x−3+2x2+3x−13−8x

1m−1n1n+1m

Solution

n−mm+n

In the following exercises, solve each equation.

1x+34=58

1z−5+1z+5=1z2−25

Solution

z=12

z2z+8−34z−8=3z2−16z−168z2+16z−64

In the following exercises, solve each rational inequality and write the solution in interval notation.

6xx−6≤2

Solution

[−3,6)

2x+3x−6>1

12+12x2≥5x

Solution

(−∞,0)∪(0,4]∪[6,∞)

In the following exercises, find R(x) given f(x)=x−4x2−3x−10 and g(x)=x−5x2−2x−8.

R(x)=f(x)−g(x)

R(x)=f(x)·g(x)

Solution

R(x)=1(x+2)(x+2)

R(x)=f(x)÷g(x)

Given the function,
R(x)=22x2+x−15, find the values of x that make the function less than or equal to 0.

Solution

(−3,52)

In the following exercises, solve.

If y varies directly with x, and x=5 when y=30, find x when y=42.

If y varies inversely with the square of x and x=3 when y=9, find y when x=4.

Solution

y=8116

Matheus can ride his bike for 30 miles with the wind in the same amount of time that he can go 21 miles against the wind. If the wind’s speed is 6 mph, what is Matheus’ speed on his bike?

Oliver can split a truckload of logs in 8 hours, but working with his dad they can get it done in 3 hours. How long would it take Oliver’s dad working alone to split the logs?

Solution

Oliver’s dad would take 445 hours to split the logs himself.

The volume of a gas in a container varies inversely with the pressure on the gas. If a container of nitrogen has a volume of 29.5 liters with 2000 psi, what is the volume if the tank has a 14.7 psi rating? Round to the nearest whole number.

The cities of Dayton, Columbus, and Cincinnati form a triangle in southern Ohio. The diagram gives the map distances between these cities in inches.

The figure is a triangle formed by Cincinnati, Dayton, and Columbus. The distance between Cincinnati and Dayton is 2.4 inches. The distance between Dayton and Columbus is 3.2 inches. The distance between Columbus and Cincinnati is 5.3 inches.

The actual distance from Dayton to Cincinnati is 48 miles. What is the actual distance between Dayton and Columbus?

Solution

The distance between Dayton and Columbus is 64 miles.

rational inequality
A rational inequality is an inequality that contains a rational expression.
zero partition number of a rational inequality
The zero partition number of a rational inequality is a number which makes the rational expression zero or undefined.

Introduction

An illustration of the crystalline structure of graphene.
Graphene is an incredibly strong and flexible material made from carbon. It can also conduct electricity. Notice the hexagonal grid pattern. (credit: “AlexanderAIUS” / Wikimedia Commons)

Imagine charging your cell phone is less than five seconds. Consider cleaning radioactive waste from contaminated water. Think about filtering salt from ocean water to make an endless supply of drinking water. Ponder the idea of bionic devices that can repair spinal injuries. These are just of few of the many possible uses of a material called graphene. Materials scientists are developing a material made up of a single layer of carbon atoms that is stronger than any other material, completely flexible, and conducts electricity better than most metals. Research into this type of material requires a solid background in mathematics, including understanding roots and radicals. In this chapter, you will learn to simplify expressions containing roots and radicals, perform operations on radical expressions and equations, and evaluate radical functions.

Simplify Expressions with Roots

Learning Objectives

By the end of this section, you will be able to:

  • Simplify expressions with roots
  • Estimate and approximate roots
  • Simplify variable expressions with roots

Before you get started, take this readiness quiz.

Simplify: ⓐ (−9)2 ⓑ −92 ⓒ (−9)3.
If you missed this problem, review Example 8 in Use a Problem Solving Strategy.

Solution

ⓐ 8; ⓑ −1; ⓒ -729

Round 3.846 to the nearest hundredth.
If you missed this problem, review Example 1 in Decimals.

Solution

3.85

Simplify: ⓐ x3·x3 ⓑ y2·y2·y2 ⓒ z3·z3·z3·z3.
If you missed this problem, review Example 1 in Properties of Exponents and Scientific Notation.

Solution

ⓐ x6; ⓑ y6; ⓒ z12

Simplify Expressions with Roots

In Foundations, we briefly looked at square roots. Remember that when a real number n is multiplied by itself, we write n2 and read it ‘n squared’. This number is called the square of n, and n is called the square root. For example,

132is read “13 squared”169 is called thesquareof 13, since132=16913 is asquare rootof 169

Square and Square Root of a number

Square

Ifn2=m,thenmis thesquareofn.

Square Root

Ifn2=m,thennis asquare rootofm.

Notice (−13)2 = 169 also, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169.

So, every positive number has two square roots—one positive and one negative. What if we only wanted the positive square root of a positive number? We use a radical sign, and write, m, which denotes the positive square root of m. The positive square root is also called the principal square root. This symbol, as well as other radicals to be introduced later, are grouping symbols.

We also use the radical sign for the square root of zero. Because 02=0, 0=0. Notice that zero has only one square root.

Square Root Notation

mis read “the square root ofm”.Ifn2=m,thenn=m,forn≥0.
The image shows the variable m inside a square root symbol. The symbol is a line that goes up along the left side and then flat above the variable. The symbol is labeled “radical sign”. The variable m is labeled “radicand”.

We know that every positive number has two square roots and the radical sign indicates the positive one. We write 169=13. If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, −169=−13.

Simplify: ⓐ 144 ⓑ −289.

Solution
ⓐ
This table demonstrates finding the square root of 144 with its corresponding explanation.
144
Since 122=144. 12


ⓑ
Demonstrates the evaluation of a negative square root, showing the expression, explanation, and final value.
−289
Since 172=289 and the negative is in front of the radical sign. −17

Simplify: ⓐ −64 ⓑ 225.

Solution

ⓐ −8 ⓑ 15

Simplify: ⓐ 100 ⓑ −121.

Solution

ⓐ 10 ⓑ −11

Can we simplify −49? Is there a number whose square is −49?

()2=−49

Any positive number squared is positive. Any negative number squared is positive. There is no real number equal to −49. The square root of a negative number is not a real number.

Simplify: ⓐ −196 ⓑ −64.

Solution
ⓐ
Explains why the square root of a negative number is not a real number.
−196
There is no real number whose square is −196. −196is not a real number.


ⓑ
Evaluation of a negative square root, showing the property and its result.
−64
The negative is in front of the radical. −8

Simplify: ⓐ −169 ⓑ −81.

Solution

ⓐ not a real number ⓑ −9

Simplify: ⓐ −49 ⓑ −121.

Solution

ⓐ −7 ⓑ not a real number

So far we have only talked about squares and square roots. Let’s now extend our work to include higher powers and higher roots.

Let’s review some vocabulary first.

We write:We say:n2nsquaredn3ncubedn4nto the fourth powern5nto the fifth power

The terms ‘squared’ and ‘cubed’ come from the formulas for area of a square and volume of a cube.

It will be helpful to have a table of the powers of the integers from −5 to 5. See Figure 1.

The figure contains two tables. The first table has 9 rows and 5 columns. The first row is a header row with the headers “Number”, “Square”, “Cube”, “Fourth power”, and “Fifth power”. The second row contains the expressions n, n squared, n cubed, n to the fourth power, and n to the fifth power. The third row contains the number 1 in each column. The fourth row contains the numbers 2, 4, 8, 16, 32. The fifth row contains the numbers 3, 9, 27, 81, 243. The sixth row contains the numbers 4, 16, 64, 256, 1024. The seventh row contains the numbers 5, 25, 125 625, 3125. The eighth row contains the expressions x, x squared, x cubed, x to the fourth power, and x to the fifth power. The last row contains the expressions x squared, x to the fourth power, x to the sixth power, x to the eighth power, and x to the tenth power. The second table has 7 rows and 5 columns. The first row is a header row with the headers “Number”, “Square”, “Cube”, “Fourth power”, and “Fifth power”. The second row contains the expressions n, n squared, n cubed, n to the fourth power, and n to the fifth power. The third row contains the numbers negative 1, 1 negative 1, 1, negative 1. The fourth row contains the numbers negative 2, 4, negative 8, 16, negative 32. The fifth row contains the numbers negative 3, 9, negative 27, 81, negative 243. The sixth row contains the numbers negative 4, 16, negative 64, 256, negative 1024. The last row contains the numbers negative 5, 25, negative 125, 625, negative 3125.

Notice the signs in the table. All powers of positive numbers are positive, of course. But when we have a negative number, the even powers are positive and the odd powers are negative. We’ll copy the row with the powers of −2 to help you see this.

The image contains a table with 2 rows and 5 columns. The first row contains the expressions n, n squared, n cubed, n to the fourth power, and n to the fifth power. The second row contains the numbers negative 2, 4, negative 8, 16, negative 32. Arrows point to the second and fourth columns with the label “Even power Positive result”. Arrows point to the first third and fifth columns with the label “Odd power Negative result”.

We will now extend the square root definition to higher roots.

nth Root of a Number

Ifbn=a,thenbis annthroot ofa.The principalnthroot ofais writtenan.nis called theindexof the radical.

Just like we use the word ‘cubed’ for b3, we use the term ‘cube root’ for a3.

We can refer to Figure 1 to help find higher roots.

43=6434=81(−2)5=−32643=4814=3−325=−2

Could we have an even root of a negative number? We know that the square root of a negative number is not a real number. The same is true for any even root. Even roots of negative numbers are not real numbers. Odd roots of negative numbers are real numbers.

Properties of an

When n is an even number and

  • a≥0, then an is a real number.
  • a<0, then an is not a real number.

When n is an odd number, an is a real number for all values of a.

We will apply these properties in the next two examples.

Simplify: ⓐ 643 ⓑ 814 ⓒ 325.

Solution
ⓐ
Demonstrates the calculation of the cube root of 64.
643
Since 43=64. 4


ⓑ
Example demonstrating the calculation of the fourth root of 81, including the mathematical expression, its result, and the supporting explanation.
814
Since (3)4=81. 3


ⓒ
Demonstration of the calculation and result of the fifth root of 32.
325
Since (2)5=32. 2

Simplify: ⓐ 273 ⓑ 2564 ⓒ 2435.

Solution

ⓐ 3 ⓑ 4 ⓒ 3

Simplify: ⓐ 10003 ⓑ 164 ⓒ 10245.

Solution

ⓐ 10 ⓑ 2 ⓒ 4

In this example be alert for the negative signs as well as even and odd powers.

Simplify: ⓐ −1253 ⓑ -164 ⓒ −2435.

Solution
ⓐ
This table explains the calculation of the cube root of -125 and shows the result.
−1253
Since (−5)3=−125. −5


ⓑ
This table demonstrates the evaluation of the fourth root of -16, explaining why it is not a real number.
−164
Think, (?)4=−16. No real number raised to the fourth power is negative. Not a real number.


ⓒ
Demonstrates the evaluation of the fifth root of -243, showing the problem, explanation, and solution.
−2435
Since (−3)5=−243. −3

Simplify: ⓐ −273 ⓑ −2564 ⓒ −325.

Solution

ⓐ −3 ⓑ not real ⓒ −2

Simplify: ⓐ −2163 ⓑ −814 ⓒ −10245.

Solution

ⓐ −6 ⓑ not real ⓒ −4

Estimate and Approximate Roots

When we see a number with a radical sign, we often don’t think about its numerical value. While we probably know that the 4=2, what is the value of 21 or 503? In some situations a quick estimate is meaningful and in others it is convenient to have a decimal approximation.

To get a numerical estimate of a square root, we look for perfect square numbers closest to the radicand. To find an estimate of 11, we see 11 is between perfect square numbers 9 and 16, closer to 9. Its square root then will be between 3 and 4, but closer to 3.

The figure contains two tables. The first table has 5 rows and 2 columns. The first row is a header row with the headers “Number” and “Square Root”. The second row has the numbers 4 and 2. The third row is 9 and 3. The fourth row is 16 and 4. The last row is 25 and 5. A callout containing the number 11 is directed between the 9 and 16 in the first column. Another callout containing the number square root of 11 is directed between the 3 and 4 of the second column. Below the table are the inequalities 9 is less than 11 is less than 16 and 3 is less than square root of 11 is less than 4. The second table has 5 rows and 2 columns. The first row is a header row with the headers “Number” and “Cube Root”. The second row has the numbers 8 and 2. The third row is 27 and 3. The fourth row is 64 and 4. The last row is 125 and 5. A callout containing the number 91 is directed between the 64 and 125 in the first column. Another callout containing the number cube root of 91 is directed between the 4 and 5 of the second column. Below the table are the inequalities 64 is less than 91 is less than 125 and 4 is less than cube root of 91 is less than 5.

Similarly, to estimate 913, we see 91 is between perfect cube numbers 64 and 125. The cube root then will be between 4 and 5.

Estimate each root between two consecutive whole numbers: ⓐ 105 ⓑ 433.

Solution

ⓐ Think of the perfect square numbers closest to 105. Make a small table of these perfect squares and their squares roots.

The square root of 105 is displayed, a mathematical expression representing the principal square root of the number 105.
A table illustrating numbers and their square roots, showing that 105 falls between 100 and 121, and therefore its square root,  R105, is between 10 and 11.
Locate 105 between two consecutive perfect squares. An inequality: 100 < 105 < 121, with 105 in red, illustrating that 105 lies between 10 squared and 11 squared, implying its square root is between 10 and 11.
105 is between their square roots. The image displays the mathematical inequality 10 < sqrt(105) < 11, indicating that the square root of 105 is between 10 and 11.

ⓑ Similarly we locate 43 between two perfect cube numbers.

The mathematical expression showing the cube root of 43 on a white background.
This table displays perfect cubes and their corresponding cube roots. The arrows highlight that the cube root of 43 is between 3 and 4, as 43 falls between 27 (3 cubed) and 64 (4 cubed).
Locate 43 between two consecutive perfect cubes. A mathematical inequality displaying '27 < 43 < 64', with the number 43 emphasized in red.
433 is between their cube roots. A mathematical inequality states that 3 is less than the cube root of 43, which is less than 4.

Estimate each root between two consecutive whole numbers:

ⓐ 38 ⓑ 933

Solution

ⓐ 6<38<7
ⓑ 4<933<5

Estimate each root between two consecutive whole numbers:

ⓐ 84 ⓑ 1523

Solution

ⓐ 9<84<10
ⓑ 5<1523<6

There are mathematical methods to approximate square roots, but nowadays most people use a calculator to find square roots. To find a square root you will use the x key on your calculator. To find a cube root, or any root with higher index, you will use the xy key.

When you use these keys, you get an approximate value. It is an approximation, accurate to the number of digits shown on your calculator’s display. The symbol for an approximation is ≈ and it is read ‘approximately’.

Suppose your calculator has a 10 digit display. You would see that

5≈2.236067978rounded to two decimal places is5≈2.24934≈3.105422799rounded to two decimal places is934≈3.11

How do we know these values are approximations and not the exact values? Look at what happens when we square them:

(2.236067978)2=5.000000002(2.24)2=5.0176(3.105422799)4=92.999999991(3.11)4=93.54951841

Their squares are close to 5, but are not exactly equal to 5. The fourth powers are close to 93, but not equal to 93.

Round to two decimal places: ⓐ 17 ⓑ 493 ⓒ 514.

Solution
ⓐ
Steps to calculate and round the square root of 17 using a calculator.
17
Use the calculator square root key. 4.123105626…
Round to two decimal places. 4.12
17≈4.12


ⓑ
Steps for calculating and rounding the cube root of 49 using a calculator.
493
Use the calculator xy key. 3.659305710…
Round to two decimal places. 3.66
493≈3.66


ⓒ
Illustrates calculating the fourth root of 51 using a calculator and rounding the result to two decimal places.
514
Use the calculator xy key. 2.6723451177…
Round to two decimal places. 2.67
514≈2.67

Round to two decimal places:

ⓐ 11 ⓑ 713 ⓒ 1274.

Solution

ⓐ ≈3.32 ⓑ ≈4.14
ⓒ ≈3.36

Round to two decimal places:

ⓐ 13 ⓑ 843 ⓒ 984.

Solution

ⓐ ≈3.61 ⓑ ≈4.38
ⓒ ≈3.15

Simplify Variable Expressions with Roots

The odd root of a number can be either positive or negative. For example,

Three equivalent expressions are written: the cube root of 4 cubed, the cube root of 64, and 4. There are arrows pointing to the 4 that is cubed in the first expression and the 4 in the last expression labeling them as “same”. Three more equivalent expressions are also written: the cube root of the quantity negative 4 in parentheses cubed, the cube root of negative 64, and negative 4. The negative 4 in the first expression and the negative 4 in the last expression are labeled as being the “same”.

But what about an even root? We want the principal root, so 6254=5.

But notice,

Three equivalent expressions are written: the fourth root of the quantity 5 to the fourth power in parentheses, the fourth root of 625, and 5. There are arrows pointing to the 5 in the first expression and the 5 in the last expression labeling them as “same”. Three more equivalent expressions are also written: the fourth root of the quantity negative 5 in parentheses to the fourth power in parentheses, the fourth root of 625, and 5. The negative 5 in the first expression and the 5 in the last expression are labeled as being the “different”.

How can we make sure the fourth root of −5 raised to the fourth power is 5? We can use the absolute value. |−5|=5. So we say that when n is even ann=|a|. This guarantees the principal root is positive.

Simplifying Odd and Even Roots

For any integer n≥2,

when the indexnis oddann=awhen the indexnis evenann=|a|

We must use the absolute value signs when we take an even root of an expression with a variable in the radical.

Simplify: ⓐ x2 ⓑ n33 ⓒ p44 ⓓ y55.

Solution

ⓐ We use the absolute value to be sure to get the positive root.

Demonstrates how the square root of x squared simplifies to the absolute value of x, explaining the underlying even index rule.
x2
Since the index n is even, ann=|a|. |x|

ⓑ This is an odd indexed root so there is no need for an absolute value sign.

Illustrates the simplification of nth roots with odd indices, showing an example and the underlying rule.
n33
Since the index n is odd, ann=a. n

ⓒ

Demonstrates the property of simplifying even roots sqrt[n](a^n) to |a| through an example.
p44
Since the index nis evenann=|a|. |p|

ⓓ

Illustration of simplifying radical expressions where the index and exponent are equal and odd.
y55
Since the index n is odd, ann=a. y

Simplify: ⓐ b2 ⓑ w33 ⓒ m44 ⓓ q55.

Solution

ⓐ |b| ⓑ w ⓒ |m| ⓓ q

Simplify: ⓐ y2 ⓑ p33 ⓒ z44 ⓓ q55.

Solution

ⓐ |y| ⓑ p ⓒ |z| ⓓ q

What about square roots of higher powers of variables? The Power Property of Exponents says (am)n=am·n. So if we square am, the exponent will become 2m.

(am)2=a2m

Looking now at the square root,

a2mSince(am)2=a2m.(am)2Sincenis evenann=|a|.|am|Soa2m=|am|.

We apply this concept in the next example.

Simplify: ⓐ x6 ⓑ y16.

Solution
ⓐ
Step-by-step simplification of the square root of x^6, illustrating the rule for even indices.
x6
Since (x3)2=x6. (x3)2
Since the index n is even an=|a|. |x3|


ⓑ
Steps for simplifying the square root of y to the power of 16, showing the application of radical rules.
y16
Since (y8)2=y16. (y8)2
Since the index n is even ann=|a|. y8
In this case the absolute value sign is not needed as y8 is positive.

Simplify: ⓐ y18 ⓑ z12.

Solution

ⓐ |y9| ⓑ z6

Simplify: ⓐ m4 ⓑ b10.

Solution

ⓐ m2 ⓑ |b5|

The next example uses the same idea for higher roots.

Simplify: ⓐ y183 ⓑ z84.

Solution
ⓐ
This table illustrates the step-by-step simplification of the cube root of y to the power of 18, showing the application of radical properties.
y183
Since (y6)3=y18. (y6)33
Since n is odd, ann=a. y6


ⓑ
Step-by-step simplification of the fourth root of z^8, illustrating the process and explaining the absence of an absolute value.
z84
Since (z2)4=z8. (z2)44
Since z2 is positive, we do not need an absolute value sign. z2

Simplify: ⓐ u124 ⓑ v153.

Solution

ⓐ |u3| ⓑ v5

Simplify: ⓐ c205 ⓑ d246

Solution

ⓐ c4 ⓑ d4

In the next example, we now have a coefficient in front of the variable. The concept a2m=|am| works in much the same way.

16r22=4|r11|because(4r11)2=16r22.

But notice 25u8=5u4 and no absolute value sign is needed as u4 is always positive.

Simplify: ⓐ 16n2 ⓑ −81c2.

Solution
ⓐ
Steps demonstrating the simplification of the mathematical expression sqrt(16n^2) by applying relevant rules.
16n2
Since (4n)2=16n2. (4n)2
Since the index n is even ann=|a|. 4|n|


ⓑ
Simplification steps for -sqrt(81c^2), detailing each transformation and the mathematical rules applied.
−81c2
Since (9c)2=81c2. −(9c)2
Since the index n is even ann=|a|. −9|c|

Simplify: ⓐ 64x2 ⓑ −100p2.

Solution

ⓐ 8|x| ⓑ −10|p|

Simplify: ⓐ 169y2 ⓑ −121y2.

Solution

ⓐ 13|y| ⓑ −11|y|

This example just takes the idea farther as it has roots of higher index.

Simplify: ⓐ 64p63 ⓑ 16q124.

Solution
ⓐ
This table illustrates the step-by-step process for simplifying the cube root of the algebraic expression 64p^6.
64p63
Rewrite 64p6 as (4p2)3. (4p2)33
Take the cube root. 4p2


ⓑ
This table illustrates the step-by-step simplification of a fourth root mathematical expression.
16q124
Rewrite the radicand as a fourth power. (2q3)44
Take the fourth root. 2|q3|

Simplify: ⓐ 27x273 ⓑ 81q284.

Solution

ⓐ 3x9 ⓑ 3|q7|

Simplify: ⓐ 125q93 ⓑ 243q255.

Solution

ⓐ 5q3 ⓑ 3q5

The next examples have two variables.

Simplify: ⓐ 36x2y2 ⓑ 121a6b8 ⓒ 64p63q93.

Solution
ⓐ
Step-by-step simplification of the square root expression √(36x²y²) to 6|xy|.
36x2y2
Since (6xy)2=36x2y2 (6xy)2
Take the square root. 6|xy|


ⓑ
Step-by-step simplification of the square root of a monomial expression, sqrt(121a^6b^8).
121a6b8
Since (11a3b4)2=121a6b8 (11a3b4)2
Take the square root. 11|a3|b4


ⓒ
Steps for simplifying a cube root of a monomial expression.
64p63q93
Since (4p21q3)3=64p63q9 (4p21q3)33
Take the cube root. 4p21q3

Simplify: ⓐ 100a2b2 ⓑ 144p12q20 ⓒ 8x30y123

Solution

ⓐ 10|ab| ⓑ 12p6q10
ⓒ 2x10y4

Simplify: ⓐ 225m2n2 ⓑ 169x10y14 ⓒ 27w36z153

Solution

ⓐ 15|mn| ⓑ 13|x5y7|
ⓒ 3w12z5

Access this online resource for additional instruction and practice with simplifying expressions with roots.

  • Simplifying Variables Exponents with Roots using Absolute Values

Key Concepts

  • Square Root Notation
    • m is read ‘the square root of m’
    • If n2 = m, then n=m, for n≥0.
      The image shows the variable m inside a square root symbol. The symbol is a line that goes up along the left side and then flat above the variable. The symbol is labeled “radical sign”. The variable m is labeled “radicand”.
    • The square root of m, m, is a positive number whose square is m.
  • nth Root of a Number
    • If bn=a, then b is an nth root of a.
    • The principal nth root of a is written an.
    • n is called the index of the radical.
  • Properties of an
    • When n is an even number and
      • a≥0, then an is a real number
      • a<0, then an is not a real number
    • When n is an odd number, an is a real number for all values of a.
  • Simplifying Odd and Even Roots
    • For any integer n≥2,
      • when n is odd ann=a
      • when n is even ann=|a|
    • We must use the absolute value signs when we take an even root of an expression with a variable in the radical.

Practice Makes Perfect

Simplify Expressions with Roots

In the following exercises, simplify.

ⓐ 64 ⓑ −81

Solution

ⓐ 8 ⓑ −9

ⓐ 169 ⓑ −100

ⓐ 196 ⓑ −1

Solution

ⓐ 14 ⓑ −1

ⓐ 144 ⓑ −121

ⓐ 49 ⓑ −0.01

Solution

ⓐ 23 ⓑ −0.1

ⓐ 64121 ⓑ −0.16

ⓐ −121 ⓑ −289

Solution

ⓐ not real number ⓑ −17

ⓐ −400 ⓑ −36

ⓐ −225 ⓑ −9

Solution

ⓐ −15 ⓑ not real number

ⓐ −49 ⓑ −256

ⓐ 2163 ⓑ 2564

Solution

ⓐ 6 ⓑ 4

ⓐ 273 ⓑ 164 ⓒ 2435

ⓐ 5123 ⓑ 814 ⓒ 15

Solution

ⓐ 8 ⓑ 3 ⓒ 1

ⓐ 1253 ⓑ 12964 ⓒ 10245

ⓐ −83 ⓑ −814 ⓒ −325

Solution

ⓐ −2 ⓑ not real ⓒ −2

ⓐ −643 ⓑ −164 ⓒ −2435

ⓐ −1253 ⓑ −12964 ⓒ −10245

Solution

ⓐ −5 ⓑ not real ⓒ −4

ⓐ −5123 ⓑ −814 ⓒ −15

Estimate and Approximate Roots

In the following exercises, estimate each root between two consecutive whole numbers.

ⓐ 70 ⓑ 713

Solution

ⓐ 8<70<9
ⓑ 4<713<5

ⓐ 55 ⓑ 1193

ⓐ 200 ⓑ 1373

Solution

ⓐ 14<200<15
ⓑ 5<1373<6

ⓐ 172 ⓑ 2003

In the following exercises, approximate each root and round to two decimal places.

ⓐ 19 ⓑ 893 ⓒ 974

Solution

ⓐ ≈4.36 ⓑ ≈4.46
ⓒ ≈3.14

ⓐ 21 ⓑ 933 ⓒ 1014

ⓐ 53 ⓑ 1473 ⓒ 4524

Solution

ⓐ ≈7.28 ⓑ ≈5.28
ⓒ ≈4.61

ⓐ 47 ⓑ 1633 ⓒ 5274

Simplify Variable Expressions with Roots

In the following exercises, simplify using absolute values as necessary.

ⓐ u55 ⓑ v88

Solution

ⓐ u ⓑ |v|

ⓐ a33 ⓑ b99

ⓐ y44 ⓑ m77

Solution

ⓐ |y| ⓑ m

ⓐ k88 ⓑ p66

ⓐ x6 ⓑ y16

Solution

ⓐ |x3| ⓑ y8

ⓐ a14 ⓑ w24

ⓐ x24 ⓑ y22

Solution

ⓐ x12 ⓑ |y11|

ⓐ a12 ⓑ b26

ⓐ x93 ⓑ y124

Solution

ⓐ x3 ⓑ |y3|

ⓐ a105 ⓑ b273

ⓐ m84 ⓑ n205

Solution

ⓐ m2 ⓑ n4

ⓐ r126 ⓑ s303

ⓐ 49x2 ⓑ −81x18

Solution

ⓐ 7|x| ⓑ −9|x9|

ⓐ 100y2 ⓑ −100m32

ⓐ 121m20 ⓑ −64a2

Solution

ⓐ 11m10 ⓑ −8|a|

ⓐ 81x36 ⓑ −25x2

ⓐ 16x84 ⓑ 64y126

Solution

ⓐ 2x2 ⓑ 2y2

ⓐ −8c93 ⓑ 125d153

ⓐ 216a63 ⓑ 32b205

Solution

ⓐ 6a2 ⓑ 2b4

ⓐ 128r147 ⓑ 81s244

ⓐ 144x2y2 ⓑ 169w8y10 ⓒ 8a51b63

Solution

ⓐ 12|xy| ⓑ 13w4|y5|
ⓒ 2a17b2

ⓐ 196a2b2 ⓑ 81p24q6 ⓒ 27p45q93

ⓐ 121a2b2 ⓑ 9c8d12 ⓒ 64x15y663

Solution

ⓐ 11|ab| ⓑ 3c4d6
ⓒ 4x5y22

ⓐ 225x2y2z2 ⓑ 36r6s20 ⓒ 125y18z273

Writing Exercises

Why is there no real number equal to −64?

Solution

Answers will vary.

What is the difference between 92 and 9?

Explain what is meant by the nth root of a number.

Solution

Answers will vary.

Explain the difference of finding the nth root of a number when the index is even compared to when the index is odd.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “simplify expressions with roots.”, “estimate and approximate roots”, and “simplify variable expressions with roots”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

square of a number
If n2 = m, then m is the square of n.
square root of a number
If n2 = m, then n is a square root of m.

Simplify Radical Expressions

Learning Objectives

By the end of this section, you will be able to:

  • Use the Product Property to simplify radical expressions
  • Use the Quotient Property to simplify radical expressions

Before you get started, take this readiness quiz.

Simplify: x9x4.
If you missed this problem, review Example 2 in Properties of Exponents and Scientific Notation.

Solution

x5

Simplify: y3y11.
If you missed this problem, review Example 2 in Properties of Exponents and Scientific Notation.

Solution

1y8

Simplify: (n2)6.
If you missed this problem, review Example 6 in Properties of Exponents and Scientific Notation.

Solution

n12

Use the Product Property to Simplify Radical Expressions

We will simplify radical expressions in a way similar to how we simplified fractions. A fraction is simplified if there are no common factors in the numerator and denominator. To simplify a fraction, we look for any common factors in the numerator and denominator.

A radical expression, an, is considered simplified if it has no factors of mn. So, to simplify a radical expression, we look for any factors in the radicand that are powers of the index.

Simplified Radical Expression

For real integers or rational numbers a and m, and n≥2,

anis considered simplified ifahas no factors ofmn

For example, 5 is considered simplified because there are no perfect square factors in 5. But 12 is not simplified because 12 has a perfect square factor of 4.

Similarly, 43 is simplified because there are no perfect cube factors in 4. But 243 is not simplified because 24 has a perfect cube factor of 8.

To simplify radical expressions, we will also use some properties of roots. The properties we will use to simplify radical expressions are similar to the properties of exponents. We know that (ab)n=anbn. The corresponding of Product Property of Roots says that abn=an·bn.

Product Property of nth Roots

If an and bn are real numbers, and n≥2 is an integer, then

abn=an·bnandan·bn=abn

We use the Product Property of Roots to remove all perfect square factors from a square root.

Simplify Square Roots Using the Product Property of Roots

Simplify: 98.

Solution
The first step in the process is to find the largest factor in the radicand that is a perfect power of the index and rewrite the radicand as a product of two factors, using that factor. We see that 49 is the largest factor of 98 that has a power of 2. In other words 49 is the largest perfect square factor of 98. We can write 98 equals 49 times 2. Always write the perfect square factor first. The square root of 98 can then be written as the square root of the quantity 49 times 2 in parentheses. The second step in the process is to use the product rule to rewrite the radical as the product of two radicals. The square root of the quantity 49 times 2 in parentheses can be written as the square root of 49 times the square root of 2. The third step is to simplify the root of the perfect power. The square root of 49 times the square root of 2 can be written as 7 times the square root of 2.

Simplify: 48.

Solution

43

Simplify: 45.

Solution

35

Notice in the previous example that the simplified form of 98 is 72, which is the product of an integer and a square root. We always write the integer in front of the square root.

Be careful to write your integer so that it is not confused with the index. The expression 72 is very different from 27.

Simplify a radical expression using the Product Property.

  1. Find the largest factor in the radicand that is a perfect power of the index. Rewrite the radicand as a product of two factors, using that factor.
  2. Use the product rule to rewrite the radical as the product of two radicals.
  3. Simplify the root of the perfect power.

We will apply this method in the next example. It may be helpful to have a table of perfect squares, cubes, and fourth powers.

Simplify: ⓐ 500 ⓑ 163 ⓒ 2434.

Solution
ⓐ
Steps demonstrating how to simplify the square root of 500 into its simplest radical form.
500
Rewrite the radicand as a product using the largest perfect square factor. 100·5
Rewrite the radical as the product of two radicals. 100·5
Simplify. 105


ⓑ
This table demonstrates the step-by-step simplification of the cube root of 16.
163
Rewrite the radicand as a product using the greatest perfect cube factor. 23=8 8·23
Rewrite the radical as the product of two radicals. 83·23
Simplify. 223


ⓒ
Step-by-step simplification of the fourth root of 243, illustrating the process of extracting perfect fourth power factors.
2434
Rewrite the radicand as a product using the greatest perfect fourth power factor.
34=81
81·34
Rewrite the radical as the product of two radicals. 814·34
Simplify. 334

Simplify: ⓐ 288 ⓑ 813 ⓒ 644.

Solution

ⓐ 122 ⓑ 333 ⓒ 244

Simplify: ⓐ 432 ⓑ 6253 ⓒ 7294.

Solution

ⓐ 123 ⓑ 553 ⓒ 394

The next example is much like the previous examples, but with variables. Don’t forget to use the absolute value signs when taking an even root of an expression with a variable in the radical.

Simplify: ⓐ x3 ⓑ x43 ⓒ x74.

Solution
ⓐ
Step-by-step simplification of the radical expression sqrt(x^3).
x3
Rewrite the radicand as a product using the largest perfect square factor. x2·x
Rewrite the radical as the product of two radicals. x2·x
Simplify. |x|x


ⓑ
Step-by-step simplification of the radical expression abdul,{
x43
Rewrite the radicand as a product using the largest perfect cube factor. x3·x3.
Rewrite the radical as the product of two radicals. x33·x3
Simplify. xx3


ⓒ
This table illustrates the step-by-step process of simplifying a radical expression.
x74
Rewrite the radicand as a product using the greatest perfect fourth power factor. x4·x34
Rewrite the radical as the product of two radicals. x44·x34
Simplify. |x|x34

Simplify: ⓐ b5 ⓑ y64 ⓒ z53

Solution

ⓐ b2b ⓑ |y|y24 ⓒ zz23

Simplify: ⓐ p9 ⓑ y85 ⓒ q136

Solution

ⓐ p4p ⓑ yy35
ⓒ q2q6

We follow the same procedure when there is a coefficient in the radicand. In the next example, both the constant and the variable have perfect square factors.

Simplify: ⓐ 72n7 ⓑ 24x73 ⓒ 80y144.

Solution
ⓐ
Illustrates the step-by-step process of simplifying the radical expression sqrt(72n^7) to 6|n^3|sqrt(2n).
72n7
Rewrite the radicand as a product using the largest perfect square factor. 36n6·2n
Rewrite the radical as the product of two radicals. 36n6·2n
Simplify. 6|n3|2n


ⓑ
Step-by-step process for simplifying the cube root of 24x^7.
24x73
Rewrite the radicand as a product using perfect cube factors. 8x6·3x3
Rewrite the radical as the product of two radicals. 8x63·3x3
Rewrite the first radicand as (2x2)3. (2x2)33·3x3
Simplify. 2x23x3


ⓒ
Step-by-step process demonstrating the simplification of the fourth root of 80y^14.
80y144
Rewrite the radicand as a product using perfect fourth power factors. 16y12·5y24
Rewrite the radical as the product of two radicals. 16y124·5y24
Rewrite the first radicand as (2y3)4. (2y3)44·5y24
Simplify. 2|y3|5y24

Simplify: ⓐ 32y5 ⓑ 54p103 ⓒ 64q104.

Solution

ⓐ 4y22y ⓑ 3p32p3
ⓒ 2q24q24

Simplify: ⓐ 75a9 ⓑ 128m113 ⓒ 162n74.

Solution

ⓐ 5a43a ⓑ 4m32m23
ⓒ 3|n|2n34

In the next example, we continue to use the same methods even though there are more than one variable under the radical.

Simplify: ⓐ 63u3v5 ⓑ 40x4y53 ⓒ 48x4y74.

Solution
ⓐ
Step-by-step process demonstrating the simplification of a complex square root expression involving variables.
63u3v5
Rewrite the radicand as a product using the largest perfect square factor. 9u2v4·7uv
Rewrite the radical as the product of two radicals. 9u2v4·7uv
Rewrite the first radicand as (3uv2)2. (3uv2)2·7uv
Simplify. 3|u|v27uv


ⓑ
This table demonstrates the step-by-step process of simplifying a cube root expression, showing each operation and its corresponding mathematical form.
40x4y53
Rewrite the radicand as a product using the largest perfect cube factor. 8x3y3·5xy23
Rewrite the radical as the product of two radicals. 8x3y33·5xy23
Rewrite the first radicand as (2xy)3. (2xy)33·5xy23
Simplify. 2xy5xy23


ⓒ
Step-by-step simplification of the radical expression fourth root of 48x^4y^7, demonstrating each algebraic transformation.
48x4y74
Rewrite the radicand as a product using the largest perfect fourth power factor. 16x4y4·3y34
Rewrite the radical as the product of two radicals. 16x4y44·3y34
Rewrite the first radicand as (2xy)4. (2xy)44·3y34
Simplify. 2|xy|3y34

Simplify: ⓐ 98a7b5 ⓑ 56x5y43 ⓒ 32x5y84.

Solution

ⓐ 7|a3|b22ab
ⓑ 2xy7x2y3 ⓒ 2|x|y22x4

Simplify: ⓐ 180m9n11 ⓑ 72x6y53 ⓒ 80x7y44.

Solution

ⓐ 6m4|n5|5mn
ⓑ 2x2y9y23 ⓒ 2|xy|5x34

Simplify: ⓐ −273 ⓑ −164.

Solution
ⓐ
Step-by-step demonstration of calculating the cube root of -27.
−273
Rewrite the radicand as a product using perfect cube factors. (−3)33
Take the cube root. −3


ⓑ
Illustrates why the fourth root of a negative number, such as ⁴√(-16), is not a real number.
−164
There is no real number n where n4=−16. Not a real number.

Simplify: ⓐ −643 ⓑ −814.

Solution

ⓐ −4 ⓑ no real number

Simplify: ⓐ −6253 ⓑ −3244.

Solution

ⓐ −553 ⓑ no real number

We have seen how to use the order of operations to simplify some expressions with radicals. In the next example, we have the sum of an integer and a square root. We simplify the square root but cannot add the resulting expression to the integer since one term contains a radical and the other does not. The next example also includes a fraction with a radical in the numerator. Remember that in order to simplify a fraction you need a common factor in the numerator and denominator.

Simplify: ⓐ 3+32 ⓑ 4−482.

Solution
ⓐ
This table illustrates the step-by-step process for simplifying the mathematical expression 3 + sqrt(32).
3+32
Rewrite the radicand as a product using the largest perfect square factor. 3+16·2
Rewrite the radical as the product of two radicals. 3+16·2
Simplify. 3+42

The terms cannot be added as one has a radical and the other does not. Trying to add an integer and a radical is like trying to add an integer and a variable. They are not like terms!

ⓑ
Step-by-step simplification of the mathematical expression (4 - sqrt(48))/2 to its simplified form 2(1 - sqrt(3)).
4−482
Rewrite the radicand as a product using the largest perfect square factor. 4−16·32
Rewrite the radical as the product of two radicals. 4−16·32
Simplify. 4−432
Factor the common factor from the numerator. 4(1−3)2
Remove the common factor, 2, from the numerator and denominator. 2·2(1−3)2
Simplify. 2(1−3)

Simplify: ⓐ 5+75 ⓑ 10−755

Solution

ⓐ 5+53 ⓑ 2−3

Simplify: ⓐ 2+98 ⓑ 6−453

Solution

ⓐ 2+72 ⓑ 2−5

Use the Quotient Property to Simplify Radical Expressions

Whenever you have to simplify a radical expression, the first step you should take is to determine whether the radicand is a perfect power of the index. If not, check the numerator and denominator for any common factors, and remove them. You may find a fraction in which both the numerator and the denominator are perfect powers of the index.

Simplify: ⓐ 4580 ⓑ 16543 ⓒ 5804.

Solution
ⓐ
This table illustrates the step-by-step process of simplifying the square root of a fraction, starting from sqrt(45/80) and ending with 3/4.
4580
Simplify inside the radical first.
Rewrite showing the common factors of the numerator and denominator. 5·95·16
Simplify the fraction by removing common factors. 916
Simplify. Note (34)2=916. 34


ⓑ
Step-by-step simplification of the cube root of a fraction by simplifying the expression inside the radical.
16543
Simplify inside the radical first.
Rewrite showing the common factors of the numerator and denominator. 2·82·273
Simplify the fraction by removing common factors. 8273
Simplify. Note (23)3=827. 23


ⓒ
This table illustrates the step-by-step process for simplifying the fourth root of 5/80, culminating in the simplified value of 1/2.
5804
Simplify inside the radical first.
Rewrite showing the common factors of the numerator and denominator. 5·15·164
Simplify the fraction by removing common factors. 1164
Simplify. Note (12)4=116. 12

Simplify: ⓐ 7548 ⓑ 542503 ⓒ 321624.

Solution

ⓐ 54 ⓑ 35 ⓒ 23

Simplify: ⓐ 98162 ⓑ 243753 ⓒ 43244.

Solution

ⓐ 79 ⓑ 25 ⓒ 13

In the last example, our first step was to simplify the fraction under the radical by removing common factors. In the next example we will use the Quotient Property to simplify under the radical. We divide the like bases by subtracting their exponents,

aman=am−n,a≠0

Simplify: ⓐ m6m4 ⓑ a8a53 ⓒ a10a24.

Solution
ⓐ
Step-by-step simplification of a radical expression involving variables with exponents.
m6m4
Simplify the fraction inside the radical first.
Divide the like bases by subtracting the exponents. m2
Simplify. |m|


ⓑ
Steps to simplify a cube root expression using the Quotient Property of exponents.
a8a53
Use the Quotient Property of exponents to simplify the fraction under the radical first. a33
Simplify. a


ⓒ
Step-by-step guide on simplifying a radical expression involving variable exponents, illustrating the application of the Quotient Property.
a10a24
Use the Quotient Property of exponents to simplify the fraction under the radical first. a84
Rewrite the radicand using perfect fourth power factors. (a2)44
Simplify. a2

Simplify: ⓐ a8a6 ⓑ x7x34 ⓒ y17y54.

Solution

ⓐ |a| ⓑ |x| ⓒ y3

Simplify: ⓐ x14x10 ⓑ m13m73 ⓒ n12n25.

Solution

ⓐ x2 ⓑ m2 ⓒ n2

Remember the Quotient to a Power Property? It said we could raise a fraction to a power by raising the numerator and denominator to the power separately.

(ab)m=ambm,b≠0

We can use a similar property to simplify a root of a fraction. After removing all common factors from the numerator and denominator, if the fraction is not a perfect power of the index, we simplify the numerator and denominator separately.

Quotient Property of Radical Expressions

If an and bn are real numbers,b≠0, and for any integer n≥2 then,

abn=anbnandanbn=abn

How to Simplify the Quotient of Radical Expressions

Simplify: 27m3196.

Solution
The first step in the process is to simplify the fraction in the radicand, if possible. In this example the quantity 27 m cubed in parentheses divided by 196 cannot be simplified. The second step in the process is to use the quotient property to rewrite the radical as the quotient of two radicals. We rewrite the square root of the quantity 27 m cubed divided by 196 in parentheses as the quotient of the square root of the quantity 27 m cubed in parentheses and the square root of 196. The third step is to simplify the radicals in the numerator and the denominator. 9 m squared and 196 are perfect squares. We rewrite the expression as the quantity square root of quantity 9 m squared in parentheses times square root of the quantity 3 m in parentheses in parentheses divided by square root of 196. The simplified version is the quantity 3 absolute value m times square root of the quantity 3 m in parentheses in parentheses divided by 14.

Simplify: 24p349.

Solution

2|p|6p7

Simplify: 48x5100.

Solution

2x23x5

Simplify a square root using the Quotient Property.

  1. Simplify the fraction in the radicand, if possible.
  2. Use the Quotient Property to rewrite the radical as the quotient of two radicals.
  3. Simplify the radicals in the numerator and the denominator.

Simplify: ⓐ 45x5y4 ⓑ 24x7y33 ⓒ 48x10y84.

Solution
ⓐ
This table illustrates the step-by-step process of simplifying the radical expression (45x^5/y^4).
45x5y4
We cannot simplify the fraction in the radicand. Rewrite using the Quotient Property. 45x5y4
Simplify the radicals in the numerator and the denominator. 9x4·5xy2
Simplify. 3x25xy2


ⓑ
Step-by-step method for simplifying a cube root expression with a fractional radicand, demonstrating the application of radical properties.
24x7y33
The fraction in the radicand cannot be simplified. Use the Quotient Property to write as two radicals. 24x73y33
Rewrite each radicand as a product using perfect cube factors. 8x6·3x3y33
Rewrite the numerator as the product of two radicals. (2x2)33·3x3y33
Simplify. 2x23x3y


ⓒ
This table illustrates the step-by-step process of simplifying a fourth root expression containing a fraction with variables, utilizing properties of radicals.
48x10y84
The fraction in the radicand cannot be simplified. 48x104y84
Use the Quotient Property to write as two radicals. Rewrite each radicand as a product using perfect fourth power factors. 16x8·3x24y84
Rewrite the numerator as the product of two radicals. (2x2)44·3x24(y2)44
Simplify. 2x23x24y2

Simplify: ⓐ 80m3n6 ⓑ 108c10d63 ⓒ 80x10y44.

Solution

ⓐ 4|m|5m|n3| ⓑ 3c34c3d2
ⓒ 2x25x24|y|

Simplify: ⓐ 54u7v8 ⓑ 40r3s63 ⓒ 162m14n124.

Solution

ⓐ 3u36uv4 ⓑ 2r53s2
ⓒ 3|m3|2m24|n3|

Be sure to simplify the fraction in the radicand first, if possible.

Simplify: ⓐ 18p5q732pq2 ⓑ 16x5y754x2y23 ⓒ 5a8b680a3b24.

Solution
ⓐ
This table demonstrates the step-by-step process of simplifying a complex square root expression using algebraic rules.
18p5q732pq2
Simplify the fraction in the radicand, if possible. 9p4q516
Rewrite using the Quotient Property. 9p4q516
Simplify the radicals in the numerator and the denominator. 9p4q4·q4
Simplify. 3p2q2q4


ⓑ
Step-by-step simplification of a cube root expression involving fractions and variables, demonstrating radical properties.
16x5y754x2y23
Simplify the fraction in the radicand, if possible. 8x3y5273
Rewrite using the Quotient Property. 8x3y53273
Simplify the radicals in the numerator and the denominator. 8x3y33·y23273
Simplify. 2xyy233


ⓒ
Step-by-step simplification of a fourth root radical expression involving fractions and variables.
5a8b680a3b24
Simplify the fraction in the radicand, if possible. a5b4164
Rewrite using the Quotient Property. a5b44164
Simplify the radicals in the numerator and the denominator. a4b44·a4164
Simplify. |ab|a42

Simplify: ⓐ 50x5y372x4y ⓑ 16x5y754x2y23 ⓒ 5a8b680a3b24.

Solution

ⓐ 5|y|x6 ⓑ 2xyy233
ⓒ |ab|a42

Simplify: ⓐ 48m7n2100m5n8 ⓑ 54x7y5250x2y23 ⓒ 32a9b7162a3b34.

Solution

ⓐ 2|m|35|n3| ⓑ 3xyx235
ⓒ 2|ab|a243

In the next example, there is nothing to simplify in the denominators. Since the index on the radicals is the same, we can use the Quotient Property again, to combine them into one radical. We will then look to see if we can simplify the expression.

Simplify: ⓐ 48a73a ⓑ −108323 ⓒ 96x743x24.

Solution
ⓐ
Step-by-step simplification of a radical expression, demonstrating the process of simplifying quotients of square roots using algebraic properties.
48a73a
The denominator cannot be simplified, so use the Quotient Property to write as one radical. 48a73a
Simplify the fraction under the radical. 16a6
Simplify. 4|a3|


ⓑ
Step-by-step simplification of a cube root expression, illustrating the application of radical properties.
−108323
The denominator cannot be simplified, so use the Quotient Property to write as one radical. −10823
Simplify the fraction under the radical. −543
Rewrite the radicand as a product using perfect cube factors. (−3)3·23
Rewrite the radical as the product of two radicals. (−3)33·23
Simplify. −323


ⓒ
Step-by-step simplification of a radical expression by applying the quotient property and factoring out perfect fourth powers.
96x743x24
The denominator cannot be simplified, so use the Quotient Property to write as one radical. 96x73x24
Simplify the fraction under the radical. 32x54
Rewrite the radicand as a product using perfect fourth power factors. 16x44·2x4
Rewrite the radical as the product of two radicals. (2x)44·2x4
Simplify. 2|x|2x4

Simplify: ⓐ 98z52z ⓑ −500323 ⓒ 486m1143m54.

Solution

ⓐ 7z2 ⓑ −523
ⓒ 3|m|2m24

Simplify: ⓐ 128m92m ⓑ −192333 ⓒ 324n742n34.

Solution

ⓐ 8m4 ⓑ −4 ⓒ 3|n|24

Access these online resources for additional instruction and practice with simplifying radical expressions.

  • Simplifying Square Root and Cube Root with Variables
  • Express a Radical in Simplified Form-Square and Cube Roots with Variables and Exponents
  • Simplifying Cube Roots

Key Concepts

  • Simplified Radical Expression
    • For real numbers a, m and n≥2
      an is considered simplified if a has no factors of mn
  • Product Property of nth Roots
    • For any real numbers, an and bn, and for any integer n≥2
      abn=an·bn and an·bn=abn
  • How to simplify a radical expression using the Product Property
    1. Find the largest factor in the radicand that is a perfect power of the index.
      Rewrite the radicand as a product of two factors, using that factor.
    2. Use the product rule to rewrite the radical as the product of two radicals.
    3. Simplify the root of the perfect power.
  • Quotient Property of Radical Expressions
    • If an and bn are real numbers, b≠0, and for any integer n≥2 then,
      abn=anbn and anbn=abn
  • How to simplify a radical expression using the Quotient Property.
    1. Simplify the fraction in the radicand, if possible.
    2. Use the Quotient Property to rewrite the radical as the quotient of two radicals.
    3. Simplify the radicals in the numerator and the denominator.

Practice Makes Perfect

Use the Product Property to Simplify Radical Expressions

In the following exercises, use the Product Property to simplify radical expressions.

27

Solution

33

80

125

Solution

55

96

147

Solution

73

450

800

Solution

202

675

ⓐ 324 ⓑ 645

Solution

ⓐ 224 ⓑ 225

ⓐ 6253 ⓑ 1286

ⓐ 644 ⓑ 2563

Solution

ⓐ 244 ⓑ 443

ⓐ 31254 ⓑ 813

In the following exercises, simplify using absolute value signs as needed.

ⓐ y11 ⓑ r53 ⓒ s104

Solution

ⓐ | y5 |y ⓑ rr23 ⓒ s2s24

ⓐ m13 ⓑ u75 ⓒ v116

ⓐ n21 ⓑ q83 ⓒ n108

Solution

ⓐ n10n ⓑ q2q23
ⓒ |n|n28

ⓐ r25 ⓑ p85 ⓒ m54

ⓐ 125r13 ⓑ 108x53 ⓒ 48y64

Solution

ⓐ 5r65r ⓑ 3x4x23
ⓒ 2|y|3y24

ⓐ 80s15 ⓑ 96a75 ⓒ 128b76

ⓐ 242m23 ⓑ 405m104 ⓒ 160n85

Solution

ⓐ 11|m11|2m ⓑ 3m25m24 ⓒ 2n5n35

ⓐ 175n13 ⓑ 512p55 ⓒ 324q74

ⓐ 147m7n11 ⓑ 48x6y73 ⓒ 32x5y44

Solution

ⓐ 7|m3n5|3mn ⓑ 2x2y26y3 ⓒ 2|xy|2x4

ⓐ 96r3s3 ⓑ 80x7y63 ⓒ 80x8y94

ⓐ 192q3r7 ⓑ 54m9n103 ⓒ 81a9b84

Solution

ⓐ 8|qr3|3qr ⓑ 3m3n32n3 ⓒ 3a2b2a4

ⓐ 150m9n3 ⓑ 81p7q83 ⓒ 162c11d124

ⓐ −8643 ⓑ −2564

Solution

ⓐ −643 ⓑ not real

ⓐ −4865 ⓑ −646

ⓐ −325 ⓑ −18

Solution

ⓐ −2 ⓑ not real

ⓐ −83 ⓑ −164

ⓐ 5+12 ⓑ 10−242

Solution

ⓐ 5+23 ⓑ 5−6

ⓐ 8+96 ⓑ 8−804

ⓐ 1+45 ⓑ 3+903

Solution

ⓐ 1+35 ⓑ 1+10

ⓐ 3+125 ⓑ 15+755

Use the Quotient Property to Simplify Radical Expressions

In the following exercises, use the Quotient Property to simplify square roots.

ⓐ 4580 ⓑ 8273 ⓒ 1814

Solution

ⓐ 34 ⓑ 23 ⓒ 13

ⓐ 7298 ⓑ 24813 ⓒ 6964

ⓐ 10036 ⓑ 813753 ⓒ 12564

Solution

ⓐ 53 ⓑ 35 ⓒ 14

ⓐ 12116 ⓑ 162503 ⓒ 321624

ⓐ x10x6 ⓑ p11p23 ⓒ q17q134

Solution

ⓐ x2 ⓑ p3 ⓒ |q|

ⓐ p20p10 ⓑ d12d75 ⓒ m12m48

ⓐ y4y8 ⓑ u21u115 ⓒ v30v126

Solution

ⓐ 1y2 ⓑ u2 ⓒ |v3|

ⓐ q8q14 ⓑ r14r53 ⓒ c21c94

96x7121

Solution

4|x3|6x11

108y449

300m564

Solution

5m23m4

125n7169

98r5100

Solution

7r22r10

180s10144

28q6225

Solution

2|q3|715

150r3256

ⓐ 75r9s8 ⓑ 54a8b33 ⓒ 64c5d44

Solution

ⓐ 5r43rs4 ⓑ 3a22a23b
ⓒ 2|c|4c4|d|

ⓐ 72x5y6 ⓑ 96r11s55 ⓒ 128u7v126

ⓐ 28p7q2 ⓑ 81s8t33 ⓒ 64p15q124

Solution

ⓐ 2|p3|7p|q| ⓑ 3s23s23t
ⓒ 2|p3|4p34|q3|

ⓐ 45r3s10 ⓑ 625u10v33 ⓒ 729c21d84

ⓐ 32x5y318x3y ⓑ 5x6y940x5y33 ⓒ 5a8b680a3b24

Solution

ⓐ 4|xy|3 ⓑ y2x32 ⓒ |ab|a42

ⓐ 75r6s848rs4 ⓑ 24x8y481x2y3 ⓒ 32m9n2162mn24

ⓐ 27p2q108p4q3 ⓑ 16c5d7250c2d23 ⓒ 2m9n7128m3n6

Solution

ⓐ 12|pq| ⓑ 2cdd235
ⓒ |mn|2

ⓐ 50r5s2128r2s6 ⓑ 24m9n7375m4n3 ⓒ 81m2n8256m1n24

ⓐ 45p95q2 ⓑ 64424 ⓒ 128x852x25

Solution

ⓐ 3p4p|q| ⓑ 224
ⓒ 2x2x5

ⓐ 80q55q ⓑ −625353 ⓒ 80m745m4

ⓐ 50m72m ⓑ 125023 ⓒ 486y92y34

Solution

ⓐ 5|m3| ⓑ 553
ⓒ 3|y|3y24

ⓐ 72n112n ⓑ 16263 ⓒ 160r105r34

Writing Exercises

Explain why x4=x2. Then explain why x16=x8.

Solution

Answers will vary.

Explain why 7+9 is not equal to 7+9.

Explain how you know that x105=x2.

Solution

Answers will vary.

Explain why −644 is not a real number but −643 is.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 3 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “use the product property to simplify radical expressions” and “use the quotient property to simplify radical expressions”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Simplify Rational Exponents

Learning Objectives

By the end of this section, you will be able to:

  • Simplify expressions with a1n
  • Simplify expressions with amn
  • Use the properties of exponents to simplify expressions with rational exponents

Before you get started, take this readiness quiz.

Add: 715+512.
If you missed this problem, review Example 5 in Fractions.

Solution

5360

Simplify: (4x2y5)3.
If you missed this problem, review Example 7 in Properties of Exponents and Scientific Notation.

Solution

64x6y15

Simplify: 5−3.
If you missed this problem, review Example 3 in Properties of Exponents and Scientific Notation.

Solution

1125

Simplify Expressions with a1n

Rational exponents are another way of writing expressions with radicals. When we use rational exponents, we can apply the properties of exponents to simplify expressions.

The Power Property for Exponents says that (am)n=am·n when m and n are whole numbers. Let’s assume we are now not limited to whole numbers.

Suppose we want to find a number p such that (8p)3=8. We will use the Power Property of Exponents to find the value of p.

(8p)3=8Multiply the exponents on the left.83p=8Write the exponent 1 on the right.83p=81Since the bases are the same, the exponents must be equal.3p=1Solve forp.p=13

So (813)3=8. But we know also (83)3=8. Then it must be that 813=83.

This same logic can be used for any positive integer exponent n to show that a1n=an.

Rational Exponent a1n

If an is a real number and n≥2, then

a1n=an

The denominator of the rational exponent is the index of the radical.

There will be times when working with expressions will be easier if you use rational exponents and times when it will be easier if you use radicals. In the first few examples, you’ll practice converting expressions between these two notations.

Write as a radical expression: ⓐ x12 ⓑ y13 ⓒ z14.

Solution

We want to write each expression in the form an.

ⓐ
Illustrates the conversion of the rational exponent x^(1/2) to its radical form, sqrt(x), explaining the rule for the radical index.
x12
The denominator of the rational exponent is 2, so
the index of the radical is 2. We do not show the
index when it is 2.
x
ⓑ
This table illustrates the conversion from fractional exponents to radical form, specifically demonstrating y^(1/3) as the cube root of y.
y13
The denominator of the exponent is 3, so the
index is 3.
y3
ⓒ
This table illustrates the conversion of a fractional exponent to its equivalent radical form, focusing on how the exponent's denominator determines the radical index.
z14
The denominator of the exponent is 4, so the
index is 4.
z4

Write as a radical expression: ⓐ t12 ⓑ m13 ⓒ r14.

Solution

ⓐ t ⓑ m3 ⓒ r4

Write as a radial expression: ⓐ b16 ⓑ z15 ⓒ p14.

Solution

ⓐ b6 ⓑ z5 ⓒ p4

In the next example, we will write each radical using a rational exponent. It is important to use parentheses around the entire expression in the radicand since the entire expression is raised to the rational power.

Write with a rational exponent: ⓐ 5y ⓑ 4x3 ⓒ 35z4.

Solution

We want to write each radical in the form a1n.

ⓐ
Shows the conversion of a square root radical expression to its equivalent exponential form, with explanatory notes.
5y
No index is shown, so it is 2.
The denominator of the exponent will be 2.
(5y)12
Put parentheses around the entire
expression 5y.
ⓑ
This table illustrates converting a radical expression to its equivalent fractional exponent form, with an explanation of the underlying process.
4x3
The index is 3, so the denominator of the
exponent is 3. Include parentheses (4x).
(4x)13
ⓒ
Illustrates converting a radical expression to its rational exponent form, with detailed explanations.
35z4
The index is 4, so the denominator of the
exponent is 4. Put parentheses only around
the 5z since 3 is not under the radical sign.
3(5z)14

Write with a rational exponent: ⓐ 10m ⓑ 3n5 ⓒ 36y4.

Solution

ⓐ (10m)12 ⓑ (3n)15
ⓒ 3(6y)14

Write with a rational exponent: ⓐ 3k7 ⓑ 5j4 ⓒ 82a3.

Solution

ⓐ (3k)17 ⓑ (5j)14
ⓒ 8(2a)13

In the next example, you may find it easier to simplify the expressions if you rewrite them as radicals first.

Simplify: ⓐ 2512 ⓑ 6413 ⓒ 25614.

Solution
ⓐ
Steps to simplify a fractional exponent: rewriting 25^(1/2) as a square root and finding its value.
2512
Rewrite as a square root. 25
Simplify. 5
ⓑ
This table illustrates the step-by-step simplification of 64 raised to the power of one-third, demonstrating its conversion to a cube root and final evaluation.
6413
Rewrite as a cube root. 643
Recognize 64 is a perfect cube. 433
Simplify. 4
ⓒ
This table illustrates the step-by-step simplification of the mathematical expression 256 to the power of 1/4, demonstrating its transformation into a fourth root and final numerical evaluation.
25614
Rewrite as a fourth root. 2564
Recognize 256 is a perfect fourth power. 444
Simplify. 4

Simplify: ⓐ 3612 ⓑ 813 ⓒ 1614.

Solution

ⓐ 6 ⓑ 2 ⓒ 2

Simplify: ⓐ 10012 ⓑ 2713 ⓒ 8114.

Solution

ⓐ 10 ⓑ 3 ⓒ 3

Be careful of the placement of the negative signs in the next example. We will need to use the property a−n=1an in one case.

Simplify: ⓐ (−16)14 ⓑ −1614 ⓒ (16)−14.

Solution
ⓐ
This table illustrates the step-by-step simplification of the expression (-16)^(1/4), demonstrating the process to determine there is no real solution.
(−16)14
Rewrite as a fourth root. −164
(−2)44
Simplify. No real solution.
ⓑ
Step-by-step simplification of the mathematical expression -16^(1/4), detailing each transformation to reach the final result.
−1614
The exponent only applies to the 16.
Rewrite as a fouth root.
−164
Rewrite 16 as 24. −244
Simplify. −2
ⓒ
Demonstrates the step-by-step simplification of (16)^(-1/4), converting it to 1/2 using exponent and root properties.
(16)−14
Rewrite using the property a−n=1an. 1(16)14
Rewrite as a fourth root. 1164
Rewrite 16 as 24. 1244
Simplify. 12

Simplify: ⓐ (−64)−12 ⓑ −6412 ⓒ (64)−12.

Solution

ⓐ No real solution ⓑ −8
ⓒ 18

Simplify: ⓐ (−256)14 ⓑ −25614 ⓒ (256)−14.

Solution

ⓐ No real solution ⓑ −4
ⓒ 14

Simplify Expressions with amn

We can look at amn in two ways. Remember the Power Property tells us to multiply the exponents and so (a1n)m and (am)1n both equal amn. If we write these expressions in radical form, we get

amn=(a1n)m=(an)mandamn=(am)1n=amn

This leads us to the following definition.

Rational Exponent amn

For any positive integers m and n,

amn=(an)mandamn=amn

Which form do we use to simplify an expression? We usually take the root first—that way we keep the numbers in the radicand smaller, before raising it to the power indicated.

Write with a rational exponent: ⓐ y3 ⓑ (2x3)4 ⓒ (3a4b)3.

Solution

We want to use amn=amn to write each radical in the form amn.

ⓐ
Conversion of radical to exponential form: √y³ is equivalent to y^(3/2). The exponent's numerator (3) comes from the radicand's exponent, and the denominator (2) is the radical's index.


ⓑ
This image demonstrates how to convert radical expressions to rational exponents. The external exponent becomes the numerator and the radical index becomes the denominator, as seen with (³√2x)⁴ = (2x)^(4/3).


ⓒ
Illustration showing the relationship between radical form and exponential form, where the numerator of a fractional exponent is the power and the denominator is the radical index.

Write with a rational exponent: ⓐ x5 ⓑ (3y4)3 ⓒ (2m3n)5.

Solution

ⓐ x52 ⓑ (3y)34 ⓒ (2m3n)52

Write with a rational exponent: ⓐ a25 ⓑ (5ab3)5 ⓒ (7xyz)3.

Solution

ⓐ a25 ⓑ (5ab)53
ⓒ (7xyz)32

Remember that a−n=1an. The negative sign in the exponent does not change the sign of the expression.

Simplify: ⓐ 12523 ⓑ 16−32 ⓒ 32−25.

Solution

We will rewrite the expression as a radical first using the defintion, amn=(an)m. This form lets us take the root first and so we keep the numbers in the radicand smaller than if we used the other form.

ⓐ
Step-by-step simplification of an expression with a fractional exponent by converting it to radical form.
12523
The power of the radical is the numerator of the exponent, 2.
The index of the radical is the denominator of the
exponent, 3.
(1253)2
Simplify. (5)2
25
ⓑ We will rewrite each expression first using a−n=1an and then change to radical form.
Steps for simplifying the exponential expression 16^(-3/2) using exponent and radical rules.
16−32
Rewrite using a−n=1an 11632
Change to radical form. The power of the radical is the
numerator of the exponent, 3. The index is the denominator
of the exponent, 2.
1(16)3
Simplify. 143
164
ⓒ
Step-by-step simplification of 32^(-2/5), illustrating conversion from negative fractional exponent to radical form and final evaluation.
32−25
Rewrite using a−n=1an. 13225
Change to radical form. 1(325)2
Rewrite the radicand as a power. 1(255)2
Simplify. 122
14

Simplify: ⓐ 2723 ⓑ 81−32 ⓒ 16−34.

Solution

ⓐ 9 ⓑ 1729 ⓒ 18

Simplify: ⓐ 432 ⓑ 27−23 ⓒ 625−34.

Solution

ⓐ 8 ⓑ 19 ⓒ 1125

Simplify: ⓐ −2532 ⓑ −25−32 ⓒ (−25)32.

Solution
ⓐ
Step-by-step simplification of a negative number raised to a fractional exponent.
−2532
Rewrite in radical form. −(25)3
Simplify the radical. −(5)3
Simplify. −125
ⓑ
Step-by-step simplification of the expression -25^(-3/2) using exponent rules, radical forms, and arithmetic operations.
−25−32
Rewrite using a−n=1an. −(12532)
Rewrite in radical form. −(1(25)3)
Simplify the radical. −(1(5)3)
Simplify. −1125
ⓒ
Evaluates the expression (-25)^(3/2), demonstrating the steps to rewrite it in radical form and concluding that it is not a real number.
(−25)32
Rewrite in radical form. (−25)3
There is no real number whose square root
is−25.
Not a real number.

Simplify: ⓐ −1632 ⓑ −16−32 ⓒ (−16)−32.

Solution

ⓐ −64 ⓑ −164 ⓒ not a real number

Simplify: ⓐ −8132 ⓑ −81−32 ⓒ (−81)−32.

Solution

ⓐ −729 ⓑ −1729 ⓒ not a real number

Use the Properties of Exponents to Simplify Expressions with Rational Exponents

The same properties of exponents that we have already used also apply to rational exponents. We will list the Properties of Exponenets here to have them for reference as we simplify expressions.

Properties of Exponents

If a and b are real numbers and m and n are rational numbers, then

Product Propertyam·an=am+nPower Property(am)n=am·nProduct to a Power(ab)m=ambmQuotient Propertyaman=am−n,a≠0Zero Exponent Definitiona0=1,a≠0Quotient to a Power Property(ab)m=ambm,b≠0Negative Exponent Propertya−n=1an,a≠0

We will apply these properties in the next example.

Simplify: ⓐ x12·x56 ⓑ (z9)23 ⓒ x13x53.

Solution

ⓐ The Product Property tells us that when we multiply the same base, we add the exponents.

This table demonstrates the step-by-step simplification of an algebraic expression involving fractional exponents by applying the rule for multiplying powers with the same base.
x12·x56
The bases are the same, so we add the
exponents.
x12+56
Add the fractions. x86
Simplify the exponent. x43

ⓑ The Power Property tells us that when we raise a power to a power, we multiply the exponents.

Steps for simplifying an exponential expression using the power of a power rule, showing the rule's application and the final result.
(z9)23
To raise a power to a power, we multiply
the exponents.
z9·23
Simplify. z6

ⓒ The Quotient Property tells us that when we divide with the same base, we subtract the exponents.

This table demonstrates the step-by-step simplification of an exponential expression using the quotient rule for exponents.
x13x53
x13x53
To divide with the same base, we subtract
the exponents.
1x53−13
Simplify. 1x43

Simplify: ⓐ x16·x43 ⓑ (x6)43 ⓒ x23x53.

Solution

ⓐ x32 ⓑ x8 ⓒ 1x

Simplify: ⓐ y34·y58 ⓑ (m9)29 ⓒ d15d65.

Solution

ⓐ y118 ⓑ m2 ⓒ 1d

Sometimes we need to use more than one property. In the next example, we will use both the Product to a Power Property and then the Power Property.

Simplify: ⓐ (27u12)23 ⓑ (m23n12)32.

Solution
ⓐ
Step-by-step simplification of an algebraic expression involving fractional exponents using exponent properties.
(27u12)23
First we use the Product to a Power
Property.
(27)23(u12)23
Rewrite 27 as a power of 3. (33)23(u12)23
To raise a power to a power, we multiply
the exponents.
(32)(u13)
Simplify. 9u13
ⓑ
This table illustrates the step-by-step simplification of an algebraic expression involving fractional exponents and the product to a power property.
(m23n12)32
First we use the Product to a Power
Property.
(m23)32(n12)32
To raise a power to a power, we multiply
the exponents.
mn34

Simplify: ⓐ (32x13)35 ⓑ (x34y12)23.

Solution

ⓐ 8x15 ⓑ x12y13

Simplify: ⓐ (81n25)32 ⓑ (a32b12)43.

Solution

ⓐ 729n35 ⓑ a2b23

We will use both the Product Property and the Quotient Property in the next example.

Simplify: ⓐ x34·x−14x−64 ⓑ (16x43y−56x−23y16)12.

Solution
ⓐ
Step-by-step simplification of a rational expression using the Product and Quotient Properties of exponents.
x34·x−14x−64
Use the Product Property in the numerator,
add the exponents.
x24x−64
Use the Quotient Property, subtract the
exponents.
x84
Simplify. x2

ⓑ Follow the order of operations to simplify inside the parenthese first.

Step-by-step simplification of an algebraic expression involving fractional exponents using properties of exponents.
(16x43y−56x−23y16)12
Use the Quotient Property, subtract the
exponents.
(16x63y66)12
Simplify. (16x2y)12
Use the Product to a Power Property,
multiply the exponents.
4xy12

Simplify: ⓐ m23·m−13m−53 ⓑ (25m16n116m23n−16)12.

Solution

ⓐ m2 ⓑ 5nm14

Simplify: ⓐ u45·u−25u−135 ⓑ (27x45y16x15y−56)13.

Solution

ⓐ u3 ⓑ 3x15y13

Access these online resources for additional instruction and practice with simplifying rational exponents.

  • Review-Rational Exponents
  • Using Laws of Exponents on Radicals: Properties of Rational Exponents

Key Concepts

  • Rational Exponent a1n
    • If an is a real number and n≥2, then a1n=an.
  • Rational Exponent amn
    • For any positive integers m and n,
      amn=(an)m and amn=amn
  • Properties of Exponents
    • If a, b are real numbers and m, n are rational numbers, then
      • Product Property am·an=am+n
      • Power Property (am)n=am·n
      • Product to a Power (ab)m=ambm
      • Quotient Property aman=am−n,a≠0
      • Zero Exponent Definition a0=1, a≠0
      • Quotient to a Power Property (ab)m=ambm,b≠0
      • Negative Exponent Property a−n=1an,a≠0

Practice Makes Perfect

Simplify expressions with a1n

In the following exercises, write as a radical expression.

ⓐ x12 ⓑ y13 ⓒ z14

Solution

ⓐ x ⓑ y3 ⓒ z4

ⓐ r12 ⓑ s13 ⓒ t14

ⓐ u15 ⓑ v19 ⓒ w120

Solution

ⓐ u5 ⓑ v9 ⓒ w20

ⓐ g17 ⓑ h15 ⓒ j125

In the following exercises, write with a rational exponent.

ⓐ x7 ⓑ y9 ⓒ f5

Solution

ⓐ x17 ⓑ y19 ⓒ f15

ⓐ r8 ⓑ s10 ⓒ t4

ⓐ 7c3 ⓑ 12d7 ⓒ 26b4

Solution

ⓐ (7c)13 ⓑ (12d)17
ⓒ 2(6b)14

ⓐ 5x4 ⓑ 9y8 ⓒ 73z5

ⓐ 21p ⓑ 8q4 ⓒ 436r6

Solution

ⓐ (21p)12 ⓑ (8q)14
ⓒ 4(36r)16

ⓐ 25a3 ⓑ 3b ⓒ 40c8

In the following exercises, simplify.

ⓐ 8112 ⓑ 12513 ⓒ 6412

Solution

ⓐ 9 ⓑ 5 ⓒ 8

ⓐ 62514 ⓑ 24315 ⓒ 3215

ⓐ 1614 ⓑ 1612 ⓒ 62514

Solution

ⓐ 2 ⓑ 4 ⓒ 5

ⓐ 6413 ⓑ 3215 ⓒ 8114

ⓐ (−216)13 ⓑ −21613 ⓒ (216)−13

Solution

ⓐ −6 ⓑ −6 ⓒ 16

ⓐ (−1000)13 ⓑ −100013 ⓒ (1000)−13

ⓐ (−81)14 ⓑ −8114 ⓒ (81)−14

Solution

ⓐ not real ⓑ −3 ⓒ 13

ⓐ (−49)12 ⓑ −4912 ⓒ (49)−12

ⓐ (−36)12 ⓑ −3612 ⓒ (36)−12

Solution

ⓐ not real ⓑ −6 ⓒ 16

ⓐ (−16)14 ⓑ −1614 ⓒ 16−14

ⓐ (−100)12 ⓑ −10012 ⓒ (100)−12

Solution

ⓐ not real ⓑ −10 ⓒ 110

ⓐ (−32)15 ⓑ (243)−15 ⓒ −12513

Simplify Expressions with amn

In the following exercises, write with a rational exponent.

ⓐ m5 ⓑ (3y3)7 ⓒ (4x5y)35

Solution

ⓐ m52 ⓑ (3y)73 ⓒ (4x5y)35

ⓐ r74 ⓑ (2pq5)3 ⓒ (12m7n)34

ⓐ u25 ⓑ (6x3)5 ⓒ (18a5b)74

Solution

ⓐ u25 ⓑ (6x)53 ⓒ (18a5b)74

ⓐ a3 ⓑ (21v4)3 ⓒ (2xy5z)24

In the following exercises, simplify.

ⓐ 6452 ⓑ 81−32 ⓒ (−27)23

Solution

ⓐ 32,768 ⓑ 1729 ⓒ 9

ⓐ 2532 ⓑ 9−32 ⓒ (−64)23

ⓐ 3225 ⓑ 27−23 ⓒ (−25)12

Solution

ⓐ 4 ⓑ 19 ⓒ not real

ⓐ 10032 ⓑ 49−52 ⓒ (−100)32

ⓐ −932 ⓑ −9−32 ⓒ (−9)32

Solution

ⓐ −27 ⓑ −127 ⓒ not real

ⓐ −6432 ⓑ −64−32 ⓒ (−64)32

Use the Laws of Exponents to Simplify Expressions with Rational Exponents

In the following exercises, simplify. Assume all variables are positive.

ⓐ c14·c58 ⓑ (p12)34 ⓒ r45r95

Solution

ⓐ c78 ⓑ p9 ⓒ 1r

ⓐ 652·612 ⓑ (b15)35 ⓒ w27w97

ⓐ y12·y34 ⓑ (x12)23 ⓒ m58m138

Solution

ⓐ y54 ⓑ x8 ⓒ 1m

ⓐ q23·q56 ⓑ (h6)43 ⓒ n35n85

ⓐ (27q32)43 ⓑ (a13b23)32

Solution

ⓐ 81q2 ⓑ a12b

ⓐ (64s37)16 ⓑ (m43n12)34

ⓐ (16u13)34 ⓑ (4p13q12)32

Solution

ⓐ 8u14 ⓑ 8p12q34

ⓐ (625n83)34 ⓑ (9x25y35)52

ⓐ r52·r−12r−32 ⓑ (36s15t−32s−95t12)12

Solution

ⓐ r72 ⓑ 6st

ⓐ a34·a−14a−104 ⓑ (27b23c−52b−73c12)13

ⓐ c53·c−13c−23 ⓑ (8x53y−1227x−43y52)13

Solution

ⓐ c2 ⓑ 2x3y

ⓐ m74·m−54m−24 ⓑ (16m15n3281m95n−12)14

Writing Exercises

Show two different algebraic methods to simplify 432. Explain all your steps.

Solution

Answers will vary.

Explain why the expression (−16)32 cannot be evaluated.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “simplify expressions with a to the power of 1 divided by n.”, “simplify expression with a to the power of m divided by n”, and “use the laws of exponents to simplify expression with rational exponents”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Add, Subtract, and Multiply Radical Expressions

Learning Objectives

By the end of this section, you will be able to:

  • Add and subtract radical expressions
  • Multiply radical expressions
  • Use polynomial multiplication to multiply radical expressions

Before you get started, take this readiness quiz.

Add: 3x2+9x−5−(x2−2x+3).
If you missed this problem, review Example 5 in Add and Subtract Polynomials.

Solution

2x2+11x−8

Simplify: (2+a)(4−a).
If you missed this problem, review Example 4 in Multiply Polynomials.

Solution

8+2a−a2

Simplify: (9−5y)2.
If you missed this problem, review Example 7 in Multiply Polynomials.

Solution

81−90y+25y2

Add and Subtract Radical Expressions

Adding radical expressions with the same index and the same radicand is just like adding like terms. We call radicals with the same index and the same radicand like radicals to remind us they work the same as like terms.

Like Radicals

Like radicals are radical expressions with the same index and the same radicand.

We add and subtract like radicals in the same way we add and subtract like terms. We know that 3x+8x is 11x. Similarly we add 3x+8x and the result is 11x.

Think about adding like terms with variables as you do the next few examples. When you have like radicals, you just add or subtract the coefficients. When the radicals are not like, you cannot combine the terms.

Simplify: ⓐ 22−72 ⓑ 5y3+4y3 ⓒ 7x4−2y4.

Solution
ⓐ
Example illustrating the subtraction of like radical expressions, with step-by-step explanation.
22−72
Since the radicals are like, we subtract the
coefficients.
−52
ⓑ
This table demonstrates adding like radicals by combining coefficients, showing an initial expression, the rationale, and the simplified result.
5y3+4y3
Since the radicals are like, we add the
coefficients.
9y3
ⓒ
Table presenting the algebraic expression: 7⁴√x - 2⁴√y.
7x4−2y4

The indices are the same but the radicals are different. These are not like radicals. Since the radicals are not like, we cannot subtract them.

Simplify: ⓐ 82−92 ⓑ 4x3+7x3 ⓒ 3x4−5y4.

Solution

ⓐ −2 ⓑ 11x3
ⓒ 3x4−5y4

Simplify: ⓐ 53−93 ⓑ 5y3+3y3 ⓒ 5m4−2m3.

Solution

ⓐ −43 ⓑ 8y3
ⓒ 5m4−2m3

For radicals to be like, they must have the same index and radicand. When the radicands contain more than one variable, as long as all the variables and their exponents are identical, the radicands are the same.

Simplify: ⓐ 25n−65n+45n ⓑ 3xy4+53xy4−43xy4.

Solution
ⓐ
Step-by-step simplification of a radical expression, showing the process of combining like radicals to reach a final value of zero.
25n−65n+45n
Since the radicals are like, we combine them. 05n
Simplify. 0
ⓑ
Example demonstrating the combination of like radicals.
3xy4+53xy4−43xy4
Since the radicals are like, we combine them. 23xy4

Simplify: ⓐ 7x−77x+47x ⓑ 45xy4+25xy4−75xy4.

Solution

ⓐ −27x ⓑ −5xy4

Simplify: ⓐ 43y−73y+23y ⓑ 67mn3+7mn3−47mn3.

Solution

ⓐ −3y ⓑ 37mn3

Remember that we always simplify radicals by removing the largest factor from the radicand that is a power of the index. Once each radical is simplified, we can then decide if they are like radicals.

Simplify: ⓐ 20+35 ⓑ 243−3753 ⓒ 12484−232434.

Solution
ⓐ
Step-by-step simplification of a radical expression involving addition.
20+35
Simplify the radicals, when possible. 4·5+35
25+35
Combine the like radicals. 55
ⓑ
Step-by-step simplification of a radical expression involving the subtraction of cube roots.
243−3753
Simplify the radicals. 83·33−1253·33
233−533
Combine the like radicals. −333
ⓒ
Step-by-step solution demonstrating the simplification and combination of radical expressions involving fourth roots.
12484−232434
Simplify the radicals. 12164·34−23814·34
12·2·34−23·3·34
34−234
Combine the like radicals. −34

Simplify: ⓐ 18+62 ⓑ 6163−22503 ⓒ 23813−12243.

Solution

ⓐ 92 ⓑ 223 ⓒ 33

Simplify: ⓐ 27+43 ⓑ 453−7403 ⓒ 121283−53543.

Solution

ⓐ 73 ⓑ −1053 ⓒ −323

In the next example, we will remove both constant and variable factors from the radicals. Now that we have practiced taking both the even and odd roots of variables, it is common practice at this point for us to assume all variables are greater than or equal to zero so that absolute values are not needed. We will use this assumption throughout the rest of this chapter.

Simplify: ⓐ 950m2−648m2 ⓑ 54n53−16n53.

Solution
ⓐ
Step-by-step guide on simplifying a mathematical expression involving radical terms, detailing the breakdown of square roots and explaining why the final terms cannot be combined.
950m2−648m2
Simplify the radicals. 925m2·2−616m2·3
9·5m·2−6·4m·3
45m2−24m3
The radicals are not like and so cannot be
combined.
ⓑ
Illustrates the step-by-step simplification of a mathematical expression involving the subtraction of two cube roots.
54n53−16n53
Simplify the radicals. 27n33·2n23−8n33·2n23
3n2n23−2n2n23
Combine the like radicals. n2n23

Simplify: ⓐ 32m7−50m7 ⓑ 135x73−40x73.

Solution

ⓐ −m32m ⓑ x25x3

Simplify: ⓐ 27p3−48p3 ⓑ 256y53−32y53.

Solution

ⓐ −p3p
ⓑ 4y4y23−2y4n23

Multiply Radical Expressions

We have used the Product Property of Roots to simplify square roots by removing the perfect square factors. We can use the Product Property of Roots ‘in reverse’ to multiply square roots. Remember, we assume all variables are greater than or equal to zero.

We will rewrite the Product Property of Roots so we see both ways together.

Product Property of Roots

For any real numbers, an and bn, and for any integer n≥2

abn=an·bnandan·bn=abn

When we multiply two radicals they must have the same index. Once we multiply the radicals, we then look for factors that are a power of the index and simplify the radical whenever possible.

Multiplying radicals with coefficients is much like multiplying variables with coefficients. To multiply 4x·3y we multiply the coefficients together and then the variables. The result is 12xy. Keep this in mind as you do these examples.

Simplify: ⓐ (62)(310) ⓑ (−543)(−463).

Solution
ⓐ
Step-by-step guide to multiplying and simplifying radical expressions, from (6√2)(3√10) to 36√5.
(62)(310)
Multiply using the Product Property. 1820
Simplify the radical. 184·5
Simplify. 18·2·5
365
ⓑ
Step-by-step multiplication and simplification of cube root expressions.
(−543)(−463)
Multiply using the Product Property. 20243
Simplify the radical. 2083·33
Simplify. 20·2·33
4033

Simplify: ⓐ (32)(230) ⓑ (2183)(−363).

Solution

ⓐ 1215 ⓑ −1843

Simplify: ⓐ (33)(36) ⓑ (−493)(363).

Solution

ⓐ 272 ⓑ −3623

We follow the same procedures when there are variables in the radicands.

Simplify: ⓐ (106p3)(43p) ⓑ (220y24)(328y34).

Solution
ⓐ
This table illustrates the step-by-step process of multiplying and simplifying a radical expression, showing the intermediate results.
(106p3)(43p)
Multiply. 4018p4
Simplify the radical. 409p4·2
Simplify. 40·3p2·2
120p22

ⓑ When the radicands involve large numbers, it is often advantageous to factor them in order to find the perfect powers.

Step-by-step solution for multiplying and simplifying radical expressions involving fourth roots.
(220y24)(328y34)
Multiply. 64·5·4·7y54
Simplify the radical. 616y44·35y4
Simplify. 6·2y35y4
Multiply. 12y35y4

Simplify: ⓐ (66x2)(830x4) ⓑ (−412y34)(−8y34).

Solution

ⓐ 288x35 ⓑ 8y6y24

Simplify: ⓐ (26y4)(1230y) ⓑ (−49a34)(327a24).

Solution

ⓐ 144y25y ⓑ −36a3a4

Use Polynomial Multiplication to Multiply Radical Expressions

In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. First we will distribute and then simplify the radicals when possible.

Simplify: ⓐ 6(2+18) ⓑ 93(5−183).

Solution
ⓐ
This table outlines the sequential steps involved in simplifying a radical expression, from initial multiplication to combining radicals.
6(2+18)
Multiply. 12+108
Simplify. 4·3+36·3
Simplify. 23+63
Combine like radicals. 83
ⓑ
This table illustrates the step-by-step simplification of an algebraic expression involving cube roots through distribution and simplification.
93(5−183)
Distribute. 593−1623
Simplify. 593−273·63
Simplify. 593−363

Simplify: ⓐ 6(1+36) ⓑ 43(−2−63).

Solution

ⓐ 18+6 ⓑ −243−233

Simplify: ⓐ 8(2−58) ⓑ 33(−93−63).

Solution

ⓐ −40+42 ⓑ −3−183

When we worked with polynomials, we multiplied binomials by binomials. Remember, this gave us four products before we combined any like terms. To be sure to get all four products, we organized our work—usually by the FOIL method.

Simplify: ⓐ (3−27)(4−27) ⓑ (x3−2)(x3+4).

Solution
ⓐ
This table illustrates the step-by-step multiplication and simplification of two binomial expressions containing square roots.
(3−27)(4−27)
Multiply. 12−67−87+4·7
Simplify. 12−67−87+28
Combine like terms. 40−147
ⓑ
This table demonstrates the step-by-step multiplication and simplification of an algebraic expression involving cube roots.
(x3−2)(x3+4)
Multiply. x23+4x3−2x3−8
Combine like terms. x23+2x3−8

Simplify: ⓐ (6−37)(3+47) ⓑ (x3−2)(x3−3).

Solution

ⓐ −66+157
ⓑ x23−5x3+6

Simplify: ⓐ (2−311)(4−11) ⓑ (x3+1)(x3+3).

Solution

ⓐ 41−1411
ⓑ x23+4x3+3

Simplify: (32−5)(2+45).

Solution
Step-by-step multiplication and simplification of two radical expressions, showing the progression from initial problem to final solution.
(32−5)(2+45)
Multiply. 3·2+1210−10−4·5
Simplify. 6+1210−10−20
Combine like terms. −14+1110

Simplify: (53−7)(3+27)

Solution

1+921

Simplify: (6−38)(26+8)

Solution

−12−203

Recognizing some special products made our work easier when we multiplied binomials earlier. This is true when we multiply radicals, too. The special product formulas we used are shown here.

Special Products

Binomial SquaresProduct of Conjugates(a+b)2=a2+2ab+b2(a+b)(a−b)=a2−b2(a−b)2=a2−2ab+b2

We will use the special product formulas in the next few examples. We will start with the Product of Binomial Squares Pattern.

Simplify: ⓐ (2+3)2 ⓑ (4−25)2.

Solution

Be sure to include the 2ab term when squaring a binomial.

ⓐ
Two mathematical expressions are shown: (a + b)' in red text, representing the square of a sum, and directly below it, (2 + sqrt(3))', a specific example of the same algebraic form.
Multiply, using the Product of Binomial Squares Pattern. Algebraic identity for a perfect square: a^2 + 2ab + b^2, with an example where a=2 and b=sqrt(3).
Simplify. A mathematical expression reads '4 + 4 square root of 3 + 3' in black text against a white background, representing an algebraic sum involving an integer and a surd.
Combine like terms. A mathematical expression '7 + 4√3' is displayed on a white background.


ⓑ
The image displays two algebraic expressions related to squaring a difference. The first expression, in red, is (a - b)^2. Below it, in black, is the specific numerical instance (4 - 2√5)^2.
Multiply, using the Product of Binomial Squares Pattern. Illustration of the perfect square trinomial identity (a-b)^2, showing a^2 - 2ab + b^2 and its numerical application with a=4 and b=2sqrt(5).
Simplify. A mathematical expression reads '16 - 16√5 + 4 × 5' against a white background.
The image shows the mathematical expression 16 - 16√5 + 20.
Combine like terms. The image shows the mathematical expression 36 - 16√5 in a bold, black font against a white background.

Simplify: ⓐ (10+2)2 ⓑ (1+36)2.

Solution

ⓐ 102+202 ⓑ 55+66

Simplify: ⓐ (6−5)2 ⓑ (9−210)2.

Solution

ⓐ 41−125
ⓑ 121−3610

In the next example, we will use the Product of Conjugates Pattern. Notice that the final product has no radical.

Simplify: (5−23)(5+23).

Solution
This image displays the difference of squares formula, (a-b)(a+b), and an application of it with radical numbers: (5 - 2√3)(5 + 2√3).
Multiply, using the Product of Conjugates Pattern. The image displays a mathematical problem with two lines of expressions. The first line, in red text, shows the difference of squares formula, a^2 - b^2. Below it, in black text, is the specific calculation 5^2 - (2√3)^2.
Simplify. A mathematical expression '25 - 4 ⋅ 3' is shown in black text on a white background.
The number '13' is visible on the left side of the image.

Simplify: (3−25)(3+25)

Solution

−11

Simplify: (4+57)(4−57).

Solution

−159

Access these online resources for additional instruction and practice with adding, subtracting, and multiplying radical expressions.

  • Multiplying Adding Subtracting Radicals
  • Multiplying Special Products: Square Binomials Containing Square Roots
  • Multiplying Conjugates

Key Concepts

  • Product Property of Roots
    • For any real numbers, an and bn, and for any integer n≥2
      abn=an·bn and an·bn=abn
  • Special Products
    Binomial SquaresProduct of Conjugates(a+b)2=a2+2ab+b2(a+b)(a−b)=a2−b2(a−b)2=a2−2ab+b2

Practice Makes Perfect

Add and Subtract Radical Expressions

In the following exercises, simplify.

ⓐ 82−52 ⓑ 5m3+2m3 ⓒ 8m4−2m4

Solution

ⓐ 32 ⓑ 7m3 ⓒ 6m4

ⓐ 72−32 ⓑ 7p3+2p3 ⓒ 5x3−3x3

ⓐ 35+65 ⓑ 9a3+3a3 ⓒ 52z4+2z4

Solution

ⓐ 95 ⓑ 12a3 ⓒ 62z4

ⓐ 45+85 ⓑ m3−4m3 ⓒ n+3n

ⓐ 32a−42a+52a ⓑ 53ab4−33ab4−23ab4

Solution

ⓐ 42a ⓑ 0

ⓐ 11b−511b+311b ⓑ 811cd4+511cd4−911cd4

ⓐ 83c+23c−93c ⓑ 24pq3−54pq3+44pq3

Solution

ⓐ 3c ⓑ 4pq3

ⓐ 35d+85d−115d ⓑ 112rs3−92rs3+32rs3

ⓐ 27−75 ⓑ 403−3203 ⓒ 12324+231624

Solution

ⓐ −23 ⓑ −253 ⓒ 324

ⓐ 72−98 ⓑ 243+813 ⓒ 12804−234054

ⓐ 48+27 ⓑ 543+1283 ⓒ 654−32804

Solution

ⓐ 73 ⓑ 723 ⓒ 354

ⓐ 45+80 ⓑ 813−1923 ⓒ 52804+734054

ⓐ 72a5−50a5 ⓑ 980p44−6405p44

Solution

ⓐ a22a ⓑ 0

ⓐ 48b5−75b5 ⓑ 864q63−3125q63

ⓐ 80c7−20c7 ⓑ 2162r104+432r104

Solution

ⓐ 2c35c ⓑ 14r22r24

ⓐ 96d9−24d9 ⓑ 5243s64+23s64

3128y2+4y162−898y2

Solution

4y2

375y2+8y48−300y2

Multiply Radical Expressions

In the following exercises, simplify.

ⓐ (−23)(318) ⓑ (843)(−4183)

Solution

ⓐ −186 ⓑ −6493

ⓐ (−45)(510) ⓑ (−293)(793)

ⓐ (56)(−12) ⓑ (−2184)(−94)

Solution

ⓐ −302 ⓑ 624

ⓐ (−27)(−214) ⓑ (−384)(−564)

ⓐ (412z3)(39z) ⓑ (53x33)(318x33)

Solution

ⓐ 72z23 ⓑ 45x223

ⓐ (32x3)(718x2) ⓑ (−620a23)(−216a33)

ⓐ (−27z3)(314z8) ⓑ (28y24)(−212y34)

Solution

ⓐ −42z52z ⓑ −8y6y4

ⓐ (42k5)(−332k6) ⓑ (−6b34)(38b34)

Use Polynomial Multiplication to Multiply Radical Expressions

In the following exercises, multiply.

ⓐ 7(5+27) ⓑ 63(4+183)

Solution

ⓐ 14+57 ⓑ 463+343

ⓐ 11(8+411) ⓑ 33(93+183)

ⓐ 11(−3+411) ⓑ 34(544+184)

Solution

ⓐ 44−311 ⓑ 324+544

ⓐ 2(−5+92) ⓑ 24(124+244)

(7+3)(9−3)

Solution

60+23

(8−2)(3+2)

ⓐ (9−32)(6+42) ⓑ (x3−3)(x3+1)

Solution

ⓐ 30+182 ⓑ x23−2x3−3

ⓐ (3−27)(5−47) ⓑ (x3−5)(x3−3)

ⓐ (1+310)(5−210) ⓑ (2x3+6)(x3+1)

Solution

ⓐ −55+1310
ⓑ 2x23+8x3+6

ⓐ (7−25)(4+95) ⓑ (3x3+2)(x3−2)

(3+10)(3+210)

Solution

23+330

(11+5)(11+65)

(27−511)(47+911)

Solution

−439−277

(46+713)(86−313)

ⓐ (3+5)2 ⓑ (2−53)2

Solution

ⓐ 14+65 ⓑ 79−203

ⓐ (4+11)2 ⓑ (3−25)2

ⓐ (9−6)2 ⓑ (10+37)2

Solution

ⓐ 87−186
ⓑ 163+607

ⓐ (5−10)2 ⓑ (8+32)2

(4+2)(4−2)

Solution

14

(7+10)(7−10)

(4+93)(4−93)

Solution

−227

(1+82)(1−82)

(12−55)(12+55)

Solution

19

(9−43)(9+43)

(3x3+2)(3x3−2)

Solution

9x23−4

(4x3+3)(4x3−3)

Mixed Practice

2327+3448

Solution

53

175k4−63k4

56162+316128

Solution

92

243+/813

12804−234054

Solution

−54

8134−4134−3134

512c4−327c6

Solution

10c23−9c33

80a5−45a5

3575−1448

Solution

23

2193−293

864q63−3125q63

Solution

17q2

1111−1011

3·21

Solution

37

(46)(−18)

(743)(−3183)

Solution

−4293

(412x5)(26x3)

(29)2

Solution

29

(−417)(−317)

(−4+17)(−3+17)

Solution

29−717

(38a24)(12a34)

(6−32)2

Solution

54−362

3(4−33)

33(293+183)

Solution

6+323

(6+3)(6+63)

Writing Exercises

Explain when a radical expression is in simplest form.

Solution

Answers will vary.

Explain the process for determining whether two radicals are like or unlike. Make sure your answer makes sense for radicals containing both numbers and variables.


ⓐ Explain why (−n)2 is always non-negative, for n≥0.
ⓑ Explain why −(n)2 is always non-positive, for n≥0.

Solution

Answers will vary.

Use the binomial square pattern to simplify (3+2)2. Explain all your steps.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 3 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “add and subtract radical expressions.”, “ multiply radical expressions”, and “use polynomial multiplication to multiply radical expressions”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

like radicals
Like radicals are radical expressions with the same index and the same radicand.

Divide Radical Expressions

Learning Objectives

By the end of this section, you will be able to:

  • Divide radical expressions
  • Rationalize a one term denominator
  • Rationalize a two term denominator

Before you get started, take this readiness quiz.

Simplify: 3048.
If you missed this problem, review Example 1 in Fractions.

Solution

58

Simplify: x2·x4.
If you missed this problem, review Example 1 in Properties of Exponents and Scientific Notation.

Solution

x6

Multiply: (7+3x)(7−3x).
If you missed this problem, review Example 8 in Multiply Polynomials.

Solution

49−9x2

Divide Radical Expressions

We have used the Quotient Property of Radical Expressions to simplify roots of fractions. We will need to use this property ‘in reverse’ to simplify a fraction with radicals.

We give the Quotient Property of Radical Expressions again for easy reference. Remember, we assume all variables are greater than or equal to zero so that no absolute value bars are needed.

Quotient Property of Radical Expressions

If an and bn are real numbers, b≠0, and for any integer n≥2 then,

abn=anbnandanbn=abn

We will use the Quotient Property of Radical Expressions when the fraction we start with is the quotient of two radicals, and neither radicand is a perfect power of the index. When we write the fraction in a single radical, we may find common factors in the numerator and denominator.

Simplify: ⓐ 72x3162x ⓑ 32x234x53.

Solution
ⓐ
Step-by-step simplification of a radical expression using algebraic properties.
72x3162x
Rewrite using the quotient property,
anbn=abn.
72x3162x
Remove common factors. 18·4·x2·x18·9·x
Simplify. 4x29
Simplify the radical. 2x3
ⓑ
Step-by-step simplification of a radical expression using the quotient property.
32x234x53
Rewrite using the quotient property,
anbn=abn.
32x24x53
Simplify the fraction under the radical. 8x33
Simplify the radical. 2x

Simplify: ⓐ 50s3128s ⓑ 56a37a43.

Solution

ⓐ 5s8 ⓑ 2a

Simplify: ⓐ 75q5108q ⓑ 72b239b53.

Solution

ⓐ 5q26 ⓑ 2b

Simplify: ⓐ 147ab83a3b4 ⓑ −250mn−232m−2n43.

Solution
ⓐ
Step-by-step simplification of a radical expression using quotient and algebraic properties.
147ab83a3b4
Rewrite using the quotient property. 147ab83a3b4
Remove common factors in the fraction. 49b4a2
Simplify the radical. 7b2a
ⓑ
Step-by-step simplification of a mathematical expression involving cube roots and fractions.
−250mn−232m−2n43
Rewrite using the quotient property. −250mn−22m−2n43
Simplify the fraction under the radical. −125m3n63
Simplify the radical. −5mn2

Simplify: ⓐ 162x10y22x6y6 ⓑ −128x2y−132x−1y23.

Solution

ⓐ 9x2y2 ⓑ −4xy

Simplify: ⓐ 300m3n73m5n ⓑ −81pq−133p−2q53.

Solution

ⓐ 10n3m ⓑ −3pq2

Simplify: 54x5y33x2y.

Solution
Steps to simplify a radical expression using quotient and product properties, showing the transformation of the mathematical expression at each stage.
54x5y33x2y
Rewrite using the quotient property. 54x5y33x2y
Remove common factors in the fraction. 18x3y2
Rewrite the radicand as a product
using the largest perfect square factor.
9x2y2⋅2x
Rewrite the radical as the product of two
radicals.
9x2y2⋅2x
Simplify. 3xy2x

Simplify: 64x4y52xy3.

Solution

4xy2x

Simplify: 96a5b42a3b.

Solution

4ab3b

Rationalize a One Term Denominator

Before the calculator became a tool of everyday life, approximating the value of a fraction with a radical in the denominator was a very cumbersome process!

For this reason, a process called rationalizing the denominator was developed. A fraction with a radical in the denominator is converted to an equivalent fraction whose denominator is an integer. Square roots of numbers that are not perfect squares are irrational numbers. When we rationalize the denominator, we write an equivalent fraction with a rational number in the denominator.

This process is still used today, and is useful in other areas of mathematics, too.

Rationalizing the Denominator

Rationalizing the denominator is the process of converting a fraction with a radical in the denominator to an equivalent fraction whose denominator is an integer.

Even though we have calculators available nearly everywhere, a fraction with a radical in the denominator still must be rationalized. It is not considered simplified if the denominator contains a radical.

Similarly, a radical expression is not considered simplified if the radicand contains a fraction.

Simplified Radical Expressions

A radical expression is considered simplified if there are

  • no factors in the radicand have perfect powers of the index
  • no fractions in the radicand
  • no radicals in the denominator of a fraction

To rationalize a denominator with a square root, we use the property that (a)2=a. If we square an irrational square root, we get a rational number.

We will use this property to rationalize the denominator in the next example.

Simplify: ⓐ 43 ⓑ 320 ⓒ 36x.

Solution

To rationalize a denominator with one term, we can multiply a square root by itself. To keep the fraction equivalent, we multiply both the numerator and denominator by the same factor.

ⓐ
A mathematical expression representing the fraction four divided by the square root of three.
Multiply both the numerator and denominator by 3. A mathematical expression illustrating the process of rationalizing a denominator. The fraction shows (4 *  3) / ( 3 *  3), with the multiplying  3 highlighted in red for clarity.
Simplify. The mathematical expression showing four times the square root of three, divided by three.

ⓑ We always simplify the radical in the denominator first, before we rationalize it. This way the numbers stay smaller and easier to work with.

The image displays a mathematical expression: the square root of the fraction 3/20.
The fraction is not a perfect square, so rewrite using the
Quotient Property.
A mathematical expression displaying a fraction where the numerator is the square root of 3 and the denominator is the square root of 20.
Simplify the denominator. A mathematical expression displaying the fraction: square root of 3 divided by 2 times the square root of 5.
Multiply the numerator and denominator by 5. A mathematical expression showing the multiplication of square roots in both the numerator and denominator, likely a step in rationalizing or simplifying an expression involving radicals.
Simplify. A mathematical expression showing the square root of 15 divided by the product of 2 and 5.
Simplify. The mathematical expression shows the square root of 15, divided by 10. It is written as a fraction with 'sqrt(15)' in the numerator and '10' in the denominator.
ⓒ
A mathematical expression showing the fraction 3 over the square root of 6x.
Multiply the numerator and denominator by 6x.    A step in simplifying an algebraic fraction, showing 3 times the square root of 6x divided by the product of the square root of 6x with itself.
Simplify. A mathematical expression showing the fraction 3 times the square root of 6x divided by 6x.
Simplify. A mathematical expression showing the square root of 6x divided by 2x.

Simplify: ⓐ 53 ⓑ 332 ⓒ 22x.

Solution

ⓐ 533 ⓑ 68 ⓒ 2xx

Simplify: ⓐ 65 ⓑ 718 ⓒ 55x.

Solution

ⓐ 655 ⓑ 146 ⓒ 5xx

When we rationalized a square root, we multiplied the numerator and denominator by a square root that would give us a perfect square under the radical in the denominator. When we took the square root, the denominator no longer had a radical.

We will follow a similar process to rationalize higher roots. To rationalize a denominator with a higher index radical, we multiply the numerator and denominator by a radical that would give us a radicand that is a perfect power of the index. When we simplify the new radical, the denominator will no longer have a radical.

For example,

Two examples of rationalizing denominators are shown. The first example is 1 divided by cube root 2. A note is made that the radicand in the denominator is 1 power of 2 and that we need 2 more to get a perfect cube. We multiply numerator and denominator by the cube root of the quantity 2 squared. The result is cube root 4 divided by cube root of quantity 2 cubed. This simplifies to cube root 4 divided by 2. The second example is 1 divided by fourth root 5. A note is made that the radicand in the denominator is 1 power of 5 and that we need 3 more to get a perfect fourth. We multiply numerator and denominator by the fourth root of the quantity 5 cubed. The result is fourth root of 125 divided by fourth root of quantity 5 to the fourth. This simplifies to fourth root 125 divided by 5.

We will use this technique in the next examples.

Simplify ⓐ 163 ⓑ 7243 ⓒ 34x3.

Solution

To rationalize a denominator with a cube root, we can multiply by a cube root that will give us a perfect cube in the radicand in the denominator. To keep the fraction equivalent, we multiply both the numerator and denominator by the same factor.

ⓐ
A mathematical expression showing the fraction 1 over the cube root of 6.
The radical in the denominator has one factor of 6.
Multiply both the numerator and denominator by 623,
which gives us 2 more factors of 6.
A mathematical expression displaying a fraction where the numerator is 1 multiplied by the cube root of 6 squared, and the denominator is the cube root of 6 multiplied by the cube root of 6 squared.
Multiply. Notice the radicand in the denominator
has 3 powers of 6.
A mathematical fraction displaying the cube root of 6 squared over the cube root of 6 cubed.
Simplify the cube root in the denominator. A mathematical expression displaying the cube root of 36 in the numerator, divided by 6 in the denominator.

ⓑ We always simplify the radical in the denominator first, before we rationalize it. This way the numbers stay smaller and easier to work with.

A mathematical expression showing the cube root of the fraction seven over twenty-four.
The fraction is not a perfect cube, so
rewrite using the Quotient Property.
A fraction with the cube root of 7 in the numerator and the cube root of 24 in the denominator, representing a mathematical expression involving radicals.
Simplify the denominator. A mathematical expression displaying the cube root of 7 divided by 2 times the cube root of 3.
Multiply the numerator and denominator       
by 323. This will give us 3 factors of 3.
A fraction with cube root expressions. The numerator is the product of cube root of 7 and cube root of 3 squared. The denominator is 2 times the product of cube root of 3 and cube root of 3 squared.
Simplify. A mathematical fraction with the cube root of 63 over 2 times the cube root of 3 cubed.
Remember, 333=3. The cube root of 63 divided by 2 times 3.
Simplify. The cube root of sixty-three divided by six.
ⓒ
A mathematical expression showing the fraction 3 over the cube root of 4x. The numerator is 3, and the denominator is the cube root symbol with 3 as the index, encompassing the term 4x.
Rewrite the radicand to show the factors. A mathematical expression showing the fraction 3 divided by the cube root of (2 squared multiplied by x).
Multiply the numerator and denominator by 2·x23.
This will get us 3 factors of 2 and 3 factors of x.
A mathematical expression showing a fraction. The numerator is 3 multiplied by the cube root of 2, then multiplied by x squared. The denominator is the cube root of (2 squared times x), multiplied by the cube root of 2, then multiplied by x squared.
Simplify. A mathematical fraction with 3 times the cube root of 2x squared in the numerator, and the cube root of 2 cubed x cubed in the denominator.
Simplify the radical in the denominator. A mathematical expression showing the fraction 3 times the cube root of 2x squared, all over 2x.

Simplify: ⓐ 173 ⓑ 5123 ⓒ 59y3.

Solution

ⓐ 4937 ⓑ 9036 ⓒ 53y233y

Simplify: ⓐ 123 ⓑ 3203 ⓒ 225n3.

Solution

ⓐ 432 ⓑ 150310 ⓒ 25n235n

Simplify: ⓐ 124 ⓑ 5644 ⓒ 28x4.

Solution

To rationalize a denominator with a fourth root, we can multiply by a fourth root that will give us a perfect fourth power in the radicand in the denominator. To keep the fraction equivalent, we multiply both the numerator and denominator by the same factor.

ⓐ
A mathematical expression showing the fraction 1 over the fourth root of 2, often represented as 1 /  th root(2). This is equivalent to 2 raised to the power of -1/4.
The radical in the denominator has one factor of 2.
Multiply both the numerator and denominator by 234,   
which gives us 3 more factors of 2.
Mathematical expression illustrating a step in rationalizing a denominator. Both the numerator and denominator are multiplied by the fourth root of 2 cubed (⁴√2³).
Multiply. Notice the radicand in the denominator
has 4 powers of 2.
A mathematical expression showing the division of the fourth root of 8 by the fourth root of 2 raised to the power of 4.
Simplify the fourth root in the denominator. A mathematical expression featuring the fourth root of 8, all divided by 2.

ⓑ We always simplify the radical in the denominator first, before we rationalize it. This way the numbers stay smaller and easier to work with.

A mathematical expression showing the fourth root of the fraction 5 divided by 64.
The fraction is not a perfect fourth power, so rewrite
using the Quotient Property.
A fraction with the fourth root of 5 in the numerator and the fourth root of 64 in the denominator.
Rewrite the radicand in the denominator to show the factors. A mathematical expression showing the fourth root of 5 divided by the fourth root of 2 to the power of 6.
Simplify the denominator. A mathematical expression showing a fraction. The numerator is the fourth root of 5. The denominator is 2 multiplied by the fourth root of 2 squared.
Multiply the numerator and denominator by 224.
This will give us 4 factors of 2.
A fraction showing an algebraic expression. The numerator is (4th root of 5) * (4th root of 2^2). The denominator is 2 * (4th root of 2^2) * (4th root of 2^2). Some elements are in red.
Simplify. A mathematical fraction is shown. The numerator is the fourth root of 5 multiplied by the fourth root of 4. The denominator is 2 multiplied by the fourth root of 2 to the power of 4.
Remember, 244=2. A mathematical expression showing a fraction. The numerator is the fourth root of 20. The denominator is 2 multiplied by 2.
Simplify. The image displays the mathematical expression: the fourth root of 20, all divided by 4.
ⓒ
A mathematical expression showing the fraction 2 divided by the fourth root of 8x. The number 2 is in the numerator, and the denominator is the fourth root symbol with 8x inside it.
Rewrite the radicand to show the factors. A mathematical fraction with 2 in the numerator and the fourth root of 2 cubed multiplied by x in the denominator.
Multiply the numerator and denominator by 2·x34.   
This will get us 4 factors of 2 and 4 factors of x.
A mathematical fraction with a numerator of 2 times the fourth root of 2 times x cubed, and a denominator of the fourth root of 2 cubed x times the fourth root of 2 times x cubed.
Simplify. A mathematical expression displaying a fraction. The numerator is 2 multiplied by the fourth root of (2x^3). The denominator is the fourth root of (2^4x^4).
Simplify the radical in the denominator. A mathematical expression displaying a fraction. The numerator is 2 multiplied by the fourth root of 2x cubed. The denominator is 2x.
Simplify the fraction. The fourth root of two x cubed, divided by x.

Simplify: ⓐ 134 ⓑ 3644 ⓒ 3125x4.

Solution

ⓐ 2743 ⓑ 1244 ⓒ 35x345x

Simplify: ⓐ 154 ⓑ 71284 ⓒ 44x4

Solution

ⓐ 12545 ⓑ 1444
ⓒ 24x34x

Rationalize a Two Term Denominator

When the denominator of a fraction is a sum or difference with square roots, we use the Product of Conjugates Pattern to rationalize the denominator.

(a−b)(a+b)(2−5)(2+5)a2−b222−(5)24−5−1

When we multiply a binomial that includes a square root by its conjugate, the product has no square roots.

Simplify: 52−3.

Solution
A mathematical expression showing the fraction 5 over 2 minus the square root of 3.
Multiply the numerator and denominator by the
conjugate of the denominator.
A fraction with 5(2 + sqrt(3)) in the numerator and (2 - sqrt(3))(2 + sqrt(3)) in the denominator, illustrating the process of rationalizing a denominator.
Multiply the conjugates in the denominator. A mathematical fraction with 5(2+sqrt(3)) in the numerator and 2^2 - (sqrt(3))^2 in the denominator.
Simplify the denominator. A mathematical expression showing the fraction 5(2 + square root of 3) divided by 4 - 3.
Simplify the denominator. A mathematical expression showing 5 multiplied by the sum of 2 and the square root of 3, all divided by 1.
Simplify. The image shows the mathematical expression 5(2 + √3).

Simplify: 31−5.

Solution

−3(1+5)4

Simplify: 24−6.

Solution

4+65

Notice we did not distribute the 5 in the answer of the last example. By leaving the result factored we can see if there are any factors that may be common to both the numerator and denominator.

Simplify: 3u−6.

Solution
A mathematical expression showing the fraction with square root of 3 as the numerator and the difference of square root of u and square root of 6 as the denominator.
Multiply the numerator and denominator by the
conjugate of the denominator.
A mathematical fraction with a numerator of sqrt(3)(sqrt(u)+sqrt(6)) and a denominator of (sqrt(u)-sqrt(6))(sqrt(u)+sqrt(6)), highlighting the (sqrt(u)+sqrt(6)) term in red.
Multiply the conjugates in the denominator. A fraction with sqrt(3)(sqrt(u)+sqrt(6)) in the numerator and (sqrt(u))^2-(sqrt(6))^2 in the denominator, illustrating a rationalization process.
Simplify the denominator. A mathematical expression showing the fraction sqrt(3)(sqrt(u) + sqrt(6)) over (u - 6).

Simplify: 5x+2.

Solution

5(x−2)x−2

Simplify: 10y−3.

Solution

10(y+3)y−3

Be careful of the signs when multiplying. The numerator and denominator look very similar when you multiply by the conjugate.

Simplify: x+7x−7.

Solution
A mathematical expression showing the fraction (square root of x + square root of 7) divided by (square root of x - square root of 7).
Multiply the numerator and denominator by the
conjugate of the denominator.
A step in rationalizing a denominator, showing multiplication by a conjugate radical expression to eliminate the square root from the denominator.
Multiply the conjugates in the denominator. A mathematical expression featuring a fraction. The numerator is (sqrt(x) + sqrt(7)) multiplied by itself, and the denominator is the difference of squares: (sqrt(x))^2 - (sqrt(7))^2.
Simplify the denominator. A mathematical expression showing a fraction. The numerator is (square root of x + square root of 7) squared, and the denominator is x - 7.

We do not square the numerator. Leaving it in factored form, we can see there are no common factors to remove from the numerator and denominator.

Simplify: p+2p−2.

Solution

(p+2)p−22

Simplify: q−10q+10

Solution

(q−10)q−102

Access these online resources for additional instruction and practice with dividing radical expressions.

  • Rationalize the Denominator
  • Dividing Radical Expressions and Rationalizing the Denominator
  • Simplifying a Radical Expression with a Conjugate
  • Rationalize the Denominator of a Radical Expression

Key Concepts

  • Quotient Property of Radical Expressions
    • If an and bn are real numbers, b≠0, and for any integer n≥2 then,
      abn=anbn and anbn=abn
  • Simplified Radical Expressions
    • A radical expression is considered simplified if there are:
      • no factors in the radicand that have perfect powers of the index
      • no fractions in the radicand
      • no radicals in the denominator of a fraction

Practice Makes Perfect

Divide Square Roots

In the following exercises, simplify.

ⓐ 12872 ⓑ 1283543

Solution

ⓐ 43 ⓑ 43

ⓐ 4875 ⓑ 813243

ⓐ 200m598m ⓑ 54y232y53

Solution

ⓐ 10m27 ⓑ 3y

ⓐ 108n7243n3 ⓑ 54y316y43

ⓐ 75r3108r7 ⓑ 24x7381x43

Solution

ⓐ 56r2 ⓑ 2x3

ⓐ 196q484q5 ⓑ 16m4354m3

ⓐ 108p5q23p3q6 ⓑ −16a4b−232a−2b3

Solution

ⓐ 6pq2 ⓑ −2a2b

ⓐ 98rs102r3s4 ⓑ −375y4z−233y−2z43

ⓐ 320mn−545m−7n3 ⓑ 16x4y−23−54x−2y43

Solution

ⓐ 8m43n4 ⓑ −2x23y2

ⓐ 810c−3d71000cd−1 ⓑ 24a7b−13−81a−2b23

56x5y42xy3

Solution

2x27y

72a3b63ab3

48a3b633a−1b33

Solution

2ab2a3

162x−3y632x3y−23

Rationalize a One Term Denominator

In the following exercises, rationalize the denominator.

ⓐ 106 ⓑ 427 ⓒ 105x

Solution

ⓐ 563 ⓑ 239 ⓒ 25xx

ⓐ 83 ⓑ 740 ⓒ 82y

ⓐ 67 ⓑ 845 ⓒ 123p

Solution

ⓐ 677 ⓑ 21015 ⓒ 43pp

ⓐ 45 ⓑ 2780 ⓒ 186q

ⓐ 153 ⓑ 5243 ⓒ 436a3

Solution

ⓐ 2535 ⓑ 4536 ⓒ 26a233a

ⓐ 133 ⓑ 5323 ⓒ 749b3

ⓐ 1113 ⓑ 7543 ⓒ 33x23

Solution

ⓐ 121311 ⓑ 2836 ⓒ 9x3x

ⓐ 1133 ⓑ 31283 ⓒ 36y23

ⓐ 174 ⓑ 5324 ⓒ 44x24

Solution

ⓐ 34347 ⓑ 4044 ⓒ 24x24x

ⓐ 144 ⓑ 9324 ⓒ 69x34

ⓐ 194 ⓑ 251284 ⓒ 627a4

Solution

ⓐ 943 ⓑ 5044 ⓒ 23a34a

ⓐ 184 ⓑ 271284 ⓒ 1664b24

Rationalize a Two Term Denominator

In the following exercises, simplify.

81−5

Solution

−2(1+5)

72−6

63−7

Solution

3(3+7)

54−11

3m−5

Solution

3(m+5)m−5

5n−7

2x−6

Solution

2(x+6)x−6

7y+3

r+5r−5

Solution

(r+5)r−52

s−6s+6

x+8x−8

Solution

(x+22)x−82

m−3m+3

Writing Exercises


ⓐ Simplify 273 and explain all your steps.
ⓑ Simplify 275 and explain all your steps.
ⓒ Why are the two methods of simplifying square roots different?

Solution

Answers will vary.

Explain what is meant by the word rationalize in the phrase, “rationalize a denominator.”

Explain why multiplying 2x−3 by its conjugate results in an expression with no radicals.

Solution

Answers will vary.

Explain why multiplying 7x3 by x3x3 does not rationalize the denominator.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “divide radical expressions.”, “rationalize a one term denominator”, and “rationalize a two term denominator”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

rationalizing the denominator
Rationalizing the denominator is the process of converting a fraction with a radical in the denominator to an equivalent fraction whose denominator is an integer.

Solve Radical Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve radical equations
  • Solve radical equations with two radicals
  • Use radicals in applications

Before you get started, take this readiness quiz.

Simplify: (y−3)2.
If you missed this problem, review Example 7 in Multiply Polynomials.

Solution

y2−6y+9

Solve: 2x−5=0.
If you missed this problem, review Example 2 in Use a General Strategy to Solve Linear Equations.

Solution

x=52

Solve n2−6n+8=0.
If you missed this problem, review Example 2 in Polynomial Equations.

Solution

n=2orn=4

Solve Radical Equations

In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation.

Radical Equation

An equation in which a variable is in the radicand of a radical expression is called a radical equation.

As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.

In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the nth power. This will eliminate the radical.

Fora≥0,(an)n=a.

How to Solve a Radical Equation

Solve: 5n−4−9=0.

Solution

Step 1 is to isolate the radical on one side of the equation. To isolate the radical add 9 to both sides. The resulting equation is square root of the quantity 5 n minus 4 in parentheses minus 9 plus 9 equals 0 plus 9. This simplifies to square root of the quantity 5 n minus 4 in parentheses equals 9. Step 2 is to raise both sides of the equation to the power of the index. Since the index of a square root is 2, we square both sides. Remember that the square of the square root of “a” is equal to “a”. The equation that results is the square of the square root of the quantity 5 n minus 4 in parentheses equals 9 squared. This simplifies to 5 n minus 4 equals 81. Step 3 is to solve the new equation. We get 5 n equals 85 and then n equals 17. Step 4 is to check the answer in the original equation. Does the square root of the quantity 5 times 17 minus 4 in parentheses minus 9 equal zero? Simplifying the left side we get square root of the quantity 85 minus 4 in parentheses minus 9 and then square root of 81 minus 9 and then 9 minus 9 which does equal 0. This verifies that the solution is n equals 17.

Solve: 3m+2−5=0.

Solution

m=233

Solve: 10z+1−2=0.

Solution

z=310

Solve a radical equation with one radical.

  1. Isolate the radical on one side of the equation.
  2. Raise both sides of the equation to the power of the index.
  3. Solve the new equation.
  4. Check the answer in the original equation.

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

Solve: 9k−2+1=0.

Solution
A mathematical equation is displayed, showing the square root of (9k - 2), with a plus 1 outside the square root, set equal to 0. The full equation is: sqrt(9k - 2) + 1 = 0.
To isolate the radical, subtract 1 to both sides. A mathematical equation: the square root of 9k minus 2, plus 1, minus 1, equals 0 minus 1. The '-1' terms are highlighted in red on both sides of the equation.
Simplify. A mathematical equation displays the square root of 9k minus 2, which equals negative 1. The equation is 'sqrt(9k - 2) = -1'.

Because the square root is equal to a negative number, the equation has no solution.

Solve: 2r−3+5=0.

Solution

no solution

Solve: 7s−3+2=0.

Solution

no solution

If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.

Binomial Squares

(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2

Don’t forget the middle term!

Solve: p−1+1=p.

Solution
A mathematical equation is displayed, showing 'sqrt(p - 1) + 1 = p'.
To isolate the radical, subtract 1 from both sides. An equation, sqrt(p-1) + 1 - 1 = p-1, featuring mathematical symbols and numbers, with the second '-1' and the 'p-1' on the right side highlighted in red.
Simplify. A mathematical equation is displayed, showing the square root of (p-1) equal to (p-1).
Square both sides of the equation. The equation shows (the square root of p-1) squared equals (p-1) squared.
Simplify, using the Product of Binomial Squares Pattern on the
right. Then solve the new equation.
A mathematical equation is displayed on a white background, which reads 'p - 1 = p^2 - 2p + 1'.
It is a quadratic equation, so get zero on one side. A quadratic equation showing zero equals the quantity p squared minus three p plus two.
Factor the right side. A mathematical equation showing 0 equals the product of (p - 1) and (p - 2).
Use the Zero Product Property. Two mathematical equations are displayed horizontally: '0 = p - 1' and '0 = p - 2'. The background is white.
Solve each equation. Two mathematical expressions are displayed on a white background: 'p = 1' on the left and 'p = 2' on the right, indicating two distinct parameter values.
Check the answers.
Step-by-step verification of the equation sqrt(p-1) + 1 = p, confirming its truth for p=1 and p=2 in these two examples.
The solutions are p=1,p=2.

Solve: x−2+2=x.

Solution

x=2,x=3

Solve: y−5+5=y.

Solution

y=5,y=6

When the index of the radical is 3, we cube both sides to remove the radical.

(a3)3=a

Solve: 5x+13+8=4.

Solution
5x+13+8=4
To isolate the radical, subtract 8 from both sides. 5x+13=−4
Cube both sides of the equation. (5x+13)3=(−4)3
Simplify. 5x+1=−64
Solve the equation. 5x=−65
x=−13
Check the answer.
Verifying that x = -13 is the correct solution for the equation cube_root(5x + 1) + 8 = 4, demonstrating step-by-step substitution and simplification to confirm the equality 4 = 4.
The solution is x=−13.

Solve: 4x−33+8=5

Solution

x=−6

Solve: 6x−103+1=−3

Solution

x=−9

Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical. We raise each side of the equation to the power of the denominator of the rational exponent. Since (am)n=am·n, we have for example,

(x12)2=x,(x13)3=x

Remember, x12=x and x13=x3.

Solve: (3x−2)14+3=5.

Solution
(3x−2)14+3=5
To isolate the term with the rational exponent,
subtract 3 from both sides.
(3x−2)14=2
Raise each side of the equation to the fourth power. ((3x−2)14)4=(2)4
Simplify. 3x−2=16
Solve the equation. 3x=18
x=6
Check the answer.
Verifying x=6 as the solution to (3x-2)^(1/4) + 3 = 5 by substituting x and simplifying the equation to confirm 5=5.
The solution is x=6.

Solve: (9x+9)14−2=1.

Solution

x=8

Solve: (4x−8)14+5=7.

Solution

x=6

Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution!

Solve: r+4−r+2=0.

Solution
r+4−r+2=0
Isolate the radical. r+4=r−2
Square both sides of the equation. (r+4)2=(r−2)2
Simplify and then solve the equation r+4=r2−4r+4
It is a quadratic equation, so get zero on
one side.
0=r2−5r
Factor the right side. 0=r(r−5)
Use the Zero Product Property. 0=r0=r−5
Solve the equation. r=0r=5
Check your answer.
Verification of solutions for the radical equation sqrt(r+4) - r + 2 = 0. It shows that r=0 is an extraneous solution, while r=5 is a valid solution. The solution is r = 5.
r=0 is an extraneous solution.

Solve: m+9−m+3=0.

Solution

m=7

Solve: n+1−n+1=0.

Solution

n=3

When there is a coefficient in front of the radical, we must raise it to the power of the index, too.

Solve: 33x−5−8=4.

Solution
33x−5−8=4
Isolate the radical term. 33x−5=12
Isolate the radical by dividing both sides by 3. 3x−5=4
Square both sides of the equation. (3x−5)2=(4)2
Simplify, then solve the new equation. 3x−5=16
3x=21
Solve the equation. x=7
Check the answer.
This image demonstrates the step-by-step verification that x=7 is a valid solution to the radical equation 3√(3x-5) - 8 = 4, successfully leading to 4=4.
The solution is x=7.

Solve: 24a+4−16=16.

Solution

a=63

Solve: 32b+3−25=50.

Solution

b=311

Solve Radical Equations with Two Radicals

If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.

In the next example, when one radical is isolated, the second radical is also isolated.

Solve: 4x−33=3x+23.

Solution
Step-by-step solution for an equation involving cube roots.
The radical terms are isolated. 4x−33=3x+23
Since the index is 3, cube both sides of the
equation.
(4x−33)3=(3x+23)3
Simplify, then solve the new equation. 4x−3=3x+2
x−3=2
x=5
The solution isx=5.
Check the answer.
We leave it to you to show that 5 checks!

Solve: 5x−43=2x+53.

Solution

x=3

Solve: 7x+13=2x−53.

Solution

x=−65

Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and raise both sides of the equation to the power of the index again.

How to Solve a Radical Equation

Solve: m+1=m+9.

Solution

Step 1 is to isolate one of the radical terms on one side of the equation. The radical on the right is isolated. Step 2 is to raise both sides of the equation to the power of the index. We square both sides. The equation that results is the square of the quantity square root of m plus 1 in parentheses equals the square of the square root of the quantity m plus 9 in parentheses. Simplify – be very careful as you multiply! This simplifies to m plus 2 times square root m plus 1 equals m plus 9. Step 3 is to repeat steps 1 and 2 again if there are any more radicals. There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical term. 2 times square root m equals 8. Here, we can easily isolate the radical by dividing both sides by 2. We get square root m equals 4. Squaring both sides we get the square of the square root of m equals 4 squared. m equals 16. Step 4 is to check the answer in the original equation. Does the square root of 16 plus 1 equal the square root of the quantity 16 plus 9? Simplifying both sides we get 4 plus 1 equals 5. This verifies that the solution is m equals 16.

Solve: 3−x=x−3.

Solution

x=4

Solve: x+2=x+16.

Solution

x=9

We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.

Solve a radical equation.

  1. Isolate one of the radical terms on one side of the equation.
  2. Raise both sides of the equation to the power of the index.
  3. Are there any more radicals?
    If yes, repeat Step 1 and Step 2 again.
    If no, solve the new equation.
  4. Check the answer in the original equation.

Be careful as you square binomials in the next example. Remember the pattern is (a+b)2=a2+2ab+b2 or (a−b)2=a2−2ab+b2.

Solve: q−2+3=4q+1.

Solution
A mathematical equation shows the square root of (q - 2) plus 3 equals the square root of (4q + 1).
The radical on the right is isolated. Square
both sides.
An algebraic equation showing the square of the expression (sqrt(q-2) + 3) on the left side, and the square of sqrt(4q+1) on the right side, with both expressions being equal.
Simplify. A mathematical equation is displayed: q - 2 + 6 * sqrt(q - 2 + 9) = 4q + 1. The equation involves a variable 'q', numbers, addition, subtraction, multiplication, and a square root.
There is still a radical in the equation so
we must repeat the previous steps. Isolate
the radical.
A mathematical equation is displayed on a white background, which reads '6 sqrt(q) - 2 = 3q - 6'.
Square both sides. It would not help to
divide both sides by 6. Remember to
square both the 6 and the q−2.
The image shows mathematical equations demonstrating algebraic expansions. The top line presents two squared expressions: (6√(q-2))^2 and ((a-b)/(3q-6))^2. The bottom line details the expansion of 6^2(√(q-2))^2 and the binomial expansion of (3q-6)^2, referencing the identity a^2 - 2ab + b^2.
Simplify, then solve the new equation. A mathematical equation is displayed, showing 36(q-2) = 9q^2 - 36q + 36 on a white background.
Distribute. A quadratic equation is displayed, showing 36q - 72 = 9q^2 - 36q + 36, likely for solving the variable 'q'.
It is a quadratic equation, so get zero on
one side.
A quadratic equation is displayed: 0 = 9q^2 - 72q + 108. The equation is presented in a clear, digital format against a white background.
Factor the right side. Algebraic factorization of a quadratic expression. The equation 0 = 9(q^2 - 8q + 12) is factored into 0 = 9(q - 6)(q - 2).
Use the Zero Product Property. Two separate sets of linear equations and their solutions are shown. On the left, q - 6 = 0 is solved to q = 6. On the right, q - 2 = 0 is solved to q = 2.
The checks are left to you. The solutions are q=6 and q=2.

Solve: x−1+2=2x+6

Solution

x=5

Solve: x+2=3x+4

Solution

x=0x=4

Use Radicals in Applications

As you progress through your college courses, you’ll encounter formulas that include radicals in many disciplines. We will modify our Problem Solving Strategy for Geometry Applications slightly to give us a plan for solving applications with formulas from any discipline.

Use a problem solving strategy for applications with formulas.

  1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
  2. Identify what we are looking for.
  3. Name what we are looking for by choosing a variable to represent it.
  4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

One application of radicals has to do with the effect of gravity on falling objects. The formula allows us to determine how long it will take a fallen object to hit the gound.

Falling Objects

On Earth, if an object is dropped from a height of h feet, the time in seconds it will take to reach the ground is found by using the formula

t=h4.

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting h=64 into the formula.

A mathematical formula displaying 't equals the square root of h, all divided by 4', which can also be written as t = vh / 4 or t = (h^0.5) / 4.
A mathematical equation: t equals the square root of 64, all divided by 4, with the number 64 highlighted in red.
Take the square root of 64. A mathematical equation shows 't = 8/4' in black text against a white background.
Simplify the fraction. The text 't=2' is displayed in dark gray characters against a plain white background.

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Marissa dropped her sunglasses from a bridge 400 feet above a river. Use the formula t=h4 to find how many seconds it took for the sunglasses to reach the river.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. the time it takes for the
sunglasses to reach the river
Step 3. Name what we are looking. Let t= time.
Step 4. Translate into an equation by writing the
appropriate formula. Substitute in the given
information.
Substitution of h=400 into the equation t = sqrt(h)/4, resulting in t = sqrt(400)/4, demonstrating the first step in solving for t.
Step 5. Solve the equation. A mathematical expression reads 't = 20/4', indicating a variable 't' is equal to the fraction 20 divided by 4.
The image displays the mathematical expression 't = 5' in black text against a plain white background, indicating a variable 't' assigned the value of five.
Step 6. Check the answer in the problem and make
sure it makes sense.
A three-step math problem demonstrating that 5 is equal to the square root of 400 divided by 4, concluding with '5 = 5' and a checkmark.
Does 5 seconds seem like a reasonable length of
time?
Yes.
Step 7. Answer the question. It will take 5 seconds for the
sunglasses to reach the river.

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula t=h4 to find how many seconds it took for the package to reach the ground.

Solution

9 seconds

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula t=h4 to find how many seconds it took for the squeegee to reach the sidewalk.

Solution

3.5 seconds

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.

Skid Marks and Speed of a Car

If the length of the skid marks is d feet, then the speed, s, of the car before the brakes were applied can be found by using the formula

s=24d

After a car accident, the skid marks for one car measured 190 feet. Use the formula s=24d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Solution
Step 1. Read the problem
Step 2. Identify what we are looking for. the speed of a car
Step 3. Name what weare looking for, Let s= the speed.
Step 4. Translate into an equation by writing
the appropriate formula. Substitute in the
given information.
A mathematical problem showing the substitution of d=190 into the equation s = sqrt(24d), resulting in s = sqrt(24(190)), with 190 highlighted in red.
Step 5. Solve the equation. The image displays the mathematical equation s = sqrt(4,560), calculating the square root of 4,560.
The variable 's' is equal to the repeating decimal 67.52777...
Round to 1 decimal place. A mathematical expression displays 's' is approximately equal to 67.5.
Checking if 67.5 equals the square root of 24 times 190. Steps show simplification: square root of 24 times 190 becomes square root of 4560, which is approximately 67.5277. A checkmark confirms the approximation.
The speed of the car before the brakes were applied
was 67.5 miles per hour.

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula s=24d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Solution

42.7 feet

The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula s=24d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Solution

54.1 feet

Access these online resources for additional instruction and practice with solving radical equations.

  • Solving an Equation Involving a Single Radical
  • Solving Equations with Radicals and Rational Exponents
  • Solving Radical Equations
  • Solve Radical Equations
  • Radical Equation Application

Key Concepts

  • Binomial Squares
    (a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2
  • Solve a Radical Equation
    1. Isolate one of the radical terms on one side of the equation.
    2. Raise both sides of the equation to the power of the index.
    3. Are there any more radicals?
      If yes, repeat Step 1 and Step 2 again.
      If no, solve the new equation.
    4. Check the answer in the original equation.
  • Problem Solving Strategy for Applications with Formulas
    1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
    2. Identify what we are looking for.
    3. Name what we are looking for by choosing a variable to represent it.
    4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.
  • Falling Objects
    • On Earth, if an object is dropped from a height of h feet, the time in seconds it will take to reach the ground is found by using the formula t=h4.
  • Skid Marks and Speed of a Car
    • If the length of the skid marks is d feet, then the speed, s, of the car before the brakes were applied can be found by using the formula s=24d.

Practice Makes Perfect

Solve Radical Equations

In the following exercises, solve.

5x−6=8

Solution

x=14

4x−3=7

5x+1=−3

Solution

no solution

3y−4=−2

2x3=−2

Solution

x=−4

4x−13=3

2m−3−5=0

Solution

m=14

2n−1−3=0

6v−2−10=0

Solution

v=17

12u+1−11=0

4m+2+2=6

Solution

m=72

6n+1+4=8

2u−3+2=0

Solution

no solution

5v−2+5=0

u−3+3=u

Solution

u=3,u=4

v−10+10=v

r−1=r−1

Solution

r=1,r=2

s−8=s−8

6x+43=4

Solution

x=10

11x+43=5

4x+53−2=−5

Solution

x=−8

9x−13−1=−5

(6x+1)12−3=4

Solution

x=8

(3x−2)12+1=6

(8x+5)13+2=−1

Solution

x=−4

(12x−5)13+8=3

(12x−3)14−5=−2

Solution

x=7

(5x−4)14+7=9

x+1−x+1=0

Solution

x=3

y+4−y+2=0

z+100−z=−10

Solution

z=21

w+25−w=−5

32x−3−20=7

Solution

x=42

25x+1−8=0

28r+1−8=2

Solution

r=3

37y+1−10=8

Solve Radical Equations with Two Radicals

In the following exercises, solve.

3u+7=5u+1

Solution

u=3

4v+1=3v+3

8+2r=3r+10

Solution

r=−2

10+2c=4c+16

5x−13=x+33

Solution

x=1

8x−53=3x+53

2x2+9x−183=x2+3x−23

Solution

x=−8,x=2

x2−x+183=2x2−3x−63

a+2=a+4

Solution

a=0

r+6=r+8

u+1=u+4

Solution

u=94

x+1=x+2

a+5−a=1

Solution

a=4

−2=d−20−d

2x+1=1+x

Solution

x=0x=4

3x+1=1+2x−1

2x−1−x−1=1

Solution

x=1x=5

x+1−x−2=1

x+7−x−5=2

Solution

x=9

x+5−x−3=2

Use Radicals in Applications

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula s=A to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Solution

8.7 feet

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula s=A to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula t=h4 to find how many seconds it took for the cell phone to reach the ground.

Solution

4.7 seconds

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula t=h4 to find how many seconds it took for the hammer to reach the river.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula s=24d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Solution

72 feet

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula s=24d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Writing Exercises

Explain why an equation of the form x+1=0 has no solution.

Solution

Answers will vary.

ⓐ Solve the equation r+4−r+2=0. ⓑ Explain why one of the “solutions” that was found was not actually a solution to the equation.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “Solve radical equations”, “solve radical equations with two radicals”, and “use radicals in applications”. The other columns are left blank so the learner can indicate their level of understanding.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

radical equation
An equation in which a variable is in the radicand of a radical expression is called a radical equation.

Use Radicals in Functions

Learning Objectives

By the end of this section, you will be able to:

  • Evaluate a radical function
  • Find the domain of a radical function
  • Graph radical functions

Before you get started, take this readiness quiz.

Solve: 1−2x≥0.
If you missed this problem, review Example 3 in Solve Linear Inequalities.

Solution

(−∞,12]

For f(x)=3x−4, evaluate f(2),f(−1),f(0).
If you missed this problem, review Example 7 in Relations and Functions.

Solution

f(2)=2,f(−1)=−7,f(0)=−4

Graph f(x)=x. State the domain and range of the function in interval notation.
If you missed this problem, review Example 6 in Graphs of Functions.

Solution

domain:[0,∞);range:[0,∞)

Evaluate a Radical Function

In this section we will extend our previous work with functions to include radicals. If a function is defined by a radical expression, we call it a radical function.

The square root function is f(x)=x.

The cube root function is f(x)=x3.

Radical Function

A radical function is a function that is defined by a radical expression.

To evaluate a radical function, we find the value of f(x) for a given value of x just as we did in our previous work with functions.

For the function f(x)=2x−1, find ⓐ f(5) ⓑ f(−2).

Solution
ⓐ
Step-by-step evaluation of the function f(x) = sqrt(2x - 1) for x=5.
f(x)=2x−1
To evaluate f(5), substitute 5 for x. f(5)=2·5−1
Simplify. f(5)=9
Take the square root. f(5)=3
ⓑ
Steps to evaluate f(x)=sqrt(2x-1) at x=-2, showing substitution and simplification to sqrt(-5).
f(x)=2x−1
To evaluate f(−2), substitute −2 for x. f(−2)=2(−2)−1
Simplify. f(−2)=−5

Since the square root of a negative number is not a real number, the function does not have a value at x=−2.

For the function f(x)=3x−2, find ⓐ f(6) ⓑ f(0).

Solution

ⓐ f(6)=4 ⓑ no value at x=0

For the function g(x)=5x+5, find ⓐ g(4) ⓑ g(−3).

Solution

ⓐ g(4)=5 ⓑ no value at f(−3)

We follow the same procedure to evaluate cube roots.

For the function g(x)=x−63, find ⓐ g(14) ⓑ g(−2).

Solution
ⓐ
This table shows the step-by-step evaluation of the function g(x) = ³√(x-6) for x=14, from substitution to the final result.
g(x)=x−63
To evaluate g(14), substitute 14 for x. g(14)=14−63
Simplify. g(14)=83
Take the cube root. g(14)=2
ⓑ
Step-by-step evaluation of the function g(x) = x−63 for x = -2.
g(x)=x−63
To evaluate g(−2), substitute −2 for x. g(−2)=−2−63
Simplify. g(−2)=−83
Take the cube root. g(−2)=−2

For the function g(x)=3x−43, find ⓐ g(4) ⓑ g(1).

Solution

ⓐ g(4)=2 ⓑ g(1)=−1

For the function h(x)=5x−23, find ⓐ h(2) ⓑ h(−5).

Solution

ⓐ h(2)=2
ⓑ h(−5)=−3

The next example has fourth roots.

For the function f(x)=5x−44, find ⓐ f(4) ⓑ f(−12)

Solution
ⓐ
Step-by-step evaluation of the function f(x) = (5x-4)^(1/4) at x=4.
f(x)=5x−44
To evaluate f(4), substitute 4 for x. f(4)=5·4−44
Simplify. f(4)=164
Take the fourth root. f(4)=2
ⓑ
Step-by-step evaluation of the function f(x) = ⁴√(5x-4) at x = -12, showing substitution and simplification.
f(x)=5x−44
To evaluate f(−12), substitute −12 for x. f(−12)=5(−12)−44
Simplify. f(−12)=−644

Since the fourth root of a negative number is not a real number, the function does not have a value at x=−12.

For the function f(x)=3x+44, find ⓐ f(4) ⓑ f(−1).

Solution

ⓐ f(4)=2 ⓑ f(−1)=1

For the function g(x)=5x+14, find ⓐ g(16) ⓑ g(3).

Solution

ⓐ g(16)=3 ⓑ g(3)=2

Find the Domain of a Radical Function

To find the domain and range of radical functions, we use our properties of radicals. For a radical with an even index, we said the radicand had to be greater than or equal to zero as even roots of negative numbers are not real numbers. For an odd index, the radicand can be any real number. We restate the properties here for reference.

Properties of an

When n is an even number and:

  • a≥0, then an is a real number.
  • a<0, then an is not a real number.

When n is an odd number, an is a real number for all values of a.

So, to find the domain of a radical function with even index, we set the radicand to be greater than or equal to zero. For an odd index radical, the radicand can be any real number.

Domain of a Radical Function

When the index of the radical is even, the radicand must be greater than or equal to zero.

When the index of the radical is odd, the radicand can be any real number.

Find the domain of the function, f(x)=3x−4. Write the domain in interval notation.

Solution

Since the function, f(x)=3x−4 has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0. We set the radicand to be greater than or equal to 0 and then solve to find the domain.

Steps to solve the inequality 3x - 4 >= 0, showing each mathematical transformation.
3x−4≥0
Solve. 3x≥4
x≥43

The domain of f(x)=3x−4 is all values x≥43 and we write it in interval notation as [43,∞).

Find the domain of the function, f(x)=6x−5. Write the domain in interval notation.

Solution

[56,∞)

Find the domain of the function, f(x)=4−5x. Write the domain in interval notation.

Solution

(−∞,45]

Find the domain of the function, g(x)=6x−1. Write the domain in interval notation.

Solution

Since the function, g(x)=6x−1 has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0.

The radicand cannot be zero since the numerator is not zero.

For 6x−1 to be greater than zero, the denominator must be positive since the numerator is positive. We know a positive divided by a positive is positive.

We set x−1>0 and solve.

This table demonstrates the process of solving a simple linear inequality, starting with the instruction and showing the step-by-step solution.

Solve.
x−1>0 x>1

Also, since the radicand is a fraction, we must realize that the denominator cannot be zero.

We solve x−1=0 to find the value that must be eliminated from the domain.

Demonstration of solving an algebraic equation and determining domain restrictions.

Solve.
x−1=0 x=1sox≠1in the domain.

Putting this together we get the domain is x>1 and we write it as (1,∞).

Find the domain of the function, f(x)=4x+3. Write the domain in interval notation.

Solution

(−3,∞)

Find the domain of the function, h(x)=9x−5. Write the domain in interval notation.

Solution

(5,∞)

The next example involves a cube root and so will require different thinking.

Find the domain of the function, f(x)=2x2+33. Write the domain in interval notation.

Solution

Since the function, f(x)=2x2+33 has a radical with an index of 3, which is odd, we know the radicand can be any real number. This tells us the domain is any real number. In interval notation, we write (−∞,∞).

The domain of f(x)=2x2+33 is all real numbers and we write it in interval notation as (−∞,∞).

Find the domain of the function, f(x)=3x2−13. Write the domain in interval notation.

Solution

(−∞,∞)

Find the domain of the function, g(x)=5x−43. Write the domain in interval notation.

Solution

(−∞,∞)

Graph Radical Functions

Before we graph any radical function, we first find the domain of the function. For the function, f(x)=x, the index is even, and so the radicand must be greater than or equal to 0.

This tells us the domain is x≥0 and we write this in interval notation as [0,∞).

Previously we used point plotting to graph the function, f(x)=x. We chose x-values, substituted them in and then created a chart. Notice we chose points that are perfect squares in order to make taking the square root easier.

The figure shows the square root function graph on the x y-coordinate plane. The x-axis of the plane runs from 0 to 7. The y-axis runs from 0 to 7. The function has a starting point at (0, 0) and goes through the points (1, 1) and (4, 2). A table is shown beside the graph with 3 columns and 5 rows. The first row is a header row with the expressions “x”, “f (x) = square root of x”, and “(x, f (x))”. The second row has the numbers 0, 0, and (0, 0). The third row has the numbers 1, 1, and (1, 1). The fourth row has the numbers 4, 2, and (4, 2). The fifth row has the numbers 9, 3, and (9, 3).

Once we see the graph, we can find the range of the function. The y-values of the function are greater than or equal to zero. The range then is [0,∞).

For the function f(x)=x+3,

ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ Since the radical has index 2, we know the radicand must be greater than or equal to zero. If x+3≥0, then x≥−3. This tells us the domain is all values x≥−3 and written in interval notation as [−3,∞).

ⓑ To graph the function, we choose points in the interval [−3,∞) that will also give us a radicand which will be easy to take the square root.

The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 3 to 3. The y-axis runs from 0 to 7. The function has a starting point at (negative 3, 0) and goes through the points (negative 2, 1) and (1, 2). A table is shown beside the graph with 3 columns and 5 rows. The first row is a header row with the expressions “x”, “f (x) = square root of the quantity x plus 3”, and “(x, f (x))”. The second row has the numbers negative 3, 0, and (negative 3, 0). The third row has the numbers negative 2, 1, and (negative 2, 1). The fourth row has the numbers 1, 2, and (1, 2). The fifth row has the numbers 6, 3, and (6, 3).

ⓒ Looking at the graph, we see the y-values of the function are greater than or equal to zero. The range then is [0,∞).

For the function f(x)=x+2, ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ domain: [−2,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 6. The y-axis runs from 0 to 8. The function has a starting point at (negative 2, 0) and goes through the points (negative 1, 1) and (2, 2).
ⓒ range: [0,∞)

For the function f(x)=x−2, ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ domain: [2,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from 0 to 8. The y-axis runs from 0 to 6. The function has a starting point at (2, 0) and goes through the points (3, 1) and (6, 2).
ⓒ range: [0,∞)

In our previous work graphing functions, we graphed f(x)=x3 but we did not graph the function f(x)=x3. We will do this now in the next example.

For the function f(x)=x3, ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ Since the radical has index 3, we know the radicand can be any real number. This tells us the domain is all real numbers and written in interval notation as (−∞,∞)

ⓑ To graph the function, we choose points in the interval (−∞,∞) that will also give us a radicand which will be easy to take the cube root.

The figure shows the cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis runs from negative 10 to 10. The function has a center point at (0, 0) and goes through the points (1, 1), (negative 1, negative 1), (8, 2), and (negative 8, negative 2). A table is shown beside the graph with 3 columns and 6 rows. The first row is a header row with the expressions “x”, “f (x) = cube root of x”, and “(x, f (x))”. The second row has the numbers negative 8, negative 2, and (negative 8, negative 2). The third row has the numbers negative 1, negative 1, and (negative 1, negative 1). The fourth row has the numbers 0, 0, and (0, 0). The fifth row has the numbers 1, 1, and (1, 1). The sixth row has the numbers 8, 2, and (8, 2).

ⓒ Looking at the graph, we see the y-values of the function are all real numbers. The range then is (−∞,∞).

For the function f(x)=−x3,

ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 2. The y-axis runs from negative 2 to 2. The function has a center point at (0, 0) and goes through the points (1, negative 1) and (negative 1, 1).
ⓒ range: (−∞,∞)

For the function f(x)=x−23,

ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 1 to 5. The y-axis runs from negative 3 to 3. The function has a center point at (2, 0) and goes through the points (1, negative 1) and (3, 2).
ⓒ range: (−∞,∞)

Access these online resources for additional instruction and practice with radical functions.

  • Domain of a Radical Function
  • Domain of a Radical Function 2
  • Finding Domain of a Radical Function

Key Concepts

  • Properties of an
    • When n is an even number and:
      a≥0, then an is a real number.
      a<0, then an is not a real number.
    • When n is an odd number, an is a real number for all values of a.
  • Domain of a Radical Function
    • When the index of the radical is even, the radicand must be greater than or equal to zero.
    • When the index of the radical is odd, the radicand can be any real number.

Practice Makes Perfect

Evaluate a Radical Function

In the following exercises, evaluate each function.

f(x)=4x−4, find ⓐ f(5) ⓑ f(0).

Solution

ⓐ f(5)=4 ⓑ no value at x=0

f(x)=6x−5, find ⓐ f(5) ⓑ f(−1).

g(x)=6x+1, find ⓐ g(4) ⓑ g(8).

Solution

ⓐ g(4)=5 ⓑ g(8)=7

g(x)=3x+1, find ⓐ g(8) ⓑ g(5).

F(x)=3−2x, find ⓐ F(1) ⓑ F(−11).

Solution

ⓐ F(1)=1 ⓑ F(−11)=5

F(x)=8−4x, find ⓐ F(1) ⓑ F(−2).

G(x)=5x−1, find ⓐ G(5) ⓑ G(2).

Solution

ⓐ G(5)=26 ⓑ G(2)=3

G(x)=4x+1, find ⓐ G(11) ⓑ G(2).

g(x)=2x−43, find ⓐ g(6) ⓑ g(−2).

Solution

ⓐ g(6)=2 ⓑ g(−2)=−2

g(x)=7x−13, find ⓐ g(4) ⓑ g(−1).

h(x)=x2−43, find ⓐ h(−2) ⓑ h(6).

Solution

ⓐ h(−2)=0 ⓑ h(6)=243

h(x)=x2+43, find ⓐ h(−2) ⓑ h(6).

For the function f(x)=2x34, find ⓐ f(0) ⓑ f(2).

Solution

ⓐ f(0)=0 ⓑ f(2)=2

For the function f(x)=3x34, find ⓐ f(0) ⓑ f(3).

For the function g(x)=4−4x4, find ⓐ g(1) ⓑ g(−3).

Solution

ⓐ g(1)=0 ⓑ g(−3)=2

For the function g(x)=8−4x4, find ⓐ g(−6) ⓑ g(2).

Find the Domain of a Radical Function

In the following exercises, find the domain of the function and write the domain in interval notation.

f(x)=3x−1

Solution

[13,∞)

f(x)=4x−2

g(x)=2−3x

Solution

(−∞,23]

g(x)=8−x

h(x)=5x−2

Solution

(2,∞)

h(x)=6x+3

f(x)=x+3x−2

Solution

(−∞,−3]∪(2,∞)

f(x)=x−1x+4

g(x)=8x−13

Solution

(−∞,∞)

g(x)=6x+53

f(x)=4x2−163

Solution

(−∞,∞)

f(x)=6x2−253

F(x)=8x+34

Solution

[−38,∞)

F(x)=10−7x4

G(x)=2x−15

Solution

(−∞,∞)

G(x)=6x−35

Graph Radical Functions

In the following exercises, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.

f(x)=x+1

Solution

ⓐ domain: [−1,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 1 to 7. The y-axis runs from negative 2 to 10. The function has a starting point at (negative 1, 0) and goes through the points (0, 1) and (3, 2).
ⓒ [0,∞)

f(x)=x−1

g(x)=x+4

Solution

ⓐ domain: [−4,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 2 to 6. The function has a starting point at (negative 4, 0) and goes through the points (negative 3, 1) and (0, 2).
ⓒ [0,∞)

g(x)=x−4

f(x)=x+2

Solution

ⓐ domain: [0,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from 0 to 8. The y-axis runs from 0 to 8. The function has a starting point at (0, 2) and goes through the points (1, 3) and (4, 4).
ⓒ [2,∞)

f(x)=x−2

g(x)=2x

Solution

ⓐ domain: [0,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from 0 to 8. The y-axis runs from 0 to 8. The function has a starting point at (0, 0) and goes through the points (1, 2) and (4, 4).
ⓒ [0,∞)

g(x)=3x

f(x)=3−x

Solution

ⓐ domain: (−∞,3]
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 6 to 4. The y-axis runs from 0 to 8. The function has a starting point at (3, 0) and goes through the points (2, 1), (negative 1, 2), and (negative 6, 3).
ⓒ [0,∞)

f(x)=4−x

g(x)=−x

Solution

ⓐ domain: [0,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from 0 to 8. The y-axis runs from negative 8 to 0. The function has a starting point at (0, 0) and goes through the points (1, negative 1) and (4, negative 2).
ⓒ (−∞,0]

g(x)=−x+1

f(x)=x+13

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The function has a center point at (negative 1, 0) and goes through the points (negative 2, negative 1) and (0, 1).
ⓒ (−∞,∞)

f(x)=x−13

g(x)=x+23

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The function has a center point at (negative 4, 0) and goes through the points (negative 3, negative 1) and (negative 1, 1).
ⓒ (−∞,∞)

g(x)=x−23

f(x)=x3+3

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 2 to 6. The function has a center point at (0, 3) and goes through the points (negative 1, 2) and (1, 4).
ⓒ (−∞,∞)

f(x)=x3−3

g(x)=x3

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The function has a center point at (0, 0) and goes through the points (1, 1) and (negative 1, negative 1).
ⓒ (−∞,∞)

g(x)=−x3

f(x)=2x3

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 4 to 4. The function has a center point at (0, 0) and goes through the points (1, 2) and (negative 1, negative 2).
ⓒ (−∞,∞)

f(x)=−2x3

Writing Exercises

Explain how to find the domain of a fourth root function.

Solution

Answers will vary.

Explain how to find the domain of a fifth root function.

Explain why y=x3 is a function.

Solution

Answers will vary.

Explain why the process of finding the domain of a radical function with an even index is different from the process when the index is odd.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “evaluate a radical function”, “find the domain of a radical function”, and “graph a radical function”. The other columns are left blank so the learner can indicate their level of understanding.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

radical function
A radical function is a function that is defined by a radical expression.

Use the Complex Number System

Learning Objectives

By the end of this section, you will be able to:

  • Evaluate the square root of a negative number
  • Add and subtract complex numbers
  • Multiply complex numbers
  • Divide complex numbers
  • Simplify powers of i

Before you get started, take this readiness quiz.

Given the numbers −4,−7,0.5–,73,3,81, list the ⓐ rational numbers, ⓑ irrational numbers, ⓒ real numbers.
If you missed this problem, review Example 9 in Decimals.

Solution

ⓐ −4,0.5–,73,3,81; ⓑ 7; ⓒ −4,−7,0.5–,73,3,81

Multiply: (x−3)(2x+5).
If you missed this problem, review Example 4 in Multiply Polynomials.

Solution

2x2−x−15

Rationalize the denominator:55−3.
If you missed this problem, review Example 8 in Multiply Polynomials.

Solution

5+152

Evaluate the Square Root of a Negative Number

Whenever we have a situation where we have a square root of a negative number we say there is no real number that equals that square root. For example, to simplify −1, we are looking for a real number x so that x2 = –1. Since all real numbers squared are positive numbers, there is no real number that equals –1 when squared.

Mathematicians have often expanded their numbers systems as needed. They added 0 to the counting numbers to get the whole numbers. When they needed negative balances, they added negative numbers to get the integers. When they needed the idea of parts of a whole they added fractions and got the rational numbers. Adding the irrational numbers allowed numbers like 5. All of these together gave us the real numbers and so far in your study of mathematics, that has been sufficient.

But now we will expand the real numbers to include the square roots of negative numbers. We start by defining the imaginary unit i as the number whose square is –1.

Imaginary Unit

The imaginary unit i is the number whose square is –1.

i2=−1ori=−1

We will use the imaginary unit to simplify the square roots of negative numbers.

Square Root of a Negative Number

If b is a positive real number, then

−b=bi

We will use this definition in the next example. Be careful that it is clear that the i is not under the radical. Sometimes you will see this written as −b=ib to emphasize the i is not under the radical. But the −b=bi is considered standard form.

Write each expression in terms of i and simplify if possible:

ⓐ −25 ⓑ −7 ⓒ −12.

Solution
ⓐ
Step-by-step simplification of the square root of -25.
−25
Use the definition of the square root of negative numbers. 25i
Simplify. 5i
ⓑ
Steps to simplify the square root of a negative number, showing the use of the imaginary unit 'i'.
−7
Use the definition of the square root of negative numbers. 7i
Simplify. Be careful that it is clear that i is not under the radical sign.
ⓒ
Step-by-step simplification of sqrt(-12) into its complex number form, 2 sqrt(3)i.
−12
Use the definition of the square root of negative numbers. 12i
Simplify 12. 23i

Write each expression in terms of i and simplify if possible:

ⓐ −81 ⓑ −5 ⓒ −18.

Solution

ⓐ 9i ⓑ 5i ⓒ 32i

Write each expression in terms of i and simplify if possible:

ⓐ −36 ⓑ −3 ⓒ −27.

Solution

ⓐ 6i ⓑ 3i ⓒ 33i

Now that we are familiar with the imaginary number i, we can expand our concept of the number system to include imaginary numbers. The complex number system includes the real numbers and the imaginary numbers. A complex number is of the form a + bi, where a, b are real numbers. We call a the real part and b the imaginary part.

Complex Number

A complex number is of the form a + bi, where a and b are real numbers.

The image shows the expression a plus b i. The number a is labeled “real part” and the number b i is labeled “imaginary part”.

A complex number is in standard form when written as a+bi, where a and b are real numbers.

If b=0, then a+bi becomes a+0·i=a, and is a real number.

If b≠0, then a+bi is an imaginary number.

If a=0, then a+bi becomes 0+bi=bi, and is called a pure imaginary number.

We summarize this here.

a+bi
b=0 a+0·ia Real number
b≠0 a+bi Imaginary number
a=0 0+bibi Pure imaginary number

The standard form of a complex number is a+bi, so this explains why the preferred form is −b=bi when b>0.

The diagram helps us visualize the complex number system. It is made up of both the real numbers and the imaginary numbers.

The table has four rows and three columns. The first row is a header and the second column entry a plus b i. In the second row is b equals zero, a plus 0 i, and “Real number”. The third row contains b is not equal to 0, a plus b i, and “Imaginary number”. The fourth row contains a = 0, 0 plus b i, and “Pure imaginary number”.

Add or Subtract Complex Numbers

We are now ready to perform the operations of addition, subtraction, multiplication and division on the complex numbers—just as we did with the real numbers.

Adding and subtracting complex numbers is much like adding or subtracting like terms. We add or subtract the real parts and then add or subtract the imaginary parts. Our final result should be in standard form.

Add: −12+−27.

Solution
Step-by-step solution for adding square roots of negative numbers, demonstrating simplification into complex numbers.
−12+−27
Use the definition of the square root of negative numbers. 12i+27i
Simplify the square roots. 23i+33i
Add. 53i

Add: −8+−32.

Solution

62i

Add: −27+−48.

Solution

73i

Remember to add both the real parts and the imaginary parts in this next example.

Simplify: ⓐ (4−3i)+(5+6i) ⓑ (2−5i)−(5−2i).

Solution
ⓐ
Step-by-step example of adding complex numbers, illustrating the use of the associative property to simplify the expression.
(4−3i)+(5+6i)
Use the Associative Property to put the real
parts and the imaginary parts together.
(4+5)+(−3i+6i)
Simplify. 9+3i
ⓑ
A step-by-step guide demonstrating the subtraction and simplification of two complex numbers.
(2−5i)−(5−2i)
Distribute. 2−5i−5+2i
Use the Associative Property to put the real
parts and the imaginary parts together.
2−5−5i+2i
Simplify. −3−3i

Simplify: ⓐ (2+7i)+(4−2i) ⓑ (8−4i)−(2−i).

Solution

ⓐ 6+5i ⓑ 6−3i

Simplify: ⓐ (3−2i)+(−5−4i) ⓑ (4+3i)−(2−6i).

Solution

ⓐ −2−6i ⓑ 2+9i

Multiply Complex Numbers

Multiplying complex numbers is also much like multiplying expressions with coefficients and variables. There is only one special case we need to consider. We will look at that after we practice in the next two examples.

Multiply: 2i(7−5i).

Solution
Step-by-step simplification of the complex number expression 2i(7-5i) into standard form (10 + 14i).
2i(7−5i)
Distribute. 14i−10i2
Simplify i2. 14i−10(−1)
Multiply. 14i+10
Write in standard form. 10+14i

Multiply: 4i(5−3i).

Solution

12+20i

Multiply: −3i(2+4i).

Solution

12−6i

In the next example, we multiply the binomials using the Distributive Property or FOIL.

Multiply: (3+2i)(4−3i).

Solution
A step-by-step guide to multiplying complex numbers using the FOIL method, detailing each calculation stage from the initial product to the simplified result.
(3+2i)(4−3i)
Use FOIL. 12−9i+8i−6i2
Simplify i2 and combine like terms. 12−i−6(−1)
Multiply. 12−i+6
Combine the real parts. 18−i

Multiply: (5−3i)(−1−2i).

Solution

−11−7i

Multiply: (−4−3i)(2+i).

Solution

−5−10i

In the next example, we could use FOIL or the Product of Binomial Squares Pattern.

Multiply: (3+2i)2

Solution
A mathematical expression showing the fraction (a + b) / (3 + 2i) squared. The numerator 'a + b' is displayed in red text.
Use the Product of Binomial Squares Pattern, (a+b)2=a2+2ab+b2. An algebraic identity a^2 + 2ab + b^2 is shown with a specific example: 3^2 + 2 * 3 * 2i + (2i)^2, demonstrating the expansion of (a+b)^2 where a=3 and b=2i.
Simplify. The mathematical expression 9 + 12i + 4i^2 is shown.
Simplify i2. A mathematical expression showing the sum of 9, 12i, and 4 multiplied by -1, which is 9 + 12i + 4(-1).
Simplify. The image shows the complex number expression 5 + 12i in black text against a white background.

Multiply using the Binomial Squares pattern: (−2−5i)2.

Solution

−21+20i

Multiply using the Binomial Squares pattern: (−5+4i)2.

Solution

9−40i

Since the square root of a negative number is not a real number, when we have the square roots of two negative numbers, we cannot use the Product Property for Radicals. In order to multiply square roots of negative numbers we should first write them as complex numbers, using −b=bi. This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

Multiply: −36·−4.

Solution

To multiply square roots of negative numbers, we first write them as complex numbers.

Step-by-step multiplication of square roots of negative numbers, demonstrating conversion to complex numbers and simplification.
−36·−4
Write as complex numbers using −b=bi. 36i·4i
Simplify. 6i·2i
Multiply. 12i2
Simplify i2 and multiply. −12

Multiply: −49·−4.

Solution

−14

Multiply: −36·−81.

Solution

−54

In the next example, each binomial has a square root of a negative number. Before multiplying, each square root of a negative number must be written as a complex number.

Multiply: (3−−12)(5+−27).

Solution

To multiply square roots of negative numbers, we first write them as complex numbers.

Detailed steps demonstrating the multiplication of two complex numbers, simplifying to their final form.
(3−−12)(5+−27)
Write as complex numbers using −b=bi. (3−23i)(5+33i)
Use FOIL. 15+93i−103i−6·3i2
Combine like terms and simplify i2. 15−3i−6·(−3)
Multiply and combine like terms. 33−3i

Multiply: (4−−12)(3−−48).

Solution

−12−223i

Multiply: (−2+−8)(3−−18).

Solution

6+122i

We first looked at conjugate pairs when we studied polynomials. We said that a pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference is called a conjugate pair and is of the form (a−b),(a+b).

A complex conjugate pair is very similar. For a complex number of the form a+bi, its conjugate is a−bi. Notice they have the same first term and the same last term, but one is a sum and one is a difference.

Complex Conjugate Pair

A complex conjugate pair is of the form a+bi,a−bi.

We will multiply a complex conjugate pair in the next example.

Multiply: (3−2i)(3+2i).

Solution
Step-by-step multiplication and simplification of complex conjugates (3 - 2i)(3 + 2i) to a real number result.
(3−2i)(3+2i)
Use FOIL. 9+6i−6i−4i2
Combine like terms and simplify i2. 9−4(−1)
Multiply and combine like terms. 13

Multiply: (4−3i)·(4+3i).

Solution

25

Multiply: (−2+5i)·(−2−5i).

Solution

29

From our study of polynomials, we know the product of conjugates is always of the form (a−b)(a+b)=a2−b2. The result is called a difference of squares. We can multiply a complex conjugate pair using this pattern.

The last example we used FOIL. Now we will use the Product of Conjugates Pattern.

The quantity a minus b in parentheses times the quantity a plus b in parentheses is written above the expression showing the product of 3 minus 2 i in parentheses and 3 plus 2 i in parentheses. In the next line a squared minus b squared is written above the expression 3 squared minus the quantity 2 i in parentheses squared. Simplifying we get 9 minus 4 i squared. This is equal to 9 minus 4 times negative 1. The final result is 13.

Notice this is the same result we found in Example 9.

When we multiply complex conjugates, the product of the last terms will always have an i2 which simplifies to −1.

(a−bi)(a+bi)a2−(bi)2a2−b2i2a2−b2(−1)a2+b2

This leads us to the Product of Complex Conjugates Pattern: (a−bi)(a+bi)=a2+b2

Product of Complex Conjugates

If a and b are real numbers, then

(a−bi)(a+bi)=a2+b2

Multiply using the Product of Complex Conjugates Pattern: (8−2i)(8+2i).

Solution
Mathematical expressions illustrating the difference of squares formula, (a - b)(a + b), and its application with complex numbers, (8 - 2i)(8 + 2i).
Use the Product of Complex Conjugates Pattern,
(a−bi)(a+bi)=a2+b2.
A mathematical image displaying a general algebraic expression 'a^2 + b^2' in red text, followed by a specific numerical instance '8^2 + 2^2' in black text below it.
Simplify the squares. The image displays the mathematical expression '64 + 4' in black text against a white background.
Add. The number 68 is prominently displayed in black text against a stark white background.

Multiply using the Product of Complex Conjugates Pattern: (3−10i)(3+10i).

Solution

109

Multiply using the Product of Complex Conjugates Pattern: (−5+4i)(−5−4i).

Solution

41

Divide Complex Numbers

Dividing complex numbers is much like rationalizing a denominator. We want our result to be in standard form with no imaginary numbers in the denominator.

How to Divide Complex Numbers

Divide: 4+3i3−4i.

Solution
Step 1 is to write both the numerator and denominator in standard form. For this example they are both in standard form. Step 2 is to multiply the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of 3 minus 4 i is 3 plus 4 i. The resulting expression is the quantity 4 plus 3 i in parentheses times the quantity 3 plus 4 i in parentheses divided by the product of 3 minus 4 i in parentheses and the quantity 3 plus 4 i in parentheses. Step 3 is to simplify and write the result in standard form. Use the pattern the quantity a plus b i in parentheses equals a squared plus b squared in the denominator. The expression for this example then becomes the quantity 12 plus 16 i plus 9 i plus 12 i squared in parentheses divided by the sum of 9 and 16. Combining like terms we get the quantity 12 plus 25 i minus 12 in parentheses divided by 25. Simplifying we get 25 i divided by 25. Write the result in standard form. The result is i.

Divide: 2+5i5−2i.

Solution

i

Divide: 1+6i6−i.

Solution

i

We summarize the steps here.

How to divide complex numbers.

  1. Write both the numerator and denominator in standard form.
  2. Multiply the numerator and denominator by the complex conjugate of the denominator.
  3. Simplify and write the result in standard form.

Divide, writing the answer in standard form: −35+2i.

Solution
Step-by-step process for dividing complex numbers by rationalizing the denominator, showing actions and corresponding mathematical expressions.
−35+2i
Multiply the numerator and denominator by the
complex conjugate of the denominator.
−3(5−2i)(5+2i)(5−2i)
Multiply in the numerator and use the Product of
Complex Conjugates Pattern in the denominator.
−15+6i52+22
Simplify. −15+6i29
Write in standard form. −1529+629i

Divide, writing the answer in standard form: 41−4i.

Solution

417+1617i

Divide, writing the answer in standard form: −2−1+2i.

Solution

25+45i

Be careful as you find the conjugate of the denominator.

Divide: 5+3i4i.

Solution
This table illustrates the step-by-step process for dividing complex numbers, transforming a complex fraction into its standard a + bi form.
5+3i4i
Write the denominator in standard form. 5+3i0+4i
Multiply the numerator and denominator by
the complex conjugate of the denominator.
(5+3i)(0−4i)(0+4i)(0−4i)
Simplify. (5+3i)(−4i)(4i)(−4i)
Multiply. −20i−12i2−16i2
Simplify the i2. −20i+1216
Rewrite in standard form. 1216−2016i
Simplify the fractions. 34−54i

Divide: 3+3i2i.

Solution

32−32i

Divide: 2+4i5i.

Solution

45−25i

Simplify Powers of i

The powers of i make an interesting pattern that will help us simplify higher powers of i. Let’s evaluate the powers of i to see the pattern.

i1i2i3i4i−1i2·ii2·i2−1·i(−1)(−1) −i1 i5i6i7i8 i4·ii4·i2i4·i3i4·i4 1·i1·i21·i31·1 ii2i31 −1−i

We summarize this now.

i1=ii5=i i2=−1i6=−1 i3=−ii7=−i i4=1i8=1

If we continued, the pattern would keep repeating in blocks of four. We can use this pattern to help us simplify powers of i. Since i4 = 1, we rewrite each power, in, as a product using i4 to a power and another power of i.

We rewrite it in the form in=(i4)q·ir, where the exponent, q, is the quotient of n divided by 4 and the exponent, r, is the remainder from this division. For example, to simplify i57, we divide 57 by 4 and we get 14 with a remainder of 1. In other words, 57=4·14+1. So we write i57=(14)14·i1 and then simplify from there.

A mathematical problem demonstrating the simplification of i to the power of 57, using long division (57 / 4 = 14 remainder 1) to determine that i^57 equals i.

Simplify: i86.

Solution
This table illustrates the step-by-step simplification of the imaginary unit i raised to the power of 86.
i86
Divide 86 by 4 and rewrite i86 in the
in=(i4)q·ir form.
(14)21·i2
A long division calculation of 86 divided by 4. The steps show that 4 goes into 8 two times, resulting in 8. Bringing down the 6, 4 goes into 6 one time, resulting in 4, with a remainder of 2. The quotient is 21 with a remainder of 2.
Simplify. (1)21·(−1)
Simplify. –1

Simplify: i75.

Solution

−i

Simplify: i92.

Solution

1

Access these online resources for additional instruction and practice with the complex number system.

  • Expressing Square Roots of Negative Numbers with i
  • Subtract and Multiply Complex Numbers
  • Dividing Complex Numbers
  • Rewriting Powers of i

Key Concepts

  • Square Root of a Negative Number
    • If b is a positive real number, then −b=bi
      a+bi
      b=0 a+0·ia Real number
      b≠0 a+bi Imaginary number
      a=0 0+bibi Pure imaginary number
    • A complex number is in standard form when written as a + bi, where a, b are real numbers.
      The diagram has a rectangle with the labels “Complex Numbers” and a plus b i. A second rectangle has the labels “Real Numbers”, a plus b i, b = 0. A third rectangle has the labels “Imaginary Numbers”, a plus b i, b not equal to 0. Arrows go from the Real Numbers rectangle and Imaginary Numbers rectangle and point toward the Complex Numbers rectangle.
  • Product of Complex Conjugates
    • If a, b are real numbers, then
      (a−bi)(a+bi)=a2+b2
  • How to Divide Complex Numbers
    1. Write both the numerator and denominator in standard form.
    2. Multiply the numerator and denominator by the complex conjugate of the denominator.
    3. Simplify and write the result in standard form.

Section Exercises

Practice Makes Perfect

Evaluate the Square Root of a Negative Number

In the following exercises, write each expression in terms of i and simplify if possible.

ⓐ −16 ⓑ −11
ⓒ −8

Solution

ⓐ 4i ⓑ 11i ⓒ 22i

ⓐ −121 ⓑ −1 ⓒ −20

ⓐ −100 ⓑ −13 ⓒ −45

Solution

ⓐ 10i ⓑ 13i ⓒ 35i

ⓐ −49 ⓑ −15 ⓒ −75

Add or Subtract Complex Numbers In the following exercises, add or subtract.

−75+−48

Solution

93i

−12+−75

−50+−18

Solution

82i

−72+−8

(1+3i)+(7+4i)

Solution

8+7i

(6+2i)+(3−4i)

(8−i)+(6+3i)

Solution

14+2i

(7−4i)+(−2−6i)

(1−4i)−(3−6i)

Solution

−2+2i

(8−4i)−(3+7i)

(6+i)−(−2−4i)

Solution

8+5i

(−2+5i)−(−5+6i)

(5−−36)+(2−−49)

Solution

7−13i

(−3+−64)+(5−−16)

(−7−−50)−(−32−−18)

Solution

25−22i

(−5+−27)−(−4−−48)

Multiply Complex Numbers

In the following exercises, multiply.

4i(5−3i)

Solution

12+20i

2i(−3+4i)

−6i(−3−2i)

Solution

−12+18i

−i(6+5i)

(4+3i)(−5+6i)

Solution

−38++9i

(−2−5i)(−4+3i)

(−3+3i)(−2−7i)

Solution

27+15i

(−6−2i)(−3−5i)

In the following exercises, multiply using the Product of Binomial Squares Pattern.

(3+4i)2

Solution

−7+24i

(−1+5i)2

(−2−3i)2

Solution

−5+12i

(−6−5i)2

In the following exercises, multiply.

−25·−36

Solution

−30

−4·−16

−9·−100

Solution

−30

−64·−9

(−2−−27)(4−−48)

Solution

−44−4i3

(5−−12)(−3+−75)

(2+−8)(−4+−18)

Solution

−20−22i

(5+−18)(−2−−50)

(2−i)(2+i)

Solution

5

(4−5i)(4+5i)

(7−2i)(7+2i)

Solution

53

(−3−8i)(−3+8i)

In the following exercises, multiply using the Product of Complex Conjugates Pattern.

(7−i)(7+i)

Solution

50

(6−5i)(6+5i)

(9−2i)(9+2i)

Solution

85

(−3−4i)(−3+4i)

Divide Complex Numbers

In the following exercises, divide.

3+4i4−3i

Solution

i

5−2i2+5i

2+i3−4i

Solution

225+1125i

3−2i6+i

32−3i

Solution

613+913i

24−5i

−43−2i

Solution

−1213−813i

−13+2i

1+4i3i

Solution

43−13i

4+3i7i

−2−3i4i

Solution

−34+12i

−3−5i2i

Simplify Powers of i

In the following exercises, simplify.

i41

Solution

i

i39

i66

Solution

−1

i48

i128

Solution

1

i162

i137

Solution

i

i255

Writing Exercises

Explain the relationship between real numbers and complex numbers.

Solution

Answers will vary.

Aniket multiplied as follows and he got the wrong answer. What is wrong with his reasoning?

−7·−7497

Why is −64=8i but −643=−4.

Solution

Answers will vary.

Explain how dividing complex numbers is similar to rationalizing a denominator.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “evaluate the square root of a negative number”, “add or subtract complex numbers”, “multiply complex numbers”, “divide complex numbers”, and “simplify powers of i”. The other columns are left blank so the learner can indicate their level of understanding.

ⓑ On a scale of 1−10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Chapter Review Exercises

Simplify Expressions with Roots

Simplify Expressions with Roots

In the following exercises, simplify.

ⓐ 225 ⓑ −16

Solution

ⓐ 15 ⓑ −4

ⓐ −169 ⓑ −8

ⓐ 83 ⓑ 814 ⓒ 2435

Solution

ⓐ 2 ⓑ 3 ⓒ 3

ⓐ −5123 ⓑ −814 ⓒ −15

Estimate and Approximate Roots

In the following exercises, estimate each root between two consecutive whole numbers.

ⓐ 68 ⓑ 843

Solution

ⓐ 8<68<9
ⓑ 4<843<5

In the following exercises, approximate each root and round to two decimal places.

ⓐ 37 ⓑ 843 ⓒ 1254

Simplify Variable Expressions with Roots

In the following exercises, simplify using absolute values as necessary.


ⓐ a33
ⓑ b77

Solution

ⓐ a ⓑ b


ⓐ a14
ⓑ w24


ⓐ m84
ⓑ n205

Solution

ⓐ m2 ⓑ n4


ⓐ 121m20
ⓑ −64a2


ⓐ 216a63
ⓑ 32b205

Solution

ⓐ 6a2 ⓑ 2b4


ⓐ 144x2y2
ⓑ 169w8y10
ⓒ 8a51b63

Simplify Radical Expressions

Use the Product Property to Simplify Radical Expressions

In the following exercises, use the Product Property to simplify radical expressions.

125

Solution

55

675

ⓐ 6253 ⓑ 1286

Solution

ⓐ 553 ⓑ 226

In the following exercises, simplify using absolute value signs as needed.


ⓐ a23
ⓑ b83
ⓒ c138


ⓐ 80s15
ⓑ 96a75
ⓒ 128b76

Solution

ⓐ 4|s7|5s ⓑ 2a3a25
ⓒ 2|b|2b6


ⓐ 96r3s3
ⓑ 80x7y63
ⓒ 80x8y94


ⓐ −325
ⓑ −18

Solution

ⓐ −2 ⓑ not real


ⓐ 8+96
ⓑ 2+402

Use the Quotient Property to Simplify Radical Expressions

In the following exercises, use the Quotient Property to simplify square roots.

ⓐ 7298 ⓑ 24813 ⓒ 6964

Solution

ⓐ 67 ⓑ 23 ⓒ 12

ⓐ y4y8 ⓑ u21u115 ⓒ v30v126

300m564

Solution

5m23m4


ⓐ 28p7q2
ⓑ 81s8t33
ⓒ 64p15q124


ⓐ 27p2q108p4q3
ⓑ 16c5d7250c2d23
ⓒ 2m9n7128m3n6

Solution

ⓐ 12|pq| ⓑ 2cd5d23
ⓒ |mn|2


ⓐ 80q55q
ⓑ −625353
ⓒ 80m745m4

Simplify Rational Exponents

Simplify expressions with a1n

In the following exercises, write as a radical expression.

ⓐ r12 ⓑ s13 ⓒ t14

Solution

ⓐ r ⓑ s3 ⓒ t4

In the following exercises, write with a rational exponent.

ⓐ 21p ⓑ 8q4 ⓒ 436r6

In the following exercises, simplify.


ⓐ 62514
ⓑ 24315
ⓒ 3215

Solution

ⓐ 5 ⓑ 3 ⓒ 2


ⓐ (−1,000)13
ⓑ −1,00013
ⓒ (1,000)−13


ⓐ (−32)15
ⓑ (243)−15
ⓒ −12513

Solution

ⓐ −2 ⓑ 13 ⓒ −5

Simplify Expressions with amn

In the following exercises, write with a rational exponent.


ⓐ r74
ⓑ (2pq5)3
ⓒ (12m7n)34

In the following exercises, simplify.


ⓐ 2532
ⓑ 9−32
ⓒ (−64)23

Solution

ⓐ 125 ⓑ 127 ⓒ 16


ⓐ −6432
ⓑ −64−32
ⓒ (−64)32

Use the Laws of Exponents to Simplify Expressions with Rational Exponents

In the following exercises, simplify.


ⓐ 652·612
ⓑ (b15)35
ⓒ w27w97

Solution

ⓐ 63 ⓑ b9 ⓒ 1w


ⓐ a34·a−14a−104
ⓑ (27​b23​c−52b−73c12)13

Add, Subtract and Multiply Radical Expressions

Add and Subtract Radical Expressions

In the following exercises, simplify.


ⓐ 72−32
ⓑ 7p3+2p3
ⓒ 5x3−3x3

Solution

ⓐ 42 ⓑ 9p3 ⓒ 2x3


ⓐ 11b−511b+311b
ⓑ 811cd4+511cd4−911cd4


ⓐ 48+27
ⓑ 543+1283
ⓒ 654−32804

Solution

ⓐ 73 ⓑ 723 ⓒ 354


ⓐ 80c7−20c7
ⓑ 2162r104+432r104

375y2+8y48−300y2

Solution

37y3

Multiply Radical Expressions

In the following exercises, simplify.


ⓐ (56)(−12)
ⓑ (−2184)(−94)


ⓐ (32x3)(718x2)
ⓑ (−620a23)(−216a33)

Solution

ⓐ 126x2x ⓑ 48a5a23

Use Polynomial Multiplication to Multiply Radical Expressions

In the following exercises, multiply.


ⓐ 11(8+411)
ⓑ 33(93+183)


ⓐ (3−27)(5−47)
ⓑ (x3−5)(x3−3)

Solution

ⓐ 71−227
ⓑ x23−8x3+15

(27−511)(47+911)


ⓐ (4+11)2
ⓑ (3−25)2

Solution

ⓐ 27+811 ⓑ 29−125

(7+10)(7−10)

(3x3+2)(3x3−2)

Solution

9x23−4

Divide Radical Expressions

Divide Square Roots

In the following exercises, simplify.


ⓐ 4875
ⓑ 813243


ⓐ 320mn−545m−7n3
ⓑ 16x4y−23−54x−2y43

Solution

ⓐ 8m43n4 ⓑ −2x23y2

Rationalize a One Term Denominator

In the following exercises, rationalize the denominator.

ⓐ 83 ⓑ 740 ⓒ 82y

ⓐ 1113 ⓑ 7543 ⓒ 33x23

Solution

ⓐ 121311 ⓑ 2836 ⓒ 9x3x

ⓐ 144 ⓑ 9324 ⓒ 69x34

Rationalize a Two Term Denominator

In the following exercises, simplify.

72−6

Solution

−7(2+6)2

5n−7

x+8x−8

Solution

(x+22)x−82

Solve Radical Equations

Solve Radical Equations

In the following exercises, solve.

4x−3=7

5x+1=−3

Solution

no solution

4x−13=3

u−3+3=u

Solution

u=3,u=4

4x+53−2=−5

(8x+5)13+2=−1

Solution

x=−4

y+4−y+2=0

28r+1−8=2

Solution

r=3

Solve Radical Equations with Two Radicals

In the following exercises, solve.

10+2c=4c+16

2x2+9x−183=x2+3x−23

Solution

x=−8,x=2

r+6=r+8

x+1−x−2=1

Solution

x=3

Use Radicals in Applications

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula s=A to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula s=24d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Solution

64.8 feet

Use Radicals in Functions

Evaluate a Radical Function

In the following exercises, evaluate each function.

g(x)=6x+1, find
ⓐ g(4)
ⓑ g(8)

G(x)=5x−1, find
ⓐ G(5)
ⓑ G(2)

Solution

ⓐ G(5)=26 ⓑ G(2)=3

h(x)=x2−43, find
ⓐ h(−2)
ⓑ h(6)

For the function
g(x)=4−4x4, find
ⓐ g(1)
ⓑ g(−3)

Solution

ⓐ g(1)=0 ⓑ g(−3)=2

Find the Domain of a Radical Function

In the following exercises, find the domain of the function and write the domain in interval notation.

g(x)=2−3x

F(x)=x+3x−2

Solution

(2,∞)

f(x)=4x2−163

F(x)=10−7x4

Solution

−∞,107

Graph Radical Functions

In the following exercises, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.

g(x)=x+4

g(x)=2x

Solution

ⓐ domain: [0,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from 0 to 8. The y-axis runs from 0 to 8. The function has a starting point at (0, 0) and goes through the points (1, 2) and (4, 4).
ⓒ range: [0,∞)

f(x)=x−13

f(x)=x3+3

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 2 to 6. The function has a center point at (0, 3) and goes through the points (negative 1, 2) and (1, 4).
ⓒ range: (−∞,∞)

Use the Complex Number System

Evaluate the Square Root of a Negative Number

In the following exercises, write each expression in terms of i and simplify if possible.


ⓐ −100
ⓑ −13
ⓒ −45

Add or Subtract Complex Numbers

In the following exercises, add or subtract.

−50+−18

Solution

82i

(8−i)+(6+3i)

(6+i)−(−2−4i)

Solution

8+5i

(−7−−50)−(−32−−18)

Multiply Complex Numbers

In the following exercises, multiply.

(−2−5i)(−4+3i)

Solution

23+14i

−6i(−3−2i)

−4·−16

Solution

−8

(5−−12)(−3+−75)

In the following exercises, multiply using the Product of Binomial Squares Pattern.

(−2−3i)2

Solution

−5+12i

In the following exercises, multiply using the Product of Complex Conjugates Pattern.

(9−2i)(9+2i)

Divide Complex Numbers

In the following exercises, divide.

2+i3−4i

Solution

225+1125i

−43−2i

Simplify Powers of i

In the following exercises, simplify.

i48

Solution

1

i255

Practice Test

In the following exercises, simplify using absolute values as necessary.

125x93

Solution

5x3

169x8y6

72x8y43

Solution

2x2y9x2y3

45x3y4180x5y2

In the following exercises, simplify. Assume all variables are positive.

ⓐ 256−14 ⓑ −4932

Solution

ⓐ 14 ⓑ −343

−45

x−14·x54x−34

Solution

x74

(8​x23​y−52x−73y12)13

48x5−75x5

Solution

−x23x

27x2−4x12+108x2

212x5·36x3

Solution

36x42

43(163−63)

(4−33)(5+23)

Solution

2−73

1283543

245xy−445x−4y3

Solution

7x2x3|y3|y

153

32+3

Solution

3(2−3)

−4·−9

−4i(−2−3i)

Solution

−12+8i

4+i3−2i

i172

Solution

1

In the following exercises, solve.

2x+5+8=6

x+5+1=x

Solution

x=4

2x2−6x−233=x2−3x+53

In the following exercise, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.

g(x)=x+2

Solution

ⓐ domain: [−2,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 6. The y-axis runs from 0 to 8. The function has a starting point at (negative 2, 0) and goes through the points (negative 1, 1) and (2, 2).
ⓒ range: [0,∞)

complex conjugate pair
A complex conjugate pair is of the form a + bi, a – bi.
complex number
A complex number is of the form a + bi, where a and b are real numbers. We call a the real part and b the imaginary part.
complex number system
The complex number system is made up of both the real numbers and the imaginary numbers.
imaginary unit
The imaginary unit i is the number whose square is –1. i2 = –1 or i=−1.
standard form
A complex number is in standard form when written as a+bi, where a, b are real numbers.

Introduction

A photo of an person’s eye fitted with a contact lens camera.
Several companies have patented contact lenses equipped with cameras, suggesting that they may be the future of wearable camera technology. (credit: “intographics”/Pixabay)

Blink your eyes. You’ve taken a photo. That’s what will happen if you are wearing a contact lens with a built-in camera. Some of the same technology used to help doctors see inside the eye may someday be used to make cameras and other devices. These technologies are being developed by biomedical engineers using many mathematical principles, including an understanding of quadratic equations and functions. In this chapter, you will explore these kinds of equations and learn to solve them in different ways. Then you will solve applications modeled by quadratics, graph them, and extend your understanding to quadratic inequalities.

Solve Quadratic Equations Using the Square Root Property

Learning Objectives

By the end of this section, you will be able to:

  • Solve quadratic equations of the form ax2=k using the Square Root Property
  • Solve quadratic equations of the form a(x–h)2=k using the Square Root Property

Before you get started, take this readiness quiz.

Simplify: 128.
If you missed this problem, review Example 1 in Simplify Radical Expressions.

Solution

82

Simplify: 325.
If you missed this problem, review Example 4 in Divide Radical Expressions.

Solution

4105

Factor: 9x2−12x+4.
If you missed this problem, review Example 1 in Factor Special Products.

Solution

3x−22

A quadratic equation is an equation of the form ax2 + bx + c = 0, where a≠0. Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax2. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the form ax2=k using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x2 = 9.

Steps to solve the quadratic equation x^2=9 by factoring, detailing each action and its corresponding mathematical expression.
x2=9
Put the equation in standard form. x2−9=0
Factor the difference of squares. (x−3)(x+3)=0
Use the Zero Product Property. x−3=0x−3=0
Solve each equation. x=3x=−3

We can easily use factoring to find the solutions of similar equations, like x2 = 16 and x2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, x=4,x=−4 and x=5,x=−5.

But what happens when we have an equation like x2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13)2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

Ifn2=m,thennis a square root ofm.

Since these equations are all of the form x2 = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.

Square Root Property

If x2 = k, then

x=korx=−korx=±k.

Notice that the Square Root Property gives two solutions to an equation of the form x2 = k, the principal square root of k and its opposite. We could also write the solution as x=±k. We read this as x equals positive or negative the square root of k.

Now we will solve the equation x2 = 9 again, this time using the Square Root Property.

Steps demonstrating how to solve the quadratic equation x^2 = 9 using the Square Root Property.
x2=9
Use the Square Root Property. x=±9
x=±3
Sox=3orx=−3.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x2 = 7.

Demonstrates solving the equation x^2=7 by applying the Square Root Property.
x2=7
Use the Square Root Property. x=7,x=−7

We cannot simplify 7, so we leave the answer as a radical.

How to solve a Quadratic Equation of the form ax2 = k Using the Square Root Property

Solve: x2−50=0.

Solution
Step one is to isolate the quadratic term and make its coefficient one. For the equation x squared minus fifty equals zero, first add fifty to both sides to get x squared by itself. The new equation is x squared equals fifty. Step two is to use Square Root Property. Remember to write the plus or minus symbol. The equation created is x equals the positive or negative square root of 50. Step three is to simplify the radical if possible. Continue to write equivalent equations. X equals the positive or negative square root of twenty five times the square root of five. X equals positive or negative five times the square root of five. Rewrite to show two solutions: x equals five times the square root of five or x equals negative five times the square root of five. Checking solutions for a quadratic equation. Substitute x equals 5 times the square root of 5 and x equals negative 5 times the square root of 5 into the original equation to verify both are valid solutions.

Solve: x2−48=0.

Solution

x=43,x=−43

Solve: y2−27=0.

Solution

y=33,y=−33

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

Solve a quadratic equation using the square root property.

  1. Isolate the quadratic term and make its coefficient one.
  2. Use Square Root Property.
  3. Simplify the radical.
  4. Check the solutions.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Solve: 3z2=108.

Solution
3z2=108
The quadratic term is isolated.
Divide by 3 to make its coefficient 1.
3z23=1083
Simplify. z2=36
Use the Square Root Property. z=±36
Simplify the radical. z=±6
Rewrite to show two solutions. z=6,z=−6
Check the solutions:

This image demonstrates how to verify the solutions z = 6 and z = -6 for the equation 3z^2 = 108, showing that both positive and negative square roots satisfy the equation.

Solve: 2x2=98.

Solution

x=7,x=−7

Solve: 5m2=80.

Solution

m=4,m=−4

The Square Root Property states ‘If x2=k,’ What will happen if k<0? This will be the case in the next example.

Solve: x2+72=0.

Solution
x2+72=0
Isolate the quadratic term. x2=−72
Use the Square Root Property. x=±−72
Simplify using complex numbers. x=±72i
Simplify the radical. x=±62i
Rewrite to show two solutions. x=62i,x=−62i
Check the solutions:

This image verifies that both x = 6√2i and x = -6√2i are valid solutions for the quadratic equation x^2 + 72 = 0 through step-by-step substitution.

Solve: c2+12=0.

Solution

c=23i,c=−23i

Solve: q2+24=0.

Solution

c=26i,c=−26i

Our method also works when fractions occur in the equation; we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Solve: 23u2+5=17.

Solution
23u2+5=17
Isolate the quadratic term. A mathematical equation is displayed on a white background, which reads '2/3 u^2 = 12'.
Multiply by 32 to make the coefficient 1. An algebraic equation: (3/2) multiplied by (2/3)u squared equals (3/2) multiplied by 12, with the fraction 3/2 highlighted in red on both sides.
Simplify. A mathematical equation is displayed, showing u squared equals eighteen (u^2 = 18). The characters are in a dark grey font against a plain white background.
Use the Square Root Property. A mathematical expression displays u equals plus or minus the square root of 18.
Simplify the radical. A mathematical equation shows 'u = plus-minus square root of 9 multiplied by 2' against a white background.
Simplify. The image shows the equation u = ±3√2, representing two possible values for u: positive three times the square root of two, and negative three times the square root of two.
Rewrite to show two solutions. Two solutions for 'u' are presented: u = 3√2 and u = -3√2, representing positive and negative square roots in a mathematical context.
Check:

Verification of two solutions, u = 3√2 and u = -3√2, for the equation (2/3)u^2 + 5 = 17, showing step-by-step calculations that confirm both values satisfy the equation.

Solve: 12x2+4=24.

Solution

x=210,x=−210

Solve: 34y2−3=18.

Solution

y=27,y=−27

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

Solve: 2x2−8=41.

Solution
A mathematical equation is displayed, reading '2x^2 - 8 = 41' on a white background.
Isolate the quadratic term. The image displays the mathematical equation 2x^2 = 49, presented in a clean, straightforward manner against a white background.
Divide by 2 to make the coefficient 1. A mathematical equation is displayed, showing '2x^2 / 2 = 49 / 2' in black text against a white background.
Simplify. A mathematical equation displays 'X squared equals 49 over 2' on a white background.
Use the Square Root Property. The mathematical equation 'x = ××±×× sqrt(49/2)' is displayed in black text on a white background.
Rewrite the radical as a fraction of square roots. A mathematical equation shows x equals plus or minus the square root of 49 divided by the square root of 2.
Rationalize the denominator. A mathematical equation shows x equals plus or minus a fraction where the numerator is the square root of 49 multiplied by the square root of 2, and the denominator is the square root of 2 multiplied by the square root of 2. The second √2 in both the numerator and denominator is in red.
Simplify. A mathematical equation shows x equals plus or minus seven times the square root of two, all divided by two.
Rewrite to show two solutions. The image displays two solutions for 'x': x = 7√2 / 2 and x = -7√2 / 2.
Check:
We leave the check for you.

Solve: 5r2−2=34.

Solution

r=655,r=−655

Solve: 3t2+6=70.

Solution

t=833,t=−833

Solve Quadratic Equations of the Form a(x − h)2 = k Using the Square Root Property

We can use the Square Root Property to solve an equation of the form a(x − h)2 = k as well. Notice that the quadratic term, x, in the original form ax2 = k is replaced with (x − h).

On the left is the equation a times x square equals k. Replacing x in this equation with the expression x minus h changes the equation. It is now a times the square of x minus h equals k.

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a, then the Square Root Property can be used on (x − h)2.

Solve: 4(y−7)2=48.

Solution
4(y−7)2=48
Divide both sides by the coefficient 4. (y−7)2=12
Use the Square Root Property on the binomial y−7=±12
Simplify the radical. y−7=±23
Solve for y. y=7±23
Rewrite to show two solutions. y=7+23, y=7−23
Check:

Step-by-step verification of the solutions y = 7 + 2√3 and y = 7 - 2√3 for the equation 4(y-7)^2 = 48, showing that both satisfy the equation and result in 48=48.

Solve: 3(a−3)2=54.

Solution

a=3+32,a=3−32

Solve: 2(b+2)2=80.

Solution

b=−2+210,b=−2−210

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Solve: (x−13)2=59.

Solution
Step-by-step solution of a quadratic equation (x - 1/3)^2 = 5/9, demonstrating the application of the Square Root Property.
(x−13)2=59
Use the Square Root Property. x−13=±59
Rewrite the radical as a fraction of square roots. x−13=±59
Simplify the radical. x−13=±53
Solve for x. x=13±53
Rewrite to show two solutions. x=13+53,x=13−53
Check:
We leave the check for you.

Solve: (x−12)2=54.

Solution

x=12+52,x=12−52

Solve: (y+34)2=716.

Solution

y=−34+74,y=−34−74

We will start the solution to the next example by isolating the binomial term.

Solve: 2(x−2)2+3=57.

Solution
Step-by-step solution for the quadratic equation 2(x-2)^2 + 3 = 57, demonstrating algebraic simplification and the square root property.
2(x−2)2+3=57
Subtract 3 from both sides to isolate the binomial term. 2(x−2)2=54
Divide both sides by 2. (x−2)2=27
Use the Square Root Property. x−2=±27
Simplify the radical. x−2=±33
Solve for x. x=2±33
Rewrite to show two solutions. x=2+33,x=2−33
Check:
We leave the check for you.

Solve: 5(a−5)2+4=104.

Solution

a=5+25,a=5−25

Solve: 3(b+3)2−8=88.

Solution

b=−3+42,b=−3−42

Sometimes the solutions are complex numbers.

Solve: (2x−3)2=−12.

Solution
Step-by-step solution of a quadratic equation using the Square Root Property, resulting in complex solutions.
(2x−3)2=−12
Use the Square Root Property. 2x−3=±−12
Simplify the radical. 2x−3=±23i
Add 3 to both sides. 2x=3±23i
Divide both sides by 2. x=3±23i2
Rewrite in standard form. x=32±23i2
Simplify. x=32±3i
Rewrite to show two solutions. x=32+3i,x=32−3i
Check:
We leave the check for you.

Solve: (3r+4)2=−8.

Solution

r=−43+22i3,r=−43−22i3

Solve: (2t−8)2=−10.

Solution

t=4+10i2,t=4−10i2

The left sides of the equations in the next two examples do not seem to be of the form a(x − h)2. But they are perfect square trinomials, so we will factor to put them in the form we need.

Solve: 4n2+4n+1=16.

Solution

We notice the left side of the equation is a perfect square trinomial. We will factor it first.

4n2+4n+1=16
Factor the perfect square trinomial. (2n+1)2=16
Use the Square Root Property. 2n+1=±16
Simplify the radical. 2n+1=±4
Solve for n. 2n=−1±4
Divide each side by 2. 2n2=−1±42
n=−1±42
Rewrite to show two solutions. n=−1+42, n=−1−42
Simplify each equation. n=32, n=−52
Check:

Two solutions for the equation 4n^2 + 4n + 1 = 16 are verified: n=3/2 and n=-5/2, both satisfying the equation and confirming 16=16.

Solve: 9m2−12m+4=25.

Solution

m=73,m=−1

Solve: 16n2+40n+25=4.

Solution

n=−34,n=−74

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

  • Solving Quadratic Equations: The Square Root Property
  • Using the Square Root Property to Solve Quadratic Equations

Key Concepts

  • Square Root Property
    • If x2=k, then x=korx=−k or x=±k

    How to solve a quadratic equation using the square root property.
    1. Isolate the quadratic term and make its coefficient one.
    2. Use Square Root Property.
    3. Simplify the radical.
    4. Check the solutions.

Practice Makes Perfect

Solve Quadratic Equations of the Form ax2 = k Using the Square Root Property

In the following exercises, solve each equation.

a2=49

Solution

a=±7

b2=144

r2−24=0

Solution

r=±26

t2−75=0

u2−300=0

Solution

u=±103

v2−80=0

4m2=36

Solution

m=±3

3n2=48

43x2=48

Solution

x=±6

53y2=60

x2+25=0

Solution

x=±5i

y2+64=0

x2+63=0

Solution

x=±37i

y2+45=0

43x2+2=110

Solution

x=±9

23y2−8=−2

25a2+3=11

Solution

a=±25

32b2−7=41

7p2+10=26

Solution

p=±477

2q2+5=30

5y2−7=25

Solution

y=±4105

3x2−8=46

Solve Quadratic Equations of the Form a(x − h)2 = k Using the Square Root Property

In the following exercises, solve each equation.

(u−6)2=64

Solution

u=14,u=−2

(v+10)2=121

(m−6)2=20

Solution

m=6±25

(n+5)2=32

(r−12)2=34

Solution

r=12±32

(x+15)2=725

(y+23)2=881

Solution

y=−23±229

(t−56)2=1125

(a−7)2+5=55

Solution

a=7±52

(b−1)2−9=39

4(x+3)2−5=27

Solution

x=−3±22

5(x+3)2−7=68

(5c+1)2=−27

Solution

c=−15±335i

(8d−6)2=−24

(4x−3)2+11=−17

Solution

x=34±72i

(2y+1)2−5=−23

m2−4m+4=8

Solution

m=2±22

n2+8n+16=27

x2−6x+9=12

Solution

x=3±23

y2+12y+36=32

25x2−30x+9=36

Solution

x=−35,x=95

9y2+12y+4=9

36x2−24x+4=81

Solution

x=−76,x=116

64x2+144x+81=25

Mixed Practice

In the following exercises, solve using the Square Root Property.

2r2=32

Solution

r=±4

4t2=16

(a−4)2=28

Solution

a=4±27

(b+7)2=8

9w2−24w+16=1

Solution

w=1,w=53

4z2+4z+1=49

a2−18=0

Solution

a=±32

b2−108=0

(p−13)2=79

Solution

p=13±73

(q−35)2=34

m2+12=0

Solution

m=±23i

n2+48=0.

u2−14u+49=72

Solution

u=7±62

v2+18v+81=50

(m−4)2+3=15

Solution

m=4±23

(n−7)2−8=64

(x+5)2=4

Solution

x=−3,x=−7

(y−4)2=64

6c2+4=29

Solution

c=±566

2d2−4=77

(x−6)2+7=3

Solution

x=6±2i

(y−4)2+10=9

Writing Exercises

In your own words, explain the Square Root Property.

Solution

Answers will vary.

In your own words, explain how to use the Square Root Property to solve the quadratic equation (x+2)2=16.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

Table checklist for evaluating understanding. It asks students to rate their ability to solve two types of quadratic equations using the Square Root Property: first, a times x squared equals k; second, a times the square of x minus h equals k. Response options are "Confidently," "with some help," or "No, I don’t get it."

Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

Solve Quadratic Equations by Completing the Square

Learning Objectives

By the end of this section, you will be able to:

  • Complete the square of a binomial expression
  • Solve quadratic equations of the form x2+bx+c=0 by completing the square
  • Solve quadratic equations of the form ax2+bx+c=0 by completing the square

Before you get started, take this readiness quiz.

Expand: (x+9)2.
If you missed this problem, review Example 8 in Multiply Polynomials.

Solution

x2+18x+81

Factor y2−14y+49.
If you missed this problem, review Example 1 in Factor Trinomials.

Solution

(y−7)2

Factor 5n2+40n+80.
If you missed this problem, review Example 6 in Factor Trinomials.

Solution

5(n+4)2

So far we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called completing the square, which is important for our work on conics later.

Complete the Square of a Binomial Expression

In the last section, we were able to use the Square Root Property to solve the equation (y − 7)2 = 12 because the left side was a perfect square.

(y−7)2=12y−7=±12y−7=±23y=7±23

We also solved an equation in which the left side was a perfect square trinomial, but we had to rewrite it the form (x−k)2 in order to use the Square Root Property.

x2−10x+25=18(x−5)2=18

What happens if the variable is not part of a perfect square? Can we use algebra to make a perfect square?

Let’s look at two examples to help us recognize the patterns.

(x+9)2(y−7)2 (x+9)(x+9)(y−7)(y−7) x2+9x+9x+81y2−7y−7y+49 x2+18x+81y2−14y+49

We restate the patterns here for reference.

Binomial Squares Pattern

If a and b are real numbers,

Quantity a plus b squared equals a squared plus 2 a b plus b2 where the binomial squared equals the first term squared plus 2 times the product of terms plus the second term squared. Quantity a minus b squared equals a squared minus 2 a b plus b2 where the binomial squared equals the first term squared minus 2 times the product of terms plus the second term squared.

We can use this pattern to “make” a perfect square.

We will start with the expression x2 + 6x. Since there is a plus sign between the two terms, we will use the (a + b)2 pattern, a2 + 2ab + b2 = (a + b)2.

The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 6x plus an unknown to allow a comparison of the corresponding terms of the expressions.

We ultimately need to find the last term of this trinomial that will make it a perfect square trinomial. To do that we will need to find b. But first we start with determining a. Notice that the first term of x2 + 6x is a square, x2. This tells us that a = x.

The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 2 x b + b squared. Note that x has been substituted for a in the second equation and compare corresponding terms.

What number, b, when multiplied with 2x gives 6x? It would have to be 3, which is 12(6). So b = 3.

The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 2 times 3 times x plus an unknown value to help compare terms.

Now to complete the perfect square trinomial, we will find the last term by squaring b, which is 32 = 9.

The perfect square expression a squared plus 2 a b plus b squared is shown above the expression x squared plus 6 x plus 9.

We can now factor.

The factored expression, the square of a plus b, is shown over the square of the expression x + 3.

So we found that adding 9 to x2 + 6x ‘completes the square’, and we write it as (x + 3)2.

Complete a square of x2+bx.

  1. Identify b, the coefficient of x.
  2. Find (12b)2, the number to complete the square.
  3. Add the (12b)2 to x2 + bx.
  4. Factor the perfect square trinomial, writing it as a binomial squared.

Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ x2−26x ⓑ y2−9y ⓒ n2+12n

Solution
ⓐ
Two algebraic expressions are displayed: 'x² - bx' in red text, followed by 'x² - 26x' in black text directly below it.
The coefficient of x is −26.
Find(12b)2.(12·(−26))2(13)2169
Add 169 to the binomial to complete the square. The image shows the algebraic expression x^2 - 26x + 169, which is a quadratic trinomial. This expression is a perfect square trinomial, factoring to (x - 13)^2.
Factor the perfect square trinomial, writing it as
a binomial squared.
A mathematical expression showing the quantity (x-13) raised to the power of 2, enclosed in parentheses.
ⓑ
Two algebraic expressions are shown: x^2 - bx in red, and y^2 - 9y in black, stacked vertically.
The coefficient of y is −9.
Find(12b)2.(12·(−9))2(−92)2814
Add 814 to the binomial to complete the square. A mathematical expression shown in black text on a white background, which reads y squared minus 9y plus 81 over 4.
Factor the perfect square trinomial, writing it as
a binomial squared.
(y - 9/2)^2
ⓒ
Two lines of algebraic expressions are shown on a white background. The top line reads 'x^2 + bx' in red text. The bottom line reads 'n^2 + (1/2)n' in black text.
The coefficient of n is 12.
Find(12b)2.(12·12)2(14)2116
Add 116 to the binomial to complete the square. A mathematical expression displays n squared plus one-half n plus one-sixteenth, set against a plain white background.
Rewrite as a binomial square. A mathematical expression showing the quantity 'n plus one-fourth' enclosed in parentheses, all raised to the power of two: (n + 1/4)^2.

Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ a2−20a ⓑ m2−5m ⓒ p2+14p

Solution

ⓐ (a−10)2 ⓑ (b−52)2
ⓒ (p+18)2

Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ b2−4b ⓑ n2+13n ⓒ q2−23q

Solution

ⓐ (b−2)2 ⓑ (n+132)2
ⓒ (q−13)2

Solve Quadratic Equations of the Form x2 + bx + c = 0 by Completing the Square

In solving equations, we must always do the same thing to both sides of the equation. This is true, of course, when we solve a quadratic equation by completing the square too. When we add a term to one side of the equation to make a perfect square trinomial, we must also add the same term to the other side of the equation.

For example, if we start with the equation x2 + 6x = 40, and we want to complete the square on the left, we will add 9 to both sides of the equation.

The image displays the quadratic equation x squared plus 6x equals 40.
An incomplete quadratic equation is shown, x squared plus 6x plus an empty blank space on the left side, which equals 40 plus another empty blank space on the right side.
A mathematical equation, x^2 + 6x + 9 = 40 + 9, with the number 9 highlighted in red, indicating it has been added to both sides to complete the square.
Add 9 to both sides to complete the square. A mathematical equation is displayed: (x + 3)'  = 49. The equation involves an algebraic expression squared equal to a numerical value, representing a quadratic equation in a specific format.

Now the equation is in the form to solve using the Square Root Property! Completing the square is a way to transform an equation into the form we need to be able to use the Square Root Property.

How to Solve a Quadratic Equation of the Form x2+bx+c=0 by Completing the Square

Solve by completing the square: x2+8x=48.

Solution
Step 1 is to isolate the variable terms on one side and the constant terms on the other. This equation, x squared plus 8 x equals 48 already has all variable terms on the left. Note that the leading coefficient is 1, so b equals 8. In step 2, find the expression one half times b, squared, the number needed to complete the square. Add this value to both sides of the equation. Take half of 8 and square it. The square of one half times 8 equals 16, so add 16 to BOTH sides of the equation. The equation becomes x squared plus 8 x plus 16 equals 48 plus 16. In step 3, factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right. Factor x squared plus 8 x plus 16 on the left side. Add 48+16 on the right side. The equation becomes the square of x plus 4 equals 64. Step 4 is to use the Square Root Property. Take the square root of both sides of the equation to yield x plus 4 equals the positive or negative square root of 64. In step 5, simplify the radical and then solve the two resulting equations. X plus 4 equals positive 8 or negative 8. If x plus 4 equals 8, then x equals 4. If x plus 4 equals negative 8, then x equals negative 12. Finally, step 6, check the solutions. Put each answer in the original equation to check. First substitute x equals 4. We need to show that 4 squared plus 8 times 4 equals 48. Simplify. The expression 4 squared plus 8 times 4 is equivalent to 16 plus 32, or 48. X equals 4 is a solution. Next substitute x equals negative 12 into the original equation, x squared plus 8 x equals 48. The square of negative 12 plus 8 times negative 12 equals 144 minus 96, or 48. X equals negative 12 is also a solution.

Solve by completing the square: x2+4x=5.

Solution

x=−5,x=1

Solve by completing the square: y2−10y=−9.

Solution

y=1,y=9

The steps to solve a quadratic equation by completing the square are listed here.

Solve a quadratic equation of the form x2+bx+c=0 by completing the square.

  1. Isolate the variable terms on one side and the constant terms on the other.
  2. Find (12·b)2, the number needed to complete the square. Add it to both sides of the equation.
  3. Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
  4. Use the Square Root Property.
  5. Simplify the radical and then solve the two resulting equations.
  6. Check the solutions.

When we solve an equation by completing the square, the answers will not always be integers.

Solve by completing the square: x2+4x=−21.

Solution
Two math expressions are shown: the general quadratic form x² + bx + c in red, and a specific quadratic equation x² + 4x = -21 in black on a white background.
The variable terms are on the left side.
Take half of 4 and square it.
An algebraic equation showing the initial step of completing the square: x^2 + 4x + [blank line with (1/2 * 4)^2 in red below it] = -21.
(12(4))2=4
Add 4 to both sides. A mathematical equation showing the step of adding 4 to both sides: x^2 + 4x + 4 = -21 + 4, as part of completing the square to solve for x.
Factor the perfect square trinomial,
writing it as a binomial squared.
The image displays the quadratic equation (x + 2)^2 = -17, which has no real solutions.
Use the Square Root Property. A mathematical equation is displayed on a white background: x + 2 = ±sqrt(-17). This equation involves a variable 'x', constants, an equality sign, and the plus-minus square root of a negative number, indicating complex solutions.
Simplify using complex numbers. Equation showing x plus 2 equals positive or negative square root of 17 times i. This represents complex solutions for x.
Subtract 2 from each side. The image displays a mathematical equation for x: x = -2 ± sqrt(17)i, showing a solution with both real and imaginary components.
Rewrite to show two solutions. Two complex conjugate roots are displayed: x = -2 + sqrt(17)i and x = -2 - sqrt(17)i.
We leave the check to you.

Solve by completing the square: y2−10y=−35.

Solution

y=5±10i

Solve by completing the square: z2+8z=−19.

Solution

z=−4+3i,z=−4−3i

In the previous example, our solutions were complex numbers. In the next example, the solutions will be irrational numbers.

Solve by completing the square: y2−18y=−6.

Solution
Two algebraic expressions are shown: one on the top reads 'x^2 - bx c' with 'c' appearing slightly separate, and the other below it reads 'y^2 - 18y = -6'.
The variable terms are on the left side.
Take half of −18 and square it.
(12(−18))2=81 A mathematical equation y^2 - 18y + blank = -6 is shown, with a red hint below the blank indicating (1/2 * -18)^2, illustrating the process of completing the square to solve a quadratic equation.
Add 81 to both sides. An equation: y^2 - 18y + 81 = -6 + 81. The value 81 is highlighted in red, indicating its addition to both sides, a common step in solving quadratic equations by completing the square.
Factor the perfect square trinomial,
writing it as a binomial squared.
A mathematical equation is displayed, showing (y-9)^2 = 75. The equation involves a variable y, subtraction, exponentiation, and equality to a constant.
Use the Square Root Property. A mathematical equation is displayed, showing y - 9 = +/- sqrt(75).
Simplify the radical. A mathematical equation shows 'y minus 9 equals plus or minus 5 times the square root of 3'.
Solve for y. A mathematical expression showing y equals 9 plus or minus 5 times the square root of 3.
Check.
Verifying solutions for the quadratic equation y² - 18y = -6 by substituting y with (9 + 5√3) and (9 - 5√3), both yielding -6 = -6.

Another way to check this would be to use a calculator. Evaluate y2−18yfor both of the solutions. The answer should be −6.

Solve by completing the square: x2−16x=−16.

Solution

x=8+43,x=8−43

Solve by completing the square: y2+8y=11.

Solution

y=−4+33,y=−4−33

We will start the next example by isolating the variable terms on the left side of the equation.

Solve by completing the square: x2+10x+4=15.

Solution
A mathematical equation is displayed on a white background. The equation reads: x squared plus 10x plus 4 equals 15.
Isolate the variable terms on the left side.
Subtract 4 to get the constant terms on the right side.
A quadratic equation displayed against a plain white background, reading 'x^2 + 10x = 11'.
Take half of 10 and square it.
(12(10))2=25 A mathematical equation illustrating a step in completing the square: x^2 - 10x + blank = 11, where the blank is represented by a fraction bar with an empty numerator and a denominator of (1/2 * 10)^2 in red.
Add 25 to both sides. A quadratic equation shown as x squared plus 10x plus 25 equals 11 plus 25, with the number 25 highlighted in red on both sides of the equation.
Factor the perfect square trinomial, writing it as
a binomial squared.
A mathematical equation shows '(x + 5)^2 = 36' centered on a white background.
Use the Square Root Property. A mathematical equation is displayed, reading 'x + 5 = +/- square root 36', showing a step in solving for x in an algebraic problem involving a square root.
Simplify the radical. A mathematical equation is displayed, showing 'x + 5 =  6' with a plus-minus symbol before the 6.
Solve for x. A mathematical equation is displayed on a white background: x = -5 ×1 6. This represents two possible values for x: x = -5 + 6 and x = -5 - 6.
Rewrite to show two solutions. The image displays two algebraic expressions for the variable x: x = -5 + 6 and x = -5 - 6, indicating two different calculations.
Solve the equations. Two mathematical equations are displayed: x=1 and x=-11, set against a plain white background.
Check:

Verification of the quadratic equation x^2 + 10x + 4 = 15 by substituting x=1 and x=-11, showing that both values yield 15=15 and are therefore correct solutions.

Solve by completing the square: a2+4a+9=30.

Solution

a=−7,a=3

Solve by completing the square: b2+8b−4=16.

Solution

b=−10,b=2

To solve the next equation, we must first collect all the variable terms on the left side of the equation. Then we proceed as we did in the previous examples.

Solve by completing the square: n2=3n+11.

Solution
A mathematical equation is displayed on a white background: n squared equals three n plus eleven (n^2 = 3n + 11).
Subtract 3n to get the variable terms on the left side. A mathematical equation is displayed on a white background, reading 'n squared minus 3n equals 11'.
Take half of −3 and square it.
(12(−3))2=94 A mathematical equation illustrating the process of completing the square: n^2 - 3n + (empty numerator) / ((1/2)*(-3))^2 = 11. The term to be added to complete the square is represented by the fraction.
Add 94 to both sides. An algebraic equation is displayed: n squared minus 3n plus 9 over 4 equals 11 plus 9 over 4. This likely represents a step in completing the square for a quadratic equation.
Factor the perfect square trinomial, writing it as
a binomial squared.
A mathematical equation shows (n - 3/2)^2 = 44/4 + 9/4. The expression on the left is a binomial squared with variable 'n' and constant 3/2. The right side is a sum of two fractions, both with a denominator of 4.
Add the fractions on the right side. A mathematical equation shown on a white background, which reads '(n - 3/2)^2 = 53/4'
Use the Square Root Property. A mathematical equation is displayed, showing 'n - 3/2 = ±√(53/4)' in black text on a white background.
Simplify the radical. A mathematical equation shows 'n minus 3 over 2 equals plus or minus the square root of 53 over 2'.
Solve for n. A mathematical equation displays 'n = 3/2 plus or minus the square root of 53, all divided by 2' in black text on a white background.
Rewrite to show two solutions. Two mathematical expressions for 'n' are shown: n = 3/2 + sqrt(53)/2, and n = 3/2 - sqrt(53)/2.
Check:
We leave the check for you!

Solve by completing the square: p2=5p+9.

Solution

p=52+612,p=52−612

Solve by completing the square: q2=7q−3.

Solution

q=72+372,q=72−372

Notice that the left side of the next equation is in factored form. But the right side is not zero. So, we cannot use the Zero Product Property since it says “If a·b=0, then a = 0 or b = 0.” Instead, we multiply the factors and then put the equation into standard form to solve by completing the square.

Solve by completing the square: (x−3)(x+5)=9.

Solution
A mathematical equation is displayed against a white background: (x - 3)(x + 5) = 9. The equation involves a variable 'x' in two binomials that are multiplied together and set equal to 9.
We multiply the binomials on the left. A quadratic equation is displayed on a white background, reading 'x^2 + 2x - 15 = 9'.
Add 15 to isolate the constant terms on the right. A mathematical equation is displayed on a white background, reading 'x^2 + 2x = 24'. The equation features a quadratic term, a linear term, and a constant.
Take half of 2 and square it.
(12·(2))2=1 An algebraic equation showing a step to complete the square. The expression is x^2 + 2x + a fractional term with a blank numerator and (1/2 * 2)^2 in the denominator, equaling 24.
Add 1 to both sides. The equation x^2 + 2x + 1 = 24 + 1, with the added '1' on each side emphasized in red.
Factor the perfect square trinomial, writing it as
a binomial squared.
A mathematical equation is displayed on a white background, reading '(x+1)^2 = 25' in black text.
Use the Square Root Property. A mathematical equation on a white background reads 'x + 1 = ±√25'.
Solve for x. A close-up shot of a math equation on a white background, which reads 'x = -1 ± 5'.
Rewrite to show two solutions. Two mathematical equations are displayed on a white background: 'x = -1 + 5' and 'x = -1 - 5'. These represent two possible solutions for the variable 'x'.
Simplify. Two mathematical expressions are displayed on a white background: 'x = 4,' and 'x = -6'.
Check:
We leave the check for you!

Solve by completing the square: (c−2)(c+8)=11.

Solution

c=−9,c=3

Solve by completing the square: (d−7)(d+3)=56.

Solution

d=11,d=−7

Solve Quadratic Equations of the Form ax2 + bx + c = 0 by Completing the Square

The process of completing the square works best when the coefficient of x2 is 1, so the left side of the equation is of the form x2 + bx + c. If the x2 term has a coefficient other than 1, we take some preliminary steps to make the coefficient equal to 1.

Sometimes the coefficient can be factored from all three terms of the trinomial. This will be our strategy in the next example.

Solve by completing the square: 3x2−12x−15=0.

Solution

To complete the square, we need the coefficient of x2 to be one. If we factor out the coefficient of x2 as a common factor, we can continue with solving the equation by completing the square.

A mathematical equation reads 3x^2 - 12x - 15 = 0.
Factor out the greatest common factor. A mathematical equation is displayed with the expression 3(x^2 - 4x - 5) = 0, indicating a quadratic equation to be solved.
Divide both sides by 3 to isolate the trinomial
with coefficient 1.
A mathematical equation shown as 3(x^2 - 4x - 5) / 3 = 0 / 3, demonstrating division by 3 on both sides of the equation.
Simplify. A quadratic equation, x^2 - 4x - 5 = 0, is displayed in a grayscale image. The equation is centered against a white background, representing a mathematical problem to be solved.
Add 5 to get the constant terms on the right side. A mathematical equation is displayed, showing 'x squared minus 4x equals 5' in a white background.
Take half of 4 and square it.
(12(−4))2=4 An algebraic equation showing x^2 - 4x plus a blank numerator over a red (1/2 * 4)^2 term, set equal to 5. This illustrates a step in completing the square for a quadratic expression.
Add 4 to both sides. A step in solving x^2 - 4x = 5 by adding 4 to both sides to complete the square, resulting in x^2 - 4x + 4 = 5 + 4.
Factor the perfect square trinomial, writing it
as a binomial squared.
The image displays the algebraic equation (x-2)^2 = 9, centered on a white background.
Use the Square Root Property. A mathematical equation is displayed on a white background: x - 2 = plus or minus the square root of 9. This equation involves a variable, arithmetic operations, and a radical expression.
Solve for x. A mathematical equation 'x - 2 equals plus or minus 3'
Rewrite to show two solutions. The image displays two simple mathematical equations: x = 2 + 3 and x = 2 - 3, separated by a comma. These equations show two possible values for x, one from addition and one from subtraction.
Simplify. Two mathematical equations are displayed against a white background: 'x = 5,' and 'x = -1'.
Check:

Checking the solutions x=5 and x=-1 for the quadratic equation 3x² - 12x - 15 = 0. Both substitutions correctly result in 0=0, confirming they are valid roots.

Solve by completing the square: 2m2+16m+14=0.

Solution

m=−7,m=−1

Solve by completing the square: 4n2−24n−56=8.

Solution

n=−2,n=8

To complete the square, the coefficient of the x2 must be 1. When the leading coefficient is not a factor of all the terms, we will divide both sides of the equation by the leading coefficient! This will give us a fraction for the second coefficient. We have already seen how to complete the square with fractions in this section.

Solve by completing the square: 2x2−3x=20.

Solution

To complete the square we need the coefficient of x2 to be one. We will divide both sides of the equation by the coefficient of x2. Then we can continue with solving the equation by completing the square.

A mathematical equation is displayed, reading '2x^2 - 3x = 20' in black text against a white background.
Divide both sides by 2 to get the
coefficient of x2 to be 1.
A mathematical equation is shown with the expression (2x^2 - 3x) / 2 = 20 / 2, where both sides of the equation are divided by 2.
Simplify. A mathematical equation is displayed, showing 'x squared minus three halves x equals ten' in black text on a white background.
Take half of −32 and square it.
(12(−32))2=916 A mathematical equation demonstrating the process of completing the square. It shows x squared minus three-halves x, plus a blank term, which is indicated by one-half times negative three-halves, all squared, equals 10.
Add 916 to both sides. A mathematical equation showing x squared minus three-halves x plus nine-sixteenths equals ten plus nine-sixteenths, with the fractions and plus signs in red.
Factor the perfect square trinomial,
writing it as a binomial squared.
A mathematical equation shows '(x - 3/4)^2 = 160/16 + 9/16' written in black characters on a white background.
Add the fractions on the right side. A mathematical equation shows (x - 3/4)^2 = 169/16.
Use the Square Root Property. A mathematical equation is displayed, showing x minus three-fourths equals plus or minus the square root of 169 over 16.
Simplify the radical. A mathematical equation shows 'x minus 3/4 equals plus or minus 13/4' on a white background. This equation represents a step in solving for x, likely from a quadratic equation.
Solve for x. A mathematical equation shows x equals three-fourths plus or minus thirteen-fourths, represented as x = 3/4 ×1 13/4.
Rewrite to show two solutions. Two mathematical equations are displayed horizontally. The first reads x equals 3 over 4 plus 13 over 4. The second reads x equals 3 over 4 minus 13 over 4.
Simplify. The image displays two values for x: x=4 and x=-5/2.
Check:
We leave the check for you!

Solve by completing the square: 3r2−2r=21.

Solution

r=−73,r=3

Solve by completing the square: 4t2+2t=20.

Solution

t=−52,t=2

Now that we have seen that the coefficient of x2 must be 1 for us to complete the square, we update our procedure for solving a quadratic equation by completing the square to include equations of the form ax2 + bx + c = 0.

Solve a quadratic equation of the form ax2+bx+c=0 by completing the square.

  1. Divide by a to make the coefficient of x2 term 1.
  2. Isolate the variable terms on one side and the constant terms on the other.
  3. Find (12·b)2, the number needed to complete the square. Add it to both sides of the equation.
  4. Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
  5. Use the Square Root Property.
  6. Simplify the radical and then solve the two resulting equations.
  7. Check the solutions.

Solve by completing the square: 3x2+2x=4.

Solution

Again, our first step will be to make the coefficient of x2 one. By dividing both sides of the equation by the coefficient of x2, we can then continue with solving the equation by completing the square.

A mathematical equation reads '3x² + 2x = 4' against a plain white background.
Divide both sides by 3 to make the
coefficient of x2 equal 1.
A mathematical equation shows (3x^2 + 2x) divided by 3, equaling 4/3. The equation is presented in black text on a white background.
Simplify. A quadratic equation displayed on a white background, reading x squared plus two-thirds x equals four-thirds.
Take half of 23 and square it.
(12·23)2=19 A quadratic equation, x squared plus two-thirds x plus a blank equals four-thirds. A red box highlights the term one-half times two-thirds, all squared, which is being added to complete the square.
Add 19 to both sides. A quadratic equation displayed as x squared plus two-thirds x plus one-ninth equals four-thirds plus one-ninth, with the one-ninth terms highlighted in red on both sides.
Factor the perfect square trinomial, writing it as
a binomial squared.
A mathematical equation shows (x + 1/3) squared equals 12/9 plus 1/9, demonstrating steps in solving a quadratic equation or simplifying expressions.
Use the Square Root Property. A mathematical equation is displayed, showing x plus one-third equals plus or minus the square root of thirteen over nine.
Simplify the radical. A mathematical equation is displayed, reading 'x + 1/3 = +/- sqrt(13)/3' on a white background. The equation involves a variable 'x', fractions, a square root, and the plus-minus symbol.
Solve for x . A mathematical equation shows 'X = -1/3 ×1 (sqrt)13 / 3' written in the center of a plain white background.
Rewrite to show two solutions. The image displays two mathematical solutions for x: X = -1/3 + sqrt(13)/3 and X = -1/3 - sqrt(13)/3, representing the roots of a quadratic equation.
Check:
We leave the check for you!

Solve by completing the square: 4x2+3x=2.

Solution

x=−38+418,x=−38−418

Solve by completing the square: 3y2−10y=−5.

Solution

y=53+103,y=53−103

Access these online resources for additional instruction and practice with completing the square.

  • Completing Perfect Square Trinomials
  • Completing the Square 1
  • Completing the Square to Solve Quadratic Equations
  • Completing the Square to Solve Quadratic Equations: More Examples
  • Completing the Square 4

Key Concepts

  • Binomial Squares Pattern
    If a and b are real numbers,
    Quantity a plus b squared equals a squared plus 2 a b plus b2 where the binomial squared equals the first term squared plus 2 times the product of terms plus the second term squared. Quantity a minus b squared equals a squared minus 2 a b plus b2 where the binomial squared equals the first term squared minus 2 times the product of terms plus the second term squared.
  • How to Complete a Square
    1. Identify b, the coefficient of x.
    2. Find (12b)2, the number to complete the square.
    3. Add the (12b)2 to x2 + bx
    4. Rewrite the trinomial as a binomial square
  • How to solve a quadratic equation of the form ax2 + bx + c = 0 by completing the square.
    1. Divide by a to make the coefficient of x2 term 1.
    2. Isolate the variable terms on one side and the constant terms on the other.
    3. Find (12·b)2, the number needed to complete the square. Add it to both sides of the equation.
    4. Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right.
    5. Use the Square Root Property.
    6. Simplify the radical and then solve the two resulting equations.
    7. Check the solutions.

Practice Makes Perfect

Complete the Square of a Binomial Expression

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ m2−24m ⓑ x2−11x ⓒ p2−13p

Solution

ⓐ (m−12)2 ⓑ (x−112)2
ⓒ (p−16)2

ⓐ n2−16n ⓑ y2+15y ⓒ q2+34q

ⓐ p2−22p ⓑ y2+5y ⓒ m2+25m

Solution

ⓐ (p−11)2 ⓑ (y+52)2
ⓒ (m+15)2

ⓐ q2−6q ⓑ x2−7x ⓒ n2−23n

Solve Quadratic Equations of the form x2 + bx + c = 0 by Completing the Square

In the following exercises, solve by completing the square.

u2+2u=3

Solution

u=−3,u=1

z2+12z=−11

x2−20x=21

Solution

x=−1,x=21

y2−2y=8

m2+4m=−44

Solution

m=−2±210i

n2−2n=−3

r2+6r=−11

Solution

r=−3±2i

t2−14t=−50

a2−10a=−5

Solution

a=5±25

b2+6b=41

x2+5x=2

Solution

x=−52±332

y2−3y=2

u2−14u+12=−1

Solution

u=1,u=13

z2+2z−5=2

r2−4r−3=9

Solution

r=−2,r=6

t2−10t−6=5

v2=9v+2

Solution

v=92±892

w2=5w−1

x2−5=10x

Solution

x=5±30

y2−14=6y

(x+6)(x−2)=9

Solution

x=−7,x=3

(y+9)(y+7)=80

(x+2)(x+4)=3

Solution

x=−5,x=−1

(x−2)(x−6)=5

Solve Quadratic Equations of the form ax2 + bx + c = 0 by Completing the Square

In the following exercises, solve by completing the square.

3m2+30m−27=6

Solution

m=−11,m=1

2x2−14x+12=0

2n2+4n=26

Solution

n=−1±14

5x2+20x=15

2c2+c=6

Solution

c=−2,c=32

3d2−4d=15

2x2+7x−15=0

Solution

x=−5,x=32

3x2−14x+8=0

2p2+7p=14

Solution

p=−74±1614

3q2−5q=9

5x2−3x=−10

Solution

x=310±19110i

7x2+4x=−3

Writing Exercises

Solve the equation x2+10x=−25

ⓐ by using the Square Root Property

ⓑ by Completing the Square

ⓒ Which method do you prefer? Why?

Solution

Answers will vary.

Solve the equation y2+8y=48 by completing the square and explain all your steps.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can complete the square of a binomial expression.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve quadratic equations of the form x squared plus b times x plus c equals 0 by completing the square.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve quadratic equations of the form a times x squared plus b times x plus c equals 0 by completing the square.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Solve Quadratic Equations Using the Quadratic Formula

Learning Objectives

By the end of this section, you will be able to:

  • Solve quadratic equations using the Quadratic Formula
  • Use the discriminant to predict the number and type of solutions of a quadratic equation
  • Identify the most appropriate method to use to solve a quadratic equation

Before you get started, take this readiness quiz.

Evaluate b2−4ab when a=3 and b=−2.
If you missed this problem, review Example 10 in Integers.

Solution

28

Simplify: 108.
If you missed this problem, review Example 1 in Simplify Radical Expressions.

Solution

63

Simplify: 50.
If you missed this problem, review Example 1 in Use the Complex Number System.

Solution

52

Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for x.

We start with the standard form of a quadratic equation and solve it for x by completing the square.

A quadratic equation, ax^2 + bx + c = 0, where 'a' cannot be zero, is shown in this image. It's a fundamental algebraic expression used to find the roots of a second-degree polynomial.
Isolate the variable terms on one side. A mathematical equation displays 'ax^2 + bx = -c' in gray text against a plain white background, representing a quadratic equation.
Make the coefficient of x2 equal to 1, by
dividing by a.
A mathematical equation showing the division of a quadratic equation by 'a', resulting in x^2 + (b/a)x = -c/a, a step in solving for x by completing the square.
Simplify. A mathematical equation is displayed, showing x squared plus b over a times x equals negative c over a, representing a step in solving a quadratic equation by completing the square.
To complete the square, find (12·ba)2 and add it to both sides of the equation.
(12ba)2=b24a2 A mathematical equation demonstrating a step in completing the square: x^2 + (b/a)x + (b^2)/(4a^2) = -c/a + (b^2)/(4a^2), with the term b^2/(4a^2) highlighted in red on both sides.
The left side is a perfect square, factor it. The mathematical equation (x + b/(2a))^2 = -c/a + b^2/(4a^2) is shown on a white background. It represents an intermediate step in solving quadratic equations.
Find the common denominator of the right
side and write equivalent fractions with
the common denominator.
A mathematical equation illustrating a step in completing the square, showing (x + b/(2a))^2 equals b^2/(4a^2) minus (c/a) multiplied by (4a/4a).
Simplify. A mathematical equation illustrating a step in solving for x using the completing the square method, which is commonly used to derive the quadratic formula.
Combine to one fraction. A mathematical equation shows (x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), representing an intermediate step in deriving the quadratic formula by completing the square.
Use the square root property. A mathematical equation shows x plus b over 2a equals plus or minus the square root of b squared minus 4ac, all divided by 4a squared, illustrating a step in deriving the quadratic formula.
Simplify the radical. A mathematical equation shown as x + b/(2a) = +/- sqrt(b^2 - 4ac) / (2a), a step in deriving the quadratic formula, isolated on a white background.
Add −b2a to both sides of the equation. The quadratic formula, showing the solutions for x are equal to negative b divided by two a, plus or minus the square root of b squared minus four a c, all divided by two a.
Combine the terms on the right side. The quadratic formula, x = (-b ×1×4b^2 - 4ac) / 2a, is displayed, used to find the roots of a quadratic equation. This fundamental algebraic equation is a key tool in mathematics for solving polynomials of degree 2.
This equation is the Quadratic Formula.

Quadratic Formula

The solutions to a quadratic equation of the form ax2 + bx + c = 0, where a≠0 are given by the formula:

x=−b±b2−4ac2a

To use the Quadratic Formula, we substitute the values of a, b, and c from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.

Notice the formula is an equation. Make sure you use both sides of the equation.

How to Solve a Quadratic Equation Using the Quadratic Formula

Solve by using the Quadratic Formula: 2x2+9x−5=0.

Solution
Step 1 is to write the quadratic equation in standard form, a times x squared plus b x plus c equals zero, and identify the values a, b, and c. The equation 2 x squared plus 9 x minus 5 equals zero is in standard form. A equals 2, b equals 9, and c equals negative 5. Step 2. Write the quadratic formula. Then substitute the values of a, b, and c. Substitute a equals 2, b equals 9, and c equals negative 5 into the equation x equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. So x equals the quotient negative 9 plus or minus the square root of the difference 9 squared minus the product 4 times 2 times negative 5 divided by the product 2 times 2. In step 3, simplify the fraction and solve for x. x equals the quotient negative 9 plus or minus the square root of the difference 81 minus negative 40 divided by 4. Simplify the radicand. x equals the quotient negative 9 plus or minus the square root of 121 divided by 4. Simplify the square root. x equals the quotient negative 9 plus or minus 11 divided by 4. Separate into two equations. The first equation is x equals the quotient negative 9 plus 11 divided by 4 which simplifies to 2 divided by 4. The first solution is x equals one half. The second equation is x equals the quotient negative 9 minus 11 divided by 4 which simplifies to negative 20 divided by 4. The second solution is x equals negative 5. The fourth, and final, step is to check the solution. Put each answer into the original equation to check. First, substitute x equals one half into the original equation, 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of one half plus nine times one half minus 5. We need to show that this expression equals 0. Simplify the square. 2 times one fourth plus nine times one half minus 5 equals one half plus 9 halves minus 5, or 10 halves minus 5. 5 minus 5 equals 0, so x equals one half is indeed a solution. Next substitute x = negative 5 into the equation 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of negative 5 plus 9 times negative 5 minus 5. We need to show that this expression equals 0. Simplify the square. 2 times 25 plus nine times negative 5 minus 5 equals 50 minus 45 minus 5, or 0. x equals negative 5 is a solution as well.

Solve by using the Quadratic Formula: 3y2−5y+2=0.

Solution

y=1,y=23

Solve by using the Quadratic Formula: 4z2+2z−6=0.

Solution

z=1,z=−32

Solve a quadratic equation using the quadratic formula.

  1. Write the quadratic equation in standard form, ax2 + bx + c = 0. Identify the values of a, b, and c.
  2. Write the Quadratic Formula. Then substitute in the values of a, b, and c.
  3. Simplify.
  4. Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “x =”.

Solve by using the Quadratic Formula: x2−6x=−5.

Solution
A mathematical equation is displayed, showing 'x squared minus 6x equals minus 5' against a white background.
Write the equation in standard form by adding
5 to each side.
A clear, focused image displays the quadratic equation x^2 - 6x + 5 = 0 presented in black text against a plain white background.
This equation is now in standard form. Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red, and a specific example x^2 - 6x + 5 = 0 in black.
Identify the values of a,  b,  c. The image displays mathematical coefficients a=1 in light blue, b=-6 in red, and c=5 in light green, set against a plain white background.
Write the Quadratic Formula. The quadratic formula, used to find the solutions for x in a quadratic equation of the form ax^2 + bx + c = 0.
Then substitute in the values of a,  b,  c. The quadratic formula, x = -(-6) ×177 sqrt((-6)^2 - 4 * 1 * 5) / (2 * 1), demonstrates the substitution of values for 'a', 'b', and 'c' to solve a quadratic equation.
Simplify. A quadratic formula step showing x equals six plus or minus the square root of thirty-six minus twenty, all divided by two.
A mathematical equation shows x equals the fraction of (6 plus or minus the square root of 16) all over 2. This is a step in solving a quadratic equation.
A mathematical equation displays x equals 6 plus or minus 4, all divided by 2.
Rewrite to show two solutions. Two mathematical expressions for 'x' are displayed: x = (6+4)/2 and x = (6-4)/2, representing two possible solutions in a mathematical context.
Simplify. Two mathematical equations are displayed horizontally: x = 10/2, and x = 2/2.
The image shows two mathematical expressions on a white background: 'x = 5,' and 'x = 1'.
Check:

Verification of solutions for the quadratic equation x^2 - 6x + 5 = 0, demonstrating that both x=5 and x=1 satisfy the equation, resulting in 0=0.

Solve by using the Quadratic Formula: a2−2a=15.

Solution

a=−3,a=5

Solve by using the Quadratic Formula: b2+24=−10b.

Solution

b=−6,b=−4

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

Solve by using the Quadratic Formula: 2x2+10x+11=0.

Solution
A quadratic equation is displayed, reading '2x^2 + 10x + 11 = 0'.
This equation is in standard form. The general form of a quadratic equation, ax^2 + bx + c = 0, is shown above a specific example: 2x^2 + 10x + 11 = 0.
Identify the values of a, b, and c. The image displays the values of three variables: a = 2, b = 10, and c = 11, written in a horizontal line with different colors for each variable and its value.
Write the Quadratic Formula. The image displays the quadratic formula, an equation used to solve quadratic equations for the variable x. It shows x equals negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.
Then substitute in the values of a, b, and c. The quadratic formula is shown with values substituted for solving an equation: x = (- (10) +/- sqrt((10)^2 - 4 * 2 * (11))) / (2 * 2). The numbers 10, 2, and 11 are color-coded in red, cyan, and yellow, respectively.
Simplify. A quadratic formula calculation shows x equals negative 10 plus or minus the square root of 100 minus 88, all divided by 4.
A mathematical equation showing x equals the fraction with a numerator of -10 plus or minus the square root of 12, all divided by 4. This is a common step in the quadratic formula.
Simplify the radical. The equation shows x equals the fraction with numerator -10 plus or minus 2 times the square root of 3, and denominator 4.
Factor out the common factor in the numerator. The equation x equals the fraction 2 multiplied by the quantity -5 plus or minus the square root of 3, all divided by 4, is displayed on a white background.
Remove the common factors. Equation showing x equals negative five plus or minus the square root of three, all divided by two.
Rewrite to show two solutions. Two mathematical expressions for x are displayed. The first is x = (-5 + sqrt(3))/2, and the second is x = (-5 - sqrt(3))/2, representing solutions to a quadratic equation.
Check:
We leave the check for you!

Solve by using the Quadratic Formula: 3m2+12m+7=0.

Solution

m=−6+153,m=−6−153

Solve by using the Quadratic Formula: 5n2+4n−4=0.

Solution

n=−2+265,n=−2−265

When we substitute a, b, and c into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

Solve by using the Quadratic Formula: 3p2+2p+9=0.

Solution
A quadratic equation displayed as 3p^2 + 2p + 9 = 0.
This equation is in standard form Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red and a specific example 3p^2 + 2p + 9 = 0 in black.
Identify the values of a,b,c. The image displays the values of three variables: a=3, b=2, and c=9, written in different colors on a white background.
Write the Quadratic Formula. The quadratic formula, an algebraic formula that provides the solution(s) for a quadratic equation, where p equals negative b plus or minus the square root of b squared minus 4ac, all divided by 2a.
Then substitute in the values of a,b,c. A mathematical expression for 'p' using the quadratic formula, featuring numbers 2, 3, and 9. It shows p = [-2 ×1 sqrt(2^2 - 4 * 3 * 9)] / (2 * 3).
Simplify. A mathematical equation shows p equals a fraction where the numerator is -2 plus or minus the square root of (4 minus 108), and the denominator is 6.
A mathematical equation showing the variable p is equal to a fraction where the numerator is -2 plus or minus the square root of -104, and the denominator is 6.
Simplify the radical using complex numbers. The image displays a mathematical equation for the variable 'p', showing it as a complex number: p = (-2  A plus-minus sign   sqrt(104) i) / 6.
Simplify the radical. The image shows the mathematical equation for p, which is equal to a fraction with -2 plus or minus 2 times the square root of 26i in the numerator, and 6 in the denominator.
Factor the common factor in the numerator. A mathematical equation displaying 'p' equals 2 multiplied by the quantity of negative 1 plus or minus the square root of 26i, all divided by 6.
Remove the common factors. A mathematical equation shows p equals the fraction with a numerator of -1 plus or minus the square root of 26 multiplied by i, and a denominator of 3.
Rewrite in standard a+bi form. The image shows the complex solution for p, expressed as p = -1/3 plus or minus (square root of 26 multiplied by i) all divided by 3, written in a clear, standard mathematical notation.
Write as two solutions. The image displays the two complex conjugate values for 'p': p = -1/3 + (sqrt(26)i)/3 and p = -1/3 - (sqrt(26)i)/3.

Solve by using the Quadratic Formula: 4a2−2a+8=0.

Solution

a=14+314i,a=14−314i

Solve by using the Quadratic Formula: 5b2+2b+4=0.

Solution

b=−15+195i,b=−15−195i

Remember, to use the Quadratic Formula, the equation must be written in standard form, ax2 + bx + c = 0. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Solve by using the Quadratic Formula: x(x+6)+4=0.

Solution

Our first step is to get the equation in standard form.

A mathematical equation is displayed, showing x multiplied by the quantity (x plus 6), plus 4, which equals 0. The equation is x(x+6) + 4 = 0.
Distribute to get the equation in standard form. A mathematical equation is displayed, showing a quadratic expression set equal to zero: x^2 + 6x + 4 = 0. The equation is rendered in a clear, digital font on a plain white background.
This equation is now in standard form Two quadratic equations are displayed: 'ax^2 + bx + c = 0' in red text, followed by 'x^2 + 6x + 4 = 0' in black text below it.
Identify the values of a,b,c. The image displays the values of three variables: a=1, b=6, and c=4, with each variable and its value presented in a distinct color, suggesting a mathematical or programming context.
Write the Quadratic Formula. The quadratic formula: x equals negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.
Then substitute in the values of a,b,c. Mathematical equation showing the quadratic formula with 'b=6' (red), 'a=1' (light blue), and 'c=4' (yellow) color-coded and substituted into the expression x = -(6) ×plusminus ×sqrt{(6)^2 - 4 * 1 * 4} / (2 * 1).
Simplify. The image shows a step in solving a quadratic equation, with x equal to a fraction where the numerator is -6 plus or minus the square root of (36 minus 16), and the denominator is 2.
A mathematical equation is displayed: x equals the fraction with a numerator of -6 plus or minus the square root of 20, and a denominator of 2.
Simplify the radical. A mathematical equation shows x equals the fraction with a numerator of -6 plus or minus 2 times the square root of 5, all divided by 2.
Factor the common factor in the numerator. A mathematical equation shows x equals 2 times the quantity of negative 3 plus or minus the square root of 5, all divided by 2. This represents a solution to a quadratic equation.
Remove the common factors. A mathematical equation is displayed, showing 'x = -3 ×1 ×5' in black font against a white background.
Write as two solutions. The image displays the solutions to a quadratic equation, showing x = -3 + sqrt(5) and x = -3 - sqrt(5).
Check:
We leave the check for you!

Solve by using the Quadratic Formula: x(x+2)−5=0.

Solution

x=−1+6,x=−1−6

Solve by using the Quadratic Formula: 3y(y−2)−3=0.

Solution

y=1+2,y=1−2

When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.

Solve by using the Quadratic Formula: 12u2+23u=13.

Solution

Our first step is to clear the fractions.

A quadratic equation is displayed on a white background, reading one-half u squared plus two-thirds u equals one-third.
Multiply both sides by the LCD, 6, to clear the fractions. A mathematical equation is displayed, showing 6 multiplied by the sum of one-half u squared and two-thirds u, equaling 6 multiplied by one-third.
Multiply. The image displays the mathematical quadratic equation 3u^2 + 4u = 2, presented in a clear, dark gray font against a plain white background.
Subtract 2 to get the equation in standard form. Two lines of equations are displayed. The first line is 'ax^2 + bx + c = 0' in red. The second line, in black, is '3u^2 + 4u - 2 = 0'.
Identify the values of a, b, and c. The image displays the equations a = 3 in light blue, b = 4 in red, and c = 2 in yellow-green, all arranged horizontally on a white background.
Write the Quadratic Formula. The image displays the quadratic formula, an algebraic formula used to solve quadratic equations of the form ax^2 + bx + c = 0, where 'u' represents the variable.
Then substitute in the values of a, b, and c. An example of the quadratic formula in use, with coefficients 'b=4', 'a=3', and 'c=-2' substituted to find the value of 'u'. The values are color-coded for clarity.
Simplify. A mathematical equation, 'u = (-4 plus or minus sqrt(16 + 24)) / 6'
A mathematical equation shows 'u equals negative four plus or minus the square root of forty, all divided by six'.
Simplify the radical. The image shows a mathematical equation where 'u' equals the fraction whose numerator is '-4 plus or minus 2 times the square root of 10' and whose denominator is '6'.
Factor the common factor in the numerator. A mathematical equation is displayed where 'u' is equal to 2 multiplied by the quantity of -2 plus or minus the square root of 10, all divided by 6, set against a plain white background.
Remove the common factors. A mathematical equation is displayed, showing u equals a fraction: negative 2 plus or minus the square root of 10, all divided by 3.
Rewrite to show two solutions. Two mathematical expressions are shown for the variable u. The first is u equals negative two plus the square root of ten, all divided by three. The second is u equals negative two minus the square root of ten, all divided by three.
Check:
We leave the check for you!

Solve by using the Quadratic Formula: 14c2−13c=112.

Solution

c=2+73,c=2−73

Solve by using the Quadratic Formula: 19d2−12d=−13.

Solution

d=9+334,d=9−334

Think about the equation (x − 3)2 = 0. We know from the Zero Product Property that this equation has only one solution,
x = 3.

We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to 0 gives just one solution. Notice that once the radicand is simplified it becomes 0 , which leads to only one solution.

Solve by using the Quadratic Formula: 4x2−20x=−25.

Solution
A quadratic equation, 4x^2 - 20x = -25, is displayed on a white background.
Add 25 to get the equation in standard form. A quadratic equation in its general form (ax² + bx + c = 0) is shown, followed by a specific example (4x² - 20x + 25 = 0).
Identify the values of a, b, and c. The image displays mathematical variable assignments: a = 4, b = 20, c = 25, shown in blue, red, and yellow text, respectively, against a white background.
Write the quadratic formula. The quadratic formula, x = (-b ×1 sqrt(b^2 - 4ac)) / 2a, is displayed in black text on a white background.
Then substitute in the values of a, b, and c. An image showing the quadratic formula x = -b ± sqrt(b^2 - 4ac) / 2a with values b=-20, a=4, and c=25 substituted, displaying an intermediate step in solving a quadratic equation.
Simplify. A mathematical equation shows x equals the fraction with a numerator of 20 plus or minus the square root of 400 minus 400, and a denominator of 8.
A mathematical equation showing x equals twenty plus or minus the square root of zero, all divided by eight.
Simplify the radical. A mathematical equation shows 'x = 20/8' written in black against a white background.
Simplify the fraction. The image displays the mathematical equation 'x = 5/2' in black text against a plain white background, showing the value of the variable x as a fraction.
Check:
We leave the check for you!

Did you recognize that 4x2 − 20x + 25 is a perfect square trinomial. It is equivalent to (2x − 5)2? If you solve
4x2 − 20x + 25 = 0 by factoring and then using the Square Root Property, do you get the same result?

Solve by using the Quadratic Formula: r2+10r+25=0.

Solution

r=−5

Solve by using the Quadratic Formula: 25t2−40t=−16.

Solution

t=45

Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant.

Discriminant

In the Quadratic Formula, x equals the quotient of negative b plus or minus the square root of b squared minus 4 times a times c and 2 a, the value under the radical, b squared minus 4 times a times c, is called the discriminant.

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.

Quadratic Equation
(in standard form)
Discriminant
b2−4ac
Value of the Discriminant Number and Type of solutions
2x2+9x−5=0 92−4·2(−5)121 + 2 real
4x2−20x+25=0 (−20)2−4·4·250 0 1 real
3p2+2p+9=0 22−4·3·9−104 − 2 complex
When the value under the radical in the Quadratic Formula, the discriminant, is positive, the equation has two real solutions. When the value under the radical in the Quadratic Formula, the discriminant, is zero, the equation has one real solution. When the value under the radical in the Quadratic Formula, the discriminant, is negative, the equation has two complex solutions.

Using the Discriminant, b2 − 4ac, to Determine the Number and Type of Solutions of a Quadratic Equation

For a quadratic equation of the form ax2 + bx + c = 0, a≠0,

  • If b2 − 4ac > 0, the equation has 2 real solutions.
  • if b2 − 4ac = 0, the equation has 1 real solution.
  • if b2 − 4ac < 0, the equation has 2 complex solutions.

Determine the number of solutions to each quadratic equation.

ⓐ 3x2+7x−9=0 ⓑ 5n2+n+4=0 ⓒ 9y2−6y+1=0.

Solution

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

ⓐ
Steps demonstrating the calculation of the discriminant for the quadratic equation 3x^2 + 7x - 9 = 0.
3x2+7x−9=0
The equation is in standard form, identify a, b, and c. a=3,b=7,c=−9
Write the discriminant. b2−4ac
Substitute in the values of a, b, and c. (7)2−4·3·(−9)
Simplify. 49+108
157

Since the discriminant is positive, there are 2 real solutions to the equation.

ⓑ
Step-by-step calculation of the discriminant for the quadratic equation 5n^2 + n + 4 = 0.
5n2+n+4=0
The equation is in standard form, identify a, b, and c. a=5,b=1,c=4
Write the discriminant. b2−4ac
Substitute in the values of a, b, and c. (1)2−4·5·4
Simplify. 1−80
−79

Since the discriminant is negative, there are 2 complex solutions to the equation.

ⓒ
Step-by-step calculation of the discriminant for the quadratic equation 9y^2 - 6y + 1 = 0.
9y2−6y+1=0
The equation is in standard form, identify a, b, and c. a=9,b=−6,c=1
Write the discriminant. b2−4ac
Substitute in the values of a, b, and c. (−6)2−4·9·1
Simplify. 36−36
0

Since the discriminant is 0, there is 1 real solution to the equation.

Determine the numberand type of solutions to each quadratic equation.

ⓐ 8m2−3m+6=0 ⓑ 5z2+6z−2=0 ⓒ 9w2+24w+16=0.

Solution

ⓐ 2 complex solutions; ⓑ 2 real solutions; ⓒ 1 real solution

Determine the number and type of solutions to each quadratic equation.

ⓐ b2+7b−13=0 ⓑ 5a2−6a+10=0 ⓒ 4r2−20r+25=0.

Solution

ⓐ 2 real solutions; ⓑ 2 complex solutions; ⓒ 1 real solution

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We summarize the four methods that we have used to solve quadratic equations below.

Methods for Solving Quadratic Equations

  1. Factoring
  2. Square Root Property
  3. Completing the Square
  4. Quadratic Formula

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is ax2=k or a(x−h)2=k we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.

Identify the most appropriate method to solve a quadratic equation.

  1. Try Factoring first. If the quadratic factors easily, this method is very quick.
  2. Try the Square Root Property next. If the equation fits the form ax2=k or a(x−h)2=k, it can easily be solved by using the Square Root Property.
  3. Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.

The next example uses this strategy to decide how to solve each quadratic equation.

Identify the most appropriate method to use to solve each quadratic equation.

ⓐ 5z2=17 ⓑ 4x2−12x+9=0 ⓒ 8u2+6u=11.

Solution

ⓐ
5z2=17

Since the equation is in the ax2=k, the most appropriate method is to use the Square Root Property.

ⓑ
4x2−12x+9=0

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

ⓒ
Demonstrates converting the quadratic equation 8u^2 + 6u = 11 to its standard form.
8u2+6u=11
Put the equation in standard form. 8u2+6u−11=0

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.

Identify the most appropriate method to use to solve each quadratic equation.

ⓐ x2+6x+8=0 ⓑ (n−3)2=16 ⓒ 5p2−6p=9.

Solution

ⓐ factoring; ⓑ Square Root Property; ⓒ Quadratic Formula

Identify the most appropriate method to use to solve each quadratic equation.

ⓐ 8a2+3a−9=0 ⓑ 4b2+4b+1=0 ⓒ 5c2=125.

Solution

ⓐ Quadratic Forumula;
ⓑ Factoring or Square Root Property ⓒ Square Root Property

Access these online resources for additional instruction and practice with using the Quadratic Formula.

  • Using the Quadratic Formula
  • Solve a Quadratic Equation Using the Quadratic Formula with Complex Solutions
  • Discriminant in Quadratic Formula

Key Concepts

  • Quadratic Formula
    • The solutions to a quadratic equation of the form ax2 + bx + c = 0, a≠0 are given by the formula:
      x=−b±b2−4ac2a
  • How to solve a quadratic equation using the Quadratic Formula.
    1. Write the quadratic equation in standard form, ax2 + bx + c = 0. Identify the values of a, b, c.
    2. Write the Quadratic Formula. Then substitute in the values of a, b, c.
    3. Simplify.
    4. Check the solutions.
  • Using the Discriminant, b2 − 4ac, to Determine the Number and Type of Solutions of a Quadratic Equation
    • For a quadratic equation of the form ax2 + bx + c = 0, a≠0,
      • If b2 − 4ac > 0, the equation has 2 real solutions.
      • if b2 − 4ac = 0, the equation has 1 real solution.
      • if b2 − 4ac < 0, the equation has 2 complex solutions.
  • Methods to Solve Quadratic Equations:
    • Factoring
    • Square Root Property
    • Completing the Square
    • Quadratic Formula
  • How to identify the most appropriate method to solve a quadratic equation.
    1. Try Factoring first. If the quadratic factors easily, this method is very quick.
    2. Try the Square Root Property next. If the equation fits the form ax2 = k or a(x − h)2 = k, it can easily be solved by using the Square Root Property.
    3. Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.

Practice Makes Perfect

Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

4m2+m−3=0

Solution

m=−1,m=34

4n2−9n+5=0

2p2−7p+3=0

Solution

p=12,p=3

3q2+8q−3=0

p2+7p+12=0

Solution

p=−4,p=−3

q2+3q−18=0

r2−8r=33

Solution

r=−3,r=11

t2+13t=−40

3u2+7u−2=0

Solution

u=−7±736

2p2+8p+5=0

2a2−6a+3=0

Solution

a=3±32

5b2+2b−4=0

x2+8x−4=0

Solution

x=−4±25

y2+4y−4=0

3y2+5y−2=0

Solution

y=−2,y=13

6x2+2x−20=0

2x2+3x+3=0

Solution

x=−34±154i

2x2−x+1=0

8x2−6x+2=0

Solution

x=38±78i

8x2−4x+1=0

(v+1)(v−5)−4=0

Solution

v=2±13

(x+1)(x−3)=2

(y+4)(y−7)=18

Solution

y=3±1932

(x+2)(x+6)=21

13m2+112m=14

Solution

m=−1,m=34

13n2+n=−12

34b2+12b=38

Solution

b=−2±226

19c2+23c=3

16c2+24c+9=0

Solution

c=−34

25d2−60d+36=0

25q2+30q+9=0

Solution

q=−35

16y2+8y+1=0

Use the Discriminant to Predict the Number of Real Solutions of a Quadratic Equation

In the following exercises, determine the number of real solutions for each quadratic equation.

ⓐ 4x2−5x+16=0 ⓑ 36y2+36y+9=0 ⓒ 6m2+3m−5=0

Solution

ⓐ no real solutions ⓑ 1
ⓒ 2

ⓐ 9v2−15v+25=0 ⓑ 100w2+60w+9=0 ⓒ 5c2+7c−10=0

ⓐ r2+12r+36=0 ⓑ 8t2−11t+5=0 ⓒ 3v2−5v−1=0

Solution

ⓐ 1 ⓑ no real solutions
ⓒ 2

ⓐ 25p2+10p+1=0 ⓑ 7q2−3q−6=0 ⓒ 7y2+2y+8=0

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.


ⓐ x2−5x−24=0
ⓑ (y+5)2=12
ⓒ 14m2+3m=11

Solution


ⓐ factor
ⓑ square root
ⓒ Quadratic Formula


ⓐ (8v+3)2=81
ⓑ w2−9w−22=0
ⓒ 4n2−10n=6


ⓐ 6a2+14a=20
ⓑ (x−14)2=516
ⓒ y2−2y=8

Solution


ⓐ Quadratic Formula
ⓑ square root
ⓒ factor


ⓐ 8b2+15b=4
ⓑ 59v2−23v=1
ⓒ (w+43)2=29

Writing Exercises

Solve the equation x2+10x=120

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?

Solution

Answers will vary.

Solve the equation 12y2+23y=24

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve quadratic equations using the quadratic formula.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can use the discriminant to predict the number of solutions of a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can identify the most appropriate method to use to solve a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

discriminant
In the Quadratic Formula, x=−b±b2−4ac2a, the quantity b2 − 4ac is called the discriminant.

Solve Equations in Quadratic Form

Learning Objectives

By the end of this section, you will be able to:

  • Solve equations in quadratic form

Before you get started, take this readiness quiz.

Factor by substitution: y4−y2−20.
If you missed this problem, review Example 13 in Factor Trinomials.

Solution

y2+4y2−5

Factor by substitution: (y−4)2+8(y−4)+15.
If you missed this problem, review Example 14 in Factor Trinomials.

Solution

(y−1)(y+1)

Simplify: ⓐ x12·x14 ⓑ (x13)2 ⓒ (x−1)2.
If you missed this problem, review Example 8 in Simplify Rational Exponents.

Solution

ⓐ x34; ⓑ x23; ⓒ x−2

Solve Equations in Quadratic Form

Sometimes when we factored trinomials, the trinomial did not appear to be in the ax2 + bx + c form. So we factored by substitution allowing us to make it fit the ax2 + bx + c form. We used the standard u for the substitution.

To factor the expression x4 − 4x2 − 5, we noticed the variable part of the middle term is x2 and its square, x4, is the variable part of the first term. (We know (x2)2=x4.) So we let u = x2 and factored.

The image displays the mathematical expression x^4 - 4x^2 - 5, a polynomial equation often seen in algebra and calculus.
The mathematical expression (x^2)^2 - 4(x^2) - 5 is shown, representing a quadratic form in terms of x^2. The 'x' variable and its exponents are highlighted in red.
Let u=x2 and substitute. The image shows the mathematical expression u^2 - 4u - 5, where 'u' is highlighted in red, indicating it as a variable.
Factor the trinomial. The mathematical expression (u + 1)(u - 5) is displayed on a white background.
Replace u with x2. A mathematical expression showing the product of two binomials: (x^2 + 1)(x^2 - 5). The variable 'x' is in red, while the numbers, operators, and parentheses are in black.

Similarly, sometimes an equation is not in the ax2 + bx + c = 0 form but looks much like a quadratic equation. Then, we can often make a thoughtful substitution that will allow us to make it fit the ax2 + bx + c = 0 form. If we can make it fit the form, we can then use all of our methods to solve quadratic equations.

Notice that in the quadratic equation ax2 + bx + c = 0, the middle term has a variable, x, and its square, x2, is the variable part of the first term. Look for this relationship as you try to find a substitution.

Again, we will use the standard u to make a substitution that will put the equation in quadratic form. If the substitution gives us an equation of the form ax2 + bx + c = 0, we say the original equation was of quadratic form.

The next example shows the steps for solving an equation in quadratic form.

How to Solve Equations in Quadratic Form

Solve: 6x4−7x2+2=0

Solution
Step 1 is to identify a substitution that will put the equation in quadratic form. Look at the equation 6 x to the fourth power minus 7 x squared plus 2 equals 0. Since the square of x squared equals x to the fourth, let u equals x squared. Step 2 is to rewrite the equation with the substitution to put it in quadratic form. Rewrite the equation to prepare for the substitution to show 6 times the square of x squared minus 7 times x squared plus 2 equals 0. Substitute u equals x squared to get the new equation 6 times u squared minus 7 u plus 2 equals 0. Step 3 is to solve the quadratic equation for u. We can solve by factoring, so rewrite the equation as the product of 2 u minus 1 and 3 u minus 2 equals 0. Use the Zero Product Property to create 2 equations. If 2 u minus 1 equals 0, then 2u equals one, so u equals one half. If 3 u minus 2 equals 0, then 3 u equals 2 and u equals two thirds. In step 4, substitute the original variable back into the results. In this case replace u with x squared. So u equals one half becomes x squared equals one half and u equals two thirds becomes x squared equals two thirds. Step 5 is to solve for the original variable, so use the Square Root Property to solve for x. If x squared equals one half, then x equals the positive or negative square root of one half. Rationalize the denominator to see that x equals the positive or negative square root of 2 divided by 2. If x squared equals two thirds, then x equals the positive or negative square root of two thirds. Rationalize the denominator to see that x equals the positive or negative square root of 6 divided by 3. In step 6, check your solutions. We will show one check here, x equals square root 2 divided by 2. Substitute this value into the original equation. 6 times the fourth power of the quotient square root 2 divided by 2 minus 7 times the square of the quotient square root of 2 divided by 2 plus 2. Does this expression equal 0? Simplify the powers. 6 times four sixteenths minus 7 times two fourths plus 2. Simplify terms. Three halves minus seven halves plus four halves equals zero. Square root 2 divided by 2 is a solution. We leave the other checks to you!

Solve: x4−6x2+8=0.

Solution

x=2,x=−2,x=2,x=−2

Solve: x4−11x2+28=0.

Solution

x=7,x=−7,x=2,x=−2

We summarize the steps to solve an equation in quadratic form.

Solve equations in quadratic form.

  1. Identify a substitution that will put the equation in quadratic form.
  2. Rewrite the equation with the substitution to put it in quadratic form.
  3. Solve the quadratic equation for u.
  4. Substitute the original variable back into the results, using the substitution.
  5. Solve for the original variable.
  6. Check the solutions.

In the next example, the binomial in the middle term, (x − 2) is squared in the first term. If we let u = x − 2 and substitute, our trinomial will be in ax2 + bx + c form.

Solve: (x−2)2+7(x−2)+12=0.

Solution
A quadratic equation (x-2)^2 + 7(x-2) + 12 = 0 is displayed on a white background, representing a math problem typically solved using substitution.
Prepare for the substitution. Equation showing the expression (x minus 2) squared plus 7 times (x minus 2) plus 12 equals zero.
Let u=x−2 and substitute. A quadratic equation displayed on a white background: u^2 + 7u + 12 = 0. The variable 'u' appears in red, while the numbers, exponents, and operators are in black.
Solve by factoring. A mathematical equation is displayed against a white background. The equation reads as follows: (u + 3)(u + 4) = 0.
Two mathematical equations are displayed: u + 3 = 0, u + 4 = 0, shown in a white background with black text.
The image displays two mathematical expressions: u = -3, and u = -4, written in a clear, sans-serif font against a plain white background.
Replace u with x−2. Two mathematical equations are displayed: x - 2 = -3 and x - 2 = -4. These are linear equations where a variable 'x' is involved in a subtraction operation resulting in a negative integer.
Solve for x. Two mathematical equations are displayed on a white background: 'X = -1' followed by a comma, and then 'X = -2'.
Check:

Verification of solutions for a quadratic equation (x-2)^2 + 7(x-2) + 12 = 0, showing that both x=-1 and x=-2 satisfy the equation.

Solve: (x−5)2+6(x−5)+8=0.

Solution

x=3,x=1

Solve: (y−4)2+8(y−4)+15=0.

Solution

y=−1,y=1

In the next example, we notice that (x)2=x. Also, remember that when we square both sides of an equation, we may introduce extraneous roots. Be sure to check your answers!

Solve: x−3x+2=0.

Solution

The x in the middle term, is squared in the first term (x)2=x. If we let u=x and substitute, our trinomial will be in ax2 + bx + c = 0 form.

A mathematical equation is displayed: x - 3 times the square root of x + 2 = 0. This is a radical equation that can be solved by substitution, letting y = square root of x.
Rewrite the trinomial to prepare for the substitution. A quadratic equation, (sqrt(x))^2 - 3sqrt(x) + 2 = 0, is shown, with the square root of x highlighted in red.
Let u=x and substitute. A white background displays the quadratic equation u^2 - 3u + 2 = 0 in horizontal alignment. The variable 'u' is rendered in red, while the numbers, operators, and equals sign are black.
Solve by factoring. A mathematical equation is displayed on a white background: (u-2)(u-1)=0.
Two mathematical equations are displayed: u - 2 = 0, u - 1 = 0, separated by a comma. The equations are presented in a clear, sans-serif font against a plain white background.
The image displays mathematical notation 'u = 2, u = 1' in black text against a white background.
Replace u with x. Two mathematical equations are displayed: square root of x equals 2, and square root of x equals 1.
Solve for x, by squaring both sides. The text 'x = 4, x = 1' is shown against a white background.
Check:

Checking the solutions x=4 and x=1 for the equation x - 3sqrt(x) + 2 = 0, demonstrating that both values correctly satisfy the equality.

Solve: x−7x+12=0.

Solution

x=9,x=16

Solve: x−6x+8=0.

Solution

x=4,x=16

Substitutions for rational exponents can also help us solve an equation in quadratic form. Think of the properties of exponents as you begin the next example.

Solve: x23−2x13−24=0.

Solution

The x13 in the middle term is squared in the first term (x13)2=x23. If we let u=x13 and substitute, our trinomial will be in ax2 + bx + c = 0 form.

A mathematical equation is displayed, reading 'x to the power of 2/3 minus 2x to the power of 1/3 minus 24 equals 0'.
Rewrite the trinomial to prepare for the substitution. A mathematical equation is displayed: (x raised to the power of 1/3) squared minus 2 times (x raised to the power of 1/3) minus 24 equals 0.
Let u=x13 and substitute. A quadratic equation is displayed, showing u^2 - 2u - 24 = 0. The variable 'u' and its coefficient '2' in the second term are highlighted in red.
Solve by factoring. A mathematical equation in factored form is shown against a white background: (u - 6)(u + 4) = 0.
Two mathematical equations are displayed: u - 6 = 0 and u + 4 = 0. These linear equations show simple algebraic expressions involving the variable 'u' and constants, both set to zero.
Two mathematical expressions are shown: u = 6, and u = -4.
Replace u with x13. Two mathematical equations are shown: x^(1/3) = 6, and x^(1/3) = -4. These represent two separate conditions for x raised to the power of one-third.
Solve for x by cubing both sides. Two algebraic equations are displayed: (x^(1/3))^3 = (6)^3 and (x^(1/3))^3 = (-4)^3, illustrating how to solve for x when involving cube roots and cubing.
Two mathematical expressions are displayed on a white background: 'x = 216,' and 'x = -64'.
Check:

This image demonstrates verifying the solutions x=216 and x=-64 for the equation x^(2/3) - 2x^(1/3) - 24 = 0 by substituting them into the equation, both resulting in 0=0.

Solve: x23−5x13−14=0.

Solution

x=−8,x=343

Solve: x12−8x14+15=0.

Solution

x=81,x=625

In the next example, we need to keep in mind the definition of a negative exponent as well as the properties of exponents.

Solve: 3x−2−7x−1+2=0.

Solution

The x−1 in the middle term is squared in the first term (x−1)2=x−2. If we let u=x−1 and substitute, our trinomial will be in ax2 + bx + c = 0 form.

A mathematical equation displays '3x^(-2) - 7x^(-1) + 2 = 0' in a simple, clear font against a white background.
Rewrite the trinomial to prepare for the substitution. A mathematical equation is displayed: 3(x⁻¹)² - 7(x⁻¹) + 2 = 0. The variable 'x' is shown in red, with an exponent of -1. The equation features a quadratic form with respect to x⁻¹.
Let u=x−1 and substitute. A quadratic equation is displayed on a white background: 3u^2 - 7u + 2 = 0. The coefficients for the u-squared and u terms, '3u' and '7u', are highlighted in red, contrasting with the black numbers and operators.
Solve by factoring. A mathematical equation where the product of two binomials, (3u - 1) and (u - 2), is set equal to zero.
Two separate linear equations, 3u - 1 = 0 and u - 2 = 0, are shown in white text against a dark background, representing a mathematical problem to solve for the variable 'u'.
Two mathematical expressions are displayed: u = 1/3, and u = 2.
Replace u with x−1. Two mathematical equations are displayed: x^-1 = 1/3, and x^-1 = 2. Both equations show an inverse power of x set equal to a constant value, with the first constant being a fraction and the second an integer.
Solve for x by taking the reciprocal since x−1=1x. Two mathematical expressions are displayed: 'x = 3,' and 'x = 1/2'. These represent potential solutions or values for the variable x.
Check:

Two algebraic problems demonstrate the verification of solutions for the equation 3x^-2 - 7x^-1 + 2 = 0, with x=3 and x=1/2 both resulting in 0=0.

Solve: 8x−2−10x−1+3=0.

Solution

x=43x=2

Solve: 6x−2−23x−1+20=0.

Solution

x=25,x=34

Access this online resource for additional instruction and practice with solving quadratic equations.

  • Solving Equations in Quadratic Form

Key Concepts

  • How to solve equations in quadratic form.
    1. Identify a substitution that will put the equation in quadratic form.
    2. Rewrite the equation with the substitution to put it in quadratic form.
    3. Solve the quadratic equation for u.
    4. Substitute the original variable back into the results, using the substitution.
    5. Solve for the original variable.
    6. Check the solutions.

Practice Makes Perfect

Solve Equations in Quadratic Form

In the following exercises, solve.

x4−7x2+12=0

Solution

x=±3,x=±2

x4−9x2+18=0

x4−13x2−30=0

Solution

x=±15,x=±2i

x4+5x2−36=0

2x4−5x2+3=0

Solution

x=±1,x=±62

4x4−5x2+1=0

2x4−7x2+3=0

Solution

x=±3,x=±22

3x4−14x2+8=0

(x−3)2−5(x−3)−36=0

Solution

x=−1,x=12

(x+2)2−3(x+2)−54=0

(3y+2)2+(3y+2)−6=0

Solution

x=−53,x=0

(5y−1)2+3(5y−1)−28=0

(x2+1)2−5(x2+1)+4=0

Solution

x=0,x=±3

(x2−4)2−4(x2−4)+3=0

2(x2−5)2−5(x2−5)+2=0

Solution

x=±222,x=±7

2(x2−5)2−7(x2−5)+6=0

x−x−20=0

Solution

x=25

x−8x+15=0

x+6x−16=0

Solution

x=4

x+4x−21=0

6x+x−2=0

Solution

x=14

6x+x−1=0

10x−17x+3=0

Solution

x=125, x=94

12x+5x−3=0

x23+9x13+8=0

Solution

x=−1,x=−512

x23−3x13=28

x23+4x13=12

Solution

x=8,x=−216

x23−11x13+30=0

6x23−x13=12

Solution

x=278,x=−6427

3x23−10x13=8

8x23−43x13+15=0

Solution

x=27512,x=125

20x23−23x13+6=0

x−8x12+7=0

Solution

x=1,x=49

2x−7x12=15

6x−2+13x−1+5=0

Solution

x=−2,x=−35

15x−2−26x−1+8=0

8x−2−2x−1−3=0

Solution

x=−2,x=43

15x−2−4x−1−4=0

Writing Exercises

Explain how to recognize an equation in quadratic form.

Solution

Answers will vary.

Explain the procedure for solving an equation in quadratic form.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve equations in quadratic form.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Solve Applications of Quadratic Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

The sum of two consecutive odd numbers is −100. Find the numbers.
If you missed this problem, review Example 5 in Use a Problem Solving Strategy.

Solution

−51,−49

Solve: 2x+1+1x−1=1x2−1.
If you missed this problem, review Example 3 in Solve Rational Equations.

Solution

x=23

Find the length of the hypotenuse of a right triangle with legs 5 inches and 12 inches.
If you missed this problem, review Example 7 in Solve a Formula for a Specific Variable.

Solution

13inches

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

  1. Factoring
  2. Square Root Property
  3. Completing the Square
  4. Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy.

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are looking for.
  3. Name what we are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
  5. Solve the equation using algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one n, then the next one is n + 2. The next one would be n + 2 + 2 or n + 4. This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

Consecutive even integersConsecutive odd integers 64,66,6877,79,81 n1steven integern1stodd integer n+22ndconsecutive even integern+22ndconsecutive odd integer n+43rdconsecutive even integern+43rdconsecutive odd integer

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

The product of two consecutive odd integers is 195. Find the integers.

Solution
This table provides a seven-step guide for solving a word problem to find two consecutive odd integers whose product is 195, detailing each stage from problem understanding to solution verification.
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for two consecutive odd integers.
Step 3. Name what we are looking for. Let n= the first odd integer.
n+2= the next odd integer
Step 4. Translate into an equation. State the problem in one sentence. “The product of two consecutive odd integers is 195.”
The product of the first odd integer and the second odd integer is 195.
Translate into an equation. n(n+2)=195n2+2n=195
Step 5. Solve the equation. Distribute.
Write the equation in standard form.
Factor.
n2+2n−195=0(n+15)(n−13)=0
Use the Zero Product Property.
Solve each equation.
n+15=0n−13=0n=−15,n=13
There are two values of n that are solutions. This will give us two pairs of consecutive odd integers for our solution.
First odd integern=13First odd integern=−15next odd integern+2next odd integern+213+2−15+215−13
Step 6. Check the answer.
Do these pairs work?
Are they consecutive odd integers?
13,15yes−13,−15yes
Is their product 195?
13·15=195yes−13(−15)=195yes
Step 7. Answer the question. Two consecutive odd integers whose product is 195 are 13, 15 and −13, −15.

The product of two consecutive odd integers is 99. Find the integers.

Solution

The two consecutive odd integers whose product is 99 are 9, 11, and −9, −11

The product of two consecutive even integers is 168. Find the integers.

Solution

The two consecutive even integers whose product is 128 are 12, 14 and −12, −14.

We will use the formula for the area of a triangle to solve the next example.

Area of a Triangle

For a triangle with base, b, and height, h, the area, A, is given by the formula A=12bh.

Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.

Recall that when we solve geometric applications, it is helpful to draw the figure.

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of 120 square feet and the architect wants the base to be 4 feet more than twice the height. Find the base and height of the window.

Solution
Step 1. Read the problem.
Draw a picture.
A geometric diagram depicts a triangle with a height 'h' drawn from its apex to its base, dividing it into two right-angled triangles. The total length of the base is labeled '2h + 4'.
Step 2. Identify what we are looking for. We are looking for the base and height.
Step 3. Name what we are looking for. Let h = the height of the triangle.
2h + 4 = the base of the triangle
Step 4. Translate into an equation.
We know the area. Write the
formula for the area of a triangle.
A=12bh
Step 5. Solve the equation.
Substitute in the values.
120=12(2h+4)h
Distribute. 120=h2+2h
This is a quadratic equation, rewrite it in standard form. h2+2h−120=0
Factor. (h−10)(h+12)=0
Use the Zero Product Property. h−10=0h+12=0
Simplify. h=10,h=−12
Since h is the height of a window, a value of h = −12 does not make sense.
The height of the triangle h=10.
The base of the triangle 2h+4.
2·10+4
24
Step 6. Check the answer.
Does a triangle with height 10 and base 24 have area 120? Yes.
Step 7. Answer the question. The height of the triangular window is 10 feet and the base is 24 feet.

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of 456 square inches.

Solution

The height of the triangle is 12 inches and the base is 76 inches.

If a triangle that has an area of 110 square feet has a base that is two feet less than twice the height, what is the length of its base and height?

Solution

The height of the triangle is 11 feet and the base is 20 feet.

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Area of a Rectangle

For a rectangle with length, L, and width, W, the area, A, is given by the formula A = LW.

Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.

Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than 3 times the width. Find the length and width. Round to the nearest tenth of a foot.

Solution
Step 1. Read the problem.
Draw a picture.
A rectangle is shown with its width labeled as 'w' and its length labeled as '3w - 1'.
Step 2. Identify what we are looking for. We are looking for the length and width.
Step 3. Name what we are looking for. Let w= the width of the rectangle.
3w−1= the length of the rectangle
Step 4. Translate into an equation.
We know the area. Write the formula for the area of a rectangle.
Area equals length multiplied by width. A = L * W.
Step 5. Solve the equation. Substitute in the values. A mathematical equation is displayed, reading 150 = (3w - 1)w. The numbers and variables are in a dark gray font against a white background.
Distribute. A mathematical equation is displayed on a white background: '150 = 3w^2 - W'.
This is a quadratic equation; rewrite it in standard form.
Solve the equation using the Quadratic Formula.
Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red, and a specific example 3w^2 - w - 150 = 0 in black, both set equal to zero.
Identify the a,b,c values. The image displays the values of three variables: a = 3, b = -1, and c = -150, written in a horizontal line with different colors for each assignment. The text is on a white background.
Write the Quadratic Formula. A mathematical formula for W, showing W equals negative b plus or minus the square root of b squared minus 4ac, all divided by 2q. This is a variation of the quadratic formula.
Then substitute in the values of a,b,c. The quadratic formula is shown calculating W: W = [-(-1) ± √((-1)² - 4·3·(-150))] / (2·3), with 'a' = 3, 'b' = -1, and 'c' = -150 highlighted.
Simplify. A mathematical equation is displayed on a white background, reading 'W = (1 plus or minus the square root of 1 + 1800) divided by 6'.
A mathematical equation displaying W equals the fraction 1 plus or minus the square root of 1801, all divided by 6.
Rewrite to show two solutions. Two mathematical expressions are shown for W: W = (1 + sqrt(1801))/6 and W = (1 - sqrt(1801))/6.
Approximate the answers using a calculator.
We eliminate the negative solution for the width.
Mathematical calculations showing the length of an object. The width 'w' is approximately 7.2. The length is calculated as 3w - 1, substituting 7.2 for 'w' to get approximately 20.6. A crossed-out value of w ≈ 6.9 is also visible.
Step 6. Check the answer.
Make sure that the answers make sense. Since the
answers are approximate, the area will not come
out exactly to 150.
Step 7. Answer the question. The width of the rectangle is
approximately 7.2 feet and the
length is approximately 20.6 feet.

The length of a 200 square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

Solution

The length of the garden is approximately 18 feet and the width 11 feet.

A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot.?

Solution

The length of the tablecloth is approximatel 11.8 feet and the width 6.8 feet.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Pythagorean Theorem

In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, a2 + b2 = c2.

Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Solution
Step 1. Read the problem. Draw a picture. An isosceles triangle with an altitude drawn from the apex to the base. One slanted side is labeled '10', and downward arrows mark the base vertices.
Step 2. Identify what we are looking for. We are looking for the height of the pole.
Step 3. Name what we are looking for. The distance from the base of the pole to either stake is the same as the height of the pole.

Let x= the height of the pole.
x= the distance from pole to stake

Each side is a right triangle. We draw a picture of one of them.
A right-angled triangle with two legs of length 'x' and a hypotenuse of length '10'.
Step 4. Translate into an equation.
We can use the Pythagorean Theorem to solve for x.
Write the Pythagorean Theorem.
a2+b2=c2
Step 5. Solve the equation. Substitute. x2+x2=102
Simplify. 2x2=100
Divide by 2 to isolate the variable. 2x22=1002
Simplify. x2=50
Use the Square Root Property. x=±50
Simplify the radical. x=±52
Rewrite to show two solutions. x=52,x=−52
If we approximate this number to the
nearest tenth with a calculator, we find
x≈7.1.
Step 6. Check the answer.
Check on your own in the Pythagorean Theorem.
Step 7. Answer the question. The pole should be about 7.1 feet tall.

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is 20 feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

Solution

The length of the flag pole’s shadow is approximately 6.3 feet and the height of the flag pole is 18.9 feet.

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

Solution

The distance between the opposite corners is approximately 7.2 feet.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, v0, propels the object up until gravity causes the object to fall back down.

Projectile motion

The height in feet, h , of an object shot upwards into the air with initial velocity, v0, after t seconds is given by the formula

h=−16t2+v0t

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.

Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for. We are looking for the number of
seconds, which is time.
Step 3. Name what we are looking for. Let t= the number of seconds.
Step 4. Translate into an equation. Use the formula. The mathematical formula h = -16t^2 + v₀t, commonly used to calculate the height (h) of a projectile at a given time (t), considering initial vertical velocity (v₀) and the effect of gravity.
Step 5. Solve the equation.
We know the velocity v0 is 130 feet per second.
The height is 260 feet. Substitute the values.
A mathematical equation is displayed, reading 260 = -16t^2 + 130t, set against a plain white background.
This is a quadratic equation, rewrite it in standard form.
Solve the equation using the Quadratic Formula.
The general quadratic equation ax^2 + bx + c = 0 is shown alongside a specific example: 16t^2 - 130t + 260 = 0.
Identify the values of a,b,c. The values a = 16, b = -130, and c = 260 are displayed in colorful text against a white background.
Write the Quadratic Formula. The quadratic formula showing the solution for t equals negative b plus or minus the square root of b squared minus four a c all divided by two a. It is used to solve quadratic equations.
Then substitute in the values of a,b,c. A mathematical equation displays the quadratic formula with specific numerical values for a, b, and c, used to solve for 't'. The formula is t = [-(-130) ×1 (sqrt((-130)^2 - 4*16*260))] / (2*16).
Simplify. A mathematical equation calculates 't' using a quadratic formula, involving the square root of 16,900 minus 16,640, added or subtracted from 130, all divided by 32.
A mathematical equation is displayed, showing 't = (130 ± √260) / 32' on a white background.
Rewrite to show two solutions. Two mathematical expressions are displayed on a white background, showing the variable 't' defined by two different fractions: t = (130 + sqrt(260)) / 32 and t = (130 - sqrt(260)) / 32.
Approximate the answer with a calculator. Text displays 't   4.6 seconds, t   3.6 seconds' on a white background.
Step 6. Check the answer.
The check is left to you.
Step 7. Answer the question. The firework will go up and then fall back
down. As the firework goes up, it will
reach 260 feet after approximately 3.6
seconds. It will also pass that height on
the way down at 4.6 seconds.

An arrow is shot from the ground into the air at an initial speed of 108 ft/s. Use the formula h = −16t2 + v0t to determine when the arrow will be 180 feet from the ground. Round the nearest tenth.

Solution

The arrow will reach 180 feet on its way up after 3 seconds and again on its way down after approximately 3.8 seconds.

A man throws a ball into the air with a velocity of 96 ft/s. Use the formula h = −16t2 + v0t to determine when the height of the ball will be 48 feet. Round to the nearest tenth.

Solution

The ball will reach 48 feet on its way up after approximately .6 second and again on its way down after approximately 5.4 seconds.

We have solved uniform motion problems using the formula D = rt in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

Image shows the template for a table with three rows and four columns. The first column is empty. The second column is labeled “Rate.” The third column is labeled “Time.” The fourth column is labeled “Distance.” The labels are written in the equation Rate times Time equals Distance. There is one extra cell at the bottom of the fourth column.

The formula D = rt assumes we know r and t and use them to find D. If we know D and r and need to find t, we would solve the equation for t and get the formula t=Dr.

Some uniform motion problems are also modeled by quadratic equations.

Professor Smith just returned from a conference that was 2,000 miles east of his home. His total time in the airplane for the round trip was 9 hours. If the plane was flying at a rate of 450 miles per hour, what was the speed of the jet stream?

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

Illustrates the setup for a word problem: calculating the speed of a jet stream affecting a plane's round trip, including variable definitions and rate formulas.
Diagram first shows motion of the plane at 450 miles per hour with an arrow to the right. The plane is traveling 2000 miles with the wind, represented by the expression 450 plus r. The jet stream motion is to the right. The round trip takes 9 hours. At the bottom of the diagram, an arrow to the left models the return motion of the plane. The plane’s velocity is 450 miles per hour, and the motion is 2000 miles against the wind modeled by the expression 450 – r.
We fill in the chart to organize the information.
We are looking for the speed of the jet stream. Let r= the speed of the jet stream.
When the plane flies with the wind, the wind increases its speed and so the rate is 450 + r.
When the plane flies against the wind, the wind decreases its speed and the rate is 450 − r.
Write in the rates.
Write in the distances.
Since D=r·t, we solve for
t and get t=Dr.
We divide the distance by
the rate in each row, and
place the expression in the
time column.
A table illustrates a rate-time-distance problem for an aircraft with headwind and tailwind. Headwind: Rate (450-r), Time (2000/(450-r)), Distance (2000). Tailwind: Rate (450+r), Time (2000/(450+r)), Distance (2000). Total time is 9.
We know the times add to 9
and so we write our equation.
2000450−r+2000450+r=9
We multiply both sides by the LCD. (450−r)(450+r)(2000450−r+2000450+r)= 9(450−r)(450+r)
Simplify. 2000(450+r)+2000(450−r)= 9(450−r)(450+r)
Factor the 2,000. 2000(450+r+450−r)= 9(4502−r2)
Solve. 2000(900)= 9(4502−r2)
Divide by 9. 2000(100)= 4502−r2
Simplify. 200000= 202500−r2
−2500=−r2
50=rThe speed of the jet stream.
Check:
Is 50 mph a reasonable speed for the jet stream? Yes.
If the plane is traveling 450 mph and the wind is 50 mph,
Tailwind 450+50=500mph2000500=4hours
Headwind 450−50=400 mph2000400=5 hours
The times add to 9 hours, so it checks.
The speed of the jet stream was 50 mph.

MaryAnne just returned from a visit with her grandchildren back east . The trip was 2400 miles from her home and her total time in the airplane for the round trip was 10 hours. If the plane was flying at a rate of 500 miles per hour, what was the speed of the jet stream?

Solution

The speed of the jet stream was 100 mph.

Gerry just returned from a cross country trip. The trip was 3000 miles from his home and his total time in the airplane for the round trip was 11 hours. If the plane was flying at a rate of 550 miles per hour, what was the speed of the jet stream?

Solution

The speed of the jet stream was 50 mph.

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 12 hours more than Press #2 to do the job and when both presses are running they can print the job in 8 hours. How long does it take for each press to print the job alone?

Solution

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Let x= the number of hours for Press #2
to complete the job.
Enter the hours per job for Press #1,
Press #2, and when they work together.
A table showing the number of hours needed to complete a job and the part of the job completed per hour for two presses (Press #1, Press #2) working individually and together.
The part completed by Press #1 plus the part
completed by Press #2 equals the
amount completed together.
Translate to an equation.
A mathematical equation illustrating a work-rate problem, showing 'Work completed by Press #1 + Press #2 = Together' with the formula 1/(x+12) + 1/x = 1/8.
Solve. A mathematical equation is displayed on a white background. The equation is a fractional expression: 1 over (x + 12) plus 1 over x equals 1 over 8.
Multiply by the LCD, 8x(x+12). A mathematical equation shown in red and black text on a white background. The equation is: 8x(x + 12)(1/(x + 12) + 1/x) = (1/8)8x(x + 12).
Simplify. The image shows the algebraic equation 8x + 8(x + 12) = x(x + 12).
A mathematical equation is displayed, reading 8x + 8x + 96 = x^2 + 12x. The equation features variables, coefficients, constants, and an exponent.
A quadratic equation is displayed: 0 = x^2 - 4x - 96, against a plain white background.
Solve. A mathematical equation shows '0 = (x - 12)(x + 8)' in a white font against a plain white background.
Two simple linear equations are displayed on a white background: 'x - 12 = 0' and 'x + 8 = 0', suggesting solutions for x.
The image shows mathematical notation with text. It displays 'x = 12,' and then 'y = 8 hours' with the 'y = 8' part struck through, suggesting a correction or change to the value of y.
Since the idea of negative hours does not make sense, we use the value x=12. A mathematical expression displays '12 + 12' on a white background, with the '1' of each '12' in red and the '2's and plus sign in black. To the right, a fully red '12' stands alone.
The image displays the text '24 hours' followed by '12 hours' on a white background, suggesting a comparison or timeline between the two durations.
Write our sentence answer. Press #1 would take 24 hours and
Press #2 would take 12 hours to do the job alone.

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours more than Press #2 to do the job and when both presses are running they can print the job in 4 hours. How long does it take for each press to print the job alone?

Solution

Press #1 would take 12 hours, and Press #2 would take 6 hours to do the job alone.

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes 3 hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in 2 hours. How long does it take for each hose to fill the hot tub?

Solution

The red hose take 6 hours and the green hose take 3 hours alone.

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

  • Quadratic Equation Word Problems
  • Applying the Quadratic Formula

Key Concepts

  • Methods to Solve Quadratic Equations
    • Factoring
    • Square Root Property
    • Completing the Square
    • Quadratic Formula
  • How to use a Problem-Solving Strategy.
    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what we are looking for.
    3. Name what we are looking for. Choose a variable to represent that quantity.
    4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.
  • Area of a Triangle
    • For a triangle with base, b, and height, h, the area, A, is given by the formula A=12bh.
      Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.
  • Area of a Rectangle
    • For a rectangle with length, L, and width, W, the area, A, is given by the formula A = LW.
      Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.
  • Pythagorean Theorem
    • In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, a2 + b2 = c2.
      Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.
  • Projectile motion
    • The height in feet, h, of an object shot upwards into the air with initial velocity, v0, after t seconds is given by the formula h = −16t2 + v0t.

Practice Makes Perfect

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve using any method.

The product of two consecutive odd numbers is 255. Find the numbers.

Solution

Two consecutive odd numbers whose product is 255 are 15 and 17, and −15 and −17.

The product of two consecutive even numbers is 360. Find the numbers.

The product of two consecutive even numbers is 624. Find the numbers.

Solution

The first and second consecutive odd numbers are 24 and 26, and −26 and −24.

The product of two consecutive odd numbers is 1,023. Find the numbers.

The product of two consecutive odd numbers is 483. Find the numbers.

Solution

Two consecutive odd numbers whose product is 483 are 21 and 23, and −21 and −23.

The product of two consecutive even numbers is 528. Find the numbers.

In the following exercises, solve using any method. Round your answers to the nearest tenth, if needed.

A triangle with area 45 square inches has a height that is two less than four times the base Find the base and height of the triangle.

Solution

The width of the triangle is 5 inches and the height is 18 inches.

The base of a triangle is six more than twice the height. The area of the triangle is 88 square yards. Find the base and height of the triangle.

The area of a triangular flower bed in the park has an area of 120 square feet. The base is 4 feet longer that twice the height. What are the base and height of the triangle?

Solution

The base is 24 feet and the height of the triangle is 10 feet.

A triangular banner for the basketball championship hangs in the gym. It has an area of 75 square feet. What is the length of the base and height , if the base is two-thirds of the height?

The length of a rectangular driveway is five feet more than three times the width. The area is 50 square feet. Find the length and width of the driveway.

Solution

The length of the driveway is 15.0 feet and the width is 3.3 feet.

A rectangular lawn has area 140 square yards. Its width that is six less than twice the length. What are the length and width of the lawn?

A rectangular table for the dining room has a surface area of 24 square feet. The length is two more feet than twice the width of the table. Find the length and width of the table.

Solution

The length of table is 8 feet and the width is 3 feet.

The new computer has a surface area of 168 square inches. If the the width is 5.5 inches less that the length, what are the dimensions of the computer?

The hypotenuse of a right triangle is twice the length of one of its legs. The length of the other leg is three feet. Find the lengths of the three sides of the triangle.

Solution

The lengths of the three sides of the triangle are 1.7, 3, and 3.5 ft.

The hypotenuse of a right triangle is 10 cm long. One of the triangle’s legs is three times the length of the other leg. Find the lengths of the two legs of the triangle. Round to the nearest tenth.

A rectangular garden will be divided into two plots by fencing it on the diagonal. The diagonal distance from one corner of the garden to the opposite corner is five yards longer than the width of the garden. The length of the garden is three times the width. Find the length of the diagonal of the garden.

Image shows a rectangular segment of grass with fence around 4 sides and across the diagonal. The vertical side of the rectangle is labeled w and the horizontal side is labeled 3 w. The diagonal fence is labeled w plus 5.
Solution

The length of the diagonal fencing is 7.3 yards.

Nautical flags are used to represent letters of the alphabet. The flag for the letter, O consists of a yellow right triangle and a red right triangle which are sewn together along their hypotenuse to form a square. The hypotenuse of the two triangles is three inches longer than a side of the flag. Find the length of the side of the flag.

Image shows a square with side lengths s. The square is divided into two triangles with a diagonal. The top triangle is red and the lower triangle is yellow. The diagonal is labeled s plus 3.

Gerry plans to place a 25-foot ladder against the side of his house to clean his gutters. The bottom of the ladder will be 5 feet from the house.How far up the side of the house will the ladder reach?

Solution

The ladder will reach 24.5 feet on the side of the house.

John has a 10-foot piece of rope that he wants to use to support his 8-foot tree. How far from the base of the tree should he secure the rope?

A firework rocket is shot upward at a rate of 640 ft/sec. Use the projectile formula h = −16t2 + v0t to determine when the height of the firework rocket will be 1200 feet.

Solution

The rocket will reach 1200 feet on its way up at 1.97 seconds and on its way down at 38.03 seconds.

An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula h = −16t2 + v0t, to determine when height of the arrow will be 400 feet.

A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial height of 8 feet. Use the formula h = −16t2 + v0t + 8 to determine how many seconds it will take for the bullet to hit the ground. (That is, when will h = 0?)

Solution

The bullet will take 70 seconds to hit the ground.

A stone is dropped from a 196-foot platform. Use the formula h = −16t2 + v0t + 196 to determine how many seconds it will take for the stone to hit the ground. (Since the stone is dropped, v0= 0.)

The businessman took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours and each way the trip was 200 miles. What was the speed of the wind that affected the plane which was flying at a speed of 120 mph?

Solution

The speed of the wind was 49 mph.

The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 300 miles. If the plane was flying at 125 mph, what was the speed of the wind that affected the plane?

Roy kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?

Solution

The speed of the current was 4.3 mph.

Rick paddled up the river, spent the night camping, and then paddled back. He spent 10 hours paddling and the campground was 24 miles away. If Rick kayaked at a speed of 5 mph, what was the speed of the current?

Two painters can paint a room in 2 hours if they work together. The less experienced painter takes 3 hours more than the more experienced painter to finish the job. How long does it take for each painter to paint the room individually?

Solution

The less experienced painter takes 6 hours and the experienced painter takes 3 hours to do the job alone.

Two gardeners can do the weekly yard maintenance in 8 minutes if they work together. The older gardener takes 12 minutes more than the younger gardener to finish the job by himself. How long does it take for each gardener to do the weekly yard maintainence individually?

It takes two hours for two machines to manufacture 10,000 parts. If Machine #1 can do the job alone in one hour less than Machine #2 can do the job, how long does it take for each machine to manufacture 10,000 parts alone?

Solution

Machine #1 takes 3.6 hours and Machine #2 takes 4.6 hours to do the job alone.

Sully is having a party and wants to fill his swimming pool. If he only uses his hose it takes 2 hours more than if he only uses his neighbor’s hose. If he uses both hoses together, the pool fills in 4 hours. How long does it take for each hose to fill the pool?

Writing Exercises

Make up a problem involving the product of two consecutive odd integers.

ⓐ Start by choosing two consecutive odd integers. What are your integers?

ⓑ What is the product of your integers?

ⓒ Solve the equation n(n + 2) = p, where p is the product you found in part (b).

ⓓ Did you get the numbers you started with?

Solution

Answers will vary.

Make up a problem involving the product of two consecutive even integers.

ⓐ Start by choosing two consecutive even integers. What are your integers?

ⓑ What is the product of your integers?

ⓒ Solve the equation n(n + 2) = p, where p is the product you found in part (b).

ⓓ Did you get the numbers you started with?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve applications of the quadratic formula.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Graph Quadratic Functions Using Properties

Learning Objectives

By the end of this section, you will be able to:

  • Recognize the graph of a quadratic function
  • Find the axis of symmetry and vertex of a parabola
  • Find the intercepts of a parabola
  • Graph quadratic functions using properties
  • Solve maximum and minimum applications

Before you get started, take this readiness quiz.

Graph the function f(x)=x2 by plotting points.
If you missed this problem, review Example 4 in Graphs of Functions.

Solution

An upward-opening parabola graphed on a Cartesian coordinate system, with its vertex at (0,0) and key points highlighted.

Solve: 2x2+3x−2=0.
If you missed this problem, review Example 2 in Polynomial Equations.

Solution

x=12,x=−2

Evaluate −b2a when a = 3 and b = −6.
If you missed this problem, review Example 10 in Integers.

Solution

1

Recognize the Graph of a Quadratic Function

Previously we very briefly looked at the function f(x)=x2, which we called the square function. It was one of the first non-linear functions we looked at. Now we will graph functions of the form f(x)=ax2+bx+c if a≠0. We call this kind of function a quadratic function.

Quadratic Function

A quadratic function, where a, b, and c are real numbers and a≠0, is a function of the form

f(x)=ax2+bx+c

We graphed the quadratic function f(x)=x2 by plotting points.

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis of the plane runs from negative 2 to 6. The parabola has a vertex at (0, 0) and also passes through the points (-2, 4), (-1, 1), (1, 1), and (2, 4). To the right of the graph is a table of values with 3 columns. The first row is a header row and labels each column, “x”, “f of x equals x squared”, and “the order pair x, f of x.” In row 2, x equals negative 3, f of x equals x squared is 9 and the ordered pair x, f of x is the ordered pair negative 3, 9. In row 3, x equals negative 2, f of x equals x squared is 4 and the ordered pair x, f of x is the ordered pair negative 2, 4. In row 4, x equals negative 1, f of x equals x squared is 1 and the ordered pair x, f of x is the ordered pair negative 1, 1. In row 5, x equals 0, f of x equals x squared is 0 and the ordered pair x, f of x is the ordered pair 0, 0. In row 6, x equals 1, f of x equals x squared is 1 and the ordered pair x, f of x is the ordered pair 1, 1. In row 7, x equals 2, f of x equals x squared is 4 and the ordered pair x, f of x is the ordered pair 2, 4. In row 8, x equals 3, f of x equals x squared is 9 and the ordered pair x, f of x is the ordered pair 3, 9.

Every quadratic function has a graph that looks like this. We call this figure a parabola.

Let’s practice graphing a parabola by plotting a few points.

Graph f(x)=x2−1.

Solution

We will graph the function by plotting points.

Choose integer values for x,
substitute them into the equation
and simplify to find f(x).

Record the values of the ordered pairs in the chart.
A table displays values for the function f(x) = x^2 - 1. For x=0, f(x)=-1. For x=1 and x=-1, f(x)=0. For x=2 and x=-2, f(x)=3.
Plot the points, and then connect
them with a smooth curve. The
result will be the graph of the
function f(x)=x2−1.
A quadratic function graphed as an upward-opening parabola on a Cartesian plane, featuring highlighted points at (-2,3), (-1,0), (0,-1), (1,0), and (2,3).

Graph f(x)=−x2..

Solution

This figure shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, 0).

Graph f(x)=x2+1.

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, −1).

All graphs of quadratic functions of the form f (x) = ax2 + bx + c are parabolas that open upward or downward. See Figure 1.

This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The general form for the equation of this graph is f of x equals a x squared plus b x plus c. The equation of this parabola is x squared plus 4 x plus 3. The leading coefficient, a, is greater than 0, so this parabola opens upward.The graph on the right shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The general form for the equation of this graph is f of x equals a x squared plus b x plus c. The equation of this parabola is negative x squared plus 4 x plus 3. The leading coefficient, a, is less than 0, so this parabola opens downward.

Notice that the only difference in the two functions is the negative sign before the quadratic term (x2 in the equation of the graph in Figure 1). When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.

Parabola Orientation

For the graph of the quadratic function f (x) = ax2 + bx + c, if

This images shows a bulleted list. The first bullet notes that, if a is greater than 0, then the parabola opens upward and shows an image of an upward-opening parabola. The second bullet notes that, if a is less than 0, then the parabola opens downward and shows an image of a downward-opening parabola.

Determine whether each parabola opens upward or downward:

ⓐ f(x)=−3x2+2x−4 ⓑ f(x)=6x2+7x−9.

Solution
ⓐ
Find the value of “a”. A mathematical image shows the general quadratic equation f(x) = ax^2 + bx + c, followed by a specific instance f(x) = -3x^2 + 2x - 4, and then explicitly states that a = -3.
Since the “a” is negative, the parabola will open downward.
ⓑ
Find the value of “a”. The image displays the general form of a quadratic function, f(x) = ax^2 + bx + c, followed by a specific example, f(x) = 6x^2 + 7x - 9, and identifies the value of 'a' as 6.
Since the “a” is positive, the parabola will open upward.

Determine whether the graph of each function is a parabola that opens upward or downward:

ⓐ f(x)=2x2+5x−2 ⓑ f(x)=−3x2−4x+7.

Solution

ⓐ up; ⓑ down

Determine whether the graph of each function is a parabola that opens upward or downward:

ⓐ f(x)=−2x2−2x−3 ⓑ f(x)=5x2−2x−1.

Solution

ⓐ down; ⓑ up

Find the Axis of Symmetry and Vertex of a Parabola

Look again at Figure 1. Do you see that we could fold each parabola in half and then one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

We show the same two graphs again with the axis of symmetry. See Figure 2.

This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The equation of this parabola is x squared plus 4 x plus 3. The vertical line passes through the point (negative 2, 0) and has the equation x equals negative 2. The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The equation of this parabola is negative x squared plus 4 x plus 3. The vertical line passes through the point (2, 0) and has the equation x equals 2.

The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of f (x) = ax2 + bx + c is x=−b2a.

So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula x=−b2a.

Compare the function f of x equals x squared plus 4 x plus 3 to the standard form of a quadratic function, f of x equals a x squared plus b x plus c. The axis of symmetry is the line x equals negative b divided by the product 2 a. Substituting for b and a yields x equals negative 4 divided by the product 2 times 1. The axis of symmetry equals negative 2. Next, compare the function f of x equals negative x squared plus 4 x plus 3 to the standard form of a quadratic function, f of x equals a x squared plus b x plus c. The axis of symmetry is the line x equals negative b divided by the product 2 a. Substituting for b and a yields x equals negative 4 divided by the product 2 times negative 1. The axis of symmetry equals 2.

Notice that these are the equations of the dashed blue lines on the graphs.

The point on the parabola that is the lowest (parabola opens up), or the highest (parabola opens down), lies on the axis of symmetry. This point is called the vertex of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its
x-coordinate is −b2a. To find the y-coordinate of the vertex we substitute the value of the x-coordinate into the quadratic function.

For the function f of x equals x squared plus 4 x plus 3, the axis of symmetry is x equals negative 2. The vertex is the point on the parabola with x-coordinate negative 2. Substitute x equals negative 2 into the function f of x equals x squared plus 4 x plus 3. F of x equals the square of negative 2 plus 4 times negative 2 plus 3, so f of x equals negative 1. The vertex is the point (negative 2, negative 1). For the function f of x equals negative x squared plus 4 x plus 3, the axis of symmetry is x equals 2. The vertex is the point on the parabola with x-coordinate 2. Substitute x equals 2 into the function f of x equals x squared plus 4 x plus 3. F of x equals 2 squared plus 4 times 2 plus 3, so f of x equals 7. The vertex is the point (2, 7).

Axis of Symmetry and Vertex of a Parabola

The graph of the function f (x) = ax2 + bx + c is a parabola where:

  • the axis of symmetry is the vertical line x=−b2a.
  • the vertex is a point on the axis of symmetry, so its x-coordinate is −b2a.
  • the y-coordinate of the vertex is found by substituting x=−b2a into the quadratic equation.

For the graph of f(x)=3x2−6x+2 find:

ⓐ the axis of symmetry ⓑ the vertex.

Solution
ⓐ
The general quadratic equation f(x) = ax^2 + bx + c is shown above a specific example: f(x) = 3x^2 - 6x + 2.
The axis of symmetry is the vertical line
x=−b2a.
Substitute the values of a,b into the
equation.
An algebraic equation X = -(-6 / (2 * 3)) is shown on a white background.
Simplify. The image displays the mathematical equation 'x = 1' in black text on a plain white background, indicating that the variable x has a value of one.
The axis of symmetry is the line x=1.
ⓑ
A mathematical equation is presented, showing a quadratic function written in standard form: f(x) = 3x^2 - 6x + 2.
The vertex is a point on the line of
symmetry, so its x-coordinate will be
x=1.
Find f(1).
The image shows the evaluation of a quadratic function at x=1, represented as f(1) = 3(1)^2 - 6(1) + 2.
Simplify. A mathematical equation is displayed: f(1) = 3 * 1 - 6 + 2, where the number '1' being multiplied by 3 is highlighted in red, indicating a substitution or specific value for calculation.
The result is the y-coordinate. The mathematical equation f(1) = -1 is displayed in black text against a plain white background.
The vertex is (1,−1).

For the graph of f(x)=2x2−8x+1 find:

ⓐ the axis of symmetry ⓑ the vertex.

Solution

ⓐ x=2; ⓑ (2, −7)

For the graph of f(x)=2x2−4x−3 find:

ⓐ the axis of symmetry ⓑ the vertex.

Solution

ⓐ x=1; ⓑ (1, −5)

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the x- and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y-intercept the value of x is zero. So to find the y-intercept, we substitute x = 0 into the function.

Let’s find the y-intercepts of the two parabolas shown in Figure 3.

This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The vertical line is an axis of symmetry for the parabola, and passes through the point (negative 2, 0). It has the equation x equals negative 2. The equation of this parabola is x squared plus 4 x plus 3. When x equals 0, f of 0 equals 0 squared plus 4 times 0 plus 3. F of 0 equals 3. The y-intercept of the graph is the point (0, 3). The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The vertical line is an axis of symmetry for the parabola and passes through the point (2, 0). It has the equation x equals 2. The equation of this parabola is negative x squared plus 4 x plus 3. When x equals 0, f of 0 equals negative 0 squared plus 4 times 0 plus 3. F of 0 equals 3. The y-intercept of the graph is the point (0, 3).

An x-intercept results when the value of f (x) is zero. To find an x-intercept, we let f (x) = 0. In other words, we will need to solve the equation 0 = ax2 + bx + c for x.

f(x)=ax2+bx+c0=ax2+bx+c

Solving quadratic equations like this is exactly what we have done earlier in this chapter!

We can now find the x-intercepts of the two parabolas we looked at. First we will find the x-intercepts of the parabola whose function is f (x) = x2 + 4x + 3.

The quadratic function f(x) = x^2 + 4x + 3 is displayed, representing a parabola in a mathematical context.
Let f(x)=0. The image shows the quadratic equation 0 = x^2 + 4x + 3, presented in a clear and concise format. The number 0 is highlighted in red, emphasizing the equation's form.
Factor. A mathematical equation shows 0 equals the product of (x + 1) and (x + 3), representing a quadratic equation in factored form.
Use the Zero Product Property. Two linear equations are shown: x + 1 = 0 and x + 3 = 0.
Solve. Two mathematical expressions are displayed: x = -1 and x = -3.
The x-intercepts are (−1,0) and (−3,0).

Now we will find the x-intercepts of the parabola whose function is f (x) = −x2 + 4x + 3.

The image displays the quadratic function f(x) = -x^2 + 4x + 3 written in a standard mathematical notation on a white background.
Let f(x)=0. A mathematical equation is displayed with a red zero, an equals sign, and the expression '-x^2 + 4x + 3' in black on a white background, representing a quadratic equation.
This quadratic does not factor, so
we use the Quadratic Formula.
The quadratic formula, x equals negative b plus or minus the square root of b squared minus 4ac, all divided by 2a, for solving quadratic equations.
a=−1,b=4,c=3 A step in solving a quadratic equation using the quadratic formula: x = (-4 plus or minus sqrt(4^2 - 4(-1)(3))) / (2(-1)).
Simplify. A mathematical equation is displayed, showing 'x equals negative 4 plus or minus the square root of 28, all divided by negative 2' on a white background.
A mathematical equation shows x = (-4 ± 2√7) / -2. This expression simplifies to x = 2 ± √7, representing two possible values for x from a quadratic formula solution.
A mathematical equation displays x equals a fraction where the numerator is -2 multiplied by the quantity 2 plus or minus the square root of 7, and the denominator is -2.
A mathematical equation displaying x = 2 ± √7, showing the value of x as two plus or minus the square root of seven.
The x-intercepts are (2+7,0) and
(2−7,0).

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph,

(2+7,0)≈(4.6,0)(2−7,0)≈(−0.6,0)

Do these results agree with our graphs? See Figure 4.

This image shows 2 graphs side-by-side. The graph on the left shows the upward-opening parabola defined by the function f of x equals x squared plus 4 x plus 3 and a dashed vertical line, x equals negative 2, graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1). The y-intercept is (0, 3) and the x-intercepts are (negative 1, 0) and (negative 3, 0). The graph on the right shows the downward-opening parabola defined by the function f of x equals negative x squared plus 4 x plus 3 and a dashed vertical line, x equals 2, graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7). The y-intercept is (0, 3) and the x-intercepts are (2 plus square root 7, 0), approximately (4.6, 0) and (2 minus square root, 0), approximately (negative 0.6, 0).

Find the Intercepts of a Parabola

To find the intercepts of a parabola whose function is f(x)=ax2+bx+c:

y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.

Find the intercepts of the parabola whose function is f(x)=x2−2x−8.

Solution
To find the y-intercept, let x=0 and
solve for f(x).
The image shows the quadratic function f(x) = x^2 - 2x - 8.
A mathematical equation shows f(0) equals 0 squared minus 2 multiplied by 0, minus 8. The '0' values in 0^2 and 2*0 are highlighted in red.
The mathematical equation f(0) = -8 is presented in a clear and concise format.
When x=0, then f(0)=−8.
The y-intercept is the point (0,−8).
To find the x-intercept, let f(x)=0 and
solve for x.
The mathematical equation f(x) = x^2 - 2x - 8 is displayed.
A mathematical equation shows '0 = x^2 - 2x - 8' against a white background.
Solve by factoring. A mathematical equation is displayed with the number 0 on the left side, an equals sign in the middle, and the expression (x - 4)(x + 2) on the right side.
Two equations are shown side-by-side: 0 = x - 4 and 0 = x + 2. These represent two simple linear equations with variable x.
The mathematical expressions '4=X' and '-2=X' are shown, indicating a contradiction where the variable X is equated to two different values simultaneously.
When f(x)=0, then x=4orx=−2.
The x-intercepts are the points (4,0) and
(−2,0).

Find the intercepts of the parabola whose function is f(x)=x2+2x−8.

Solution

y-intercept: (0, −8) x-intercepts (−4,0),(2,0)

Find the intercepts of the parabola whose function is f(x)=x2−4x−12.

Solution

y-intercept: (0, −12) x-intercepts (−2,0),(6,0)

In this chapter, we have been solving quadratic equations of the form ax2 + bx + c = 0. We solved for x and the results were the solutions to the equation.

We are now looking at quadratic functions of the form f (x) = ax2 + bx + c. The graphs of these functions are parabolas. The x-intercepts of the parabolas occur where f (x) = 0.

For example:

Quadratic equationQuadratic function x2−2x−15=0(x−5)(x+3)=0x−5=0x+3=0x=5x=−3Letf(x)=0.f(x)=x2−2x−150=x2−2x−150=(x−5)(x+3)x−5=0x+3=0x=5x=−3(5,0)and(−3,0)x-intercepts

The solutions of the quadratic function are the x values of the x-intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the functions give the x-intercepts of the graphs, the number of x-intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic function of the form ax2+bx+c=0. Now we can use the discriminant to tell us how many x-intercepts there are on the graph.

This image shows three graphs side-by-side. The graph on the left shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies below the x-axis and the parabola crosses the x-axis at two different points. If b squared minus 4 a c is greater than 0, then the quadratic equation a x squared plus b x plus c equals 0 has two solutions, and the graph of the parabola has 2 x-intercepts. The graph in the middle shows a downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies on the x-axis, the only point of intersection between the parabola and the x-axis. If b squared minus 4 a c equals 0, then the quadratic equation a x squared plus b x plus c equals 0 has one solution, and the graph of the parabola has 1 x-intercept. The graph on the right shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies above the x-axis and the parabola does not cross the x-axis. If b squared minus 4 a c is less than 0, then the quadratic equation a x squared plus b x plus c equals 0 has no solutions, and the graph of the parabola has no x-intercepts.

Before you to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Find the intercepts of the parabola for the function f(x)=5x2+x+4.

Solution
The image displays the quadratic function f(x) = 5x^2 + x + 4.
To find the y-intercept, let x=0 and
solve for f(x).
A mathematical equation shows f(0) = 5 * 0^2 + 0 + 4, illustrating the substitution of 0 into a function, with the substituted zeros highlighted in red.
The mathematical equation f(0) = 4 is displayed in black text against a plain white background.
When x=0, then f(0)=4.
The y-intercept is the point (0,4).
To find the x-intercept, let f(x)=0 and
solve for x.
A mathematical equation is displayed, showing f(x) = 5x^2 + x + 4 on a white background. This is a quadratic function, representing a parabola when graphed.
A mathematical equation is displayed: 0 = 5x^2 + x + 4.
Find the value of the discriminant to
predict the number of solutions which is
also the number of x-intercepts.
b2−4ac12−4·5·41−80−79
Since the value of the discriminant is
negative, there is no real solution to the
equation.
There are no x-intercepts.

Find the intercepts of the parabola whose function is f(x)=3x2+4x+4.

Solution

y-intercept: (0, 4) no x-intercept

Find the intercepts of the parabola whose function is f(x)=x2−4x−5.

Solution

y-intercept: (0, −5) x-intercepts (−1, 0),(5, 0)

Graph Quadratic Functions Using Properties

Now we have all the pieces we need in order to graph a quadratic function. We just need to put them together. In the next example we will see how to do this.

How to Graph a Quadratic Function Using Properties

Graph f (x) = x2 −6x + 8 by using its properties.

Solution

Step 1 is to determine whether the parabola opens upward or downward. Loot at the leading coefficient, a, in the equation. If f of x equals x squared minus 6 x plus 8, then a equals 1. Since a is positive, the parabola opens upward. Step 2 is to find the axis of symmetry. The axis of symmetry is the line x equals negative b divided by the product 2 a. For the function f of x equals x squared minust 6 x plus 8, the axis of symmetry is negative b divided by the product 2 a. x equals the opposite of negative 6 divided by the product 2 times 1. X equals 3. In step 3, find the vertex. The vertex is on the axis of symmetry. Substitute x equals 3 into the function. F of x equals x squared minus 6 x plus 8. F of 3 equals 3 squared minus 6 times 3 plus 8. F of 3 equals negative 1. The vertex is the point (3, negative 1). Step 4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry. We first find f of 0 to find the y-intercept. F of x equals x squared plus 6 x plus 8, so f of 0 equals 0 squared plus 6 times 0 plus 8. F of 0 equals 8. The y-intercept is the point (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x-value 6. The point symmetric to the y-intercept is the point (6, 8). Step 5 is to find the x-intercepts. Find additional points if needed. We solve f of x equals 0. We can solve this quadratic equation by factoring. To find the x-intercepts, set f of x equal to 0. F of x equals x squared minus 6 x plus 8. 0 equals x squared minus 6 x plus 8. 0 equals the product of x minus 2 and x minus 4. So x equals 2 or x equals 4. The x-intercepts are the points (2, 0) and (4, 0). The final step, step 6, is to graph the parabola. We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these 5 points to sketch the parabola. An image shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 10. The y-axis of the plane runs from negative 3 to 7. The parabola has a vertex at (3, negative 1). Other points plotted include the x-intercepts, (2, 0) and (4, 0), the y-intercept, (0, 8), and the point (6, 8) that is symmetric to the y-intercept across the axis of symmetry.

Graph f (x) = x2 + 2x − 8 by using its properties.

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The axis of symmetry, x equals negative 1, is graphed as a dashed line. The parabola has a vertex at (negative 1, negative 9). The y-intercept of the parabola is the point (0, negative 8). The x-intercepts of the parabola are the points (negative 4, 0) and (4, 0).

Graph f (x) = x2 − 8x + 12 by using its properties.

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 15. The axis of symmetry, x equals 4, is graphed as a dashed line. The parabola has a vertex at (4, negative 4). The y-intercept of the parabola is the point (0, 12). The x-intercepts of the parabola are the points (2, 0) and (6, 0).

We list the steps to take in order to graph a quadratic function here.

To graph a quadratic function using properties.

  1. Determine whether the parabola opens upward or downward.
  2. Find the equation of the axis of symmetry.
  3. Find the vertex.
  4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
  5. Find the x-intercepts. Find additional points if needed.
  6. Graph the parabola.

We were able to find the x-intercepts in the last example by factoring. We find the x-intercepts in the next example by factoring, too.

Graph f (x) = –x2 + 6x − 9 by using its properties.

Solution
Two mathematical equations are displayed: a general quadratic function f(x) = ax^2 + bx + c, and a specific instance f(x) = -x^2 + 6x - 9.
Since a is −1, the parabola opens downward.
A red, inverted U-shaped arrow with both ends pointing downwards, symbolizing a cycle, feedback loop, or a two-way process.
To find the equation of the axis of symmetry, use
x=−b2a.
The image displays a mathematical equation on a white background. The equation shown is x = -b / 2a, which represents the x-coordinate of the vertex of a parabola in standard form.
A mathematical expression reads X = -6 / (2(-1)), showing a fraction with a negative sign, numerator 6, and denominator 2 multiplied by -1 in parentheses.
The image displays the mathematical equation X=3 against a white background.
The axis of symmetry is x=3.
The vertex is on the line x=3.
A Cartesian coordinate system shows a vertical dashed line passing through x = 3. The x-axis ranges from -10 to 10, and the y-axis ranges from -10 to 10, with increments of 1 for both axes.
Find f(3). The quadratic function f(x) = -x^2 + 6x - 9 is displayed on a white background.
A mathematical equation is displayed on a white background: f(3) = -3^2 + 6 * 3 - 9. The numbers '3' within the terms '-3^2' and '6' are highlighted in red, indicating a substitution or evaluation.
A mathematical equation is displayed, showing 'f(3) = -9 + 18 - 9' in the center of a white background. This is a function evaluation, and the right side simplifies to 0.
The mathematical equation f(3) = 0 is displayed on a white background.
The vertex is (3,0).
A graph displays a vertical dashed line passing through x=3 on a coordinate plane. The x and y axes range from -10 to 10, with a blue dot marking the line's intersection at (3,0).
The y-intercept occurs when x=0. Find f(0). A mathematical equation for a quadratic function is displayed: f(x) = -x^2 + 6x - 9.
Substitute x=0. Substitution into a quadratic function: The equation f of x equals negative x squared plus six x minus nine is evaluated at x equals zero, demonstrating how to find the value of the function when zero is input.
Simplify. The image displays the mathematical expression 'f(0) = -9' centered on a white background.
The y-intercept is (0,−9).
The point (0,−9) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is (6,−9). A coordinate plane displays a dashed vertical line at x=3. Two blue points are plotted: one at (0, -9) and another at (6, -9).
Point symmetric to the y-intercept is (6,−9)
The x-intercept occurs when f(x)=0. The image displays the quadratic function f(x) = -x^2 + 6x - 9 on a white background.
Find f(x)=0. A quadratic equation, 0 = -x^2 + 6x - 9, is displayed on a white background, representing a mathematical expression.
Factor the GCF. The image shows the mathematical equation '0 = -(x^2 - 6x + 9)' centered on a white background, representing an algebraic expression likely for solving for x.
Factor the trinomial. The image displays the mathematical equation 0 = -(x - 3)^2, presented against a plain white background.
Solve for x. The number 0 equals 3, displayed as '0 = 3' in a simple, faded black font against a white background.
Connect the points to graph the parabola. Graph of a downward-opening parabola with vertex (3,0) and axis of symmetry x=3. The curve passes through (0,-9) and (6,-9).

Graph f (x) = −3x2 + 12x − 12 by using its properties.

Solution

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 15 to 10. The parabola has a vertex at (2, 0). The y-intercept (0, negative 12) is plotted as well as the axis of symmetry, x equals 2.

Graph f (x) = 4x2 + 24x + 36 by using its properties.

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 30 to 20. The y-axis of the plane runs from negative 10 to 40. The parabola has a vertex at (negative 3, 0). The y-intercept (0, 36) is plotted as well as the axis of symmetry, x equals negative 3.

For the graph of f (x) = −x2 + 6x − 9, the vertex and the x-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation 0 = −x2 + 6x − 9 is 0, so there is only one solution. That means there is only one x-intercept, and it is the vertex of the parabola.

How many x-intercepts would you expect to see on the graph of f (x) = x2 + 4x + 5?

Graph f (x) = x2 + 4x + 5 by using its properties.

Solution
Two mathematical equations are displayed: the general form of a quadratic function, f(x) = ax^2 + bx + c, and a specific instance, f(x) = x^2 + 4x + 5.
Since a is 1, the parabola opens upward.
Two red arrows form a U-shape, pointing upwards, suggesting a connection or uplift.
To find the axis of symmetry, find x=−b2a. The mathematical formula for finding the x-coordinate of the vertex of a parabola, x = -b / 2a, is displayed on a white background.
A mathematical equation is displayed on a white background, reading 'x = -4 / (2)1'.
The image shows x=-2.
The equation of the axis of symmetry is x=−2.
A coordinate plane with an x-axis ranging from -9 to 9 and a y-axis ranging from -9 to 9. A dashed vertical line is graphed at x = -2, extending across the entire y-range.
The vertex is on the line x=−2.
Find f(x) when x=−2. The image displays the quadratic function f(x) = x^2 + 4x + 5 written in black text against a white background.
A mathematical equation is displayed, showing the function f evaluated at -2: f(-2) = (-2)^2 + 4(-2) + 5, which involves squaring, multiplication, and addition.
A mathematical equation is displayed on a white background, reading 'f(-2) = 4 - 8 + 5' in black text.
The image displays the mathematical equation f(-2) = 1 centered on a white background.
The vertex is (−2,1).
A coordinate plane shows a vertical dashed line passing through x = -2. A blue point is plotted on this line at (-2, 1). The x-axis and y-axis both range from -9 to 9.
The y-intercept occurs when x=0. A mathematical equation is displayed on a white background: f(x) = x^2 + 4x + 5.
Find f(0). The image displays the mathematical expression f(0) = 5 in black text against a plain white background, indicating that the function f evaluates to 5 when its input is 0.
Simplify. The image displays the mathematical expression f(0) = 5 in black text against a plain white background, indicating that the function f evaluates to 5 when its input is 0.
The y-intercept is (0,5).
The point (−4,5) is two units to the left of the line of
symmetry.
The point two units to the right of the line of
symmetry is (0,5).
A graph on a coordinate plane shows a dashed vertical line at x = -2. Three points are plotted: (-4, 5), (-2, 1), and (0, 5).
Point symmetric to the y-intercept is (−4,5).
The x-intercept occurs when f(x)=0. A graph with three points: (-4, 5), (-2, 1), and (0, 5). A dashed vertical line is at x = -2, passing through the point (-2, 1) on the coordinate plane.
Find f(x)=0. The image shows the quadratic equation 0 = x^2 + 4x + 5.
Test the discriminant.
The mathematical expression b^2 - 4ac, which is the discriminant from the quadratic formula.
A mathematical expression displaying '4^2 - 4 * 1 * 5' against a white background.
A vertical image displays the numbers '15-20' in black text against a plain white background, appearing as if on a vertical sign or banner.
-4
Since the value of the discriminant is negative, there is
no real solution and so no x-intercept.
Connect the points to graph the parabola. You may
want to choose two more points for greater accuracy.
A parabola opens upwards with vertex at (-2, 1) and axis of symmetry at x = -2. The curve passes through points (-4, 5) and (0, 5).

Graph f (x) = x2 − 2x + 3 by using its properties.

Solution


This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 4. The y-axis of the plane runs from negative 1 to 5. The parabola has a vertex at (1, 2). The y-intercept (0, 3) is plotted as is the line of symmetry, x equals 1.

Graph f (x) = −3x2 − 6x − 4 by using its properties.

Solution


This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 2. The y-axis of the plane runs from negative 5 to 1. The parabola has a vertex at (negative 1, negative 2). The y-intercept (0, negative 4) is plotted as is the line of symmetry, x equals negative 1.

Finding the y-intercept by finding f (0) is easy, isn’t it? Sometimes we need to use the Quadratic Formula to find the x-intercepts.

Graph f (x) = 2x2 − 4x − 3 by using its properties.

Solution
Two mathematical equations are displayed: a general form of a quadratic function, f(x) = ax^2 + bx + c, and a specific example, f(x) = 2x^2 - 4x - 3, on a white background.
Since a is 2, the parabola opens upward. A vibrant red U-shaped graphic featuring two distinct arrowheads, both pointing directly upwards, signifying an upward trend, connection, or flow. A clear visual for growth or movement.
To find the equation of the axis of symmetry, use
x=−b2a.
A mathematical formula is displayed as 'x = -b / 2a' against a white background.
A mathematical equation is displayed on a white background: X = -4 / (2 * 2). The formula simplifies to X = -4/4, which means X = -1.
The equation 'X = 1' is displayed on a plain white background.
The equation of the axis of
symmetry is x=1.
The vertex is on the line x=1. A mathematical expression displays the quadratic function f(x) = 2x^2 - 4x - 3, presented in a bold, sans-serif font on a white background.
Find f(1). The image shows the equation f(x) = 2(1)^2 - 4(1) - 3.
The image shows the mathematical equation f(1) = 2 - 4 - 3 in black text on a white background.
A mathematical expression reads f(1) = -5, representing a function f evaluated at 1 resulting in a value of -5.
The vertex is (1,−5).
The y-intercept occurs when x=0. The image displays the quadratic function f(x) = 2x^2 - 4x - 3.
Find f(0). The image shows the evaluation of a quadratic function at x=0, specifically f(0) = 2(0)^2 - 4(0) - 3, simplifying to -3.
Simplify. The mathematical equation 'f(0) = -3' is displayed on a plain white background.
The y-intercept is (0,−3).
The point (0,−3) is one unit to the left of the line of
symmetry.
Point symmetric to the
y-intercept is (2,−3)
The point one unit to the right of the line of
symmetry is (2,−3).
The x-intercept occurs when y=0. A mathematical equation for a quadratic function, f(x) = 2x^2 - 4x - 3, displayed in black text on a white background.
Find f(x)=0. A mathematical equation is displayed, reading '0 = 2x^2 - 4x - 3' on a white background.
Use the Quadratic Formula. The quadratic formula, used to find the roots of a quadratic equation, displayed as x = (-b +/- sqrt(b^2 - 4ac)) / (2a).
Substitute in the values of a,b, and c. An instance of the quadratic formula used to solve for x, showing the values -4, 2, and 3 substituted into the standard algebraic expression.
Simplify. A quadratic formula step showing X equals negative 4 plus or minus the square root of 16 plus 24, all divided by 4.
Simplify inside the radical. A mathematical equation is displayed on a white background: x = (4 ± √40) / 4. This represents the solution to a quadratic equation, showing the plus or minus operation for the square root of 40.
Simplify the radical. The image displays the mathematical equation for x: x = (4 plus or minus 2 times the square root of 10) / 4.
Factor the GCF. A mathematical equation showing x equals two times two plus or minus the square root of ten, all divided by four. This simplifies to x = (2 ±  10)/2.
Remove common factors. A mathematical equation showing x equals 2 plus or minus the square root of 10, all divided by 2.
Write as two equations. Two solutions for x are displayed: x equals (2 plus the square root of 10) all over 2, and x equals (2 minus the square root of 10) all over 2.
Approximate the values. Two mathematical approximations are displayed: x is approximately equal to 2.5, and x is approximately equal to -0.6. The text is clear and centrally placed on a plain white background.
The approximate values of the
x-intercepts are (2.5,0) and
(−0.6,0).
Graph the parabola using the points found. A parabola is graphed on a coordinate plane, with x and y axes ranging from -9 to 9. The parabola opens upwards, with its vertex at (1, -5). A dashed line at x=1 represents the axis of symmetry.

Graph f (x) = 5x2 + 10x + 3 by using its properties.

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis of the plane runs from negative 4 to 4. The axis of symmetry, x equals negative 1, is graphed as a dashed line. The parabola has a vertex at (negative 1, negative 2). The y-intercept of the parabola is the point (0, 3). The x-intercepts of the parabola are approximately (negative 1.6, 0) and (negative 0.4, 0).

Graph f (x) = −3x2 − 6x + 5 by using its properties.

Solution

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The axis of symmetry, x equals negative 1, is graphed as a dashed line. The parabola has a vertex at (negative 1, 8). The y-intercept of the parabola is the point (0, 5). The x-intercepts of the parabola are approximately (negative 2.6, 0) and (0.6, 0).

Solve Maximum and Minimum Applications

Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic function. The y-coordinate of the vertex is the minimum value of a parabola that opens upward. It is the maximum value of a parabola that opens downward. See Figure 5.

This figure shows 2 graphs side-by-side. The left graph shows a downward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label maximum. The right graph shows an upward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label minimum.

Minimum or Maximum Values of a Quadratic Function

The y-coordinate of the vertex of the graph of a quadratic function is the

  • minimum value of the quadratic equation if the parabola opens upward.
  • maximum value of the quadratic equation if the parabola opens downward.

Find the minimum or maximum value of the quadratic function f(x)=x2+2x−8.

Solution
A mathematical equation is displayed on a white background, reading 'f(x) = x^2 + 2x - 8'.
Since a is positive, the parabola opens upward.
The quadratic equation has a minimum.
Find the equation of the axis of symmetry. Mathematical formula: x = -b/2a, representing the axis of symmetry or the x-coordinate of the vertex of a parabola.
The image displays a mathematical equation, x = -2 / (2 * 1), indicating a step in solving for x where the numerator is -2 and the denominator is the product of 2 and 1.
A white background with the equation 'X = -1' written in black in the upper left corner.
The equation of the axis of
symmetry is x=−1.
The vertex is on the line x=−1. A mathematical equation is displayed against a white background, showing the function f(x) = x^2 + 2x - 8. The text is in a dark grey or black font.
Find f(−1). A mathematical equation shows the evaluation of a function f(x) at x=-1, with the expression f(-1) = (-1)^2 + 2(-1) - 8. The number -1 is highlighted in red to emphasize the substitution.
A mathematical expression shows f(-1) equals 1 minus 2 minus 8, displayed in black text against a white background.
The mathematical equation f(-1) = -9 is displayed on a white background.
The vertex is (−1,−9).
Since the parabola has a minimum, the y-coordinate of
the vertex is the minimum y-value of the quadratic
equation.
The minimum value of the quadratic is −9 and it
occurs when x=−1.
A graph of a parabola opening upwards, with its vertex at (-1, -9). A dashed vertical line at x = -1 indicates the axis of symmetry. The x and y axes both range from -9 to 9.
Show the graph to verify the result.

Find the maximum or minimum value of the quadratic function f(x)=x2−8x+12.

Solution

The minimum value of the quadratic function is −4 and it occurs when x = 4.

Find the maximum or minimum value of the quadratic function f(x)=−4x2+16x−11.

Solution

The maximum value of the quadratic function is 5 and it occurs when x = 2.

We have used the formula

h(t)=−16t2+v0t+h0

to calculate the height in feet, h , of an object shot upwards into the air with initial velocity, v0, after t seconds .

This formula is a quadratic function, so its graph is a parabola. By solving for the coordinates of the vertex (t, h), we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.

The quadratic equation h(t) = −16t2 + 176t + 4 models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

ⓐ How many seconds will it take the volleyball to reach its maximum height? ⓑ Find the maximum height of the volleyball.

Solution
This table presents a quadratic function and observations regarding its properties, such as its downward opening and maximum value.
h(t)=−16t2+176t+4
Since a is negative, the parabola opens downward.
The quadratic function has a maximum.

ⓐ
This table demonstrates finding the axis of symmetry through calculation and explains related concepts like the vertex and maximum value.
Find the equation of the axis of symmetry. t=−b2a t=−1762(−16) t=5.5
The equation of the axis of symmetry is t=5.5.
The vertex is on the line t=5.5. The maximum occurs when t=5.5 seconds.

ⓑ
This table demonstrates evaluating a quadratic function h(t) at t=5.5 and identifying the resulting vertex (5.5, 488).
Find h(5.5).



Use a calculator to simplify.
h(t)=−16t2+176t+4 h(t)=−16(5.5)2+176(5.5)+4 h(t)=488
The vertex is (5.5,488).

Since the parabola has a maximum, the h-coordinate of the vertex is the maximum value of the quadratic function.

The maximum value of the quadratic is 488 feet and it occurs when t = 5.5 seconds.

After 5.5 seconds, the volleyball will reach its maximum height of 488 feet.

Solve, rounding answers to the nearest tenth.

The quadratic function h(t) = −16t2 + 128t + 32 is used to find the height of a stone thrown upward from a height of 32 feet at a rate of 128 ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height?

Solution

It will take 4 seconds for the stone to reach its maximum height of 288 feet.

A path of a toy rocket thrown upward from the ground at a rate of 208 ft/sec is modeled by the quadratic function of h(t) = −16t2 + 208t. When will the rocket reach its maximum height? What will be the maximum height?

Solution

It will 6.5 seconds for the rocket to reach its maximum height of 676 feet.

Access these online resources for additional instruction and practice with graphing quadratic functions using properties.

  • Quadratic Functions: Axis of Symmetry and Vertex
  • Finding x- and y-intercepts of a Quadratic Function
  • Graphing Quadratic Functions
  • Solve Maxiumum or Minimum Applications
  • Quadratic Applications: Minimum and Maximum

Key Concepts

  • Parabola Orientation
    • For the graph of the quadratic function f(x)=ax2+bx+c, if
      • a > 0, the parabola opens upward.
      • a < 0, the parabola opens downward.
  • Axis of Symmetry and Vertex of a Parabola The graph of the function f(x)=ax2+bx+c is a parabola where:
    • the axis of symmetry is the vertical line x=−b2a.
    • the vertex is a point on the axis of symmetry, so its x-coordinate is −b2a.
    • the y-coordinate of the vertex is found by substituting x=−b2a into the quadratic equation.
  • Find the Intercepts of a Parabola
    • To find the intercepts of a parabola whose function is f(x)=ax2+bx+c:
      y-interceptx-interceptsLetx=0and solve forf(x).Letf(x)=0and solve forx.
  • How to graph a quadratic function using properties.
    1. Determine whether the parabola opens upward or downward.
    2. Find the equation of the axis of symmetry.
    3. Find the vertex.
    4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
    5. Find the x-intercepts. Find additional points if needed.
    6. Graph the parabola.
  • Minimum or Maximum Values of a Quadratic Equation
    • The y-coordinate of the vertex of the graph of a quadratic equation is the
    • minimum value of the quadratic equation if the parabola opens upward.
    • maximum value of the quadratic equation if the parabola opens downward.

Practice Makes Perfect

Recognize the Graph of a Quadratic Function

In the following exercises, graph the functions by plotting points.

f(x)=x2+3

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, 3).

f(x)=x2−3

y=−x2+1

Solution

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, 1).

f(x)=−x2−1

For each of the following exercises, determine if the parabola opens up or down.

ⓐ f(x)=−2x2−6x−7 ⓑ f(x)=6x2+2x+3

Solution

ⓐ down ⓑ up

ⓐ f(x)=4x2+x−4 ⓑ f(x)=−9x2−24x−16

ⓐ f(x)=−3x2+5x−1 ⓑ f(x)=2x2−4x+5

Solution

ⓐ down ⓑ up

ⓐ f(x)=x2+3x−4 ⓑ f(x)=−4x2−12x−9

Find the Axis of Symmetry and Vertex of a Parabola

In the following functions, find ⓐ the equation of the axis of symmetry and ⓑ the vertex of its graph.

f(x)=x2+8x−1

Solution

ⓐ x=−4; ⓑ (−4, −17)

f(x)=x2+10x+25

f(x)=−x2+2x+5

Solution

ⓐ x=1; ⓑ (1,6)

f(x)=−2x2−8x−3

Find the Intercepts of a Parabola

In the following exercises, find the intercepts of the parabola whose function is given.

f(x)=x2+7x+6

Solution

y-intercept: (0, 6); x-intercept (−1, 0),(−6, 0)

f(x)=x2+10x−11

f(x)=x2+8x+12

Solution

y-intercept: (0, 12); x-intercept (−2, 0),(−6, 0)

f(x)=x2+5x+6

f(x)=−x2+8x−19

Solution

y-intercept: (0, −19); x-intercept: none

f(x)=−3x2+x−1

f(x)=x2+6x+13

Solution

y-intercept: (0, 13); x-intercept: none

f(x)=x2+8x+12

f(x)=4x2−20x+25

Solution

y-intercept: (0,25); x-intercept (52,0)

f(x)=−x2−14x−49

f(x)=−x2−6x−9

Solution

y-intercept: (0,−9); x-intercept (−3, 0)

f(x)=4x2+4x+1

Graph Quadratic Functions Using Properties

In the following exercises, graph the function by using its properties.

f(x)=x2+6x+5

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 3, negative 4). The y-intercept, point (0, 5), is plotted as are the x-intercepts, (negative 5, 0) and (negative 1, 0).

f(x)=x2+4x−12

f(x)=x2+4x+3

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1). The y-intercept, point (0, 3), is plotted as are the x-intercepts, (negative 3, 0) and (negative 1, 0).

f(x)=x2−6x+8

f(x)=9x2+12x+4

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis of the plane runs from negative 4 to 4. The parabola has a vertex at (negative 2 thirds, 0). The y-intercept, point (0, 4), is plotted. The axis of symmetry, x equals negative 2 thirds, is plotted as a dashed vertical line.

f(x)=−x2+8x−16

f(x)=−x2+2x−7

Solution

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 15 to 10. The parabola has a vertex at (1, negative 6). The y-intercept, point (0, negative 7), is plotted. The axis of symmetry, x equals 1, is plotted as a dashed vertical line.

f(x)=5x2+2

f(x)=2x2−4x+1

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (1, negative 1). The y-intercept, point (0, 1), is plotted as are the x-intercepts, approximately (0.3, 0) and (1.7, 0). The axis of symmetry is the vertical line x equals 1, plotted as a dashed line.

f(x)=3x2−6x−1

f(x)=2x2−4x+2

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (1, 0). This point is the only x-intercept. The y-intercept, point (0, 2), is plotted. The axis of symmetry is the vertical line x equals 1, plotted as a dashed line.

f(x)=−4x2−6x−2

f(x)=−x2−4x+2

Solution

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, 6). The y-intercept, point (0, 2), is plotted as are the x-intercepts, approximately (negative 4.4, 0) and (0.4, 0). The axis of symmetry is the vertical line x equals 2, plotted as a dashed line.

f(x)=x2+6x+8

f(x)=5x2−10x+8

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (1, 3). The y-intercept, point (0, 8), is plotted; there are no x-intercepts. The axis of symmetry is the vertical line x equals 1, plotted as a dashed line.

f(x)=−16x2+24x−9

f(x)=3x2+18x+20

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 3, negative 7). The x-intercepts are plotted at the approximate points (negative 4.5, 0) and (negative 1.5, 0). The axis of symmetry is the vertical line x equals negative 3, plotted as a dashed line.

f(x)=−2x2+8x−10

Solve Maximum and Minimum Applications

In the following exercises, find the maximum or minimum value of each function.

f(x)=2x2+x−1

Solution

The minimum value is −98 when x=−14.

y=−4x2+12x−5

y=x2−6x+15

Solution

The minimum value is 6 when x = 3.

y=−x2+4x−5

y=−9x2+16

Solution

The maximum value is 16 when x = 0.

y=4x2−49

In the following exercises, solve. Round answers to the nearest tenth.

An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic function h(t) = −16t2 + 168t + 45 find how long it will take the arrow to reach its maximum height, and then find the maximum height.

Solution

In 5.3 sec the arrow will reach maximum height of 486 ft.

A stone is thrown vertically upward from a platform that is 20 feet height at a rate of 160 ft/sec. Use the quadratic function h(t) = −16t2 + 160t + 20 to find how long it will take the stone to reach its maximum height, and then find the maximum height.

A ball is thrown vertically upward from the ground with an initial velocity of 109 ft/sec. Use the quadratic function h(t) = −16t2 + 109t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.

Solution

In 3.4 seconds the ball will reach its maximum height of 185.6 feet.

A ball is thrown vertically upward from the ground with an initial velocity of 122 ft/sec. Use the quadratic function h(t) = −16t2 + 122t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.

A computer store owner estimates that by charging x dollars each for a certain computer, he can sell 40 − x computers each week. The quadratic function R(x) = −x2 +40x is used to find the revenue, R, received when the selling price of a computer is x, Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

Solution

A selling price of $20 per computer will give the maximum revenue of $400.

A retailer who sells backpacks estimates that by selling them for x dollars each, he will be able to sell 100 − x backpacks a month. The quadratic function R(x) = −x2 +100x is used to find the R, received when the selling price of a backpack is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

A retailer who sells fashion boots estimates that by selling them for x dollars each, he will be able to sell 70 − x boots a week. Use the quadratic function R(x) = −x2 +70x to find the revenue received when the average selling price of a pair of fashion boots is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue per day.

Solution

A selling price of $35 per pair of boots will give a maximum revenue of $1,225.

A cell phone company estimates that by charging x dollars each for a certain cell phone, they can sell 8 − x cell phones per day. Use the quadratic function R(x) = −x2 +8x to find the revenue received per day when the selling price of a cell phone is x. Find the selling price that will give them the maximum revenue per day, and then find the amount of the maximum revenue.

A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation A(x)=x120−x2 gives the area of the corral, A, for the length, x, of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.

Solution

The length of one side along the river is 120 feet and the maximum are is 7,200 square feet.

A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic function A(x)=x50−x2 gives the area, A, of the dog run for the length, x, of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.

A land owner is planning to build a fenced in rectangular patio behind his garage, using his garage as one of the “walls.” He wants to maximize the area using 80 feet of fencing. The quadratic function A(x) = x(80 − 2x) gives the area of the patio, where x is the width of one side. Find the maximum area of the patio.

Solution

The maximum area of the patio is 800 feet.

A family of three young children just moved into a house with a yard that is not fenced in. The previous owner gave them 300 feet of fencing to use to enclose part of their backyard. Use the quadratic function A(x)=x150−x2 determine the maximum area of the fenced in yard.

Writing Exercise

How do the graphs of the functions f(x)=x2 and f(x)=x2−1 differ? We graphed them at the start of this section. What is the difference between their graphs? How are their graphs the same?

Solution

Answers will vary.

Explain the process of finding the vertex of a parabola.

Explain how to find the intercepts of a parabola.

Solution

Answers will vary.

How can you use the discriminant when you are graphing a quadratic function?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can recognize the graph of a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can find the axis of symmetry and vertex of a parabola.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can find the intercepts of a parabola.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can graph quadratic equations in two variables.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve maximum and minimum applications.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

quadratic function
A quadratic function, where a, b, and c are real numbers and a≠0, is a function of the form f(x)=ax2+bx+c.

Graph Quadratic Functions Using Transformations

Learning Objectives

By the end of this section, you will be able to:

  • Graph quadratic equations of the form f(x)=x2+k
  • Graph quadratic functions of the form f(x)=(x-h)2
  • Graph quadratic functions of the form f(x)=ax2
  • Graph quadratic functions using transformations
  • Find a quadratic function from its graph

Before you get started, take this readiness quiz.

Graph the function f(x)=x2 by plotting points.
If you missed this problem, review Example 4 in Graphs of Functions.

Solution

A parabola opening upwards, with its vertex at the origin (0,0) and symmetrical about the y-axis, is displayed on a Cartesian coordinate system with grid lines from -8 to 8 on both axes.

Factor completely: y2−14y+49.
If you missed this problem, review Example 2 in Factor Special Products.

Solution

y−72

Factor completely: 2x2−16x+32.
If you missed this problem, review Example 4 in Factor Special Products.

Solution

2(x−4)2

Graph Quadratic Functions of the form f(x)=x2+k

In the last section, we learned how to graph quadratic functions using their properties. Another method involves starting with the basic graph of f(x)=x2 and ‘moving’ it according to information given in the function equation. We call this graphing quadratic functions using transformations.

In the first example, we will graph the quadratic function f(x)=x2 by plotting points. Then we will see what effect adding a constant, k, to the equation will have on the graph of the new function f(x)=x2+k.

Graph f(x)=x2,g(x)=x2+2, and h(x)=x2−2 on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.

Solution

Plotting points will help us see the effect of the constants on the basic f(x)=x2 graph. We fill in the chart for all three functions.

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals x squared plus 2, the ordered pair (x, g of x), h of x equals x squared minus 2, and the ordered pair (x, h of x). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the f of x equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, f of x) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The g of x equals x squared plus 2 column contains the expressions 9 plus 2, 4 plus 2, 1 plus 2, 0 plus 2, 1 plus 2, 4 plus 2, and 9 plus 2. The (x, g of x) column has the ordered pairs of (negative 3, 11), (negative 2, 6), (negative 1, 3), (0, 2), (1, 3), (2, 6), and (3, 11). In the h of x equals x squared minus 2 column, the expressions given are 9 minus 2, 4 minus 2, 1 minus 2, 0 minus 2, 1 minus 2, 4 minus 2, and 9 minus 2. In last column, (x, h of x), contains the ordered pairs (negative 3, 7), (negative 2, 2), (negative 1, negative 1), (0, negative 2), (1, negative 1), (2, 2), and (3, 7).

The g(x) values are two more than the f(x) values. Also, the h(x) values are two less than the f(x) values. Now we will graph all three functions on the same rectangular coordinate system.

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle is the graph of f of x equals x squared has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top parabola has been moved up 2 units, and the bottom has been moved down 2 units.

The graph of g(x)=x2+2 is the same as the graph of f(x)=x2 but shifted up 2 units.

The graph of h(x)=x2−2 is the same as the graph of f(x)=x2 but shifted down 2 units.

ⓐ Graph f(x)=x2,g(x)=x2+1, and h(x)=x2−1 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Solution


ⓐ
This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle graph is of f of x equals x squared has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 1 unit, and the bottom has been moved down 1 unit.
ⓑ The graph of g(x)=x2+1 is the same as the graph of f(x)=x2 but shifted up 1 unit. The graph of h(x)=x2−1 is the same as the graph of f(x)=x2 but shifted down 1 unit.

ⓐ Graph f(x)=x2,g(x)=x2+6, and h(x)=x2−6 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Solution


ⓐ
This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 6 units, and the bottom has been moved down 6 units.
ⓑ The graph of h(x)=x2+6 is the same as the graph of f(x)=x2 but shifted up 6 units. The graph of h(x)=x2−6 is the same as the graph of f(x)=x2 but shifted down 6 units.

The last example shows us that to graph a quadratic function of the form f(x)=x2+k, we take the basic parabola graph of f(x)=x2 and vertically shift it up (k>0) or shift it down (k<0).

This transformation is called a vertical shift.

Graph a Quadratic Function of the form f(x)=x2+k Using a Vertical Shift

The graph of f(x)=x2+k shifts the graph of f(x)=x2 vertically k units.

  • If k > 0, shift the parabola vertically up k units.
  • If k < 0, shift the parabola vertically down |k| units.

Now that we have seen the effect of the constant, k, it is easy to graph functions of the form f(x)=x2+k. We just start with the basic parabola of f(x)=x2 and then shift it up or down.

It may be helpful to practice sketching f(x)=x2 quickly. We know the values and can sketch the graph from there.

This figure shows an upward-opening parabola on the x y-coordinate plane, with vertex (0, 0). Other points on the curve are located at (negative 4, 16), (negative 3, 9), (negative 2, 4), (negative 1, 1), (1, 1), (2, 4), (3, 9), and (4, 16).

Once we know this parabola, it will be easy to apply the transformations. The next example will require a vertical shift.

Graph f(x)=x2−3 using a vertical shift.

Solution
This table illustrates steps and their visual outcomes for graphing transformations of functions, specifically showing shifts of f(x) = x^2.
We first draw the graph of f(x)=x2 on
the grid.
This figure shows an upward-opening parabola on the x y-coordinate plane with a vertex of (0, 0) with other points on the curve located at (negative 1, 1) and (1, 1). It is the graph of f of x equals x squared.
Determine k. Two mathematical equations are displayed: f(x) = x^2 + k and f(x) = x^2 - 3 (highlighted in red). The image appears to be for a math problem determining the value of k.
The mathematical equation k = -3 is displayed in black text on a plain white background, centrally positioned within the frame.
Shift the graph f(x)=x2 down 3. This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The top curve is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The bottom curve has been moved down 3 units.

Graph f(x)=x2−5 using a vertical shift.

Solution


This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The top curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The bottom curve has been moved down 5 units.

Graph f(x)=x2+7 using a vertical shift.

Solution

This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The bottom curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 7 units.

Graph Quadratic Functions of the form f(x)=(x−h)2

In the first example, we graphed the quadratic function f(x)=x2 by plotting points and then saw the effect of adding a constant k to the function had on the resulting graph of the new function f(x)=x2+k.

We will now explore the effect of subtracting a constant, h, from x has on the resulting graph of the new function f(x)=(x−h)2.

Graph f(x)=x2,g(x)=(x−1)2, and h(x)=(x+1)2 on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.

Solution

Plotting points will help us see the effect of the constants on the basic f(x)=x2 graph. We fill in the chart for all three functions.

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals the quantity of x minus 1 squared, the ordered pair (x, g of x), h of x equals the quantity of x plus 1 squared, and the ordered pair (x, h of x). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the f of x equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, f of x) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The g of x equals the quantity of x minus 1 squared column contains the values of 16, 9, 4, 1, 0, 1, and 4. The (x, g of x) column has the ordered pairs of (negative 3, 1), (negative 2, 9), (negative 1, 4), (0, 1), (1, 0), (2, 1), and (3, 4). In the h of x equals the quantity of x plus 1 squared, the values given are 4, 1, 0, 1, 4, 9, and 16. In last column, (x, h of x), contains the ordered pairs (negative 3, 4), (negative 2, 1), (negative 1, 0), (0, 4), (1, negative 1), (2, 9), and (3, 16).

The g(x) values and the h(x) values share the common numbers 0, 1, 4, 9, and 16, but are shifted.

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 1 unit, and the right curve has been moved to the right 1 unit.

The figure says on the first line that the graph of g of x equals the quantity x minus 1 square is the same as the graph of f of x equals x squared but shifted right 1 unit. The second line states that the graph of h of x equals the quantity x plus 1 squared is the same as the graph of f of x equals x squared but shifted left 1 unit. The third line of the figure says g of x equals the quantity x minus 1 squared with an arrow underneath it pointing to the right with 1 unit written beside it. Finally, it gives h of x equals the quantity of x plus 1 squared with an arrow underneath it pointing to the left with 1 unit written beside it.

ⓐ Graph f(x)=x2,g(x)=(x+2)2, and h(x)=(x−2)2 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Solution


ⓐ
This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 2 units, and the right curve has been moved to the right 2 units.
ⓑ The graph of g(x)=(x+2)2 is the same as the graph of f(x)=x2 but shifted left 2 units. The graph of h(x)=(x−2)2 is the same as the graph of f(x)=x2 but shift right 2 units.

ⓐ Graph f(x)=x2,g(x)=x2+5, and h(x)=x2−5 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Solution


ⓐ
This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 5 units, and the right curve has been moved to the right 5 units.
ⓑ The graph of g(x)=(x+5)2 is the same as the graph of f(x)=x2 but shifted left 5 units. The graph of h(x)=(x−5)2 is the same as the graph of f(x)=x2 but shifted right 5 units.

The last example shows us that to graph a quadratic function of the form f(x)=(x−h)2, we take the basic parabola graph of f(x)=x2 and shift it left (h > 0) or shift it right (h < 0).

This transformation is called a horizontal shift.

Graph a Quadratic Function of the form f(x)=(x−h)2 Using a Horizontal Shift

The graph of f(x)=(x−h)2 shifts the graph of f(x)=x2 horizontally h units.

  • If h > 0, shift the parabola horizontally right h units.
  • If h < 0, shift the parabola horizontally left |h| units.

Now that we have seen the effect of the constant, h, it is easy to graph functions of the form f(x)=(x−h)2. We just start with the basic parabola of f(x)=x2 and then shift it left or right.

The next example will require a horizontal shift.

Graph f(x)=(x−5)2 using a horizontal shift.

Solution
We first draw the graph of f(x)=x2 on
the grid.
A coordinate plane displays the graph of a parabola y=x^2. The curve opens upwards, symmetric about the y-axis, with its vertex located at the origin (0,0).
Determine h. Two mathematical equations are shown: f(x) = (x - h) squared in black and f(x) = (x - 5) squared in red, demonstrating a quadratic function and its specific form.
The image displays the equation 'h = 5' in a light gray font against a clean white background.
Shift the graph f(x)=x2 to the right 5 units. Graph of y=x^2 (blue) and y=(x-5)^2 (red), demonstrating a horizontal translation of 5 units to the right on the coordinate plane.

Graph f(x)=(x−4)2 using a horizontal shift.

Solution

This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The left curve is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The right curve has been moved right 4 units.

Graph f(x)=(x+6)2 using a horizontal shift.

Solution

This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The right curve is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 6 units.

Now that we know the effect of the constants h and k, we will graph a quadratic function of the form f(x)=(x−h)2+k by first drawing the basic parabola and then making a horizontal shift followed by a vertical shift. We could do the vertical shift followed by the horizontal shift, but most students prefer the horizontal shift followed by the vertical.

Graph f(x)=(x+1)2−2 using transformations.

Solution

This function will involve two transformations and we need a plan.

Let’s first identify the constants h, k.

F of x equals the quantity x plush 1 squared minus 2 is given on the top line with f of x equals the quanitity x minus h squared minis k on the second line. The given equation was changed to f of x equals the quantity of x minus negative 1 squared plush negative 2 on the third line. The final line says h equals negative 1 and k equals negative 2.

The h constant gives us a horizontal shift and the k gives us a vertical shift.

F of x equals x squared is given with an arrow coming from it pointing to f of x equals the quantity x plus 1 squared with an arrow coming from it pointing to f of x equals the quantity x plus 1 squared minus 2. The next lines say h equals negative 1 which means shift left 1 unit and k equals negative 2 which means shift down 2 units.

We first draw the graph of f(x)=x2 on the grid.

The figure says on the first line that the graph of f of x equals the quantity x plus 1 squared is the same as the graph of f of x equals x squared but shifted left 1 unit. The second line states that the graph of f of x equals the quantity x plus 1 squared minus 2 is the same as the graph of f of x equals the quantity x plus 1 squared but shifted down 2 units.

The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of f of x equals x squared left 1, we move to the next graph, which shows the original f of x equals x squared and then another curve moved left one unit to produce f of x equals the quantity of x plus 1 squared. By moving f of x equals the quantity of x plus 1 squared down 1, we move to the final graph, which shows the original f of x equals x squared and the f of x equals the quantity of x plus 1, then another curve moved down 1 to produce f of x equals the quantity of x plus 1 squared minus 2.

Graph f(x)=(x+2)2−3 using transformations.

Solution

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). Then, the original function is moved 2 units to the left to produce f of x equals the quantity of x plus 2 squared. The final curve is produced by moving down 3 units to produce f of x equals the quantity of x plus 2 squared minus 3.

Graph f(x)=(x−3)2+1 using transformations.

Solution

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). Then, the original function is moved 3 units to the right to produce f of x equals the quantity of x minus 3 squared. The final curve is produced by moving up 1 unit to produce f of x equals the quantity of x minus 3squared plus 1.

Graph Quadratic Functions of the Form f(x)=ax2

So far we graphed the quadratic function f(x)=x2 and then saw the effect of including a constant h or k in the equation had on the resulting graph of the new function. We will now explore the effect of the coefficient a on the resulting graph of the new function f(x)=ax2.

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals 2 times x squared, the ordered pair (x, g of x), h of x equals one-half times x squared, and the ordered pair (x, h of x). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the f of x equals x squared column, the values are 4, 1, 0, 1, and 4. In the (x, f of x) column, the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The g of x equals 2 times x squared column contains the expressions 2 times 4, 2 times 1, 2 times 0, 2 times 1, and 2 times 4. The (x, g of x) column has the ordered pairs of (negative 2, 8), (negative 1, 2), (0, 0), (1, 2), and (2,8). In the h of x equals one-half times x squared, the expressions given are one-half times 4, one-half times 1, one-half times 0, one-half times 1, and one-half times 4. In last column, (x, h of x), contains the ordered pairs (negative 2, 2), (negative 1, one-half), (0, 0), (1, one-half), and (2, 2).

If we graph these functions, we can see the effect of the constant a, assuming a > 0.

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The slimmer curve of g of x equals 2 times x square has a vertex at (0,0) and other points of (negative 1, one-half) and (1, one-half). The wider curve, h of x equals one-half x squared, has a vertex at (0,0) and other points of (negative 2, 2) and (2,2).

To graph a function with constant a it is easiest to choose a few points on f(x)=x2 and multiply the y-values by a.

Graph of a Quadratic Function of the form f(x)=ax2

The coefficient a in the function f(x)=ax2 affects the graph of f(x)=x2 by stretching or compressing it.

  • If 0<|a|<1, the graph of f(x)=ax2 will be “wider” than the graph of f(x)=x2.
  • If |a|>1, the graph of f(x)=ax2 will be “skinnier” than the graph of f(x)=x2.

Graph f(x)=3x2.

Solution

We will graph the functions f(x)=x2 and g(x)=3x2 on the same grid. We will choose a few points on f(x)=x2 and then multiply the y-values by 3 to get the points for g(x)=3x2.

The table depicts the effect of constants on the basic function of x squared. The table has 3 columns labeled x, f of x equals x squared with the ordered pair (x, f of x), and g of x equals 3 times x squared with the ordered pair (x, g of x). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the f of x equals x squared with the ordered pair (x, f of x), the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The g of x equals 3 times x squared with the ordered pair (x, g of x) column has the ordered pairs of (negative 2, 12) because 3 times 4 equals 12, (negative 1, 3) because 3 times 1 equals 3, (0, 0) because 3 times 0 equals 0, (1, 3) because 3 times 1 equals 3, and (2,12) because 3 times 4 equals 12. The graph beside the table shows 2 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points given on the curve are located at (negative 2, 4) (negative 1, 1), (1, 1), and (2,4). The slimmer curve of g of x equals 3 times x squared has a vertex at (0,0) and other points given of (negative 2, 12), (negative 1, 3), (1, 3), and (2,12).

Graph f(x)=−3x2.

Solution


The graph shows the upward-opening parabola on the x y-coordinate plane of f of x equals x squared that has a vertex of (0, 0). Other points given on the curve are located at (negative 2, 4) (negative 1, 1), (1, 1), and (2,4). Also shown is a downward-opening parabola of f of x equals negative 3 times x squared. It has a vertex of (0,0) with other points at (negative 1, negative 3) and (1, negative 3)

Graph f(x)=2x2.

Solution

This figure shows 2 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The slimmer curve of f of x equals 2 times x square has a vertex at (0,0) and other points of (negative 1, one-half) and (1, one-half).

Graph Quadratic Functions Using Transformations

We have learned how the constants a, h, and k in the functions, f(x)=x2+k,f(x)=(x−h)2, and f(x)=ax2 affect their graphs. We can now put this together and graph quadratic functions f(x)=ax2+bx+c by first putting them into the form f(x)=a(x−h)2+k by completing the square. This form is sometimes known as the vertex form or standard form.

We must be careful to both add and subtract the number to the SAME side of the function to complete the square. We cannot add the number to both sides as we did when we completed the square with quadratic equations.

This figure shows the difference when completing the square with a quadratic equation and a quadratic function. For the quadratic equation, start with x squared plus 8 times x plus 6 equals zero. Subtract 6 from both sides to get x squared plus 8 times x equals negative 6 while leaving space to complete the square. Then, complete the square by adding 16 to both sides to get x squared plush 8 times x plush 16 equals negative 6 plush 16. Factor to get the quantity x plus 4 squared equals 10. For the quadratic function, start with f of x equals x squared plus 8 times x plus 6. The second line shows to leave space between the 8 times x and the 6 in order to complete the square. Complete the square by adding 16 and subtracting 16 on the same side to get f of x equals x squared plus 8 times x plush 16 plus 6 minus 16. Factor to get f of x equals the quantity of x plush 4 squared minus 10.

When we complete the square in a function with a coefficient of x2 that is not one, we have to factor that coefficient from just the x-terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the x-terms.

Once we get the constant we want to complete the square, we must remember to multiply it by that coefficient before we then subtract it.

Rewrite f(x)=−3x2−6x−1 in the f(x)=a(x−h)2+k form by completing the square.

Solution
A mathematical equation is presented, showing the function f(x) equal to -3x^2 - 6x - 1. This quadratic equation is displayed in a clear, digital font on a white background.
Separate the x terms from the constant. A mathematical equation is displayed on a white background, reading 'f(x) = -3x^2 - 6x - 1' in black text.
Factor the coefficient of x2, −3. The image displays the quadratic function f(x) = -3(x^2 + 2x) - 1.
Prepare to complete the square. The equation f(x) = -3(x^2 + 2x) - 1 is shown, with a blank space after '2x' indicating an incomplete expression, possibly for completing the square.
Take half of 2 and then square it to complete the
square. (12·2)2=1
The constant 1 completes the square in the
parentheses, but the parentheses is multiplied by
−3. So we are really adding −3 We must then
add 3 to not change the value of the function.
A mathematical equation f(x) = -3(x^2 + 2x + 1) - 1 + 3. It demonstrates how adding 1 inside the parenthesis, multiplied by -3, means you must add 3 outside to balance the expression.
Rewrite the trinomial as a square and subtract the
constants.
A mathematical equation is displayed against a white background. The equation reads: f(x) = -3(x + 1)^2 + 2. It represents a quadratic function in vertex form.
The function is now in the f(x)=a(x−h)2+k
form.
Two mathematical equations are displayed: the vertex form of a quadratic function, f(x) = a(x - h) ^2 + k, followed by a specific example, f(x) = -3(x + 1) ^2 + 2.

Rewrite f(x)=−4x2−8x+1 in the f(x)=a(x−h)2+k form by completing the square.

Solution

f(x)=−4(x+1)2+5

Rewrite f(x)=2x2−8x+3 in the f(x)=a(x−h)2+k form by completing the square.

Solution

f(x)=2(x−2)2−5

Once we put the function into the f(x)=(x−h)2+k form, we can then use the transformations as we did in the last few problems. The next example will show us how to do this.

Graph f(x)=x2+6x+5 by using transformations.

Solution

Step 1. Rewrite the function in f(x)=a(x−h)2+k vertex form by completing the square.

A mathematical equation is displayed, showing a quadratic function: f(x) = x^2 + 6x + 5. The function is written in a standard algebraic notation with variables and coefficients.
Separate the x terms from the constant. A mathematical equation is displayed, showing a quadratic function: f(x) = x^2 + 6x + 5. The equation is rendered in black text on a white background, appearing as a standard algebraic expression.
Take half of 6 and then square it to complete the square.
(12·6)2=9
We both add 9 and subtract 9 to not change the value of the function. A mathematical equation for f(x) is shown as f(x) = x^2 + 6x + 9 + 5 - 9, with the numbers +9 and -9 highlighted in red.
Rewrite the trinomial as a square and subtract the constants. A mathematical equation is displayed on a white background: f(x) = (x + 3)^2 - 4. This represents a quadratic function in vertex form.
The function is now in the f(x)=(x−h)2+k form. Two mathematical equations are displayed: the general vertex form of a parabola, f(x) = (x - h)^2 + k, in red, and a specific instance, f(x) = (x + 3)^2 - 4, in black.

Step 2: Graph the function using transformations.

Looking at the h, k values, we see the graph will take the graph of f(x)=x2 and shift it to the left 3 units and down 4 units.

F of x equals x squared is given with an arrow coming from it pointing to f of x equals the quantity x plus 3 squared with an arrow coming from it pointing to f of x equals the quantity x plus 3 squared minus 4. The next lines say h equals negative 3 which means shift left 3 unit and k equals negative 4 which means shift down 4 units

We first draw the graph of f(x)=x2 on the grid.

To graph f of x equals the quantity x plus 3 squared, shift the graph of f of x equals x squares to the left 3 units. To graph f of x equals the quantity x plus 3 squared minus 4, shift the graph the quantity x plus 3 squared down 4 units.

The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of f of x equals x squared left 3, we move to the next graph, which shows the original f of x equals x squared and then another curve moved left 3 units to produce f of x equals the quantity of x plus 3 squared. By moving f of x equals the quantity of x plus 3 squared down 2, we move to the final graph, which shows the original f of x equals x squared and the f of x equals the quantity of x plus 3 squared, then another curve moved down 4 to produce f of x equals the quantity of x plus 1 squared minus 4.

Graph f(x)=x2+2x−3 by using transformations.

Solution


This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The curve to the left has been moved 1 unit to the left to produce f of x equals the quantity of x plus 1 squared. The third graph has been moved down 4 units to produce f of x equals the quantity of x plus 1 squared minus 4.

Graph f(x)=x2−8x+12 by using transformations.

Solution


This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The curve to the right has been moved 4 units to the right to produce f of x equals the quantity of x minus 4 squared. The third graph has been moved down 4 units to produce f of x equals the quantity of x minus 4 squared minus 4.

We list the steps to take to graph a quadratic function using transformations here.

Graph a quadratic function using transformations.

  1. Rewrite the function in f(x)=a(x−h)2+k form by completing the square.
  2. Graph the function using transformations.

Graph f(x)=−2x2−4x+2 by using transformations.

Solution

Step 1. Rewrite the function in f(x)=a(x−h)2+k vertex form by completing the square.

The quadratic function f(x) = -2x^2 - 4x + 2 is displayed on a white background.
Separate the x terms from the constant. A mathematical equation for a quadratic function is displayed, reading f(x) = -2x^2 - 4x + 2.
We need the coefficient of x2 to be one.
We factor −2 from the x-terms.
A mathematical equation is displayed, showing a function f(x) = -2(x^2 + 2x) + 2. This is a quadratic function written in a form that can be simplified or used for completing the square.
Take half of 2 and then square it to complete the square.
(12·2)2=1
We add 1 to complete the square in the parentheses, but the parentheses is multiplied by −2. Se we are really adding −2. To not change the value of the function we add 2. An algebraic expression f(x) = -2(x^2 + 2x + 1) + 2 + 2, highlighting the addition of 1 inside the parenthesis and the compensating addition of 2 outside, as shown by the red numbers and arrow.
Rewrite the trinomial as a square and subtract the constants. The image displays the equation of a quadratic function in vertex form: f(x) = -2(x + 1)^2 + 4.
The function is now in the f(x)=a(x−h)2+k form. Two equations for quadratic functions, with the general vertex form f(x) = a(x - h)^2 + k shown in red, and a specific example f(x) = -2(x + 1)^2 + 4 in black.

Step 2. Graph the function using transformations.

F of x equals x squared is given with an arrow coming from it pointing to f of x equals negative 2 times x squared with an arrow coming from it pointing to f of x equals negative 2 times the quantity x plus 1 squared. An arrow come from it to point to f of x equals negative 2 times the quantity x plus 1 squared plus 4. The next line says a equals negative 2 which means multiply the y-values by negative 2, then h equals negative 1 which means shift left 1 unit and k equals 4 which means shift up 4 units

We first draw the graph of f(x)=x2 on the grid.

To graph f of x equals negative 2 times x squared, multiply the y-values in parabola of f of x equals x squared by negative 2. To graph f of x equals negative 2 times the quantity x plus 1 squared, shift the graph of f of x equals negative 2 times x squared to the left 1 unit. To graph f of x equals negative 2 times the quantity x plus 1 squared plus 4, shift the graph of f of x equals negative 2 times the quantity x plus 1 squared up 4 units.

The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By multiplying by negative 2, move to the next graph showing the original f of x equals x squared and the new slimmer and flipped graph of f of x equals negative 2 x squared. By shifting that graph of f of x equals negative 2 times x squared left 1, we move to the next graph, which shows the original f of x equals x squared, f of x equals negative 2 x squared, and then another curve moved left 1 unit to produce f of x equals negative 2 times the quantity of x plus 1 squared. By moving f of x equals negative 2 times the quantity of x plus 1 squared up 4, we move to the final graph, which shows the original f of x equals x squared, f of x equals negative 2 x squared, and the f of x equals negative 2 times the quantity of x plus 1 squared, then another curve moved up 4 to produce f of x equals negative 2 times the quantity of x plus 1 squared plus 4.

Graph f(x)=−3x2+12x−4 by using transformations.

Solution


This figure shows a downward-opening parabola on the x y-coordinate plane with a vertex of (2,8) and other points of (1,5) and (3,5).

Graph f(x)=−2x2+12x−9 by using transformations.

Solution


This figure shows a downward-opening parabola on the x y-coordinate plane with a vertex of (3, 9) and other points of (1, 1) and (5, 1).

Now that we have completed the square to put a quadratic function into f(x)=a(x−h)2+k form, we can also use this technique to graph the function using its properties as in the previous section.

If we look back at the last few examples, we see that the vertex is related to the constants h and k.

The first graph shows an upward-opening parabola on the x y-coordinate plane with a vertex of (negative 3, negative 4) with other points of (0, negative 5) and (0, negative 1). Underneath the graph, it shows the standard form of a parabola, f of x equals the quantity x minus h squared plus k, with the equation of the parabola f of x equals the quantity of x plus 3 squared minus 4 where h equals negative 3 and k equals negative 4. The second graph shows a downward-opening parabola on the x y-coordinate plane with a vertex of (negative 1, 4) and other points of (0,2) and (negative 2,2). Underneath the graph, it shows the standard form of a parabola, f of x equals a times the quantity x minus h squared plus k, with the equation of the parabola f of x equals negative 2 times the quantity of x plus 1 squared plus 4 where h equals negative 1 and k equals 4.

In each case, the vertex is (h, k). Also the axis of symmetry is the line x = h.

We rewrite our steps for graphing a quadratic function using properties for when the function is in f(x)=a(x−h)2+k form.

Graph a quadratic function in the form f(x)=a(x−h)2+k using properties.

  1. Rewrite the function in f(x)=a(x−h)2+k form.
  2. Determine whether the parabola opens upward, a > 0, or downward, a < 0.
  3. Find the axis of symmetry, x = h.
  4. Find the vertex, (h, k).
  5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
  6. Find the x-intercepts.
  7. Graph the parabola.

ⓐ Rewrite f(x)=2x2+4x+5 in f(x)=a(x−h)2+k form and ⓑ graph the function using properties.

Solution
Rewrite the function in f(x)=a(x−h)2+k
form by completing the square.
f(x)=2x2+4x+5
f(x)=2(x2+2x)+5
f(x)=2(x2+2x+1)+5−2
f(x)=2(x+1)2+3
Identify the constants a,h,k. a=2h=−1k=3
Since a=2, the parabola opens upward. A red U-shaped arrow with arrowheads pointing upwards and outwards at both ends, indicating a curve, cycle, or expansion.
The axis of symmetry is x=h. The axis of symmetry is x=−1.
The vertex is (h,k). The vertex is (−1,3).
Find the y-intercept by finding f(0). f(0)=2⋅02+4⋅0+5
f(0)=5
y-intercept (0,5)
Find the point symmetric to (0,5) across the
axis of symmetry.
(−2,5)
Find the x-intercepts. The discriminant negative, so there are
no x-intercepts. Graph the parabola.
A parabola graphed on a coordinate plane. The vertex is at (-1, 3), and points (-2, 5) and (0, 5) are marked. A dashed line at x=-1 represents the axis of symmetry.

ⓐ Rewrite f(x)=3x2−6x+5 in f(x)=a(x−h)2+k form and ⓑ graph the function using properties.

Solution


ⓐ f(x)=3(x−1)2+2
ⓑ
The graph shown is an upward facing parabola with vertex (1, 2) and y-intercept (0, 5). The axis of symmetry is shown, x equals 1.

ⓐ Rewrite f(x)=−2x2+8x−7 in f(x)=a(x−h)2+k form and ⓑ graph the function using properties.

Solution


ⓐ f(x)=−2(x−2)2+1
ⓑ
The graph shown is a downward facing parabola with vertex (2, 1) and x-intercepts (1, 0) and (3, 0). The axis of symmetry is shown, x equals 2.

Find a Quadratic Function from its Graph

So far we have started with a function and then found its graph.

Now we are going to reverse the process. Starting with the graph, we will find the function.

Determine the quadratic function whose graph is shown.

The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0, 7).
Solution
Steps for deriving the equation of a quadratic function in vertex form, given its vertex and y-intercept.
Since it is quadratic, we start with the f(x)=a(x−h)2+kform.
The vertex, (h,k), is (−2,−1) so h=−2 and k=−1. f(x)=a(x−(−2))2−1
To find a, we use the y-intercept, (0,7).
So f(0)=7. 7=a(0+2)2−1
Solve for a. 7=4a−1
8=4a
2=a
Write the function. f(x)=a(x−h)2+k
Substitute in h=−2,k=−1 and a=2. f(x)=2(x+2)2−1

Write the quadratic function in f(x)=a(x−h)2+k form whose graph is shown.

The graph shown is an upward facing parabola with vertex (3, negative 4) and y-intercept (0, 5).
Solution

f(x)=(x−3)2−4

Determine the quadratic function whose graph is shown.

The graph shown is an upward facing parabola with vertex (negative 3, negative 1) and y-intercept (0, 8).
Solution

f(x)=(x+3)2−1

Access these online resources for additional instruction and practice with graphing quadratic functions using transformations.

  • Function Shift Rules Applied to Quadratic Functions
  • Changing a Quadratic from Standard Form to Vertex Form
  • Using Transformations to Graph Quadratic Functions
  • Finding Quadratic Equation in Vertex Form from Graph

Key Concepts

  • Graph a Quadratic Function of the form f(x)=x2+k Using a Vertical Shift
    • The graph of f(x)=x2+k shifts the graph of f(x)=x2 vertically k units.
      • If k > 0, shift the parabola vertically up k units.
      • If k < 0, shift the parabola vertically down |k| units.
  • Graph a Quadratic Function of the form f(x)=(x−h)2 Using a Horizontal Shift
    • The graph of f(x)=(x−h)2 shifts the graph of f(x)=x2 horizontally h units.
      • If h > 0, shift the parabola horizontally left h units.
      • If h < 0, shift the parabola horizontally right |h| units.
  • Graph of a Quadratic Function of the form f(x)=ax2
    • The coefficient a in the function f(x)=ax2 affects the graph of f(x)=x2 by stretching or compressing it.
      If 0<|a|<1, then the graph of f(x)=ax2 will be “wider” than the graph of f(x)=x2.
      If |a|>1, then the graph of f(x)=ax2 will be “skinnier” than the graph of f(x)=x2.
  • How to graph a quadratic function using transformations
    1. Rewrite the function in f(x)=a(x−h)2+k form by completing the square.
    2. Graph the function using transformations.
  • Graph a quadratic function in the vertex form f(x)=a(x−h)2+k using properties
    1. Rewrite the function in f(x)=a(x−h)2+k form.
    2. Determine whether the parabola opens upward, a > 0, or downward, a < 0.
    3. Find the axis of symmetry, x = h.
    4. Find the vertex, (h, k).
    5. Find they-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
    6. Find the x-intercepts, if possible.
    7. Graph the parabola.

Practice Makes Perfect

Graph Quadratic Functions of the form f(x)=x2+k

In the following exercises, ⓐ graph the quadratic functions on the same rectangular coordinate system and ⓑ describe what effect adding a constant, k, to the function has on the basic parabola.

f(x)=x2,g(x)=x2+4, and h(x)=x2−4.

Solution


ⓐ
This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top curve has been moved up 4 units, and the bottom has been moved down 4 units.
ⓑ The graph of g(x)=x2+4 is the same as the graph of f(x)=x2 but shifted up 4 units. The graph of h(x)=x2−4 is the same as the graph of f(x)=x2 but shift down 4 units.

f(x)=x2,g(x)=x2+7, and h(x)=x2−7.

In the following exercises, graph each function using a vertical shift.

f(x)=x2+3

Solution

This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (0, 3) and other points (7, 2) and (7, negative 2).

f(x)=x2−7

g(x)=x2+2

Solution

This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (0, 2) and other points (negative 2, 6) and (2, 6).

g(x)=x2+5

h(x)=x2−4

Solution

This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (0, negative 4) and other points (negative 2, 0) and (2, 0).

h(x)=x2−5

Graph Quadratic Functions of the form f(x)=(x−h)2

In the following exercises, ⓐ graph the quadratic functions on the same rectangular coordinate system and ⓑ describe what effect adding a constant, h, inside the parentheses has

f(x)=x2,g(x)=(x−3)2, and h(x)=(x+3)2.

Solution


ⓐ
This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The graph to the right is shifted 3 units to the right to produce g of x equals the quantity of x minus 3 squared. The graph the left is shifted 3 units to the left to produce h of x equals the quantity of x plus 3 squared.
ⓑ The graph of g(x)=(x−3)2 is the same as the graph of f(x)=x2 but shifted right 3 units. The graph of h(x)=(x+3)2 is the same as the graph of f(x)=x2 but shifted left 3 units.

f(x)=x2,g(x)=(x+4)2, and h(x)=(x−4)2.

In the following exercises, graph each function using a horizontal shift.

f(x)=(x−2)2

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (2, 0) and other points (0, 4) and (4, 4).

f(x)=(x−1)2

f(x)=(x+5)2

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (negative 5, 0) and other points (negative 7, 4) and (negative 3, 4).

f(x)=(x+3)2

f(x)=(x−5)2

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (5, 0) and other points (3, 4) and (7, 4).

f(x)=(x+2)2

In the following exercises, graph each function using transformations.

f(x)=(x+2)2+1

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (negative 2, 1) and other points (negative 4, 5) and (0, 5).

f(x)=(x+4)2+2

f(x)=(x−1)2+5

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (1, 5) and other points (negative 1, 9) and (3, 9).

f(x)=(x−3)2+4

f(x)=(x+3)2−1

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (negative 3, 1) and other points (negative 4, 0) and (negative 2, 0).

f(x)=(x+5)2−2

f(x)=(x−4)2−3

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (4, negative 2) and other points (3, negative 2) and (5, negative 2).

f(x)=(x−6)2−2

Graph Quadratic Functions of the form f(x)=ax2

In the following exercises, graph each function.

f(x)=−2x2

Solution


This figure shows a downward-opening parabolas on the x y-coordinate plane. It has a vertex of (0, 0) and other points (negative 1, negative 2) and (1, negative 2).

f(x)=4x2

f(x)=−4x2

Solution


This figure shows a downward-opening parabolas on the x y-coordinate plane. It has a vertex of (0, 0) and other points (negative 1, negative 4) and (1, negative 4).

f(x)=−x2

f(x)=12x2

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (0, 0) and other points (negative 2, 2) and (2, 2).

f(x)=13x2

f(x)=14x2

Solution


This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (0, 0) and other points (2, 1) and (negative 2, 1).

f(x)=−12x2

Graph Quadratic Functions Using Transformations

In the following exercises, rewrite each function in the f(x)=a(x−h)2+k form by completing the square.

f(x)=−3x2−12x−5

Solution

f(x)=−3(x+2)2+7

f(x)=2x2−12x+7

f(x)=3x2+6x−1

Solution

f(x)=3(x+1)2−4

f(x)=−4x2−16x−9

In the following exercises, ⓐ rewrite each function in f(x)=a(x−h)2+k form and ⓑ graph it by using transformations.

f(x)=x2+6x+5

Solution

ⓐ f(x)=(x+3)2−4
ⓑ
This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (negative 3, 3), y-intercept of (0, 5), and axis of symmetry shown at x equals negative 3.

f(x)=x2+4x−12

f(x)=x2+4x+3

Solution

ⓐ f(x)=(x+2)2−1
ⓑ
This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (negative 2, negative 1), y-intercept of (0, 3), and axis of symmetry shown at x equals negative 2.

f(x)=x2−6x+8

f(x)=x2−6x+15

Solution

ⓐ f(x)=(x−3)2+6
ⓑ
This figure shows an upward-opening parabolas on the x y-coordinate plane. It has a vertex of (3, 6), y-intercept of (0, 10), and axis of symmetry shown at x equals 3.

f(x)=x2+8x+10

f(x)=−x2+8x−16

Solution

ⓐ f(x)=−(x−4)2+0
ⓑ
This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (4, 0), y-intercept of (0, negative 16), and axis of symmetry shown at x equals 4.

f(x)=−x2+2x−7

f(x)=−x2−4x+2

Solution

ⓐ f(x)=−(x+2)2+6
ⓑ
This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, 6), y-intercept of (0, 2), and axis of symmetry shown at x equals negative 2.

f(x)=−x2+4x−5

f(x)=5x2−10x+8

Solution

ⓐ f(x)=5(x−1)2+3
ⓑ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, 3), y-intercept of (0, 8), and axis of symmetry shown at x equals 1.

f(x)=3x2+18x+20

f(x)=2x2−4x+1

Solution

ⓐ f(x)=2(x−1)2−1
ⓑ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 1), y-intercept of (0, 1), and axis of symmetry shown at x equals 1.

f(x)=3x2−6x−1

f(x)=−2x2+8x−10

Solution

ⓐ f(x)=−2(x−2)2−2
ⓑ
This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (2, negative 2), y-intercept of (0, negative 10), and axis of symmetry shown at x equals 2.

f(x)=−3x2+6x+1

In the following exercises, ⓐ rewrite each function in f(x)=a(x−h)2+k form and ⓑ graph it using properties.

f(x)=2x2+4x+6

Solution

ⓐ f(x)=2(x+1)2+4
ⓑ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, 4), y-intercept of (0, 6), and axis of symmetry shown at x equals negative 1.

f(x)=3x2−12x+7

f(x)=−x2+2x−4

Solution

ⓐ f(x)=−(x−1)2−3
ⓑ
This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 3), y-intercept of (0, negative 4), and axis of symmetry shown at x equals 1.

f(x)=−2x2−4x−5

Matching

In the following exercises, match the graphs to one of the following functions: ⓐ f(x)=x2+4 ⓑ f(x)=x2−4 ⓒ f(x)=(x+4)2 ⓓ f(x)=(x−4)2 ⓔ f(x)=(x+4)2−4 ⓕ f(x)=(x+4)2+4 ⓖ f(x)=(x−4)2−4 ⓗ f(x)=(x−4)2+4

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 4, 0) and other points (negative 4, 4) and (negative 2, 4).
Solution

ⓒ

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (0, negative 4) and other points (negative 2, 0) and (2, 0).
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 4, negative 4) and other points (negative 4, 0) and (negative 2, 0).
Solution

ⓔ

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 4, 4) and other points (negative 6, 8) and (negative 2, 8).
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, 0) and other points (2, 4) and (2, 4).
Solution

ⓓ

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (0, 4) and other points (negative 2, 8) and (2, 8).
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 4) and other points (2,0) and (6,0).
Solution

ⓖ

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, 4) and other points (2,8) and (6,8).

Find a Quadratic Function from its Graph

In the following exercises, write the quadratic function in f(x)=a(x−h)2+k form whose graph is shown.

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, negative 5) and y-intercept (0, negative 4).
Solution

f(x)=(x+1)2−5

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2,4) and y-intercept (0, 8).
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 3) and y-intercept (0, negative 1).
Solution

f(x)=2(x−1)2−3

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, negative 5) and y-intercept (0, negative 3).

Writing Exercise

Graph the quadratic function f(x)=x2+4x+5 first using the properties as we did in the last section and then graph it using transformations. Which method do you prefer? Why?

Solution

Answers will vary.

Graph the quadratic function f(x)=2x2−4x−3 first using the properties as we did in the last section and then graph it using transformations. Which method do you prefer? Why?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This figure is a list to assess your understanding of the concepts presented in this section. It has 4 columns labeled I can…, Confidently, With some help, and No-I don’t get it! Below I can…, there is graph Quadratic Functions of the form f of x equals x squared plus k; graph Quadratic Functions of the form f of x equals the quantity x minus h squared; graph Quadratic functions of the form f of x equals a times x squared; graph Quadratic Functions Using Transformations; find a Quadratic Function from its Graph. The other columns are left blank for you to check you understanding.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Solve Quadratic Inequalities

Learning Objectives

By the end of this section, you will be able to:

  • Solve quadratic inequalities graphically
  • Solve quadratic inequalities algebraically

Before you get started, take this readiness quiz.

Solve: 2x−3=0.
If you missed this problem, review Example 2 in Use a General Strategy to Solve Linear Equations.

Solution

x=32

Solve: 2y2+y=15.
If you missed this problem, review Example 2 in Polynomial Equations.

Solution

y=−3,y=52

Solve 1x2+2x−8>0
If you missed this problem, review Example 3 in Solve Rational Inequalities.

Solution

−∞,−4∪2,∞

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different.

We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations.

We will solve quadratic inequalities two ways—both graphically and algebraically.

Solve Quadratic Inequalities Graphically

A quadratic equation is in standard form when written as ax2 + bx + c = 0. If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

Quadratic Inequality

A quadratic inequality is an inequality that contains a quadratic expression.

The standard form of a quadratic inequality is written:

ax2+bx+c<0ax2+bx+c≤0 ax2+bx+c>0ax2+bx+c≥0

The graph of a quadratic function f(x) = ax2 + bx + c = 0 is a parabola. When we ask when is ax2 + bx + c < 0, we are asking when is f(x) < 0. We want to know when the parabola is below the x-axis.

When we ask when is ax2 + bx + c > 0, we are asking when is f(x) > 0. We want to know when the parabola is above the x-axis.

The first graph is an upward facing parabola, f of x, on an x y-coordinate plane. To the left of the function, f of x is greater than 0. Between the x-intercepts, f of x is less than 0. To the right of the function, f of x is greater than 0. The second graph is a downward-facing parabola, f of x, on an x y coordinate plane. To the left of the function, f of x is less than 0. Between the x-intercepts, f of x is greater than 0. To the right of the function, f of x is less than 0.

How to Solve a Quadratic Inequality Graphically

Solve x2−6x+8<0 graphically. Write the solution in interval notation.

Solution
The figure is a table with 3 columns. The first column is Step 1: Write the quadratic inequality in standard form. The second column says the inequality is in standard form. The third column says x squared minus 6 times x plus 8 less than 0. The figure is a table with 3 columns. The first column says Step 2-Graph the function f of x equals a times x squared plus b times x plus c using properties or transformations. The second column gives instructions and the third column shows the work for step 3 as follows. We will graph using properties. The function is f of x equals x squared minus 6 times x plus 8 where a equals 1, b equals negative 6, and c equals 8. Look at a in the function f of x equals x squared minus 6 times x plus 8. Since a is positive, the parabola opens upward. The equation of the axis of symmetry is the line x equals negative b divided by 2 times a, so x equals negative negative 6 divided by 2 times 1. X equals 3. The axis of symmetry is the line x equals 3. The vertex is on the axis of symmetry. Substitute x equals 3 into the function, so f of 3 equals 3 squared minus 6 times 3 plus 8. F of 3 equals negative 1, so the vertex is (3, negative 1). We find f of 0 in order to find the y-intercept, so f of 0 equals 0 squared minus 6 times 0 plus 8. F of 0 equals 8, so the y intercept is (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. Point symmetric to y-intercept is (6, 8). We solve f of x equals 0 in order to find the x-intercepts. We can solve this quadratic equation by factoring. 0 equals x squared minus 6 times x plus 8, 0 equals the quantity x minus 2 times the quantity x minus 4, x equals 2 or x equals 4. The x-intercepts are (2, 0) and (4, 0). We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these 5 points to sketch the parabola shown that is upward-facing with the points found through this process. The figure is a table with 3 columns. The first column says Step 3- Determine the solution from the graph. The second column gives instructions. X squared minus 6 x plus 8 less than 0. The inequality asks for the values of x which make the function less than 0. Which values of x make the parabola below the x-axis. We do not include the values 2, 4 as the inequality is strictly less than. The third column says The solution, in interval notation, is (2, 4).

ⓐ Solve x2+2x−8<0 graphically and ⓑ write the solution in interval notation.

Solution


ⓐ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, negative 9), y-intercept of (0, 8), and axis of symmetry shown at x equals negative 2.
ⓑ (−4,2)

ⓐ Solve x2−8x+12≥0 graphically and ⓑ write the solution in interval notation.

Solution


ⓐ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 4) and x-intercepts of (2, 0) and (6, 0).
ⓑ (−∞,2]∪[6,∞)

We list the steps to take to solve a quadratic inequality graphically.

Solve a quadratic inequality graphically.

  1. Write the quadratic inequality in standard form.
  2. Graph the function f(x)=ax2+bx+c.
  3. Determine the solution from the graph.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the x-axis but note how the position of the parabola affects the solution.

Solve −x2−8x−12≤0 graphically. Write the solution in interval notation.

Solution
The quadratic inequality in standard form. −x2−8x−12≤0
Graph the function f(x)=−x2−8x−12. The parabola opens downward.
A red U-shaped arrow pointing downwards on both ends, suggesting a flow or return, against a plain white background.
Find the line of symmetry. x=−b2a
x=−−82(−1)
x=−4
Find the vertex. f(x)=−x2−8x−12
f(−4)=−(−4)2−8(−4)−12
f(−4)=−16+32−12
f(−4)=4
Vertex (−4,4)
Find the x-intercepts. Let f(x)=0. f(x)=−x2−8x−12
0=−x2−8x−12
Factor.
Use the Zero Product Property.
0=−1(x+6)(x+2)
x=−6x=−2
Graph the parabola. x-intercepts (−6,0),(−2,0)
A graph shows a downward-opening parabola on a Cartesian coordinate system. The parabola has its vertex at (-4, 4) and intersects the x-axis at (-6, 0) and (-2, 0).
Determine the solution from the graph.
We include the x-intercepts as the inequality
is “less than or equal to.”
(−∞,−6]∪[−2,∞)

ⓐ Solve −x2−6x−5>0 graphically and ⓑ write the solution in interval notation.

Solution


ⓐ
A downward-facing parabola on the x y-coordinate plane. It has a vertex of (negative 3, 4), a y-intercept at (0, negative 5), and an axis of symmetry shown at x equals negative 3.
ⓑ (−5,−1)

ⓐ Solve −x2+10x−16≤0 graphically and ⓑ write the solution in interval notation.

Solution


ⓐ
A downward-facing parabola on the x y-coordinate plane. It has a vertex of (5, 9), a y-intercept at (0, negative 16), and an axis of symmetry of x equals 5.
ⓑ (−∞,2]∪[8,∞)

Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the zero partition numbers for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the zero partition numbers to divide the number line into intervals and then determine whether the quadratic expression will be positive or negative in the interval. We then determine the solution for the inequality.

How To Solve Quadratic Inequalities Algebraically

Solve x2−x−12≥0 algebraically. Write the solution in interval notation.

Solution
This figure is a table giving the instructions for solving x squared minus x minus 12 greater than or equal to 0 algebraically. It consists of 3 columns where the instructions are given in the first column, the explanation in the second, and the work in the third. Step 1 is to write the quadratic inequality in standard form. The quadratic inequality in already in standard form, so x squared minus x minus 12 greater than or equal to 0. Step 2 is to determine the zero partition number -- the solutions to the related quadratic equation. To do this, change the inequality sign to an equal sign and then solve the equation. x squared minus x minus 12 equals 0 factors to the quantity x plus 3 times the quantity x minus 4 equals 0. Then, x plus 3 equals 0 and x minus 4 equals 0 to give x equals negative 3 and x equals 4. Step 3 is to use the zero partition numbers to divide the number line into intervals. Use negative 3 and 4 to divide the number line into intervals. A number line is shown that includes from left to right the values of negative 3, 0, and 4, with dotted lines on negative 3 and 4. Step 4 says above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality. X equals negative 5, x equals 0, and x equals 5 are chosen to test. The expression negative x squared minus x minus 12 is given with negative 5 squared minus negative 5 minus 12 underneath, which gives 18. The expression negative x squared minus x minus 12 is given with 0 squared minus 0 minus 12 underneath, which gives 12. The expression negative x squared minus x minus 12 is given with 5 squared minus 5 minus 12 underneath, which gives 8. For Step 5, determine the intervals where the inequality is correct. Write the solution in interval notation. x squared minus x minus 12 greater than or equal to 0 is shown. The inequality is positive in the first and last intervals and equals 0 at the points negative 4, 3 . The solution, in interval notation, is (negative infinity, negative 3] in union with [4, infinity).

Solve x2+2x−8≥0 algebraically. Write the solution in interval notation.

Solution

(−∞,−4]∪[2,∞)

Solve x2−2x−15≤0 algebraically. Write the solution in interval notation.

Solution

[−3,5]

In this example, since the expression x2−x−12 factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

The figure shows the expression x squared minus x minus 12 factored to the quantity of x plus 3 times the quantity of x minus 4. The image shows a number line showing dotted lines on negative 3 and 4. It shows the signs of the quantity x plus 3 to be negative, positive, positive, and the signs of the quantity x minus 4 to be negative, negative, positive. Under the number line, it shows the quantity x plus 3 times the quantity x minus 4 with the signs positive, negative, positive.

The result is the same as we found using the other method.

We summarize the steps here.

Solve a quadratic inequality algebraically.

  1. Write the quadratic inequality in standard form.
  2. Determine the zero partition numbers—the solutions to the related quadratic equation.
  3. Use the zero partition numbers to divide the number line into intervals.
  4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
  5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Solve -x2+6x−7≥0 algebraically. Write the solution in interval notation.

Solution
Write the quadratic inequality in standard form. −x2+6x−7≥0
Multiply both sides of the inequality by −1.
Remember to reverse the inequality sign.
x2−6x+7≤0
Determine the zero partition numbers by solving
the related quadratic equation.
x2−6x+7=0
Write the Quadratic Formula. x=−b±b2−4ac2a
Then substitute in the values of a,b,c. x=−(−6)±(−6)2−4⋅1⋅(7)2⋅1
Simplify. x=6±82
Simplify the radical. x=6±222
Remove the common factor, 2. x=2(3±2)2
x=3±2
x=3+2x=3−2
x≈1.6x≈4.4
Use the zero partition numbers to divide the
number line into intervals.
Test numbers from each interval
in the original inequality.
Sign analysis of the quadratic expression -x^2 + 6x - 7 on a number line, showing it's positive between 3-sqrt(2) (approx 1.6) and 3+sqrt(2) (approx 4.4), and negative elsewhere.
Determine the intervals where the
inequality is correct. Write the solution
in interval notation.
−x2+6x−7≥0 in the middle interval
[3−2,3+2]

Solve −x2+2x+1≥0 algebraically. Write the solution in interval notation.

Solution

[−1−2,−1+2]

Solve −x2+8x−14<0 algebraically. Write the solution in interval notation.

Solution

(−∞,4−2)∪(4+2,∞)

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation ax2 + bx + c = 0. These two solutions then gave us either the two x-intercepts for the graph or the two zero partition numbers to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

For a quadratic equation of the form ax2 + bx + c = 0, a≠0.

The figure is a table with 3 columns. Column 1 is labeled discriminant, column 2 is Number/Type of solution, and column 3 is Typical Graph. Reading across the columns, if b squared minus 4 times a times c is greater than 0, there will be 2 real solutions because there are 2 x-intercepts on the graph. The image of a typical graph an upward or downward parabola with 2 x-intercepts. If the discriminant b squared minus 4 times a times c is equals to 0, then there is 1 real solution because there is 1 x-intercept on the graph. The image of the typical graph is an upward- or downward-facing parabola that has a vertex on the x-axis instead of crossing through it. If the discriminant b squared minus 4 times a times c is less than 0, there are 2 complex solutions because there is no x-intercept. The image of the typical graph shows an upward- or downward-facing parabola that does not cross the x-axis.

The last row of the table shows us when the parabolas never intersect the x-axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

Solve, writing any solution in interval notation:

ⓐ x2−3x+4>0 ⓑ x2−3x+4≤0

Solution
ⓐ
Write the quadratic inequality in standard form. x2−3x+4>0
Determine the zero partition numbers by solving
the related quadratic equation.
x2−3x+4=0
Write the Quadratic Formula. x=−b±b2−4ac2a
Then substitute in the values of a,b,c. x=−(−3)±(−3)2−4⋅1⋅(4)2⋅1
Simplify. x=3±−72
Simplify the radicand. x=3±7i2
The complex solutions tell us the
parabola does not intercept the x-axis.
Also, the parabola opens upward. This
tells us that the parabola is completely above the x-axis.
Complex solutions
A white background features a horizontal line with arrows pointing left and right. Above the center of this line, a curved arrow forms an upward-pointing U-shape, with both ends curving upwards and terminating in arrowheads.

We are to find the solution to x2−3x+4>0. Since for all values of x the graph is above the x-axis, all values of x make the inequality true. In interval notation we write (−∞,∞).

ⓑ
Steps and corresponding mathematical expressions for solving a quadratic inequality, including standard form and the related quadratic equation.
Write the quadratic inequality in standard form. x2−3x+4≤0
Determine the zero partition numbers by solving the related quadratic equation x2−3x+4=0

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the x-axis—no part of it is below the x-axis.

We are to find the solution to x2−3x+4≤0. Since for all values of x the graph is never below the x-axis, no values of x make the inequality true. There is no solution to the inequality.

Solve and write any solution in interval notation:
ⓐ −x2+2x−4≤0 ⓑ −x2+2x−4≥0

Solution

ⓐ (−∞,∞)
ⓑ no solution

Solve and write any solution in interval notation:
ⓐ x2+3x+3<0 ⓑ x2+3x+3>0

Solution

ⓐ no solution
ⓑ (−∞,∞)

Key Concepts

  • Solve a Quadratic Inequality Graphically
    1. Write the quadratic inequality in standard form.
    2. Graph the function f(x)=ax2+bx+c using properties or transformations.
    3. Determine the solution from the graph.
  • How to Solve a Quadratic Inequality Algebraically
    1. Write the quadratic inequality in standard form.
    2. Determine the zero partition numbers -- the solutions to the related quadratic equation.
    3. Use the zero partition numbers to divide the number line into intervals.
    4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
    5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Section Exercises

Practice Makes Perfect

Solve Quadratic Inequalities Graphically

In the following exercises, ⓐ solve graphically and ⓑ write the solution in interval notation.

x2+6x+5>0

Solution


ⓐ
The graph shown is an upward-facing parabola with vertex (negative 3, negative 4) and y-intercept (0,5).
ⓑ (−∞,−5)∪(−1,∞)

x2+4x−12<0

x2+4x+3≤0

Solution


ⓐ
The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0,3).
ⓑ [−3,−1]

x2−6x+8≥0

−x2−3x+18≤0

Solution


ⓐ
The graph shown is a downward-facing parabola with vertex (negative 1 and 5 tenths, 20) and y-intercept (0, 18).
ⓑ (−∞,−6]∪[3,∞)

−x2+2x+24<0

−x2+x+12≥0

Solution


ⓐ
The graph shown is a downward facing parabola with a y-intercept of (0, 12) and x-intercepts (negative 3, 0) and (4, 0).
ⓑ [−3,4]

−x2+2x+15>0

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

x2+3x−4≥0

Solution

(−∞,−4]∪[1,∞)

x2+x−6≤0

x2−7x+10<0

Solution

(2,5)

x2−4x+3>0

x2+8x>−15

Solution

(−∞,−5)∪(−3,∞)

x2+8x<−12

x2−4x+2≤0

Solution

[2−2,2+2]

−x2+8x−11<0

x2−10x>−19

Solution

(−∞,5−6)∪(5+6,∞)

x2+6x<−3

−6x2+19x−10≥0

Solution

23, 52

−3x2−4x+4≤0

−2x2+7x+4≥0

Solution

[−12,4]

2x2+5x−12>0

x2+3x+5>0

Solution

(−∞,∞).

x2−3x+6≤0

−x2+x−7>0

Solution

no solution

−x2−4x−5<0

−2x2+8x−10<0

Solution

(−∞,∞).

−x2+2x−7≥0

Writing Exercises

Explain zero partition numbers and how they are used to solve quadratic inequalities algebraically.

Solution

Answers will vary.

Solve x2+2x≥8 both graphically and algebraically. Which method do you prefer, and why?

Describe the steps needed to solve a quadratic inequality graphically.

Solution

Answers will vary.

Describe the steps needed to solve a quadratic inequality algebraically.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This figure is a list to assess your understanding of the concepts presented in this section. It has 4 columns labeled I can…, Confidently, With some help, and No-I don’t get it! Below I can…, there is solve quadratic inequalities graphically and solve quadratic inequalities algebraically. The other columns are left blank for you to check you understanding.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Chapter Review Exercises

Solve Quadratic Equations Using the Square Root Property

Solve Quadratic Equations of the form ax2 = k Using the Square Root Property

In the following exercises, solve using the Square Root Property.

y2=144

Solution

y=±12

n2−80=0

4a2=100

Solution

a=±5

2b2=72

r2+32=0

Solution

r=±42i

t2+18=0

23w2−20=30

Solution

w=±53

5c2+3=19

Solve Quadratic Equations of the Form a(x−h)2=k Using the Square Root Property

In the following exercises, solve using the Square Root Property.

(p−5)2+3=19

Solution

p=1,p=9

(u+1)2=45

(x−14)2=316

Solution

x=14±34

(y−23)2=29

(n−4)2−50=150

Solution

n=4±102

(4c−1)2=−18

n2+10n+25=12

Solution

n=−5±23

64a2+48a+9=81

Solve Quadratic Equations by Completing the Square

Solve Quadratic Equations Using Completing the Square

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

x2+22x

Solution

(x+11)2

m2−8m

a2−3a

Solution

(a−32)2

b2+13b

In the following exercises, solve by completing the square.

d2+14d=−13

Solution

d=−13,−1

y2−6y=36

m2+6m=−109

Solution

m=−3±10i

t2−12t=−40

v2−14v=−31

Solution

v=7±32

w2−20w=100

m2+10m−4=−13

Solution

m=−9,−1

n2−6n+11=34

a2=3a+8

Solution

a=32±412

b2=11b−5

(u+8)(u+4)=14

Solution

u=−6±32

(z−10)(z+2)=28

Solve Quadratic Equations of the form ax2 + bx + c = 0 by Completing the Square

In the following exercises, solve by completing the square.

3p2−18p+15=15

Solution

p=0,6

5q2+70q+20=0

4y2−6y=4

Solution

y=−12,2

2x2+2x=4

3c2+2c=9

Solution

c=−13±273

4d2−2d=8

2x2+6x=−5

Solution

x=32±12i

2x2+4x=−5

Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

4x2−5x+1=0

Solution

x=14,1

7y2+4y−3=0

r2−r−42=0

Solution

r=−6,7

t2+13t+22=0

4v2+v−5=0

Solution

v=−54,1

2w2+9w+2=0

3m2+8m+2=0

Solution

m=−4±103

5n2+2n−1=0

6a2−5a+2=0

Solution

a=512±2312i

4b2−b+8=0

u(u−10)+3=0

Solution

u=5±22

5z(z−2)=3

18p2−15p=−120

Solution

p=4±65

25q2+310q=110

4c2+4c+1=0

Solution

c=−12

9d2−12d=−4

Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation

In the following exercises, determine the number of solutions for each quadratic equation.


ⓐ 9x2−6x+1=0
ⓑ 3y2−8y+1=0
ⓒ 7m2+12m+4=0
ⓓ 5n2−n+1=0

Solution

ⓐ 1 ⓑ 2 ⓒ 2 ⓓ 0


ⓐ 5x2−7x−8=0
ⓑ 7x2−10x+5=0
ⓒ 25x2−90x+81=0
ⓓ 15x2−8x+4=0

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.


ⓐ 16r2−8r+1=0
ⓑ 5t2−8t+3=9
ⓒ 3(c+2)2=15

Solution

ⓐ factor ⓑ Quadratic Formula ⓒ square root


ⓐ 4d2+10d−5=21
ⓑ 25x2−60x+36=0
ⓒ 6(5v−7)2=150

Solve Equations in Quadratic Form

Solve Equations in Quadratic Form

In the following exercises, solve.

x4−14x2+24=0

Solution

x=±2,±23

x4+4x2−32=0

4x4−5x2+1=0

Solution

x=±1,±12

(2y+3)2+3(2y+3)−28=0

x+3x−28=0

Solution

x=16

6x+5x−6=0

x23−10x13+24=0

Solution

x=64,216

x+7x12+6=0

8x−2−2x−1−3=0

Solution

x=−2,43

Solve Applications of Quadratic Equations

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve by using the method of factoring, the square root principle, or the Quadratic Formula. Round your answers to the nearest tenth, if needed.

Find two consecutive odd numbers whose product is 323.

Find two consecutive even numbers whose product is 624.

Solution

Two consecutive even numbers whose product is 624 are 24 and 26, and −24 and −26.

A triangular banner has an area of 351 square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.

Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is 70 square inches. Find the height and width of the case.

Solution

The height is 14 inches and the width is 10 inches.

A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is 5 feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.

A rectangle is shown is a right triangle in the corner. The hypotenuse of the triangle is 5 feet, the longer leg is 2 times s and the shorter leg is s.

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.

Solution

The length of the diagonal is 3.6 feet.

The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.

For Sophia’s graduation party, several tables of the same width will be arranged end to end to give serving table with a total area of 75 square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth . Round answer to the nearest tenth.

Solution

The width of the serving table is 4.7 feet and the length is 16.1 feet.
Four tables arranged end-to-end are shown. Together, they have an area of 75 feet. The short side measures w and the long side measures 3 times w plus 2.

A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula h = −16t2 + v0t to determine when the ball will be 384 feet from the ground. Round to the nearest tenth.

The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 360 miles. If the plane was flying at 150 mph, what was the speed of the wind that affected the plane?

Solution

The speed of the wind was 30 mph.

Ezra kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?

Two handymen can do a home repair in 2 hours if they work together. One of the men takes 3 hours more than the other man to finish the job by himself. How long does it take for each handyman to do the home repair individually?

Solution

One man takes 3 hours and the other man 6 hours to finish the repair alone.

Graph Quadratic Functions Using Properties

Recognize the Graph of a Quadratic Function

In the following exercises, graph by plotting point.

Graph y=x2−2

Graph y=−x2+3

Solution


This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (3, 0) and other points of (negative 2, negative 1) and (2, negative 1).

In the following exercises, determine if the following parabolas open up or down.


ⓐ y=−3x2+3x−1
ⓑ y=5x2+6x+3


ⓐ y=x2+8x−1
ⓑ y=−4x2−7x+1

Solution

ⓐ up ⓑ down

Find the Axis of Symmetry and Vertex of a Parabola

In the following exercises, find ⓐ the equation of the axis of symmetry and ⓑ the vertex.

y=−x2+6x+8

y=2x2−8x+1

Solution

x=2;(2,−7)

Find the Intercepts of a Parabola

In the following exercises, find the x- and y-intercepts.

y=x2−4x−5

y=x2−8x+15

Solution

y:(0,15)x:(3,0),(5,0)

y=x2−4x+10

y=−5x2−30x−46

Solution

y:(0,−46)x:none

y=16x2−8x+1

y=x2+16x+64

Solution

y:(0,64)x:(−8,0)

Graph Quadratic Functions Using Properties

In the following exercises, graph by using its properties.

y=x2+8x+15

y=x2−2x−3

Solution


This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 4) and a y-intercept of (0, negative 3).

y=−x2+8x−16

y=4x2−4x+1

Solution


This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (one-half, 0) and a y-intercept of (0, 1).

y=x2+6x+13

y=−2x2−8x−12

Solution


This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, negative 4) and a y-intercept of (0, negative 12).

Solve Maximum and Minimum Applications

In the following exercises, find the minimum or maximum value.

y=7x2+14x+6

y=−3x2+12x−10

Solution

The maximum value is 2 when x = 2.

In the following exercises, solve. Rounding answers to the nearest tenth.

A ball is thrown upward from the ground with an initial velocity of 112 ft/sec. Use the quadratic equation h = −16t2 + 112t to find how long it will take the ball to reach maximum height, and then find the maximum height.

A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using 180 feet of fencing on three sides of the yard. The quadratic equation A(x)=x90−x2 gives the area, A, of the yard for the length, x, of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.

An odd-shaped figure is given. 3 sides of a rectangle are attached to the right side of the figure.
Solution

The length adjacent to the building is 90 feet giving a maximum area of 4,050 square feet.

Graph Quadratic Functions Using Transformations

Graph Quadratic Functions of the form f(x)=x2+k

In the following exercises, graph each function using a vertical shift.

g(x)=x2+4

h(x)=x2−3

Solution


This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 3, 0) and other points of (negative 1, negative 2) and (1, negative 2).

In the following exercises, graph each function using a horizontal shift.

f(x)=(x+1)2

g(x)=(x−3)2

Solution


This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (3, 0) and other points of (2, 1) and (4,1).

In the following exercises, graph each function using transformations.

f(x)=(x+2)2+3

f(x)=(x+3)2−2

Solution


This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 3, negative 2) and other points of (negative 5, 2) and (negative 1, 2).

f(x)=(x−1)2+4

f(x)=(x−4)2−3

Solution


This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 3) and other points of (3, negative 2) and (5, negative 2).

Graph Quadratic Functions of the form f(x)=ax2

In the following exercises, graph each function.

f(x)=2x2

f(x)=−x2

Solution


This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (0, 0) and other points of (negative 1, negative 1) and (1, negative 1).

f(x)=12x2

Graph Quadratic Functions Using Transformations

In the following exercises, rewrite each function in the f(x)=a(x−h)2+k form by completing the square.

f(x)=2x2−4x−4

Solution

f(x)=2(x−1)2−6

f(x)=3x2+12x+8

In the following exercises, ⓐ rewrite each function in f(x)=a(x−h)2+k form and ⓑ graph it by using transformations.

f(x)=3x2−6x−1

Solution

ⓐ f(x)=3(x−1)2−4
ⓑ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 4) and other points of (0, negative 1) and (2, negative 1).

f(x)=−2x2−12x−5

f(x)=2x2+4x+6

Solution

ⓐ f(x)=2(x+1)2+4
ⓑ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, 4) and other points of (negative 2, 6) and (0, 6).

f(x)=3x2−12x+7

In the following exercises, ⓐ rewrite each function in f(x)=a(x−h)2+k form and ⓑ graph it using properties.

f(x)=−3x2−12x−5

Solution

ⓐ f(x)=−3(x+2)2+7
ⓑ
This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, 7) and other points of (negative 4, negative 5) and (0, negative 5).

f(x)=2x2−12x+7

Find a Quadratic Function from its Graph

In the following exercises, write the quadratic function in f(x)=a(x−h)2+k form.

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, negative 1) and other points of (negative 2, negative 4) and (0, negative 4).
Solution

f(x)=(x+1)2−5

This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2, 4) and other points of (0, 8) and (4, 8).

Solve Quadratic Inequalities

Solve Quadratic Inequalities Graphically

In the following exercises, solve graphically and write the solution in interval notation.

x2−x−6>0

Solution


ⓐ
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (one-half, negative 6 and one-fourth) and other points of (0, negative 6) and (1, negative 6).
ⓑ (−∞,−2)∪(3,∞)

x2+4x+3≤0

−x2−x+2≥0

Solution


ⓐ
This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative one-half, 2 and one-fourth) and other points of (negative 2, 0) and (1, 0).
ⓑ [−2,1]

−x2+2x+3<0

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

x2−6x+8<0

Solution

(2,4)

x2+x>12

x2−6x+4≤0

Solution

[3−5,3+5]

2x2+7x−4>0

−x2+x−6>0

Solution

no solution

x2−2x+4≥0

Practice Test

Use the Square Root Property to solve the quadratic equation 3(w+5)2=27.

Solution

w=−2,w=−8

Use Completing the Square to solve the quadratic equation a2−8a+7=23.

Use the Quadratic Formula to solve the quadratic equation 2m2−5m+3=0.

Solution

m=1,m=32

Solve the following quadratic equations. Use any method.

2x(3x−2)−1=0

94y2−3y+1=0

Solution

y=23

Use the discriminant to determine the number and type of solutions of each quadratic equation.

6p2−13p+7=0

3q2−10q+12=0

Solution

2 complex

Solve each equation.

4x4−17x2+4=0

y23+2y13−3=0

Solution

y=1,y=−27

For each parabola, find ⓐ which direction it opens, ⓑ the equation of the axis of symmetry, ⓒ the vertex, ⓓ the x- and y-intercepts, and e) the maximum or minimum value.

y=3x2+6x+8

y=−x2−8x−16

Solution

ⓐ down ⓑ x=−4
ⓒ (−4,0) ⓓ y:(0,−16);x:(−4,0)
ⓔ maximum value of 0 when x=−4

Graph each quadratic function using intercepts, the vertex, and the equation of the axis of symmetry.

f(x)=x2+6x+9

f(x)=−2x2+8x+4

Solution

This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (2, 12) and other points of (0, 4) and (4, 4).

In the following exercises, graph each function using transformations.

f(x)=(x+3)2+2

f(x)=x2−4x−1

Solution


This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2, negative 5) and other points of (0, negative 1) and (4, negative 1).
f(x)=2(x−1)2−6

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

x2−6x−8≤0

2x2+x−10>0

Solution

(−∞,−52)∪(2,∞)

Model the situation with a quadratic equation and solve by any method.

Find two consecutive even numbers whose product is 360.

The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)

Solution

The diagonal is 4.4 units long.

A water balloon is launched upward at the rate of 86 ft/sec. Using the formula h = −16t2 + 86t find how long it will take the balloon to reach the maximum height, and then find the maximum height. Round to the nearest tenth.

quadratic inequality
A quadratic inequality is an inequality that contains a quadratic expression.

Introduction

A photo of rows of vegetables growing in water in a greenhouse.
Hydroponic systems allow botanists to grow crops without land. (credit: “Izhamwong”/Wikimedia Commons)

As the world population continues to grow, food supplies are becoming less able to meet the increasing demand. At the same time, available resources of fertile soil for growing plants is dwindling. One possible solution—grow plants without soil. Botanists around the world are expanding the potential of hydroponics, which is the process of growing plants without soil. To provide the plants with the nutrients they need, the botanists keep careful growth records. Some growth is described by the types of functions you will explore in this chapter—exponential and logarithmic. You will evaluate and graph these functions, and solve equations using them.

Finding Composite and Inverse Functions

Learning Objectives

By the end of this section, you will be able to:

  • Find and evaluate composite functions
  • Determine whether a function is one-to-one
  • Find the inverse of a function

Before you get started, take this readiness quiz.

If f(x)=2x−3 and g(x)=x2+2x−3, find f(4).
If you missed this problem, review Example 7 in Relations and Functions.

Solution

ⓐ f4=5; ⓑ gf4=32

Solve for x, 3x+2y=12.
If you missed this problem, review Example 4 in Solve a Formula for a Specific Variable.

Solution

x=−23y+4

Simplify: 5(x+4)5−4.
If you missed this problem, review Example 2 in Fractions.

Solution

x

In this chapter, we will introduce two new types of functions, exponential functions and logarithmic functions. These functions are used extensively in business and the sciences as we will see.

Find and Evaluate Composite Functions

Before we introduce the functions, we need to look at another operation on functions called composition. In composition, the output of one function is the input of a second function. For functions f and g, the composition is written f∘g and is defined by (f∘g)(x)=f(g(x)).

We read f(g(x)) as “f of g of x.”

This figure shows x as the input to a box denoted as function g with g of x as the output of the box. Then, g of x is the input to a box denoted as function f with f of g of x as the output of the box.

To do a composition, the output of the first function, g(x), becomes the input of the second function, f, and so we must be sure that it is part of the domain of f.

Composition of Functions

The composition of functions f and g is written f∘g and is defined by

(f∘g)(x)=f(g(x))

We read f(g(x)) as f of g of x.

We have actually used composition without using the notation many times before. When we graphed quadratic functions using translations, we were composing functions. For example, if we first graphed g(x)=x2 as a parabola and then shifted it down vertically four units, we were using the composition defined by (f∘g)(x)=f(g(x)) where f(x)=x−4.

This figure shows x as the input to a box denoted as g of x equals x squared with x squared as the output of the box. Then, x squared is the input to a box denoted as f of x equals x minus 4 with f of g of x equals x squared minus 4 as the output of the box.

The next example will demonstrate that (f∘g)(x), (g∘f)(x) and (f·g)(x) usually result in different outputs.

For functions f(x)=4x−5 and g(x)=2x+3, find: ⓐ (f∘g)(x), ⓑ (g∘f)(x), and ⓒ (f·g)(x).

Solution
ⓐ
Use the definition of (f∘g)(x). The image shows the definition of a composite function, stating that (f o g)(x) is equal to f(g(x)). This mathematical notation explains how to apply one function to the result of another function.
The text 'Substitute 2x + 3 for g(x)' is displayed in black font, with '2x + 3' highlighted in red font, against a white background. Mathematical notation demonstrating the composition of two functions, (f o g)(x), which is further expressed as f(2x + 3), implying that the inner function g(x) is equal to 2x + 3.
A mathematical problem asking to find f(2x + 3) given that f(x) = 4x - 5. The expression '2x + 3' is highlighted in red, as is the 'x' in 'f(x)' and '4x' to denote variable substitution. The image shows a mathematical equation for the composition of two functions, f and g, denoted as (f o g)(x) = 4(2x + 3) - 5. The expression 2x+3 is highlighted in red.
Distribute. A mathematical equation shows the composition of two functions, (f ∘ g)(x), which equals the expression 8x + 12 - 5. This simplifies to (f ∘ g)(x) = 8x + 7.
Simplify. The image shows the mathematical expression for a composite function, stating that (f o g)(x) = 8x + 7, where 'f' and 'g' are functions and 'o' denotes their composition.


ⓑ
Use the definition of (f∘g)(x). The definition of a composite function: (g o f)(x) = g(f(x)).
Substitute 4x - 5 for f(x). The image displays the composition of functions, showing (g o f)(x) as g(4x - 5).
Find g(4x - 5) where g(x) = 2x + 3. A mathematical equation showing the composition of functions (g o f)(x) = 2(4x - 5) + 3, with the term 4x - 5 highlighted in red.
Distribute. A mathematical equation shows a composite function (g o f)(x) equals 8x - 10 + 3.
Simplify. The image shows the composition of two functions (g o f)(x) equal to 8x - 7.

Notice the difference in the result in part ⓐ and part ⓑ.

ⓒ Notice that (f·g)(x) is different than (f∘g)(x). In part ⓐ we did the composition of the functions. Now in part ⓒ we are not composing them, we are multiplying them.

Use the definition of(f·g)(x).(f·g)(x)=f(x)·g(x)Substitutef(x)=4x−5andg(x)=2x+3.(f·g)(x)=(4x−5)·(2x+3)Multiply.(f·g)(x)=8x2+2x−15

For functions f(x)=3x−2 and g(x)=5x+1, find ⓐ (f∘g)(x) ⓑ (g∘f)(x) ⓒ (f·g)(x).

Solution

ⓐ 15x+1 ⓑ 15x−9
ⓒ 15x2−7x−2

For functions f(x)=4x−3, and g(x)=6x−5, find ⓐ (f∘g)(x), ⓑ (g∘f)(x), and ⓒ (f·g)(x).

Solution

ⓐ 24x−23 ⓑ 24x−23
ⓒ 24x2−38x+15

In the next example we will evaluate a composition for a specific value.

For functions f(x)=x2−4, and g(x)=3x+2, find: ⓐ (f∘g)(−3), ⓑ (g∘f)(−1), and ⓒ (f∘f)(2).

Solution
ⓐ
Use the definition of (f∘g)(−3). A mathematical expression showing the composition of two functions, f and g, evaluated at -3. The notation (f o g)(-3) is explicitly shown to be equal to f(g(-3)).
The image displays a mathematical problem asking to find the value of the function g at x = -3, given that g(x) = 3x + 2. A mathematical equation shows the composition of functions (f o g)(-3) expanded as f(3 * (-3) + 2). The number -3 is highlighted in red within the expanded expression, indicating its role in the calculation.
Simplify. A mathematical equation showing the composition of functions, stating that (f composed with g) evaluated at -3 is equal to f evaluated at -7, written as (f ∘ g)(-3) = f(-7).
The image shows a mathematical problem that asks to 'Find f(-7) where f(x) = x^2 - 4.' The image shows the mathematical expression for a composite function: (f o g)(-3) = (-7)^2 - 4. The value -7 is highlighted in red, indicating a substitution or an intermediate step in the calculation.
Simplify. The image shows the mathematical expression for a composite function: (f o g)(-3) = 45.


ⓑ
Use the definition of (g∘f)(−1). Composite function notation showing that (g composed with f) of negative 1 equals g of f of negative 1.
A mathematics problem asks to find the value of f(-1) given the function f(x) = x^2 - 4. The text is clear and readable on a white background. A mathematical equation demonstrating function composition, (g composed with f) at -1, which simplifies to g evaluated at the expression (-1)^2 - 4. The -1 in the exponent is highlighted.
Simplify. A mathematical equation showing the composition of two functions, (g o f)(-1) = g(-3).
A math problem states: 'Find g(-3) where g(x) = 3x + 2.' The number -3 and the variable x in g(x) and 3x are highlighted in red. The image shows the mathematical equation for a composite function evaluation, specifically (g  f)(-1) = 3(-3) + 2, where -3 is highlighted in red.
Simplify. A mathematical expression showing the composition of two functions: (g o f)(-1) = -7. The expression is rendered in black text on a white background, representing a function evaluation resulting in a negative integer.


ⓒ
Use the definition of (f∘f)(2). (f o f)(2) = f(f(2)) is shown, illustrating the composition of function f with itself, evaluated at x=2.
A mathematical problem asking to find the value of f(2) given the function f(x) = x^2 - 4. The number 2 and the variable x in f(x) are highlighted in red. A mathematical equation shows the composite function (f o f)(2) is equal to f(2^2 - 4). The number 2 in the exponent of 2^2 is highlighted in red, indicating a specific part of the expression.
Simplify. A mathematical equation shows the composition of a function with itself evaluated at 2, equaling the function evaluated at 0: (f o f)(2) = f(0).
The image shows the mathematical problem: Find f(0) where f(x) = x^2 - 4. A mathematical equation showing the composition of a function with itself evaluated at 2, setting (f o f)(2) equal to 0 squared minus 4. The number 0 is highlighted in red.
Simplify. A mathematical expression reads (f o f)(2) = -4, representing the composite function f of f evaluated at 2 equals -4. The notation is clear and centrally positioned.

For functions f(x)=x2−9, and g(x)=2x+5, find ⓐ (f∘g)(−2), ⓑ (g∘f)(−3), and ⓒ (f∘f)(4).

Solution

ⓐ –8 ⓑ 5 ⓒ 40

For functions f(x)=x2+1, and g(x)=3x−5, find ⓐ (f∘g)(−1), ⓑ (g∘f)(2), and ⓒ (f∘f)(−1).

Solution

ⓐ 65 ⓑ 10 ⓒ 5

Determine Whether a Function is One-to-One

When we first introduced functions, we said a function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x-value is matched with only one y-value.

We used the birthday example to help us understand the definition. Every person has a birthday, but no one has two birthdays and it is okay for two people to share a birthday. Since each person has exactly one birthday, that relation is a function.

This figure shows two tables. To the left is the table labeled Name, which from top to bottom reads Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, and Danny. The table on the right is labeled Birthday, which from top to bottom reads January 12, February 3, April 25, May 10, May 23, July 24, August 2, and September 15. There are arrows going from Alison to April 25, Penelope to May 23, June to August 2, Gregory to September 15, Geoffrey to January 12, Lauren to May 10, Stephen to July 24, Alice to February 3, Liz to July 24, and Danny to no birthday.

A function is one-to-one if each value in the range has exactly one element in the domain. For each ordered pair in the function, each y-value is matched with only one x-value.

Our example of the birthday relation is not a one-to-one function. Two people can share the same birthday. The range value August 2 is the birthday of Liz and June, and so one range value has two domain values. Therefore, the function is not one-to-one.

One-to-One Function

A function is one-to-one if each value in the range corresponds to one element in the domain. For each ordered pair in the function, each y-value is matched with only one x-value. There are no repeated y-values.

For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.

ⓐ {(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)} and ⓑ {(0,0),(1,1),(4,2),(9,3),(16,4)}.

Solution

ⓐ
{(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}

Each x-value is matched with only one y-value. So this relation is a function.

But each y-value is not paired with only one x-value, (−3,27) and (3,27), for example. So this function is not one-to-one.

ⓑ
{(0,0),(1,1),(4,2),(9,3),(16,4)}

Each x-value is matched with only one y-value. So this relation is a function.

Since each y-value is paired with only one x-value, this function is one-to-one.

For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

ⓐ {(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)} ⓑ {(−4,8),(−2,4),(−1,2),(0,0),(1,2),(2,4),(4,8)}

Solution

ⓐ One-to-one function
ⓑ Function; not one-to-one

For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

ⓐ {(27,−3),(8,−2),(1,−1),(0,0),(1,1),(8,2),(27,3)} ⓑ {(7,−3),(−5,−4),(8,0),(0,0),(−6,4),(−2,2),(−1,3)}

Solution

ⓐ Not a function
ⓑ Function; not one-to-one

To help us determine whether a relation is a function, we use the vertical line test. A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. Also, if any vertical line intersects the graph in more than one point, the graph does not represent a function.

The vertical line is representing an x-value and we check that it intersects the graph in only one y-value. Then it is a function.

To check if a function is one-to-one, we use a similar process. We use a horizontal line and check that each horizontal line intersects the graph in only one point. The horizontal line is representing a y-value and we check that it intersects the graph in only one x-value. If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function. This is the horizontal line test.

Horizontal Line Test

If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function.

We can test whether a graph of a relation is a function by using the vertical line test. We can then tell if the function is one-to-one by applying the horizontal line test.

Determine ⓐ whether each graph is the graph of a function and, if so, ⓑ whether it is one-to-one.

This first graph shows a straight line passing through (0, 2) and (3, 0). This second shows a parabola opening up with vertex at (0, negative 1).
Solution

ⓐ
This figure shows a straight line passing through (0, 2) and (3, 0), with a red vertical line that only passes through one point and a blue horizontal line that only passes through one point.

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. Since any horizontal line intersects the graph in at most one point, the graph is the graph of a one-to-one function.

ⓑ
This figure shows a parabola opening up with vertex at (0, negative 1), with a red vertical line that only passes through one point and a blue horizontal line that passes through two points.

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. The horizontal line shown on the graph intersects it in two points. This graph does not represent a one-to-one function.

Determine whether each graph is the graph of a function and, if so, whether it is one-to-one.

Graph a shows a parabola opening to the right with vertex at (negative 1, 0). Graph b shows an exponential function that does not cross the x axis and that passes through (0, 1) before increasing rapidly.
Solution

ⓐ Not a function ⓑ One-to-one function

Determine whether each graph is the graph of a function and, if so, whether it is one-to-one.

Graph a shows a parabola opening up with vertex at (0, 3). Graph b shows a straight line passing through (0, negative 2) and (2, 0).
Solution

ⓐ Function; not one-to-one ⓑ One-to-one function

Find the Inverse of a Function

Let’s look at a one-to one function, f, represented by the ordered pairs {(0,5),(1,6),(2,7),(3,8)}. For each x-value, f adds 5 to get the y-value. To ‘undo’ the addition of 5, we subtract 5 from each y-value and get back to the original x-value. We can call this “taking the inverse of f” and name the function f−1.

This figure shows the set (0, 5), (1, 6), (2, 7) and (3, 8) on the left side of an oval. The oval contains the numbers 0, 1, 2, and 3. There are black arrows from these numbers that point to the numbers 5, 6, 7, and 8, respectively in a second oval to the right of the first. Above this, there is a black arrow labeled “f add 5” coming from the left oval to the right oval. There are red arrows from the numbers 5, 6, 7, and 8 in the right oval to the numbers 0, 1, 2, and 3, respectively, in the left oval. Below this, we have a red arrow labeled “f with a superscript negative 1” and “subtract 5”. To the right of this, we have the set (5, 0), (6, 1), (7, 2) and (8, 3).

Notice that that the ordered pairs of f and f−1 have their x-values and y-values reversed. The domain of f is the range of f−1 and the domain of f−1 is the range of f.

Inverse of a Function Defined by Ordered Pairs

If f(x) is a one-to-one function whose ordered pairs are of the form (x,y), then its inverse function f−1(x) is the set of ordered pairs (y,x).

In the next example we will find the inverse of a function defined by ordered pairs.

Find the inverse of the function {(0,3),(1,5),(2,7),(3,9)}. Determine the domain and range of the inverse function.

Solution

This function is one-to-one since every x-value is paired with exactly one y-value.

To find the inverse we reverse the x-values and y-values in the ordered pairs of the function.
A function, its inverse, and the domain and range of the inverse function represented as sets.
Function {(0,3),(1,5),(2,7),(3,9)}
Inverse Function {(3,0),(5,1),(7,2),(9,3)}
Domain of Inverse Function {3,5,7,9}
Range of Inverse Function {0,1,2,3}

Find the inverse of {(0,4),(1,7),(2,10),(3,13)}. Determine the domain and range of the inverse function.

Solution

Inverse function: {(4,0),(7,1),(10,2),(13,3)}. Domain: {4,7,10,13}. Range: {0,1,2,3}.

Find the inverse of {(−1,4),(−2,1),(−3,0),(−4,2)}. Determine the domain and range of the inverse function.

Solution

Inverse function: {(4,−1),(1,−2),(0,−3),(2,−4)}. Domain: {0,1,2,4}. Range: {−4,−3,−2,−1}.

We just noted that if f(x) is a one-to-one function whose ordered pairs are of the form (x,y), then its inverse function f−1(x) is the set of ordered pairs (y,x).

So if a point (a,b) is on the graph of a function f(x), then the ordered pair (b,a) is on the graph of f−1(x). See Figure 1.

This figure shows the line y equals x with points (3,1) and (1,3) on either side of the line. These two points are connected by a dashed blue line segment.

The distance between any two pairs (a,b) and (b,a) is cut in half by the line y=x. So we say the points are mirror images of each other through the line y=x.

Since every point on the graph of a function f(x) is a mirror image of a point on the graph of f−1(x), we say the graphs are mirror images of each other through the line y=x. We will use this concept to graph the inverse of a function in the next example.

Graph, on the same coordinate system, the inverse of the one-to one function shown.

This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1,0) then to (0,2) and then to (3, 4).
Solution

We can use points on the graph to find points on the inverse graph. Some points on the graph are: (−5,−3),(−3,−1),(−1,0),(0,2),(3,4).

So, the inverse function will contain the points: (−3,−5),(−1,−3),(0,−1),(2,0),(4,3).
This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1, 0) then to (0,2) and then to (3, 4). Then there is a dashed line to denote y equals x. There is also a line from (negative 3, negative 5) to (negative 1, negative 3) then to (0, negative 1), then to (2, 0) and then to (4, 3).

Notice how the graph of the original function and the graph of the inverse functions are mirror images through the line y=x.

Graph, on the same coordinate system, the inverse of the one-to one function.

The graph shows a line from (negative 3, negative 4) to (negative 2, negative 2) then to (0, negative 1), then to (1, 2) and then to (4, 3). The graph shows a line from (negative 3, 4) to (0, 3) then to (1, 2) and then to (4, 1).
Solution

This figure shows a line from (negative 4, negative 3) to (negative 2, negative 2) then to (negative 1, 0) then to (2, 1) and then to (3, 4).

Graph, on the same coordinate system, the inverse of the one-to one function.

A graph on a Cartesian coordinate plane shows a piecewise linear function connecting four points: (-3, 4), (0, 3), (1, 2), and (4, 1).
Solution

Graph extends from negative 4 to 4 on both axes. Points plotted are (negative 3, 4), (0, 3), (1, 2), and (4, 1). Line segments connect points.

When we began our discussion of an inverse function, we talked about how the inverse function ‘undoes’ what the original function did to a value in its domain in order to get back to the original x-value.

This figure shows x as the input to a box denoted as function f with f of x as the output of the box. Then, f of x is the input to a box denoted as function f superscript negative 1 with f superscript negative 1 of f of x equals x as the output of the box.

Inverse Functions

f−1(f(x))=x,for allxin the domain offf(f−1(x))=x,for allxin the domain off−1

We can use this property to verify that two functions are inverses of each other.

Verify that f(x)=5x−1 and g(x)=x+15 are inverse functions.

Solution

The functions are inverses of each other if g(f(x))=x and f(g(x))=x.

Checking if g and f are inverse functions with the expression g(f(x)) =? x.
Substitute 5x−1 for f(x). The image shows a mathematical equation in black text on a white background: g(5x - 1) =? x. The question mark above the equals sign indicates that the equation is being posed as a question or hypothesis.
A mathematical problem asks to find the expression for g(5x - 1) given the function g(x) = (x + 1)/5, with the variable 'x' highlighted in red in the original function definition. A mathematical equation where the expression ((5x - 1) + 1) / 5 is being evaluated to see if it equals x, indicated by a question mark over the equals sign.
Simplify. A mathematical equation shows 5x divided by 5, followed by a question mark over an equals sign, and then the variable x, asking if 5x/5 equals x.
Simplify. A basic principle of logic and mathematics, 'X=X', is shown with a checkmark, emphasizing its self-evident truth.
A mathematical expression showing the composition of two functions, f(g(x)), questioning if it equals x. This setup is used to determine if f and g are inverse functions.
Substitute x+15 for g(x). A mathematical equation is displayed on a white background. It shows 'f(x+1)/5' equals 'x'.
A mathematical problem asks to find the value of the function f((x+1)/5) given that f(x) = 5x - 1. The input to the function f is shown as a fraction with 'x+1' in red in the numerator and '5' in red in the denominator. The image displays the equation 5((x+1)/5) - 1 =? x. Upon simplification, the 5s cancel out, resulting in (x+1) - 1, which equals x. Thus, the statement x = x is true, making the equation valid.
Simplify. A mathematical equation 'x + 1 - 1 = x' is displayed, with a question mark placed above the equals sign, posing whether the equality is true or false.
Simplify. The equation X = X with a checkmark, symbolizing a correct or self-evident statement.

Since both g(f(x))=x and f(g(x))=x are true, the functions f(x)=5x−1 and g(x)=x+15 are inverse functions. That is, they are inverses of each other.

Verify that the functions are inverse functions.

f(x)=4x−3 and g(x)=x+34.

Solution

g(f(x))=x, and f(g(x))=x, so they are inverses.

Verify that the functions are inverse functions.

f(x)=2x+6 and g(x)=x−62.

Solution

g(f(x))=x, and f(g(x))=x, so they are inverses.

We have found inverses of function defined by ordered pairs and from a graph. We will now look at how to find an inverse using an algebraic equation. The method uses the idea that if f(x) is a one-to-one function with ordered pairs (x,y), then its inverse function f−1(x) is the set of ordered pairs (y,x).

If we reverse the x and y in the function and then solve for y, we get our inverse function.

How to Find the inverse of a One-to-One Function

Find the inverse of f(x)=4x+7.

Solution
Step 1 is to substitute y for f of x. To do so, we replace f of x with y. Hence, f of x equals 4 x plus 7 becomes y equals 4 x plus 7. Step 2 is to interchange the variables x and y. To do so, we replace x with y and then y with x. Hence, we obtain x equals 4y plus 7. Step 3 is to solve for y. To do so, we subtract 7 from each side and then divide by 4. Hence, we have x minus 7 equals 4y and then the quantity x minus 7 divided by 4 equals y. Step 4 is to substitute f superscript negative 1 of x for y. To do so, we replace y with f superscript negative 1 of x. Hence, the quantity x minus 7 divided by 4 equals f superscript negative 1 of x. Step 5 is to verify that the functions are inverses. To do so, we show that f superscript negative 1 of f of x equals x and that f of f superscript negative 1of x equals x. Hence, we ask whether f inverse of 4x plus 7 equals x. This becomes a question of whether 4 x plus 7 minus 7 all divided by 4 equals x. This becomes a question of whether 4x divided by 4 equals x. This is true. To show the other side, we examine whether f of f inverse of x equals x. This becomes a question of whether f of the quantity x minus 7 divided by 4 equals x. This becomes a question of whether 4 times the quantity x minus 7 divided by 4 equals x. This becomes a question of whether x minus 7 plus 7 equals x. This is true.

Find the inverse of the function f(x)=5x−3.

Solution

f−1(x)=x+35

Find the inverse of the function f(x)=8x+5.

Solution

f−1(x)=x−58

We summarize the steps below.

How to Find the inverse of a One-to-One Function

  1. Substitute y for f(x).
  2. Interchange the variables x and y.
  3. Solve for y.
  4. Substitute f−1(x) for y.
  5. Verify that the functions are inverses.

How to Find the Inverse of a One-to-One Function

Find the inverse of f(x)=2x−35.

Solution
Step-by-step process for finding the inverse of the function f(x) = (2x-3)^(1/5) and verifying the inverse relationship.
f(x)=2x−35
Substitute y for f(x). y=2x−35
Interchange the variables x and y. x=2y−35
Solve for y. (x)5=(2y−35)5
x5=2y−3
x5+3=2y
x5+32=y
Substitute f−1(x) for y. f−1(x)=x5+32
Verify that the functions are inverses.
f−1(f(x))=?x f(f−1(x))=?x
f−1(2x−35)=?x f(x5+32)=?x
(2x−35)5+32=?x 2(x5+32)−35=?x
2x−3+32=?x x5+3−35=?x
2x2=?x x55=?x
x=x✓ x=x✓

Find the inverse of the function f(x)=3x−25.

Solution

f−1(x)=x5+23

Find the inverse of the function f(x)=6x−74.

Solution

f−1(x)=x4+76

Key Concepts

  • Composition of Functions: The composition of functions f and g, is written f∘g and is defined by
    (f∘g)(x)=f(g(x))

    We read f(g(x)) as f of g of x.
  • Horizontal Line Test: If every horizontal line, intersects the graph of a function in at most one point, it is a one-to-one function.
  • Inverse of a Function Defined by Ordered Pairs: If f(x) is a one-to-one function whose ordered pairs are of the form (x,y), then its inverse function f−1(x) is the set of ordered pairs (y,x).
  • Inverse Functions: For every x in the domain of one-to-one function f and f−1,
    f−1(f(x))=xf(f−1(x))=x
  • How to Find the Inverse of a One-to-One Function:
    1. Substitute y for f(x).
    2. Interchange the variables x and y.
    3. Solve for y.
    4. Substitute f−1(x) for y.
    5. Verify that the functions are inverses.

Practice Makes Perfect

Find and Evaluate Composite Functions

In the following exercises, find ⓐ (f ∘ g)(x), ⓑ (g ∘ f)(x), and ⓒ (f · g)(x).

f(x)=4x+3 and g(x)=2x+5

Solution

ⓐ 8x+23 ⓑ 8x+11 ⓒ
8x2+26x+15

f(x)=3x−1 and g(x)=5x−3

f(x)=6x−5 and g(x)=4x+1

Solution

ⓐ 24x+1 ⓑ 24x−19
ⓒ 24x2−14x−5

f(x)=2x+7 and g(x)=3x−4

f(x)=3x and g(x)=2x2−3x

Solution

ⓐ 6x2−9x ⓑ 18x2−9x
ⓒ 6x3−9x2

f(x)=2x and g(x)=3x2−1

f(x)=2x−1 and g(x)=x2+2

Solution

ⓐ 2x2+3 ⓑ 4x2−4x+3
ⓒ 2x3−x2+4x−2

f(x)=4x+3 and g(x)=x2−4

In the following exercises, find the values described.

For functions f(x)=2x2+3 and g(x)=5x−1, find ⓐ (f∘g)(−2) ⓑ (g∘f)(−3) ⓒ (f∘f)(−1)

Solution

ⓐ 245 ⓑ 104 ⓒ 53

For functions f(x)=5x2−1 and g(x)=4x−1, find ⓐ (f∘g)(1) ⓑ (g∘f)(−1) ⓒ (f∘f)(2)

For functions f(x)=2x3 and g(x)=3x2+2, find ⓐ (f∘g)(−1) ⓑ (g∘f)(1) ⓒ (g∘g)(1)

Solution

ⓐ 250 ⓑ 14 ⓒ 77

For functions f(x)=3x3+1 and g(x)=2x2−3, find ⓐ (f∘g)(−2) ⓑ (g∘f)(−1) ⓒ (g∘g)(1)

Determine Whether a Function is One-to-One

In the following exercises, determine if the set of ordered pairs represents a function and if so, is the function one-to-one.

{(−3,9),(−2,4),(−1,1),(0,0),
(1,1),(2,4),(3,9)}

Solution

Function; not one-to-one

{(9,−3),(4,−2),(1,−1),(0,0),
(1,1),(4,2),(9,3)}

{(−3,−5),(−2,−3),(−1,−1),
(0,1),(1,3),(2,5),(3,7)}

Solution

One-to-one function

{(5,3),(4,2),(3,1),(2,0),
(1,−1),(0,−2),(−1,−3)}

In the following exercises, determine whether each graph is the graph of a function and if so, is it one-to-one.

ⓐ
This figure shows a graph of a circle with center at the origin and radius 3.
ⓑ
This figure shows a graph of a parabola opening upward with vertex at (0k, 2).

Solution

ⓐ Not a function ⓑ Function; not one-to-one

ⓐ
This figure shows a parabola opening to the right with vertex at (negative 2, 0).
ⓑ
This figure shows a graph of a polynomial with odd order, so that it starts in the third quadrant, increases to the origin and then continues increasing through the first quadrant.

ⓐ
This figure shows a graph of a curve that starts at (negative 6 negative 2) increases to the origin and then continues increasing slowly to (6, 2).
ⓑ
This figure shows a parabola opening upward with vertex at (0, negative 4).

Solution

ⓐ One-to-one function
ⓑ Function; not one-to-one

ⓐ
This figure shows a straight line segment decreasing from (negative 4, 6) to (2, 0), after which it increases from (2, 0) to (6, 4).
ⓑ
This figure shows a circle with radius 4 and center at the origin.

In the following exercises, find the inverse of each function. Determine the domain and range of the inverse function.

{(2,1),(4,2),(6,3),(8,4)}

Solution

Inverse function: {(1,2),(2,4),(3,6),(4,8)}. Domain: {1,2,3,4}. Range: {2,4,6,8}.

{(6,2),(9,5),(12,8),(15,11)}

{(0,−2),(1,3),(2,7),(3,12)}

Solution

Inverse function: {(−2,0),(3,1),(7,2),(12,3)}. Domain: {−2,3,7,12}. Range: {0,1,2,3}.

{(0,0),(1,1),(2,4),(3,9)}

{(−2,−3),(−1,−1),(0,1),(1,3)}

Solution

Inverse function: {(−3,−2),(−1,−1),(1,0),(3,1)}. Domain: {−3,−1,1,3}. Range: {−2,−1,0,1}.

{(5,3),(4,2),(3,1),(2,0)}

In the following exercises, graph, on the same coordinate system, the inverse of the one-to-one function shown.


This figure shows a series of line segments from (negative 4, negative 3) to (negative 3, 0) then to (negative 1, 2) and then to (3, 4).

Solution

This figure shows a series of line segments from (negative 3, negative 4) to (0, negative 3) then to (2, negative 1), and then to (4, 3).


This figure shows a series of line segments from (negative 4, negative 4) to (negative 3, 1) then to (0, 2) and then to (2, 4).


This figure shows a series of line segments from (negative 4, 4) to (0, 3) then to (3, 2) and then to (4, negative 1).

Solution

This figure shows a series of line segments from (negative 1, 4) to (2, 3) then to (3, 0), and then to (4, negative 4).


This figure shows a series of line segments from (negative 4, negative 4) to (negative 1, negative 3) then to (0, 1), then to (1, 3), and then to (4, 4).

In the following exercises, determine whether or not the given functions are inverses.

f(x)=x+8 and g(x)=x−8

Solution

g(f(x))=x, and f(g(x))=x, so they are inverses.

f(x)=x−9 and g(x)=x+9

f(x)=7x and g(x)=x7

Solution

g(f(x))=x, and f(g(x))=x, so they are inverses.

f(x)=x11 and g(x)=11x

f(x)=7x+3 and g(x)=x−37

Solution

g(f(x))=x, and f(g(x))=x, so they are inverses.

f(x)=5x−4 and g(x)=x−45

f(x)=x+2 and g(x)=x2−2(x>0)

Solution

g(f(x))=x, and f(g(x))=x, so they are inverses (for nonnegative x).

f(x)=x−43 and g(x)=x3+4

In the following exercises, find the inverse of each function.

f(x)=x−12

Solution

f−1(x)=x+12

f(x)=x+17

f(x)=9x

Solution

f−1(x)=x9

f(x)=8x

f(x)=x6

Solution

f−1(x)=6x

f(x)=x4

f(x)=6x−7

Solution

f−1(x)=x+76

f(x)=7x−1

f(x)=−2x+5

Solution

f−1(x)=x−5−2

f(x)=−5x−4

f(x)=x2+6, x≥0

Solution

f−1(x)=x−6

f(x)=x2−9, x≥0

f(x)=x3−4

Solution

f−1(x)=x+43

f(x)=x3+6

f(x)=1x+2

Solution

f−1(x)=1x−2

f(x)=1x−6

f(x)=x−2, x≥2

Solution

f−1(x)=x2+2, x≥0

f(x)=x+8, x≥−8

f(x)=x−33

Solution

f−1(x)=x3+3

f(x)=x+53

f(x)=9x−54, x≥59

Solution

f−1(x)=x4+59, x≥0

f(x)=8x−34, x≥38

f(x)=−3x+55

Solution

f−1(x)=x5−5−3

f(x)=−4x−35

Writing Exercises

Explain how the graph of the inverse of a function is related to the graph of the function.

Solution

Answers will vary.

Explain how to find the inverse of a function from its equation. Use an example to demonstrate the steps.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Find and evaluate composite functions, determine whether a function is one-to-one, and find the inverse of a function. The rest of the cells are blank.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

one-to-one function
A function is one-to-one if each value in the range has exactly one element in the domain. For each ordered pair in the function, each y-value is matched with only one x-value.

Evaluate and Graph Exponential Functions

Learning Objectives

By the end of this section, you will be able to:

  • Graph exponential functions
  • Solve Exponential equations
  • Use exponential models in applications

Before you get started, take this readiness quiz.

Simplify: (x3x2).
If you missed this problem, review Example 2 in Properties of Exponents and Scientific Notation.

Solution

x

Evaluate: ⓐ 20 ⓑ (13)0.
If you missed this problem, review Example 3 in Properties of Exponents and Scientific Notation.

Solution

ⓐ 1; ⓑ 1

Evaluate: ⓐ 2−1 ⓑ (13)−1.
If you missed this problem, review Example 4 in Properties of Exponents and Scientific Notation.

Solution

ⓐ 12; ⓑ 3

Graph Exponential Functions

The functions we have studied so far do not give us a model for many naturally occurring phenomena. From the growth of populations and the spread of viruses to radioactive decay and compounding interest, the models are very different from what we have studied so far. These models involve exponential functions.

An exponential function is a function of the form f(x)=ax where a>0 and a≠1.

Exponential Function

An exponential function, where a>0 and a≠1, is a function of the form

f(x)=ax

Notice that in this function, the variable is the exponent. In our functions so far, the variables were the base.

This figure shows three functions: f of x equals negative 3x plus 4, which is marked as linear; f of x equals 2x squared plus 5x minus 3, which is marked as quadratic; and f of x equals 6 to the x power, which is marked exponential. For the functions marked linear and quadratic, x is the base. For the function marked exponential, x is the exponent for the base 6.

Our definition says a≠1. If we let a=1, then f(x)=ax becomes f(x)=1x. Since 1x=1 for all real numbers, f(x)=1. This is the constant function.

Our definition also says a>0. If we let a base be negative, say −4, then f(x)=(−4)x is not a real number when x=12.

f(x)=(−4)x f(12)=(−4)12 f(12)=−4not a real number

In fact, f(x)=(−4)x would not be a real number any time x is a fraction with an even denominator. So our definition requires a>0.

By graphing a few exponential functions, we will be able to see their unique properties.

On the same coordinate system graph f(x)=2x and g(x)=3x.

Solution

We will use point plotting to graph the functions.

This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 3 to the x power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 3 to the negative 2 power equals 1 divided by 3 squared which equals 1 over 9, (negative 2, 1 over 9). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 3 to the negative 1 power equals 1 divided by 3 to the first power which equals 1 over 3, (negative 1, 1 over 3). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 3 to the 1 power equals 9, (1, 3). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 3 to the 2 power equals 9, (2, 9). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 3 to the 3 power equals 27, (3, 27).

This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3).

Graph: f(x)=4x.

Solution


This figure shows a curve that slopes swiftly upward from just above (negative 3, 0) through (0, 1) up to (1, 4).

Graph: g(x)=5x.

Solution


This figure shows a curve that slopes swiftly upward from just above (negative 3, 0) through (0, 1) up to (1, 5).

If we look at the graphs from the previous Example and Try Its, we can identify some of the properties of exponential functions.

The graphs of f(x)=2x and g(x)=3x, as well as the graphs of f(x)=4x and g(x)=5x, all have the same basic shape. This is the shape we expect from an exponential function where a>1.

We notice, that for each function, the graph contains the point (0,1). This make sense because a0=1 for any a.

The graph of each function, f(x)=ax also contains the point (1,a). The graph of f(x)=2x contained (1,2) and the graph of g(x)=3x contained (1,3). This makes sense as a1=a.

Notice too, the graph of each function f(x)=ax also contains the point (−1,1a). The graph of f(x)=2x contained (−1,12) and the graph of g(x)=3x contained (−1,13). This makes sense as a−1=1a.

What is the domain for each function? From the graphs we can see that the domain is the set of all real numbers. There is no restriction on the domain. We write the domain in interval notation as (−∞,∞).

Look at each graph. What is the range of the function? The graph never hits the x-axis. The range is all positive numbers. We write the range in interval notation as (0,∞).

Whenever a graph of a function approaches a line but never touches it, we call that line an asymptote. For the exponential functions we are looking at, the graph approaches the x-axis very closely but will never cross it, we call the line y=0, the x-axis, a horizontal asymptote.

Properties of the Graph of f(x)=ax when a>1

Domain (−∞,∞)
Range (0,∞)
x-intercept None
y-intercept (0,1)
Contains (1,a),(−1,1a)
Asymptote x-axis, the line y=0
This figure shows a curve that slopes upward from (negative 1, 1 over a) through (0, 1), up to (1, a).

Our definition of an exponential function f(x)=ax says a>0, but the examples and discussion so far has been about functions where a>1. What happens when 0<a<1? The next example will explore this possibility.

On the same coordinate system, graph f(x)=(12)x and g(x)=(13)x.

Solution

We will use point plotting to graph the functions.

This table has seven rows and five columns. The first row is header row and reads x, f of x, equals 1 over 2 to the x power, (x, f of x), g of x equals 1 over 3 to the x power, and (x, g of x). The second row reads negative 2, 1 over 2 to the negative 2 power equals 2 squared which equals 4, (negative 2, 4), 3 to the negative 2 power equals 3 squared which equals 9, (negative 2, 9). The third row reads negative 1, 1 over 2 to the negative 1 power equals 2 to the first power which equals 2, (negative 1, 2), 1 over 3 to the negative 1 power equals 3 to the first power which equals 3, (negative 1, 3). The fourth row reads 0, 1 over 2 to the 0 power equals 1, (0, 1), 1 over 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 1 over 2 to the 1 power equals 1 over 2, (1, 1 over 2), 1 over 3 to the 1 power equals 1 over 3, (1, 1 over 3). The sixth row reads 2, 1 over 2 to the 2 power equals 1 over 4, (2, 1 over 4), 1 over 3 to the 2 power equals 1 over 9, (2, 1 over 9). The seventh row reads 3, 1 over 2 to the 3 power equals 1 over 8, (3, 1 over 8), 1 over 3 to the 3 power equals 1 over 27, (3, 1 over 27).

This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 2), (0, 1), and (1, 1 over 2). The second curve is marked in red and passes through the points (negative 1, 3), (0, 1), and (1, 1 over 3).

Graph: f(x)=(14)x.

Solution


This figure shows a curve that passes through (negative 1, 4), (0, 1) to a point just above (3, 0).

Graph: g(x)=(15)x.

Solution


This figure shows a curve that passes through (negative 1, 5), (0, 1) to a point just above (3, 0).

Now let’s look at the graphs from the previous Example and Try Its so we can now identify some of the properties of exponential functions where 0<a<1.

The graphs of f(x)=(12)x and g(x)=(13)x as well as the graphs of f(x)=(14)x and g(x)=(15)x all have the same basic shape. While this is the shape we expect from an exponential function where 0<a<1, the graphs go down from left to right while the previous graphs, when a>1, went from up from left to right.

We notice that for each function, the graph still contains the point (0, 1). This make sense because a0=1 for any a.

As before, the graph of each function, f(x)=ax, also contains the point (1,a). The graph of f(x)=(12)x contained (1,12) and the graph of g(x)=(13)x contained (1,13). This makes sense as a1=a.

Notice too that the graph of each function, f(x)=ax, also contains the point (−1,1a). The graph of f(x)=(12)x contained (−1,2) and the graph of g(x)=(13)x contained (−1,3). This makes sense as a−1=1a.

What is the domain and range for each function? From the graphs we can see that the domain is the set of all real numbers and we write the domain in interval notation as (−∞,∞). Again, the graph never hits the x-axis. The range is all positive numbers. We write the range in interval notation as (0,∞).

We will summarize these properties in the chart below. Which also include when a>1.

Properties of the Graph of f(x)=ax

when a>1 when 0<a<1
Domain (−∞,∞) Domain (−∞,∞)
Range (0,∞) Range (0,∞)
x-intercept none x-intercept none
y-intercept (0,1) y-intercept (0,1)
Contains (1,a),(−1,1a) Contains (1,a),(−1,1a)
Asymptote x-axis, the line y=0 Asymptote x-axis, the line y=0
Basic shape increasing Basic shape decreasing
This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).

It is important for us to notice that both of these graphs are one-to-one, as they both pass the horizontal line test. This means the exponential function will have an inverse. We will look at this later.

When we graphed quadratic functions, we were able to graph using translation rather than just plotting points. Will that work in graphing exponential functions?

On the same coordinate system graph f(x)=2x and g(x)=2x+1.

Solution

We will use point plotting to graph the functions.

This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 2 to the x plus 1 power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 2 to the negative 2 plus 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 2, 1 over 2). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 2 to the negative 1 plus 1 power equals 2 to the 0 power which equals 1, (negative 1, 1). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 2 to the 0 plus 1 power equals 2 to the 1 power which equals 2, (0, 2). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 2 to the 1 plus 1 power equals 2 to the second power which equals 4, (1, 4). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 2 to the 2 plus 1 power equals 2 to the third power which equals 8, (2, 8). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 2 to the 3 plus 1 power equals 2 to the fourth power which equals 16, (3, 16).

This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1), (0, 2) and (1, 4).

On the same coordinate system, graph: f(x)=2x and g(x)=2x−1.

Solution


This figure shows the graphs of two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second function g of x equals 2 to the x minus 1 power is marked in red and passes through the points (0, 1 over 2), (1, 1), and (2, 2).

On the same coordinate system, graph: f(x)=3x and g(x)=3x+1.

Solution


This figure shows the graphs of two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The second function g of x equals 3 to the x plus 1 power is marked in red and passes through the points (negative 2, 1 over 3), (negative 1, 1), and (0, 3).

Looking at the graphs of the functions f(x)=2x and g(x)=2x+1 in the last example, we see that adding one in the exponent caused a horizontal shift of one unit to the left. Recognizing this pattern allows us to graph other functions with the same pattern by translation.

Let’s now consider another situation that might be graphed more easily by translation, once we recognize the pattern.

On the same coordinate system graph f(x)=3x and g(x)=3x−2.

Solution

We will use point plotting to graph the functions.

This table has five rows and six columns. The first row is header row and reads x, f of x equals 3 to the x power, (x, f of x), g of x equals 3 to the x power minus 2, and (x, g of x). The second row reads negative 2, 3 to the negative 2 power equals 1 over 9, (negative 2, 1 over 9), 3 to the negative 2 power minus 2 equals 1 over 9 minus 2 which equals negative 17 over 9, (negative 2, negative 17 over 9). The third row reads negative 1, 3 to the negative 1 power equals 1 over 3, (negative 1, 1 over 3), 3 to the negative 1 power minus 2 equals 1 over 3 minus 2 which equals negative 5 over 3, (negative 1, negative 5 over 3). The fourth row reads 0, 3 to the 0 power equals 1, (0, 1), 3 to the 0 power minus 2 equals 1 minus 2 which equals negative 1, (0, negative 1). The fifth row reads 1, 3 to the 1 power equals 3, (1, 3), 3 to the 1 power minus 2 equals 3 minus 2 which equals 1, (1, 1). The sixth row reads 2, 3 squared equals 9, (2, 9), 3 squared minus 2 equals 9 minus 2 which equals 7, (2, 7).

This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second curve is marked in red and passes through the points (negative 1, negative 5 over 3), (0, negative 1), and (1, 1).

On the same coordinate system, graph: f(x)=3x and g(x)=3x+2.

Solution


This figure shows the graphs of two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The second function g of x equals 3 to the x power plus 2 is marked in red and passes through the points (negative 1, 7 over 3), (0, 3) and (1, 5).

On the same coordinate system, graph: f(x)=4x and g(x)=4x−2.

Solution


This figure shows the graphs of two functions. The first function f of x equals 4 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 4), (0, 1) and (1, 4). The second function g of x equals 4 to the x power minus 2 is marked in red and passes through the points (negative 1, negative 7 over 4), (0, negative 1), and (1, 2).

Looking at the graphs of the functions f(x)=3x and g(x)=3x−2 in the last example, we see that subtracting 2 caused a vertical shift of down two units. Notice that the horizontal asymptote also shifted down 2 units. Recognizing this pattern allows us to graph other functions with the same pattern by translation.


All of our exponential functions have had either an integer or a rational number as the base. We will now look at an exponential function with an irrational number as the base.

Before we can look at this exponential function, we need to define the irrational number, e. This number is used as a base in many applications in the sciences and business that are modeled by exponential functions. The number is defined as the value of (1+1n)n as n gets larger and larger. We say, as n approaches infinity, or increases without bound. The table shows the value of (1+1n)n for several values of n.

n (1+1n)n
1 2
2 2.25
5 2.48832
10 2.59374246
100 2.704813829…
1,000 2.716923932…
10,000 2.718145927…
100,000 2.718268237…
1,000,000 2.718280469…
1,000,000,000 2.718281827…

If carried out to even larger values of n, we get

e≈2.718281828

The number e is like the number π in that we use a symbol to represent it because its decimal representation never stops or repeats. The irrational number e is called the natural base.

Natural Base e

The number e is defined as the value of (1+1n)n, as n increases without bound. We say, as n approaches infinity,

e≈2.718281828...

The exponential function whose base is e, f(x)=ex is called the natural exponential function.

Natural Exponential Function

The natural exponential function is an exponential function whose base is e

f(x)=ex

The domain is (−∞,∞) and the range is (0,∞).

Let’s graph the function f(x)=ex on the same coordinate system as g(x)=2x and h(x)=3x.

This figure shows the graphs of three functions. The first function, f of x equals 2 to the x, is marked in red and passes through the points (negative 1, negative 1 over 2), (0, negative 1), and (2, 1). The second function, f of x equals 3 to the x power, is marked in green and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The third function, f of x equals e to the x power, is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over e), (0, 1) and (0, e).

Notice that the graph of f(x)=ex is “between” the graphs of g(x)=2x and h(x)=3x. Does this make sense as 2<e<3?

Solve Exponential Equations

Equations that include an exponential expression ax are called exponential equations. To solve them we use a property that says as long as a>0 and a≠1, if ax=ay then it is true that x=y. In other words, in an exponential equation, if the bases are equal then the exponents are equal.

One-to-One Property of Exponential Equations

For a>0 and a≠1,

Ifax=ay,thenx=y.

To use this property, we must be certain that both sides of the equation are written with the same base.

How to Solve an Exponential Equation

Solve: 32x−5=27.

Solution
Step 1 is to write both sides of the equation with the same base. This means that, since the left side has base 3, we write the right side with base 3. Hence, 27 equals 3 to the third power. We have 3 to the 2x minus 5 power equals 27, which we write as 3 to the 2x minus 5 power equals 3 cubed. Step 2 is to write a new equation by setting the exponents equal. This means that, since the bases are the same, the exponents must be equal. Hence, 2x minus 5 equals 3. Step 3 is to solve the equation. This means that we add 5 to each side and divide by 2. Hence, 2x equals 8, which means that x equals 4. Step 3 is to check the solution. This means that we substitute x equals 4 into the original equation. We start with 3 to the 2 x minus 5 power equals 27. We want to know whether 3 to the 2 times 4 minus 5 power equals 27. This becomes a question of whether 3 cubed equals 27, which of course is true.

Solve: 33x−2=81.

Solution

x=2

Solve: 7x−3=7.

Solution

x=4

The steps are summarized below.

How to Solve an Exponential Equation

  1. Write both sides of the equation with the same base, if possible.
  2. Write a new equation by setting the exponents equal.
  3. Solve the equation.
  4. Check the solution.

In the next example, we will use our properties on exponents.

Solve ex2e3=e2x.

Solution
ex2e3=e2x
Use the Property of Exponents: aman=am−n. ex2−3=e2x
Write a new equation by setting the exponents
equal.
x2−3=2x
Solve the equation. x2−2x−3=0
(x−3)(x+1)=0
x=3,x=−1
Check the solutions.
Step-by-step verification that e^(x^2) / e^3 = e^(2x) for x=3 and x=-1, demonstrating equality through algebraic simplification of exponents.

Solve: ex2ex=e2.

Solution

x=−1,x=2

Solve: ex2ex=e6.

Solution

x=−2,x=3

Use Exponential Models in Applications

Exponential functions model many situations. If you own a bank account, you have experienced the use of an exponential function. There are two formulas that are used to determine the balance in the account when interest is earned. If a principal, P, is invested at an interest rate, r, for t years, the new balance, A, will depend on how often the interest is compounded. If the interest is compounded n times a year we use the formula A=P(1+rn)nt. If the interest is compounded continuously, we use the formula A=Pert. These are the formulas for compound interest.

Compound Interest

For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is:

A=P(1+rn)ntwhen compoundedntimes a year. A=Pertwhen compounded continuously.

As you work with the Interest formulas, it is often helpful to identify the values of the variables first and then substitute them into the formula.

A total of $10,000 was invested in a college fund for a new grandchild. If the interest rate is 5%, how much will be in the account in 18 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

ⓒ compound continuously

Solution
Key variables and instructions for a financial calculation problem.
A=?
Identify the values of each variable in the formulas. P=$10,000
Remember to express the percent as a decimal. r=0.05
t=18years
ⓐ
This table demonstrates the step-by-step calculation of compound interest, showing the formula, substitution, and final amount.
For quarterly compounding, n=4. There are 4 quarters in a year. A=P(1+rn)nt
Substitute the values in the formula. A=10,000(1+0.054)4·18
Compute the amount. Be careful to consider the order of operations as you enter the expression into your calculator. A=$24,459.20
ⓑ
Illustrates the step-by-step calculation of compound interest using the formula A = P(1 + r/n)^(nt).
For monthly compounding, n=12. There are 12 months in a year. A=P(1+rn)nt
Substitute the values in the formula. A=10,000(1+0.0512)12·18
Compute the amount. A=$24,550.08
ⓒ
Table demonstrating the steps and formula for calculating continuous compound interest.
For compounding continuously, A=Pert
Substitute the values in the formula. A=10,000e0.05·18
Compute the amount. A=$24,596.03

Angela invested $15,000 in a savings account. If the interest rate is 4%, how much will be in the account in 10 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

ⓒ compound continuously

Solution

ⓐ $22,332.96
ⓑ $22,362.49 ⓒ $22,377.37

Allan invested $10,000 in a mutual fund. If the interest rate is 5%, how much will be in the account in 15 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

ⓒ compound continuously

Solution

ⓐ $21,071.81 ⓑ $21,137.04
ⓒ $21,170.00

Other topics that are modeled by exponential functions involve growth and decay. Both also use the formula A=Pert we used for the growth of money. For growth and decay, generally we useA0, as the original amount instead of calling it P, the principal. We see that exponential growth has a positive rate of growth and exponential decay has a negative rate of growth.

Exponential Growth and Decay

For an original amount, A0, that grows or decays at a rate, r, for a certain time, t, the final amount, A, is:

A=A0ert

Exponential growth is typically seen in the growth of populations of humans or animals or bacteria. Our next example looks at the growth of a virus.

Chris is a researcher at the Center for Disease Control and Prevention and he is trying to understand the behavior of a new and dangerous virus. He starts his experiment with 100 of the virus that grows continously at a rate of 25% per hour. He will check on the virus in 24 hours. How many viruses will he find?

Solution
Demonstrates step-by-step calculation using the exponential growth formula A = A0e^rt, including variable identification and result computation for a quantity.
Identify the values of each variable in the formulas. A=?
Be sure to put the percent in decimal form. A0=100
Be sure the units match—the rate is per hour and the time is in hours. r=0.25/hour
t=24hours
Substitute the values in the formula: A=A0ert. A=100e0.25·24
Compute the amount. A=40,342.88
Round to the nearest whole virus. A=40,343
The researcher will find 40,343 viruses.

Another researcher at the Center for Disease Control and Prevention, Lisa, is studying the growth of a bacteria. She starts her experiment with 50 of the bacteria that grows at a rate of 15% per hour. She will check on the bacteria every 8 hours. How many bacteria will she find in 8 hours?

Solution

She will find 166 bacteria.

Milan, a biologist is observing the growth pattern of a virus. They start with 100 of the virus that grows at a rate of 10% per hour. They will check on the virus in 24 hours. How many viruses will they find?

Solution

They will find 1,102 viruses.

Access these online resources for additional instruction and practice with evaluating and graphing exponential functions.

  • Graphing Exponential Functions
  • Solving Exponential Equations
  • Applications of Exponential Functions
  • Continuously Compound Interest
  • Radioactive Decay and Exponential Growth

Key Concepts

  • Properties of the Graph of f(x)=ax:
    when a>1 when 0<a<1
    Domain (−∞,∞) Domain (−∞,∞)
    Range (0,∞) Range (0,∞)
    x-intercept none x-intercept none
    y-intercept (0,1) y-intercept (0,1)
    Contains (1,a),(−1,1a) Contains (1,a),(−1,1a)
    Asymptote x-axis, the line y=0 Asymptote x-axis, the line y=0
    Basic shape increasing Basic shape decreasing

    This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).
  • One-to-One Property of Exponential Functions:
    For a>0 and a≠1,
    Ifax=ay,thenx=y.
  • How to Solve an Exponential Equation
    1. Write both sides of the equation with the same base, if possible.
    2. Write a new equation by setting the exponents equal.
    3. Solve the equation.
    4. Check the solution.
  • Compound Interest: For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is
    A=P(1+rn)ntwhen compoundedntimes a year. A=Pertwhen compounded continuously.
  • Exponential Growth and Decay: For an original amount, A0 that grows or decays at a rate, r, for a certain time t, the final amount,A, is A=A0ert.

Practice Makes Perfect

Graph Exponential Functions

In the following exercises, graph each exponential function.

f(x)=2x

Solution

This figure shows a curve that passes through (negative 1, 1 over 2) through (0, 1) to (1, 2).

g(x)=3x

f(x)=6x

Solution

This figure shows a curve that passes through (negative 1, 1 over 6) through (0, 1) to (1, 6).

g(x)=7x

f(x)=(1.5)x

Solution

This figure shows a curve that passes through (negative 1, 2 over 3) through (0, 1) to (1, 3 over 2).

g(x)=(2.5)x

f(x)=(12)x

Solution

This figure shows a curve that passes through (negative 1, 2) through (0, 1) to (1, 1 over 2).

g(x)=(13)x

f(x)=(16)x

Solution

This figure shows a curve that passes through (negative1, 6) through (0, 1) to (1, 1 over 6).

g(x)=(17)x

f(x)=(0.4)x

Solution

This figure shows a curve that passes through (negative 1, 5 over 2) through (0, 1) to (1, 2 over 5).

g(x)=(0.6)x

In the following exercises, graph each function in the same coordinate system.

f(x)=4x,g(x)=4x−1

Solution

This figure shows two functions. The first function f of x equals 4 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 4), (0, 1) and (1, 4). The second function g of x equals 4 to the x minus 1 power is marked in red and passes through the points (0, 1 over 4), (1, 1) and (2, 4).

f(x)=3x,g(x)=3x−1

f(x)=2x,g(x)=2x−2

Solution

This figure shows two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second function g of x equals 2 to the x minus 2 power is marked in red and passes through the points (0, 1 over 4), (1, 1 over 2), and (2, 1).

f(x)=2x,g(x)=2x+2

f(x)=3x,g(x)=3x+2

Solution

This figure shows two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second function g of x equals 3 to the x power plus 2 is marked in red and passes through the points (negative 2, 1), (negative 1, 3), and (0, 5).

f(x)=4x,g(x)=4x+2

f(x)=2x,g(x)=2x+1

Solution

This figure shows two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second function g of x equals 2 to the x power plus 1 is marked in red and passes through the points (negative 1, 1), (0, 2), and (1, 4).

f(x)=2x,g(x)=2x−1

In the following exercises, graph each exponential function.

f(x)=3x+2

Solution

This figure shows an exponential curve that passes through (negative 3, 1 over 3), (negative 2, 1), and (0, 9).

f(x)=3x−2

f(x)=2x+3

Solution

This figure shows an exponential that passes through (negative 1, 7 over 2), (0, 4), and (1, 5).

f(x)=2x−3

f(x)=(12)x−4

Solution

This figure shows an exponential that passes through (2, 4), (3, 2), and (4, 1).

f(x)=(12)x−3

f(x)=ex+1

Solution

This figure shows an exponential that passes through (1, 1 plus 1 over e), (0, 2), and (1, e).

f(x)=ex−2

f(x)=−2x

Solution

This figure shows an exponential that passes through (negative 1, negative 1 over 2), (0, negative 1), and (1, 2).

f(x)=2−x−1−1

Solve Exponential Equations

In the following exercises, solve each equation.

23x−8=16

Solution

x=4

22x−3=32

3x+3=9

Solution

x=−1

3x2=81

4x2=4

Solution

x=−1,x=1

4x=32

4x+2=64

Solution

x=1

4x+3=16

2x2+2x=12

Solution

x=−1

3x2−2x=13

e3x·e4=e10

Solution

x=2

e2x·e3=e9

ex2e2=ex

Solution

x=−1,x=2

ex2e3=e2x

In the following exercises, match the graphs to one of the following functions: ⓐ 2x ⓑ 2x+1 ⓒ 2x−1 ⓓ 2x+2 ⓔ 2x−2 ⓕ 3x


This figure shows an exponential that passes through (1, 1 over 3), (0, 1), and (1, 3).

Solution

ⓕ


This figure shows an exponential that passes through (negative 2, 1 over 2), (negative 1, 1), and (0, 2).


This figure shows an exponential that passes through (1, 1 over 2), (0, 1), and (1, 2).

Solution

ⓐ


This figure shows an exponential that passes through (0, 1 over 2), (1, 1), and (2, 2).


This figure shows an exponential that passes through (negative 1, 3 over 2), (0, negative 1), and (1, 0).

Solution

ⓔ


This figure shows an exponential that passes through (negative 1, 5 over 2), (0, 3), and (1, 4).

Use exponential models in applications

In the following exercises, use an exponential model to solve.

Edgar accumulated $5,000 in credit card debt. If the interest rate is 20% per year, and he does not make any payments for 2 years, how much will he owe on this debt in 2 years by each method of compounding? ⓐ compound quarterly ⓑ compound monthly ⓒ compound continuously

Solution

ⓐ $7,387.28 ⓑ $7,434.57 ⓒ $7,459.12

Cynthia invested $12,000 in a savings account. If the interest rate is 6%, how much will be in the account in 10 years by each method of compounding? ⓐ compound quarterly
ⓑ compound monthly ⓒ compound continuously

Rochelle deposits $5,000 in an IRA. What will be the value of her investment in 25 years if the investment is earning 8% per year and is compounded continuously?

Solution

$36,945.28

Nazerhy deposits $8,000 in a certificate of deposit. The annual interest rate is 6% and the interest will be compounded quarterly. How much will the certificate be worth in 10 years?

A researcher at the Center for Disease Control and Prevention is studying the growth of a bacteria. He starts his experiment with 100 of the bacteria that grows at a rate of 6% per hour. He will check on the bacteria every 8 hours. How many bacteria will he find in 8 hours?

Solution

162 bacteria

A biologist is observing the growth pattern of a virus. She starts with 50 of the virus that grows at a rate of 20% per hour. She will check on the virus in 24 hours. How many viruses will she find?

In the last ten years the population of Indonesia has grown at a rate of 1.12% per year to 258,316,051. If this rate continues, what will be the population in 10 more years?

Solution

288,929,825

In the last ten years the population of Brazil has grown at a rate of 0.9% per year to 205,823,665. If this rate continues, what will be the population in 10 more years?

Writing Exercises

Explain how you can distinguish between exponential functions and polynomial functions.

Solution

Answers will vary.

Compare and contrast the graphs of y=x2 and y=2x.

What happens to an exponential function as the values of x decreases? Will the graph ever cross the
x-axis? Explain.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Graph exponential functions, solve exponential equations, and use exponential models in applications.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

asymptote
A line which a graph of a function approaches closely but never touches.
exponential function
An exponential function, where a>0 and a≠1, is a function of the form f(x)=ax.
natural base
The number e is defined as the value of (1+1n)n, as n gets larger and larger. We say, as n increases without bound, e≈2.718281828...
natural exponential function
The natural exponential function is an exponential function whose base is e: f(x)=ex. The domain is (−∞,∞) and the range is (0,∞).

Evaluate and Graph Logarithmic Functions

Learning Objectives

By the end of this section, you will be able to:

  • Convert between exponential and logarithmic form
  • Evaluate logarithmic functions
  • Graph Logarithmic functions
  • Solve logarithmic equations
  • Use logarithmic models in applications

Before you get started, take this readiness quiz.

Solve: x2=81.
If you missed this problem, review Example 3 in Polynomial Equations.

Solution

x=9,x=−9

Evaluate: 3−2.
If you missed this problem, review Example 4 in Properties of Exponents and Scientific Notation.

Solution

19

Solve: 24=3x−5.
If you missed this problem, review Example 2 in Use a General Strategy to Solve Linear Equations.

Solution

x=7

We have spent some time finding the inverse of many functions. It works well to ‘undo’ an operation with another operation. Subtracting ‘undoes’ addition, multiplication ‘undoes’ division, taking the square root ‘undoes’ squaring.

As we studied the exponential function, we saw that it is one-to-one as its graphs pass the horizontal line test. This means an exponential function does have an inverse. If we try our algebraic method for finding an inverse, we run into a problem.

Rewrite withy=f(x).Interchange the variablesxandy.f(x)=axy=axx=aySolve fory.Oops! We have no way to solve fory!

To deal with this we define the logarithm function with base a to be the inverse of the exponential function f(x)=ax. We use the notation f−1(x)=logax and say the inverse function of the exponential function is the logarithmic function.

Logarithmic Function

The function f(x)=logax is the logarithmic function with base a, where a>0,x>0, and a≠1.

y=logaxis equivalent tox=ay

Convert Between Exponential and Logarithmic Form

Since the equations y=logax and x=ay are equivalent, we can go back and forth between them. This will often be the method to solve some exponential and logarithmic equations. To help with converting back and forth let’s take a close look at the equations. See Figure 1. Notice the positions of the exponent and base.

This figure shows the expression y equals log sub a of x, where y is the exponent and a is the base. Next to this expression we have x equals a to the y, where again y is the exponent and a is the base.

If we realize the logarithm is the exponent it makes the conversion easier. You may want to repeat, “base to the exponent give us the number.”

Convert to logarithmic form: ⓐ 23=8, ⓑ 512=5, and ⓒ (12)4=116.

Solution
In part (a) we have 2 to the 3 power equals 8, where the 2 is red and the 3 is blue. Following this, we have blue y equals log sub red a of x. Then 3 equals log sub 2 of 8. Hence, if 2 cubed equals 8, then 3 equals log sub 2 of 8. In part (b) we have 5 to the 1 over 2 power equals square root of 5, where the 5 is red and the 1 over 2 is blue. Following this, we have blue y equals log sub red a of x. Then 1 over 2 equals log sub 5 of the square root of 5. Hence, if 5 to the 1 over 2 power equals the square root of 5, then 1 over 2 equals log sub 5 of the square root of 5. In part (c) we have 1 over 2 to the x power equals 1 over 16, where the 1 over 2 is red and the x is blue. Following this, we have blue y equals log sub red a of x. Then x equals log sub 1 over 2 of 1 over 16. Hence, if 1 over 2 to the x power equals 1 over 16, then x equals log sub 1 over 2 of 1 over 16.

Convert to logarithmic form: ⓐ 32=9 ⓑ 712=7 ⓒ (13)x=127

Solution

ⓐ log39=2
ⓑ log77=12 ⓒ log13127=x

Convert to logarithmic form: ⓐ 43=64 ⓑ 413=43 ⓒ (12)x=132

Solution

ⓐ log464=3
ⓑ log443=13 ⓒ log12132=x

In the next example we do the reverse—convert logarithmic form to exponential form.

Convert to exponential form: ⓐ 2=log864, ⓑ 0=log41, and ⓒ −3=log1011000.

Solution
In part (a) we have 2 equals log sub 8 of 64, where the 2 is blue and the 8 is red. Following this, we have x equals red a to the blue y power. Then 64 equals 8 squared. Hence, if 2 equals log sub 8 of 64, then 64 equals 8 squared. In part (b) we have 0 equals log sub 4 of 1, where the 0 is blue and the 4 is red. Following this, we have x equals red a to the blue y power. Then 1 equals 4 to the zero power. Hence, if 0 equals log sub 4 of 1, then 1 equals 4 to the zero power. In part (c) we have negative 3 equals log sub 10 of 1 over 1000, where the negative 3 is blue and the 10 is red. Following this, we have x equals red a to the blue y power. Then 1 over 1000 equals 10 to the negative three power. Hence, if negative 3 equals log sub 10 of 1 over 1000, then 1 over 1000 equals 10 to the negative 3 power.

Convert to exponential form: ⓐ 3=log464 ⓑ 0=logx1 ⓒ −2=log101100

Solution

ⓐ 64=43
ⓑ 1=x0 ⓒ 1100=10−2

Convert to exponential form: ⓐ 3=log327 ⓑ 0=log31 ⓒ −1=log10110

Solution

ⓐ 27=33 ⓑ 1=30
ⓒ 110=10−1

Evaluate Logarithmic Functions

We can solve and evaluate logarithmic equations by using the technique of converting the equation to its equivalent exponential equation.

Find the value of x: ⓐ logx36=2, ⓑ log4x=3, and ⓒ log1218=x.

Solution
ⓐ
Steps to solve the logarithmic equation log_x 36 = 2, demonstrating conversion to exponential form and applying base rules.
logx36=2
Convert to exponential form. x2=36
Solve the quadratic. x=6,x=−6
The base of a logarithmic function must be positive, so we eliminate x=−6. x=6Therefore,log636=2.
ⓑ
Steps to solve the logarithmic equation log₄x = 3 by converting it to exponential form.
log4x=3
Convert to exponential form. 43=x
Simplify. x=64Therefore,log464=3.
ⓒ
Step-by-step solution for evaluating log_(1/2)(1/8) by converting to exponential form.
log1218=x
Convert to exponential form. (12)x=18
Rewrite 18 as (12)3. (12)x=(12)3
With the same base, the exponents must be equal. x=3Therefore,log1218=3

Find the value of x: ⓐ logx64=2 ⓑ log5x=3 ⓒ log1214=x

Solution


ⓐ x=8 ⓑ x=125 ⓒ x=2

Find the value of x: ⓐ logx81=2 ⓑ log3x=5 ⓒ log13127=x

Solution


ⓐ
x=9 ⓑ x=243 ⓒ x=3

When see an expression such as log327, we can find its exact value two ways. By inspection we realize it means “3 to what power will be 27”? Since 33=27, we know log327=3. An alternate way is to set the expression equal to x and then convert it into an exponential equation.

Find the exact value of each logarithm without using a calculator: ⓐ log525, ⓑ log93, and ⓒ log2116.

Solution
ⓐ
Demonstrates two methods to evaluate log_5(25): a direct question/answer approach and a step-by-step algebraic solution, clarifying the process of solving logarithms.
log525
5 to what power will be 25? log525=2
Or
Set the expression equal to x. log525=x
Change to exponential form. 5x=25
Rewrite 25 as 52. 5x=52
With the same base the exponents must be equal. x=2Therefore,log525=2.
ⓑ
Steps for solving the logarithmic expression log_9 3 by converting it to exponential form.
log93
Set the expression equal to x. log93=x
Change to exponential form. 9x=3
Rewrite 9 as 32. (32)x=31
Simplify the exponents. 32x=31
With the same base the exponents must be equal. 2x=1
Solve the equation. x=12Therefore,log93=12.
ⓒ
This table demonstrates the step-by-step process for evaluating the logarithmic expression log
log2116
Set the expression equal to x. log2116=x
Change to exponential form. 2x=116
Rewrite 16 as 24. 2x=124
2x=2−4
With the same base the exponents must be equal. x=−4Therefore,log2116=−4.

Find the exact value of each logarithm without using a calculator: ⓐ log12144 ⓑ log42 ⓒ log2132

Solution

ⓐ
2 ⓑ 12 ⓒ −5

Find the exact value of each logarithm without using a calculator: ⓐ log981 ⓑ log82 ⓒ log319

Solution

ⓐ 2 ⓑ 13 ⓒ −2

Graph Logarithmic Functions

To graph a logarithmic function y=logax, it is easiest to convert the equation to its exponential form, x=ay. Generally, when we look for ordered pairs for the graph of a function, we usually choose an x-value and then determine its corresponding y-value. In this case you may find it easier to choose y-values and then determine its corresponding x-value.

Graph y=log2x.

Solution

To graph the function, we will first rewrite the logarithmic equation, y=log2x, in exponential form, 2y=x.

We will use point plotting to graph the function. It will be easier to start with values of y and then get x.

y 2y=x (x,y)
−2 2−2=122=14 (14,-2)
−1 2−1=121=12 (12,−1)
0 20=1 (1,0)
1 21=2 (2,1)
2 22=4 (4,2)
3 23=8 (8,3)
This figure shows the logarithmic curve going through the points (1 over 2, negative 1), (1, 0), and (2, 1).

Graph: y=log3x.

Solution


This figure shows the logarithmic curve going through the points (1 over 3, negative 1), (1, 0), and (3, 1).

Graph: y=log5x.

Solution


This figure shows the logarithmic curve going through the points (1 over 5, negative 1), (1, 0), and (5, 1).

The graphs of y=log2x,y=log3x, and y=log5x are the shape we expect from a logarithmic function where a>1.

We notice that for each function the graph contains the point (1,0). This make sense because 0=loga1 means a0=1 which is true for any a.

The graph of each function, also contains the point (a,1). This makes sense as 1=logaa means a1=a. which is true for any a.

Notice too, the graph of each function y=logax also contains the point (1a,−1). This makes sense as −1=loga1a means a−1=1a, which is true for any a.

Look at each graph again. Now we will see that many characteristics of the logarithm function are simply ’mirror images’ of the characteristics of the corresponding exponential function.

What is the domain of the function? The graph never hits the y-axis. The domain is all positive numbers. We write the domain in interval notation as (0,∞).

What is the range for each function? From the graphs we can see that the range is the set of all real numbers. There is no restriction on the range. We write the range in interval notation as (−∞,∞).

When the graph approaches the y-axis so very closely but will never cross it, we call the line x=0, the y-axis, a vertical asymptote.

Properties of the Graph of y=logax when a>1

Domain (0,∞)
Range (−∞,∞)
x-intercept (1,0)
y-intercept None
Contains (a,1),(1a,−1)
Asymptote y-axis
This figure shows the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1).

Our next example looks at the graph of y=logax when 0<a<1.

Graph y=log13x.

Solution

To graph the function, we will first rewrite the logarithmic equation, y=log13x, in exponential form, (13)y=x.

We will use point plotting to graph the function. It will be easier to start with values of y and then get x.

y (13)y=x (x,y)
−2 (13)−2=32=9 (9,−2)
−1 (13)−1=31=3 (3,−1)
0 (13)0=1 (1,0)
1 (13)1=13 (13,1)
2 (13)2=19 (19,2)
3 (13)3=127 (127,3)
This figure shows the logarithmic curve going through the points (1 over 3, 1), (1, 0), and (3, negative 1).

Graph: y=log12x.

Solution


This figure shows the logarithmic curve going through the points (1 over 2, 1), (1, 0), and (2, negative 1).

Graph: y=log14x.

Solution


This figure shows the logarithmic curve going through the points (1 over 4, 1), (1, 0), and (4, negative 1).

Now, let’s look at the graphs y=log12x,y=log13x and y=log14x, so we can identify some of the properties of logarithmic functions where 0<a<1.

The graphs of all have the same basic shape. While this is the shape we expect from a logarithmic function where 0<a<1.

We notice, that for each function again, the graph contains the points,(1,0),(a,1),(1a,−1). This make sense for the same reasons we argued above.

We notice the domain and range are also the same—the domain is (0,∞) and the range is (−∞,∞). The y-axis is again the vertical asymptote.

We will summarize these properties in the chart below. Which also include when a>1.

Properties of the Graph of y=logax

when a>1 when 0<a<1
Domain (0,∞) Domain (0,∞)
Range (−∞,∞) Range (−∞,∞)
x-intercept (1,0) x-intercept (1,0)
y-intercept none y-intercept None
Contains (a,1),(1a,−1) Contains (a,1),(1a,−1)
Asymptote y-axis Asymptote y-axis
Basic shape increasing Basic shape Decreasing
This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1).

We talked earlier about how the logarithmic function f−1(x)=logax is the inverse of the exponential function f(x)=ax. The graphs in Figure 2 show both the exponential (blue) and logarithmic (red) functions on the same graph for both a>1 and 0<a<1.

This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). It also shows the exponential curve going through the points (1, 1 over a), (0, 1), and (1, a) along with the line y equals x. The logarithmic curve is a mirror image of the exponential curve across the y equals x line. This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1). It also shows the exponential curve going through the points (negative 1, 1 over a), (0, 1), and (1, a) along with the line y equals x. The logarithmic curve is a mirror image of the exponential curve across the y equals x line.

Notice how the graphs are reflections of each other through the line y=x. We know this is true of inverse functions. Keeping a visual in your mind of these graphs will help you remember the domain and range of each function. Notice the x-axis is the horizontal asymptote for the exponential functions and the y-axis is the vertical asymptote for the logarithmic functions.

Solve Logarithmic Equations

When we talked about exponential functions, we introduced the number e. Just as e was a base for an exponential function, it can be used a base for logarithmic functions too. The logarithmic function with base e is called the natural logarithmic function. The function f(x)=logex is generally written f(x)=lnx and we read it as “el en of x.”

Natural Logarithmic Function

The function f(x)=lnx is the natural logarithmic function with base e, where x>0.

y=lnxis equivalent tox=ey

When the base of the logarithm function is 10, we call it the common logarithmic function and the base is not shown. If the base a of a logarithm is not shown, we assume it is 10.

Common Logarithmic Function

The function f(x)=logx is the common logarithmic function with base10, where x>0.

y=logxis equivalent tox=10y
It will be important for you to use your calculator to evaluate both common and natural logarithms. Find the log and ln keys on your calculator.

To solve logarithmic equations, one strategy is to change the equation to exponential form and then solve the exponential equation as we did before. As we solve logarithmic equations, y=logax, we need to remember that for the base a, a>0 and a≠1. Also, the domain is x>0. Just as with radical equations, we must check our solutions to eliminate any extraneous solutions.

Solve: ⓐ loga49=2 and ⓑ lnx=3.

Solution
ⓐ
Step-by-step solution demonstrating how to solve the logarithmic equation log_a 49 = 2, from rewriting to verifying the base.
loga49=2
Rewrite in exponential form. a2=49
Solve the equation using the square root property. a=±7
The base cannot be negative, so we eliminate a=−7. a=7,a=−7
Check.
a=7loga49=2log749=?272=?4949=49✓
ⓑ
This table illustrates the step-by-step process of solving a logarithmic equation, including rewriting it in exponential form and verifying the solution.
lnx=3
Rewrite in exponential form. e3=x
Check.
x=e3lnx=3lne3=?3e3=e3✓

Solve: ⓐ loga121=2 ⓑ lnx=7

Solution


ⓐ
a=11
ⓑ x=e7

Solve: ⓐ loga64=3 ⓑ lnx=9

Solution


ⓐ
a=4
ⓑ x=e9

Solve: ⓐ log2(3x−5)=4 and ⓑ lne2x=4.

Solution
ⓐ
Step-by-step solution and verification of the logarithmic equation log₂(3x - 5) = 4, demonstrating the conversion to exponential form and algebraic solving.
log2(3x−5)=4
Rewrite in exponential form. 24=3x−5
Simplify. 16=3x−5
Solve the equation. 21=3x
7=x
Check.
x=7log2(3x−5)=4log2(3⋅7−5)=?4log2(16)=?424=?1616=16✓
ⓑ
Step-by-step solution and check for the logarithmic equation ln(e^2x) = 4.
lne2x=4
Rewrite in exponential form. e4=e2x
Since the bases are the same the exponents are equal. 4=2x
Solve the equation. 2=x
Check.
x=2lne2x=4 lne2·2=?4 lne4=?4 e4=e4✓

Solve: ⓐ log2(5x−1)=6 ⓑ lne3x=6

Solution


ⓐ
x=13
ⓑ x=2

Solve: ⓐ log3(4x+3)=3 ⓑ lne4x=4

Solution


ⓐ
x=6
ⓑ x=1

Use Logarithmic Models in Applications

There are many applications that are modeled by logarithmic equations. We will first look at the logarithmic equation that gives the decibel (dB) level of sound. Decibels range from 0, which is barely audible to 160, which can rupture an eardrum. The 10−12 in the formula represents the intensity of sound that is barely audible.

Decibel Level of Sound

The loudness level, D, measured in decibels, of a sound of intensity, I, measured in watts per square inch is

D=10log(I10−12)

Extended exposure to noise that measures 85 dB can cause permanent damage to the inner ear which will result in hearing loss. What is the decibel level of music coming through ear phones with intensity 10−2 watts per square inch?

Solution
The image displays the formula D = 10 log(I / 10^-12), which calculates the decibel level (D) from a given sound intensity (I).
Substitute in the intensity level, I. An equation: D = 10 log(10^-2 / 10^-12), demonstrating a base-10 logarithm of a fraction of powers of ten, with 10^-2 highlighted in red.
Simplify. A mathematical equation is displayed: D = 10 log(10^10).
Since log1010=10. A mathematical equation shows D equals 10 times 10
Multiply. The image displays a clear mathematical expression: D = 100. The text is in a bold, sans-serif font against a plain white background, occupying the top left portion of the frame.
The decibel level of music coming through earphones is 100 dB.

What is the decibel level of one of the new quiet dishwashers with intensity 10−7 watts per square inch?

Solution

The quiet dishwashers have a decibel level of 50 dB.

What is the decibel level heavy city traffic with intensity 10−3 watts per square inch?

Solution

The decibel level of heavy traffic is 90 dB.

The magnitude R of an earthquake is measured by a logarithmic scale called the Richter scale. The model is R=logI, where I is the intensity of the shock wave. This model provides a way to measure earthquake intensity.

Earthquake Intensity

The magnitude R of an earthquake is measured by R=logI, where I is the intensity of its shock wave.

In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. Over 80% of the city was destroyed by the resulting fires. In 2014, Los Angeles experienced a moderate earthquake that measured 5.1 on the Richter scale and caused $108 million dollars of damage. Compare the intensities of the two earthquakes.

Solution

To compare the intensities, we first need to convert the magnitudes to intensities using the log formula. Then we will set up a ratio to compare the intensities.

This table illustrates the step-by-step calculation to compare the intensities of the 1906 and 2014 earthquakes using logarithmic and exponential conversions, culminating in their intensity ratio.
Convert the magnitudes to intensities. R=logI
1906 earthquake 7.8=logI
Convert to exponential form. I=107.8
2014 earthquake 5.1=logI
Convert to exponential form. I=105.1
Form a ratio of the intensities. Intensityfor1906Intensityfor2014
Substitute in the values. 107.8105.1
Divide by subtracting the exponents. 102.7
Evaluate. 501
The intensity of the 1906 earthquake was about 501 times the intensity of the 2014 earthquake.

In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. In 1989, the Loma Prieta earthquake also affected the San Francisco area, and measured 6.9 on the Richter scale. Compare the intensities of the two earthquakes.

Solution

The intensity of the 1906 earthquake was about 8 times the intensity of the 1989 earthquake.

In 2014, Chile experienced an intense earthquake with a magnitude of 8.2 on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured 5.1 on the Richter scale. Compare the intensities of the two earthquakes.

Solution

The intensity of the earthquake in Chile was about 1,259 times the intensity of the earthquake in Los Angeles.

Access these online resources for additional instruction and practice with evaluating and graphing logarithmic functions.

  • Re-writing logarithmic equations in exponential form
  • Simplifying Logarithmic Expressions
  • Graphing logarithmic functions
  • Using logarithms to calculate decibel levels

Key Concepts

  • Properties of the Graph of y=logax:
    when a>1 when 0<a<1
    Domain (0,∞) Domain (0,∞)
    Range (−∞,∞) Range (−∞,∞)
    x-intercept (1,0) x-intercept (1,0)
    y-intercept none y-intercept none
    Contains (a,1),(1a,−1) Contains (a,1),(1a,−1)
    Asymptote y-axis Asymptote y-axis
    Basic shape increasing Basic shape decreasing

    This figure shows that, for a greater than 1, the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1). This figure shows that, for a greater than 0 and less than 1, the logarithmic curve going through the points (a, 1), (1, 0), and (1 over a, negative 1).
  • Decibel Level of Sound: The loudness level, D, measured in decibels, of a sound of intensity, I, measured in watts per square inch is D=10log(I10−12).
  • Earthquake Intensity: The magnitude R of an earthquake is measured by R=logI, where I is the intensity of its shock wave.

Practice Makes Perfect

Convert Between Exponential and Logarithmic Form

In the following exercises, convert from exponential to logarithmic form.

42=16

25=32

Solution

log232=5

33=27

53=125

Solution

log5125=3

103=1000

10−2=1100

Solution

log1100=−2

x12=3

x13=63

Solution

logx63=13

32x=324

17x=175

Solution

log17175=x

(14)2=116

(13)4=181

Solution

log13181=4

3−2=19

4−3=164

Solution

log4164=−3

ex=6

e3=x

Solution

lnx=3

In the following exercises, convert each logarithmic equation to exponential form.

3=log464

6=log264

Solution

64=26

4=logx81

5=logx32

Solution

32=x5

0=log121

0=log71

Solution

1=70

1=log33

1=log99

Solution

9=91

−4=log10110,000

3=log101,000

Solution

1,000=103

5=logex

x=loge43

Solution

43=ex

Evaluate Logarithmic Functions

In the following exercises, find the value of x in each logarithmic equation.

logx49=2

logx121=2

Solution

x=11

logx27=3

logx64=3

Solution

x=4

log3x=4

log5x=3

Solution

x=125

log2x=−6

log3x=−5

Solution

x=1243

log14116=x

log1319=x

Solution

x=2

log1464=x

log1981=x

Solution

x=−2

In the following exercises, find the exact value of each logarithm without using a calculator.

log749

log636

Solution

2

log41

log51

Solution

0

log164

log273

Solution

13

log122

log124

Solution

−2

log2116

log3127

Solution

−3

log4116

log9181

Solution

−2

Graph Logarithmic Functions

In the following exercises, graph each logarithmic function.

y=log2x

y=log4x

Solution

This figure shows the logarithmic curve going through the points (1 over 4, negative 1), (1, 0), and (4, 1).

y=log6x

y=log7x

Solution

This figure shows that the logarithmic curve going through the points (1 over 7, negative 1), (1, 0), and (7, 1).

y=log1.5x

y=log2.5x

Solution

This figure shows the logarithmic curve going through the points (2 over 5, negative 1), (1, 0), and (2.5, 1).

y=log13x

y=log15x

Solution

This figure shows the logarithmic curve going through the points (1 over 5, 1), (1, 0), and (5, negative 1).

y=log0.4x

y=log0.6x

Solution

This figure shows the logarithmic curve going through the points (3 over 5, 1), (1, 0), and (5 over 3, negative 1).

Solve Logarithmic Equations

In the following exercises, solve each logarithmic equation.

loga16=2

loga81=2

Solution

a=9

loga8=3

loga27=3

Solution

a=3

loga32=2

loga24=3

Solution

a=233

lnx=5

lnx=4

Solution

x=e4

log2(5x+1)=4

log2(6x+2)=5

Solution

x=5

log3(4x−3)=2

log3(5x−4)=4

Solution

x=17

log4(5x+6)=3

log4(3x−2)=2

Solution

x=6

lne4x=8

lne2x=6

Solution

x=3

logx2=2

log(x2−25)=2

Solution

x=−55,x=55

log2(x2−4)=5

log3(x2+2)=3

Solution

x=−5,x=5

Use Logarithmic Models in Applications

In the following exercises, use a logarithmic model to solve.

What is the decibel level of normal conversation with intensity 10−6 watts per square inch?

What is the decibel level of a whisper with intensity 10−10 watts per square inch?

Solution

A whisper has a decibel level of 20 dB.

What is the decibel level of the noise from a motorcycle with intensity 10−2 watts per square inch?

What is the decibel level of the sound of a garbage disposal with intensity 10−2 watts per square inch?

Solution

The sound of a garbage disposal has a decibel level of 100 dB.

In 2014, Chile experienced an intense earthquake with a magnitude of 8.2 on the Richter scale. In 2010, Haiti also experienced an intense earthquake which measured 7.0 on the Richter scale. Compare the intensities of the two earthquakes.

The Los Angeles area experiences many earthquakes. In 1994, the Northridge earthquake measured magnitude of 6.7 on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured 5.1 on the Richter scale. Compare the intensities of the two earthquakes.

Solution

The intensity of the 1994 Northridge earthquake in the Los Angeles area was about 40 times the intensity of the 2014 earthquake.

Writing Exercises

Explain how to change an equation from logarithmic form to exponential form.

Explain the difference between common logarithms and natural logarithms.

Solution

Answers will vary.

Explain why logaax=x.

Explain how to find the log732 on your calculator.

Solution

Answers will vary.

Self Check

ⓐ
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and five columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Convert between exponential and logarithmic form, evaluate logarithmic functions, graph logarithmic functions, solve logarithmic equations, and use logarithmic models in applications. The rest of the cells are blank.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

common logarithmic function
The function f(x)=logx is the common logarithmic function with base10, where x>0.
y=logxis equivalent tox=10y
logarithmic function
The function f(x)=logax is the logarithmic function with base a, where a>0,x>0, and a≠1.
y=logaxis equivalent tox=ay
natural logarithmic function
The function f(x)=lnx is the natural logarithmic function with base e, where x>0.
y=lnxis equivalent tox=ey

Use the Properties of Logarithms

Learning Objectives

By the end of this section, you will be able to:

  • Use the properties of logarithms
  • Use the Change of Base Formula

Before you get started, take this readiness quiz.

Evaluate: ⓐ a0 ⓑ a1.
If you missed this problem, review Example 3 in Properties of Exponents and Scientific Notation.

Solution

ⓐ 1; ⓑ a

Write with a rational exponent: x2y3.
If you missed this problem, review Example 2 in Simplify Rational Exponents.

Solution

x2y13

Round to three decimal places: 2.5646415.
If you missed this problem, review Example 1 in Decimals.

Solution

2.565

Use the Properties of Logarithms

Now that we have learned about exponential and logarithmic functions, we can introduce some of the properties of logarithms. These will be very helpful as we continue to solve both exponential and logarithmic equations.

The first two properties derive from the definition of logarithms. Since a0=1, we can convert this to logarithmic form and get loga1=0. Also, since a1=a, we get logaa=1.

Properties of Logarithms

loga1=0logaa=1

In the next example we could evaluate the logarithm by converting to exponential form, as we have done previously, but recognizing and then applying the properties saves time.

Evaluate using the properties of logarithms: ⓐ log81 and ⓑ log66.

Solution
ⓐ
Table demonstrating the evaluation of log_8 1, using the logarithmic property log_a 1 = 0 to show the result is 0.
log81
Use the property, loga1=0. 0log81=0

ⓑ
log66Use the property,logaa=1.1log66=1

Evaluate using the properties of logarithms: ⓐ log131 ⓑ log99.

Solution

ⓐ 0 ⓑ 1

Evaluate using the properties of logarithms: ⓐ log51 ⓑ log77.

Solution

ⓐ 0 ⓑ 1

The next two properties can also be verified by converting them from exponential form to logarithmic form, or the reverse.

The exponential equation alogax=x converts to the logarithmic equation logax=logax, which is a true statement for positive values for x only.

The logarithmic equation logaax=x converts to the exponential equation ax=ax, which is also a true statement.

These two properties are called inverse properties because, when we have the same base, raising to a power “undoes” the log and taking the log “undoes” raising to a power. These two properties show the composition of functions. Both ended up with the identity function which shows again that the exponential and logarithmic functions are inverse functions.

Inverse Properties of Logarithms

For a>0,x>0 and a≠1,

alogax=xlogaax=x

In the next example, apply the inverse properties of logarithms.

Evaluate using the properties of logarithms: ⓐ 4log49 and ⓑ log335.

Solution
ⓐ
Demonstration of the logarithmic property a^(log_a x) = x using an example.
4log49
Use the property, alogax=x. 94log49=9
ⓑ
This table illustrates a logarithmic expression and a related property, showing a step in its calculation.
log335
Use the property, alogax=x. 5log335=5

Evaluate using the properties of logarithms: ⓐ 5log515 ⓑ log774.

Solution

ⓐ 15 ⓑ 4

Evaluate using the properties of logarithms: ⓐ 2log28 ⓑ log2215.

Solution

ⓐ 8 ⓑ 15

There are three more properties of logarithms that will be useful in our work. We know exponential functions and logarithmic function are very interrelated. Our definition of logarithm shows us that a logarithm is the exponent of the equivalent exponential. The properties of exponents have related properties for exponents.

In the Product Property of Exponents, am·an=am+n, we see that to multiply the same base, we add the exponents. The Product Property of Logarithms, logaM·N=logaM+logaN tells us to take the log of a product, we add the log of the factors.

Product Property of Logarithms

If M>0,N>0,a>0 and a≠1, then,

loga(M·N)=logaM+logaN

The logarithm of a product is the sum of the logarithms.

We use this property to write the log of a product as a sum of the logs of each factor.

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible: ⓐ log37x and ⓑ log464xy.

Solution
ⓐ
Demonstrates expanding a logarithmic expression log_3(7x) into log_3(7) + log_3(x) using the Product Property of logarithms.
log37x
Use the Product Property, loga(M·N)=logaM+logaN. log37+log3x
log37x=log37+log3x
ⓑ
This table demonstrates the step-by-step expansion of a logarithmic expression, log4(64xy), using the product property of logarithms, simplifying it to 3 + log4(x) + log4(y).
log464xy
Use the Product Property, loga(M·N)=logaM+logaN. log464+log4x+log4y
Simplify by evaluating log464. 3+log4x+log4y
log464xy=3+log4x+log4y

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

ⓐ log33x ⓑ log28xy

Solution

ⓐ 1+log3x
ⓑ 3+log2x+log2y

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

ⓐ log99x ⓑ log327xy

Solution

ⓐ 1+log9x
ⓑ 3+log3x+log3y

Similarly, in the Quotient Property of Exponents, aman=am−n, we see that to divide the same base, we subtract the exponents. The Quotient Property of Logarithms, logaMN=logaM−logaN tells us to take the log of a quotient, we subtract the log of the numerator and denominator.

Quotient Property of Logarithms

If M>0,N>0,a>0 and a≠1, then,

logaMN=logaM−logaN

The logarithm of a quotient is the difference of the logarithms.

Note that logaM−logaN≠loga(M−N).

We use this property to write the log of a quotient as a difference of the logs of each factor.

Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
ⓐ log557 and ⓑ logx100

Solution
ⓐ
Step-by-step simplification of the logarithmic expression log5(5/7) using the Quotient Property.
log557
Use the Quotient Property, logaMN=logaM−logaN. log55−log57
Simplify. 1−log57
log557=1−log57
ⓑ
Step-by-step expansion of log(x/100) using the logarithm quotient property and simplification.
logx100
Use the Quotient Property, logaMN=logaM−logaN. logx−log100
Simplify. logx−2
logx100=logx−2

Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.

ⓐ log434 ⓑ logx1000

Solution

ⓐ log43−1 ⓑ logx−3

Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.

ⓐ log254 ⓑ log10y

Solution

ⓐ log25−2 ⓑ 1−logy

The third property of logarithms is related to the Power Property of Exponents, (am)n=am·n, we see that to raise a power to a power, we multiply the exponents. The Power Property of Logarithms, logaMp=plogaM tells us to take the log of a number raised to a power, we multiply the power times the log of the number.

Power Property of Logarithms

If M>0,a>0,a≠1 and p is any real number then,

logaMp=plogaM

The log of a number raised to a power is the product of the power times the log of the number.

We use this property to write the log of a number raised to a power as the product of the power times the log of the number. We essentially take the exponent and throw it in front of the logarithm.

Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
ⓐ log543 and ⓑ logx10

Solution
ⓐ
Illustration of the Power Property of logarithms with an example.
log543
Use the Power Property, logaMp=plogaM. 3log54
log543=3log54
ⓑ
Application of the Power Property of Logarithms to transform log(x^10) into 10log(x).
logx10
Use the Power Property, logaMp=plogaM. 10logx
logx10=10logx

Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.

ⓐ log754 ⓑ logx100

Solution

ⓐ 4log75 ⓑ 100·logx

Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.

ⓐ log237 ⓑ logx20

Solution

ⓐ 7log23 ⓑ 20·logx

We summarize the Properties of Logarithms here for easy reference. While the natural logarithms are a special case of these properties, it is often helpful to also show the natural logarithm version of each property.

Properties of Logarithms

If M>0,N>0,a>0,a≠1 and p is any real number then,

Property Base a Base e
loga1=0 ln1=0
logaa=1 lne=1
Inverse Properties alogax=xlogaax=x elnx=x lnex=x
Product Property of Logarithms loga(M·N)=logaM+logaN ln(M·N)=lnM+lnN
Quotient Property of Logarithms logaMN=logaM−logaN lnMN=lnM−lnN
Power Property of Logarithms logaMp=plogaM lnMp=plnM

Now that we have the properties we can use them to “expand” a logarithmic expression. This means to write the logarithm as a sum or difference and without any powers.

We generally apply the Product and Quotient Properties before we apply the Power Property.

Use the Properties of Logarithms to expand the logarithm log4(2x3y2). Simplify, if possible.

Solution
Step-by-step expansion of the logarithmic expression log4(2x3y2) using logarithm properties.
log4(2x3y2)
Use the Product Property, logaM·N=logaM+logaN. log42+log4x3+log4y2
Use the Power Property, logaMp=plogaM, on the last two terms. log42+3log4x+2log4y
Simplify. 12+3log4x+2log4y
log4(2x3y2)=12+3log4x+2log4y

Use the Properties of Logarithms to expand the logarithm log2(5x4y2). Simplify, if possible.

Solution

log25+4log2x+2log2y

Use the Properties of Logarithms to expand the logarithm log3(7x5y3). Simplify, if possible.

Solution

log37+5log3x+3log3y

When we have a radical in the logarithmic expression, it is helpful to first write its radicand as a rational exponent.

Use the Properties of Logarithms to expand the logarithm log2x33y2z4. Simplify, if possible.

Solution
Step-by-step expansion of a complex logarithmic expression using properties of logarithms.
log2x33y2z4
Rewrite the radical with a rational exponent. log2(x33y2z)14
Use the Power Property, logaMp=plogaM. 14log2(x33y2z)
Use the Quotient Property, logaM·N=logaM−logaN. 14(log2(x3)−log2(3y2z))
Use the Product Property, logaM·N=logaM+logaN, in the second term. 14(log2(x3)−(log23+log2y2+log2z))
Use the Power Property, logaMp=plogaM, inside the parentheses. 14(3log2x−(log23+2log2y+log2z))
Simplify by distributing. 14(3log2x−log23−2log2y−log2z)
log2x33y2z4=14(3log2x−log23−2log2y−log2z)

Use the Properties of Logarithms to expand the logarithm log4x42y3z25. Simplify, if possible.

Solution

15(4log4x−12−3log4y−2log4z)

Use the Properties of Logarithms to expand the logarithm log3x25yz3. Simplify, if possible.

Solution

13(2log3x−log35−log3y−log3z)

The opposite of expanding a logarithm is to condense a sum or difference of logarithms that have the same base into a single logarithm. We again use the properties of logarithms to help us, but in reverse.

To condense logarithmic expressions with the same base into one logarithm, we start by using the Power Property to get the coefficients of the log terms to be one and then the Product and Quotient Properties as needed.

Use the Properties of Logarithms to condense the logarithm log43+log4x−log4y. Simplify, if possible.

Solution
Simplifying a logarithmic expression using the Product and Quotient Properties of logarithms, step-by-step.
The log expressions all have the same base, 4. log43+log4x−log4y
The first two terms are added, so we use the Product Property, logaM+logaN=logaM·N. log43x−log4y
Since the logs are subtracted, we use the Quotient Property, logaM−logaN=logaMN. log43xy
log43+log4x−log4y=log43xy

Use the Properties of Logarithms to condense the logarithm log25+log2x−log2y. Simplify, if possible.

Solution

log25xy

Use the Properties of Logarithms to condense the logarithm log36−log3x−log3y. Simplify, if possible.

Solution

log36xy

Use the Properties of Logarithms to condense the logarithm 2log3x+4log3(x+1). Simplify, if possible.

Solution
Steps to combine a sum of logarithmic expressions into a single logarithm using the Power and Product Properties.
The log expressions have the same base, 3. 2log3x+4log3(x+1)
Use the Power Property, logaM+logaN=logaM·N. log3x2+log3(x+1)4
The terms are added, so we use the Product Property, logaM+logaN=logaM·N. log3x2(x+1)4
2log3x+4log3(x+1)=log3x2(x+1)4

Use the Properties of Logarithms to condense the logarithm 3log2x+2log2(x−1). Simplify, if possible.

Solution

log2x3(x−1)2

Use the Properties of Logarithms to condense the logarithm 2logx+2log(x+1). Simplify, if possible.

Solution

logx2(x+1)2

Use the Change-of-Base Formula

To evaluate a logarithm with any other base, we can use the Change-of-Base Formula. We will show how this is derived.

This table illustrates the step-by-step derivation of the logarithm change of base formula using algebraic manipulation.
Suppose we want to evaluate logaM. logaM
Let y=logaM. y=logaM
Rewrite the expression in exponential form. ay=M
Take the logb of each side. logbay=logbM
Use the Power Property. ylogba=logbM
Solve for y. y=logbMlogba
Substitute y=logaM. logaM=logbMlogba

The Change-of-Base Formula introduces a new base b. This can be any base b we want where b>0,b≠1. Because our calculators have keys for logarithms base 10 and base e, we will rewrite the Change-of-Base Formula with the new base as 10 or e.

Change-of-Base Formula

For any logarithmic bases a,b and M>0,

logaM=logbMlogbalogaM=logMlogalogaM=lnMlna new basebnew base 10new basee

When we use a calculator to find the logarithm value, we usually round to three decimal places. This gives us an approximate value and so we use the approximately equal symbol (≈).

Rounding to three decimal places, approximate log435.

Solution
The mathematical expression log base 4 of 35 is displayed in black text on a white background.
Use the Change-of-Base Formula. The image displays the logarithm change of base formula: log subscript 'a' of M equals the fraction of log subscript 'b' of M over log subscript 'b' of 'a'.
Identify a and M. Choose 10 for b. The image displays the logarithm change of base formula, showing that log base 4 of 35 is equal to the ratio of log 35 to log 4, illustrating how to convert logarithms to a different base.
Enter the expression log35log4 in the calculator
using the log button for base 10. Round to three decimal places.
The image displays the logarithm expression 'log base 4 of 35 is approximately equal to 2.565' in black text on a white background.

Rounding to three decimal places, approximate log342.

Solution

3.402

Rounding to three decimal places, approximate log546.

Solution

2.379

Access these online resources for additional instruction and practice with using the properties of logarithms.

  • Using Properties of Logarithms to Expand Logs
  • Using Properties of Logarithms to Condense Logs
  • Change of Base

Key Concepts

  • Properties of Logarithms
    loga1=0logaa=1
  • Inverse Properties of Logarithms
    • For a>0,x>0 and a≠1
      alogax=xlogaax=x
  • Product Property of Logarithms
    • If M>0,N>0,a>0 and a≠1, then,
      logaM·N=logaM+logaN

      The logarithm of a product is the sum of the logarithms.
  • Quotient Property of Logarithms
    • If M>0,N>0,a>0 and a≠1, then,
      logaMN=logaM−logaN

      The logarithm of a quotient is the difference of the logarithms.
  • Power Property of Logarithms
    • If M>0,a>0,a≠1 and p is any real number then,
      logaMp=plogaM

      The log of a number raised to a power is the product of the power times the log of the number.
  • Properties of Logarithms Summary
    If M>0,a>0,a≠1 and p is any real number then,
    Property Base a Base e
    loga1=0 ln1=0
    logaa=1 lne=1
    Inverse Properties alogax=x logaax=x elnx=x lnex=x
    Product Property of Logarithms loga(M·N)=logaM+logaN ln(M·N)=lnM+lnN
    Quotient Property of Logarithms logaMN=logaM−logaN lnMN=lnM−lnN
    Power Property of Logarithms logaMp=plogaM lnMp=plnM
  • Change-of-Base Formula
    For any logarithmic bases a and b, and M>0,
    logaM=logbMlogbalogaM=logMlogalogaM=lnMlna new basebnew base 10new basee

Practice Makes Perfect

Use the Properties of Logarithms

In the following exercises, use the properties of logarithms to evaluate.

ⓐ log41 ⓑ log88

ⓐ log121 ⓑ lne

Solution

ⓐ 0 ⓑ 1

ⓐ 3log36 ⓑ log227

ⓐ 5log510 ⓑ log4410

Solution

ⓐ 10 ⓑ 10

ⓐ 8log87 ⓑ log66−2

ⓐ 6log615 ⓑ log88−4

Solution

ⓐ 15 ⓑ −4

ⓐ 10log5 ⓑ log10−2

ⓐ 10log3 ⓑ log10−1

Solution

ⓐ 3 ⓑ −1

ⓐ eln4 ⓑ lne2

ⓐ eln3 ⓑ lne7

Solution

ⓐ 3 ⓑ 7

In the following exercises, use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

log46x

log58y

Solution

log58+log5y

log232xy

log381xy

Solution

4+log3x+log3y

log100x

log1000y

Solution

3+logy

In the following exercises, use the Quotient Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

log338

log656

Solution

log65−1

log416y

log5125x

Solution

3−log5x

logx10

log10,000y

Solution

4−logy

lne33

lne416

Solution

4−ln16

In the following exercises, use the Power Property of Logarithms to expand each. Simplify if possible.

log3x2

log2x5

Solution

5log2x

logx−2

logx−3

Solution

−3logx

log4x

log5x3

Solution

13log5x

lnx3

lnx43

Solution

43lnx

In the following exercises, use the Properties of Logarithms to expand the logarithm. Simplify if possible.

log5(4x6y4)

log2(3x5y3)

Solution

log23+5log2x+3log2y

log3(2x2)

log5(214y3)

Solution

14log521+3log5y

log3xy2z2

log54ab3c4d2

Solution

log54+log5a+3log5b
+4log5c−2log5d

log4x16y4

log3x2327y4

Solution

23log3x−3−4log3y

log22x+y2z2

log33x+2y25z2

Solution

12log3(3x+2y2)−log35−2log3z

log25x32y2z44

log53x24y3z3

Solution

13(log53+2log5x−log54
−3log5y−log5z)

In the following exercises, use the Properties of Logarithms to condense the logarithm. Simplify if possible.

log64+log69

log4+log25

Solution

2

log280−log25

log336−log34

Solution

2

log34+log3(x+1)

log25−log2(x−1)

Solution

log25x−1

log73+log7x−log7y

log52−log5x−log5y

Solution

log52xy

4log2x+6log2y

6log3x+9log3y

Solution

log3x6y9

log3(x2−1)−2log3(x−1)

log(x2+2x+1)−2log(x+1)

Solution

0

4logx−2logy−3logz

3lnx+4lny−2lnz

Solution

lnx3y4z2

13logx−3log(x+1)

2log(2x+3)+12log(x+1)

Solution

log(2x+3)2·x+1

Use the Change-of-Base Formula

In the following exercises, use the Change-of-Base Formula, rounding to three decimal places, to approximate each logarithm.

log342

log546

Solution

2.379

log1287

log1593

Solution

1.674

log217

log321

Solution

5.542

Writing Exercises

Write the Product Property in your own words. Does it apply to each of the following? loga5x,loga(5+x). Why or why not?

Write the Power Property in your own words. Does it apply to each of the following? logaxp,(logax)r. Why or why not?

Solution

Answers will vary.

Use an example to show that
log(a+b)≠loga+logb.

Explain how to find the value of log715 using your calculator.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has three rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads use the properties of logarithms and use the change of base formula. The rest of the cells are blank.

ⓑ On a scale of 1−10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Solve Exponential and Logarithmic Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve logarithmic equations using the properties of logarithms
  • Solve exponential equations using logarithms
  • Use exponential models in applications

Before you get started, take this readiness quiz.

Solve: x2=16.
If you missed this problem, review Example 3 in Polynomial Equations.

Solution

x=4,x=−4

Solve: x2−5x+6=0.
If you missed this problem, review Example 2 in Polynomial Equations.

Solution

x=2,x=3

Solve: x(x+6)=2x+5.
If you missed this problem, review Example 4 in Polynomial Equations.

Solution

x=−5,x=1

Solve Logarithmic Equations Using the Properties of Logarithms

In the section on logarithmic functions, we solved some equations by rewriting the equation in exponential form. Now that we have the properties of logarithms, we have additional methods we can use to solve logarithmic equations.

If our equation has two logarithms we can use a property that says that if logaM=logaN then it is true that M=N. This is the One-to-One Property of Logarithmic Equations.

One-to-One Property of Logarithmic Equations

For M>0,N>0,a>0, and a≠1 is any real number:

IflogaM=logaN,thenM=N.

To use this property, we must be certain that both sides of the equation are written with the same base.

Remember that logarithms are defined only for positive real numbers. Check your results in the original equation. You may have obtained a result that gives a logarithm of zero or a negative number.

Solve: 2log5x=log581.

Solution
Step-by-step solution for the logarithmic equation 2log₅x = log₅81, demonstrating property application and solution validation.
2log5x=log581
Use the Power Property. log5x2=log581
Use the One-to-One Property, if logaM=logaN, then M=N x2=81.
Solve using the Square Root Property. x=±9
We eliminate x=−9 as we cannot take the logarithm of a negative number. x=9,x=−9
Check.
x=92log5x=log5812log59=?log581log592=?log581log581=log581✓

Solve: 2log3x=log336

Solution

x=6

Solve: 3logx=log64

Solution

x=4

Another strategy to use to solve logarithmic equations is to condense sums or differences into a single logarithm.

Solve: log3x+log3(x−8)=2.

Solution
Step-by-step solution of a logarithmic equation, illustrating property application and validation of solutions.
log3x+log3(x−8)=2
Use the Product Property, logaM+logaN=logaM⋅N. log3x(x−8)=2
Rewrite in exponential form. 32=x(x−8)
Simplify. 9=x2−8x
Subtract 9 from each side. 0=x2−8x−9
Factor. 0=(x−9)(x+1)
Use the Zero-Product Property. x−9=0,x+1=0
Solve each equation. x=9,x=−1
Check.
x=−1log3x+log3(x−8)=2log3(−1)+log3(−1−8)=?2
We cannot take the log of a negative number.
x=9log3x+log3(x−8)=2log39+log3(9−8)=?22+0=?22=2✓

Solve: log2x+log2(x−2)=3

Solution

x=4

Solve: log2x+log2(x−6)=4

Solution

x=8

When there are logarithms on both sides, we condense each side into a single logarithm. Remember to use the Power Property as needed.

Solve: log4(x+6)−log4(2x+5)=−log4x.

Solution
Step-by-step solution of a logarithmic equation, demonstrating the application of various logarithmic and algebraic properties.
log4(x+6)−log4(2x+5)=−log4x
Use the Quotient Property on the left side and the Power Property on the right. log4(x+62x+5)=log4x−1
Rewrite x−1=1x. log4(x+62x+5)=log41x
Use the One-to-One Property, if logaM=logaN, then M=N. x+62x+5=1x
Solve the rational equation. x(x+6)=2x+5
Distribute. x2+6x=2x+5
Write in standard form. x2+4x−5=0
Factor. (x+5)(x−1)=0
Use the Zero-Product Property. x+5=0,x−1=0
Solve each equation. x=−5,x=1
Check.
We leave the check for you.

Solve: log(x+2)−log(4x+3)=−logx.

Solution

x=3

Solve: log(x−2)−log(4x+16)=log1x.

Solution

x=8

Solve Exponential Equations Using Logarithms

In the section on exponential functions, we solved some equations by writing both sides of the equation with the same base. Next we wrote a new equation by setting the exponents equal.

It is not always possible or convenient to write the expressions with the same base. In that case we often take the common logarithm or natural logarithm of both sides once the exponential is isolated.

Solve 5x=11. Find the exact answer and then approximate it to three decimal places.

Solution
Illustrates solving the exponential equation 5^x = 11 using logarithms, detailing the exact and approximate solutions.
5x=11
Since the exponential is isolated, take the logarithm of both sides.
Use the Power Property to get the x as a factor, not an exponent.

Solve for x. Find the exact answer.

Approximate the answer.
log5x=log11xlog5=log11 x=log11log5 x≈1.490
Since 51=5 and 52=25, does it makes sense that 51.490≈11?

Solve 7x=43. Find the exact answer and then approximate it to three decimal places.

Solution

x=log43log7≈1.933

Solve 8x=98. Find the exact answer and then approximate it to three decimal places.

Solution

x=log98log8≈2.205

When we take the logarithm of both sides we will get the same result whether we use the common or the natural logarithm (try using the natural log in the last example. Did you get the same result?) When the exponential has base e, we use the natural logarithm.

Solve 3ex+2=24. Find the exact answer and then approximate it to three decimal places.

Solution
Step-by-step solution for the exponential equation 3e^(x+2) = 24, showing each mathematical transformation to find x.
3ex+2=24
Isolate the exponential by dividing both sides by 3. ex+2=8
Take the natural logarithm of both sides. lnex+2=ln8
Use the Power Property to get the x as a factor, not an exponent. (x+2)lne=ln8
Use the property lne=1 to simplify. x+2=ln8
Solve the equation. Find the exact answer. x=ln8−2
Approximate the answer. x≈0.079

Solve 2ex−2=18. Find the exact answer and then approximate it to three decimal places.

Solution

x=ln9+2≈4.197

Solve 5e2x=25. Find the exact answer and then approximate it to three decimal places.

Solution

x=ln52≈0.805

Use Exponential Models in Applications

In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more options to solve these equations, we are able to solve more applications.

We will again use the Compound Interest Formulas and so we list them here for reference.

Compound Interest

For a principal, P, invested at an interest rate, r, for t years, the new balance, A is:

A=P(1+rn)ntwhen compoundedntimes a year. A=Pertwhen compounded continuously.

Jermael’s parents put $10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth $50,000 when he turns 18. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal?

Solution
Detailed steps to calculate the continuous compounding growth rate 'r' when an initial principal of $10,000 grows to $50,000 over 17 years, yielding approximately 9.5%.
A=$50,000
P=$10,000
Identify the variables in the formula r=?
t=17years
A=Pert
Substitute the values into the formula. 50,000=10,000er·17
Solve for r. Divide each side by 10,000. 5=e17r
Take the natural log of each side. ln5=lne17r
Use the Power Property. ln5=17rlne
Simplify. ln5=17r
Divide each side by 17. ln517=r
Approximate the answer. r≈0.095
Convert to a percentage. r≈9.5%
They need the rate of growth to be approximately 9.5%.

Hector invests $10,000 at age 21. He hopes the investments will be worth $150,000 when he turns 50. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal?

Solution

r≈9.3%

Rachel invests $15,000 at age 25. She hopes the investments will be worth $90,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

Solution

r≈11.9%

We have seen that growth and decay are modeled by exponential functions. For growth and decay we use the formula A=A0ekt. Exponential growth has a positive rate of growth or growth constant, k, and exponential decay has a negative rate of growth or decay constant, k.

Exponential Growth and Decay

For an original amount, A0, that grows or decays at a rate, k, for a certain time, t, the final amount, A, is:

A=A0ekt

We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations.

Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

Solution

This problem requires two main steps. First we must find the unknown rate, k. Then we use that value of k to help us find the unknown number of bacteria.

Step-by-step solution for the exponential growth rate 'k' and prediction of future quantity 'A' using an exponential model.
Identify the variables in the formula. A=300A0=100k=?t=3hoursA=A0ekt
Substitute the values in the formula. 300=100ek·3
Solve for k. Divide each side by 100. 3=e3k
Take the natural log of each side. ln3=lne3k
Use the Power Property. ln3=3klne
Simplify. ln3=3k
Divide each side by 3. ln33=k
Approximate the answer. k≈0.366
We use this rate of growth to predict the number of bacteria there will be in 24 hours. A=?A0=100k=ln33t=24hoursA=A0ekt
Substitute in the values. A=100eln33·24
Evaluate. A≈656,100
At this rate of growth, they can expect 656,100 bacteria.

Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

Solution

There will be 62,500 bacteria.

Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment?

Solution

There will be 47,700 bacteria.

Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the half-life of the substance.

Similar to the previous example, we can use the given information to determine the constant of decay, and then use that constant to answer other questions.

The half-life of radium-226 is 1,590 years. How much of a 100 mg sample will be left in 500 years?

Solution

This problem requires two main steps. First we must find the decay constant k. If we start with 100-mg, at the half-life there will be 50-mg remaining. We will use this information to find k. Then we use that value of k to help us find the amount of sample that will be left in 500 years.

Step-by-step calculation to find the decay rate (k) and predict future values using an exponential decay formula.
Identify the variables in the formula. A=50A0=100k=?t=1590yearsA=A0ekt
Substitute the values in the formula. 50=100ek·1590
Solve for k. Divide each side by 100. 0.5=e1590k
Take the natural log of each side. ln0.5=lne1590k
Use the Power Property. ln0.5=1590klne
Simplify. ln0.5=1590k
Divide each side by 1590. ln0.51590=kexact answer
We use this rate of growth to predict the amount that will be left in 500 years. A=?A0=100k=ln0.51590t=500yearsA=A0ekt
Substitute in the values. A=100eln0.51590·500
Evaluate. A≈80.4mg
In 500 years there would be approximately 80.4 mg remaining.

The half-life of magnesium-27 is 9.45 minutes. How much of a 10-mg sample will be left in 6 minutes?

Solution

There will be 6.44 mg left.

The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days?

Solution

There will be 31.5 mg left.

Access these online resources for additional instruction and practice with solving exponential and logarithmic equations.

  • Solving Logarithmic Equations
  • Solving Logarithm Equations
  • Finding the rate or time in a word problem on exponential growth or decay
  • Finding the rate or time in a word problem on exponential growth or decay

Key Concepts

  • One-to-One Property of Logarithmic Equations: For M>0,N>0,a>0, and a≠1 is any real number:
    IflogaM=logaN,thenM=N.
  • Compound Interest:
    For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is:
    A=P(1+rn)ntwhen compoundedntimes a year. A=Pertwhen compounded continuously.
  • Exponential Growth and Decay: For an original amount, A0 that grows or decays at a rate, r, for a certain time t, the final amount, A, is A=A0ert.

Section Exercises

Practice Makes Perfect

Solve Logarithmic Equations Using the Properties of Logarithms

In the following exercises, solve for x.

log464=2log4x

log49=2logx

Solution

x=7

3log3x=log327

3log6x=log664

Solution

x=4

log5(4x−2)=log510

log3(x2+3)=log34x

Solution

x=1, x=3

log3x+log3x=2

log4x+log4x=3

Solution

x=8

log2x+log2(x−3)=2

log3x+log3(x+6)=3

Solution

x=3

logx+log(x+3)=1

logx+log(x−15)=2

Solution

x=20

log(x+4)−log(5x+12)=−logx

log(x−1)−log(x+3)=log1x

Solution

x=3

log5(x+3)+log5(x−6)=log510

log5(x+1)+log5(x−5)=log57

Solution

x=6

log3(2x−1)=log3(x+3)+log33

log(5x+1)=log(x+3)+log2

Solution

x=53

Solve Exponential Equations Using Logarithms

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

3x=89

2x=74

Solution

x=log74log2≈6.209

5x=110

4x=112

Solution

x=log112log4≈3.404

ex=16

ex=8

Solution

x=ln8≈2.079

(12)x=6

(13)x=8

Solution

x=log8log13≈−1.893

4ex+1=16

3ex+2=9

Solution

x=ln3−2≈−0.901

6e2x=24

2e3x=32

Solution

x=ln163≈0.924

14ex=3

13ex=2

Solution

x=ln6≈1.792

ex+1+2=16

ex−1+4=12

Solution

x=ln8+1≈3.079

In the following exercises, solve each equation.

33x+1=81

64x−17=216

Solution

x=5

ex2e14=e5x

ex2ex=e20

Solution

x=−4,x=5

loga64=2

loga81=4

Solution

a=3

lnx=−8

lnx=9

Solution

x=e9

log5(3x−8)=2

log4(7x+15)=3

Solution

x=7

lne5x=30

lne6x=18

Solution

x=3

3logx=log125

7log3x=log3128

Solution

x=2

log6x+log6(x−5)=log624

log9x+log9(x−4)=log912

Solution

x=6

log2(x+2)−log2(2x+9)=−log2x

log6(x+1)−log6(4x+10)=log61x

Solution

x=5

In the following exercises, solve for x, giving an exact answer as well as an approximation to three decimal places.

6x=91

(12)x=10

Solution

x=log10log12≈−3.322

7ex−3=35

8ex+5=56

Solution

x=ln7−5≈−3.054

Use Exponential Models in Applications

In the following exercises, solve.

Sung Lee invests $5,000 at age 18. He hopes the investments will be worth $10,000 when he turns 25. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation?

Alice invests $15,000 at age 30 from the signing bonus of her new job. She hopes the investments will be worth $30,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

Solution

6.9%

Coralee invests $5,000 in an account that compounds interest monthly and earns 7%. How long will it take for her money to double?

Simone invests $8,000 in an account that compounds interest quarterly and earns 5%. How long will it take for his money to double?

Solution

13.9 years

Researchers recorded that a certain bacteria population declined from 100,000 to 100 in 24 hours. At this rate of decay, how many bacteria will there be in 16 hours?

Researchers recorded that a certain bacteria population declined from 800,000 to 500,000 in 6 hours after the administration of medication. At this rate of decay, how many bacteria will there be in 24 hours?

Solution

122,070 bacteria

A virus takes 6 days to double its original population (A=2A0). How long will it take to triple its population?

A bacteria doubles its original population in 24 hours (A=2A0). How big will its population be in 72 hours?

Solution

8 times as large as the original population

Carbon-14 is used for archeological carbon dating. Its half-life is 5,730 years. How much of a 100-gram sample of Carbon-14 will be left in 1000 years?

Radioactive technetium-99m is often used in diagnostic medicine as it has a relatively short half-life but lasts long enough to get the needed testing done on the patient. If its half-life is 6 hours, how much of the radioactive material form a 0.5 ml injection will be in the body in 24 hours?

Solution

0.03 ml

Writing Exercises

Explain the method you would use to solve these equations: 3x+1=81, 3x+1=75. Does your method require logarithms for both equations? Why or why not?

What is the difference between the equation for exponential growth versus the equation for exponential decay?

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads solve logarithmic equations using the properties of logarithms, solve exponential equations using logarithms, and use exponential models in applications. The rest of the cells are blank.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Finding Composite and Inverse Functions

Find and Evaluate Composite Functions

In the following exercises, for each pair of functions, find ⓐ (f ∘ g)(x), ⓑ (g ∘ f)(x), and ⓒ (f · g)(x).

f(x)=7x−2 and
g(x)=5x+1

f(x)=4x and
g(x)=x2+3x

Solution

ⓐ 4x2+12x ⓑ 16x2+12x ⓒ 4x3+12x2

In the following exercises, evaluate the composition.

For functions
f(x)=3x2+2 and
g(x)=4x−3, find
ⓐ (f∘g)(−3)
ⓑ (g∘f)(−2)
ⓒ (f∘f)(−1)

For functions
f(x)=2x3+5 and
g(x)=3x2−7, find
ⓐ (f∘g)(−1)
ⓑ (g∘f)(−2)
ⓒ (g∘g)(1)

Solution

ⓐ −123 ⓑ 356 ⓒ 41

Determine Whether a Function is One-to-One

In the following exercises, for each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

{(−3,−5),(−2,−4),(−1,−3),(0,−2),
(−1,−1),(−2,0),(−3,1)}

{(−3,0),(−2,−2),(−1,0),(0,1),
(1,2),(2,1),(3,−1)}

Solution

Function; not one-to-one

{(−3,3),(−2,1),(−1,−1),(0,−3),
(1,−5),(2,−4),(3,−2)}

In the following exercises, determine whether each graph is the graph of a function and if so, is it one-to-one.

ⓐ
This figure shows a line from (negative 6, negative 2) up to (negative 1, 3) and then down from there to (6, negative 4).
ⓑ
This figure shows a line from (6, 5) down to (0, negative 1) and then down from there to (5, negative 6).

Solution

ⓐ Function; not one-to-one ⓑ Not a function

ⓐ
This figure shows a curved line from (negative 6, negative 2) up to the origin and then continuing up from there to (6, 2).
ⓑ
This figure shows a circle of radius 2 with center at the origin.

Find the Inverse of a Function

In the following exercise, find the inverse of the function. Determine the domain and range of the inverse function.

{(−3,10),(−2,5),(−1,2),(0,1)}

Solution

Inverse function: {(10,−3),(5,−2),(2,−1),(1,0)}. Domain: {1,2,5,10}. Range: {−3,−2,−1,0}.

In the following exercise, graph the inverse of the one-to-one function shown.

This figure shows a line segment from (negative 4, negative 2) up to (negative 2, 1) then up to (2, 2) and then up to (3, 4).

In the following exercises, verify that the functions are inverse functions.

f(x)=3x+7 and
g(x)=x−73

Solution

g(f(x))=x, and f(g(x))=x, so they are inverses.

f(x)=2x+9 and
g(x)=x+92

In the following exercises, find the inverse of each function.

f(x)=6x−11

Solution

f−1(x)=x+116

f(x)=x3+13

f(x)=1x+5

Solution

f−1(x)=1x−5

f(x)=x−15

Evaluate and Graph Exponential Functions

Graph Exponential Functions

In the following exercises, graph each of the following functions.

f(x)=4x

Solution

This figure shows an exponential line passing through the points (negative 1, 1 over 4), (0, 1), and (1, 4).

f(x)=(15)x

g(x)=(0.75)x

Solution

This figure shows an exponential line passing through the points (negative 1, 4 over 3), (0, 1), and (1, 3 over 4).

g(x)=3x+2

f(x)=(2.3)x−3

Solution

This figure shows an exponential line passing through the points (negative 1, negative 59 over 23), (0, negative 2), and (1, negative7 over 10).

f(x)=ex+5

f(x)=−ex

Solution

This figure shows an exponential line passing through the points (negative 1, negative 1 over e), (0, negative 1), and (1, negative e).

Solve Exponential Equations

In the following exercises, solve each equation.

35x−6=81

2x2=16

Solution

x=−2,x=2

9x=27

5x2+2x=15

Solution

x=−1

e4x·e7=e19

ex2e15=e2x

Solution

x=−3,x=5

Use Exponential Models in Applications

In the following exercises, solve.

Felix invested $12,000 in a savings account. If the interest rate is 4% how much will be in the account in 12 years by each method of compounding?

ⓐ compound quarterly
ⓑ compound monthly
ⓒ compound continuously.

Sayed deposits $20,000 in an investment account. What will be the value of his investment in 30 years if the investment is earning 7% per year and is compounded continuously?

Solution

$163,323.40

A researcher at the Center for Disease Control and Prevention is studying the growth of a bacteria. She starts her experiment with 150 of the bacteria that grows at a rate of 15% per hour. She will check on the bacteria every 24 hours. How many bacteria will he find in 24 hours?

In the last five years the population of the United States has grown at a rate of 0.7% per year to about 318,900,000. If this rate continues, what will be the population in 5 more years?

Solution

330,259,000

Evaluate and Graph Logarithmic Functions

Convert Between Exponential and Logarithmic Form

In the following exercises, convert from exponential to logarithmic form.

54=625

10−3=11,000

Solution

log11,000=−3

6315=635

ey=16

Solution

ln16=y

In the following exercises, convert each logarithmic equation to exponential form.

7=log2128

5=log100,000

Solution

100000=105

4=lnx

Evaluate Logarithmic Functions

In the following exercises, solve for x.

logx125=3

Solution

x=5

log7x=−2

log12116=x

Solution

x=4

In the following exercises, find the exact value of each logarithm without using a calculator.

log232

log81

Solution

0

log319

Graph Logarithmic Functions

In the following exercises, graph each logarithmic function.

y=log5x

Solution

This figure shows a logarithmic line passing through the points (1 over 5, negative 1), (1, 0), and (5, 1).

y=log14x

y=log0.8x

Solution

This figure shows a logarithmic line passing through the points (4 over 5, 1), (1, 0), and (5 over 4, negative 1).

Solve Logarithmic Equations

In the following exercises, solve each logarithmic equation.

loga36=5

lnx=−3

Solution

x=e−3

log2(5x−7)=3

lne3x=24

Solution

x=8

log(x2−21)=2

Use Logarithmic Models in Applications

What is the decibel level of a train whistle with intensity 10−3 watts per square inch?

Solution

90 dB

Use the Properties of Logarithms

Use the Properties of Logarithms

In the following exercises, use the properties of logarithms to evaluate.

ⓐ log71 ⓑ log1212

ⓐ 5log513 ⓑ log33−9

Solution

ⓐ 13 ⓑ −9

ⓐ 10log5 ⓑ log10−3

ⓐ eln8 ⓑ lne5

Solution

ⓐ 8 ⓑ 5

In the following exercises, use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

log4(64xy)

log10,000m

Solution

4+logm

In the following exercises, use the Quotient Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

log749y

lne52

Solution

5−ln2

In the following exercises, use the Power Property of Logarithms to expand each logarithm. Simplify, if possible.

logx−9

log4z7

Solution

17log4z

In the following exercises, use properties of logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

log3(4x7y8)

log58a2b6cd3

Solution

log58+2log5a+6log5b
+log5c−3log5d

ln3x2−y2z4

log67x26y3z53

Solution

13(log67+2log6x−1−3log6y
−5log6z)

In the following exercises, use the Properties of Logarithms to condense the logarithm. Simplify if possible.

log256−log27

3log3x+7log3y

Solution

log3x3y7

log5(x2−16)−2log5(x+4)

14logy−2log(y−3)

Solution

logy4(y−3)2

Use the Change-of-Base Formula

In the following exercises, rounding to three decimal places, approximate each logarithm.

log597

log316

Solution

5.047

Solve Exponential and Logarithmic Equations

Solve Logarithmic Equations Using the Properties of Logarithms

In the following exercises, solve for x.

3log5x=log5216

log2x+log2(x−2)=3

Solution

x=4

log(x−1)−log(3x+5)=−logx

log4(x−2)+log4(x+5)=log48

Solution

x=3

ln(3x−2)=ln(x+4)+ln2

Solve Exponential Equations Using Logarithms

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

2x=101

Solution

x=log101log2≈6.658

ex=23

(13)x=7

Solution

x=log7log13≈−1.771

7ex+3=28

ex−4+8=23

Solution

x=ln15+4≈6.708

Use Exponential Models in Applications

Jerome invests $18,000 at age 17. He hopes the investments will be worth $30,000 when he turns 26. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation?

Elise invests $4500 in an account that compounds interest monthly and earns 6%. How long will it take for her money to double?

Solution

11.6 years

Researchers recorded that a certain bacteria population grew from 100 to 300 in 8 hours. At this rate of growth, how many bacteria will there be in 24 hours?

Mouse populations can double in 8 months (A=2A0). How long will it take for a mouse population to triple?

Solution

12.7 months

The half-life of radioactive iodine is 60 days. How much of a 50 mg sample will be left in 40 days?

Practice Test

For the functions, f(x)=6x+1 and g(x)=8x−3, find ⓐ (f∘g)(x), ⓑ (g∘f)(x), and ⓒ (f·g)(x).

Solution

ⓐ 48x−17 ⓑ 48x+5
ⓒ 48x2−10x−3

Determine if the following set of ordered pairs represents a function and if so, is the function one-to-one. {(−2,2),(−1,−3),(0,1),(1,−2),(2,−3)}

Determine whether each graph is the graph of a function and if so, is it one-to-one.

ⓐ
This figure shows a parabola opening to the right with vertex (negative 3, 0).
ⓑ
This figure shows an exponential line passing through the points (negative 1, 1 over 2), (0, 1), and (1, 2).

Solution

ⓐ Not a function ⓑ One-to-one function

Graph, on the same coordinate system, the inverse of the one-to-one function shown.

This figure shows a line segment passing from the point (negative 3, 3) to (negative 1, 2) to (0, negative 2) to (2, negative 4).

Find the inverse of the function f(x)=x5−9.

Solution

f−1(x)=x+95

Graph the function g(x)=2x−3.

Solve the equation 22x−4=64.

Solution

x=5

Solve the equation ex2e4=e3x.

Megan invested $21,000 in a savings account. If the interest rate is 5%, how much will be in the account in 8 years by each method of compounding?
ⓐ compound quarterly
ⓑ compound monthly
ⓒ compound continuously.

Solution

ⓐ $31,250.74 ⓑ $31,302.29 ⓒ $31,328.32

Convert the equation from exponential to logarithmic form: 10−2=1100.

Convert the equation from logarithmic equation to exponential form: 3=log7343

Solution

343=73

Solve for x: log5x=−3

Evaluate log111.

Solution

0

Evaluate log4164.

Graph the function
y=log3x.

Solution


This figure shows a logarithmic line passing through (1 over 3, 1), (1, 0), and (3, 1).

Solve for x:
log(x2−39)=1

What is the decibel level of a small fan with intensity 10−8 watts per square inch?

Solution

40 dB

Evaluate each. ⓐ 6log617
ⓑ log99−3

In the following exercises, use properties of logarithms to write each expression as a sum of logarithms, simplifying if possible.

log525ab

Solution

2+log5a+log5b

lne128

log25x316y2z74

Solution

14(log25+3log2x−4−2log2y
−7log2z)

In the following exercises, use the Properties of Logarithms to condense the logarithm, simplifying if possible.

5log4x+3log4y

16logx−3log(x+5)

Solution

logx6(x+5)3

Rounding to three decimal places, approximate log473.

Solve for x:
log7(x+2)+log7(x−3)=log724

Solution

x=6

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

(15)x=9

5ex−4=40

Solution

x=ln8+4≈6.079

Jacob invests $14,000 in an account that compounds interest quarterly and earns 4%. How long will it take for his money to double?

Researchers recorded that a certain bacteria population grew from 500 to 700 in 5 hours. At this rate of growth, how many bacteria will there be in 20 hours?

Solution

1,921 bacteria

A certain beetle population can double in 3 months (A=2A0). How long will it take for that beetle population to triple?

Introduction

A photo of a rocket ship being launched into space.
Aerospace engineers use rockets such as this one to launch people and objects into space. (credit: WikiImages/Pixabay)

Five, Four. Three. Two. One. Lift off. The rocket launches off the ground headed toward space. Unmanned spaceships, and spaceships in general, are designed by aerospace engineers. These engineers are investigating reusable rockets that return safely to Earth to be used again. Someday, rockets may carry passengers to the International Space Station and beyond. One essential math concept for aerospace engineers is that of conics. In this chapter, you will learn about conics, including circles, parabolas, ellipses, and hyperbolas. Then you will use what you learn to investigate systems of nonlinear equations.

Distance and Midpoint Formulas; Circles

Learning Objectives

By the end of this section, you will be able to:

  • Use the Distance Formula
  • Use the Midpoint Formula
  • Write the equation of a circle in standard form
  • Graph a circle

Before you get started, take this readiness quiz.

Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches.
If you missed this problem, review Example 7 in Solve a Formula for a Specific Variable.

Solution

20inches

Factor: x2−18x+81.
If you missed this problem, review Example 2 in Factor Special Products.

Solution

x−92

Solve by completing the square: x2−12x−12=0.
If you missed this problem, review Example 2 in Solve Quadratic Equations Using the Quadratic Formula.

Solution

x=6±43

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

This figure shows two cones placed point to point. They are labeled nappes.

There are four conics—the circle, parabola, ellipse, and hyperbola. The next figure shows how the plane intersecting the double cone results in each curve.

Each of these four figures shows a double cone intersected by a plane. In the first figure, the plane is perpendicular to the axis of the cones and intersects the bottom cone to form a circle. In the second figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it intersects the base as well. Thus, the curve formed by the intersection is open at both ends. This is labeled parabola. In the third figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it does not intersect the base of the cone. Thus, the curve formed by the intersection is a closed loop, labeled ellipse. In the fourth figure, the plane is parallel to the axis, intersecting both cones. This is labeled hyperbola.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

Use the Distance Formula

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Use the rectangular coordinate system to find the distance between the points (6,4) and (2,1).

Solution
Plot the two points. Connect the two points
with a line.
Draw a right triangle as if you were going to
find slope.
A graph on a coordinate plane shows two points, (2,1) and (6,4). A right triangle connects them, illustrating the 'rise' (vertical change) and 'run' (horizontal change) used to find the distance 'd' between the points.
Find the length of each leg. A right triangle has legs of length 3 (rise) and 4 (run), and its hypotenuse is labeled 'd'. This diagram can be used to calculate 'd' using the Pythagorean theorem (3^2 + 4^2 = d^2).
Use the Pythagorean Theorem to find d, the
distance between the two points.
a2+b2=c2
Substitute in the values. 32+42=d2
Simplify. 9+16=d2
25=d2
Use the Square Root Property. d=5d=−5
Since distance, d is positive, we can eliminate
d=−5.
The distance between the points (6,4) and
(2,1) is 5.

Use the rectangular coordinate system to find the distance between the points (6,1) and (2,−2).

Solution

d=5

Use the rectangular coordinate system to find the distance between the points (5,3) and (−3,−3).

Solution

d=10

Figure shows a graph with a right triangle. The hypotenuse connects two points, (2, 1) and (6, 4). These are respectively labeled (x1, y1) and (x2, y2). The rise is y2 minus y1, which is 4 minus 1 equals 3. The run is x2 minus x1, which is 6 minus 2 equals 4.

The method we used in the last example leads us to the formula to find the distance between the two points (x1,y1) and (x2,y2).

When we found the length of the horizontal leg we subtracted 6−2 which is x2−x1.

When we found the length of the vertical leg we subtracted 4−1 which is y2−y1.

If the triangle had been in a different position, we may have subtracted x1−x2 or y1−y2. The expressions x2−x1 and x1−x2 vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say |x2−x1| and |y2−y1|.

In the Pythagorean Theorem, we substitute the general expressions |x2−x1| and |y2−y1| rather than the numbers.

Derivation of the distance formula from the Pythagorean theorem, detailing each algebraic step.
a2+b2=c2
Substitute in the values. (|x2−x1|)2+(|y2−y1|)2=d2
Squaring the expressions makes them positive, so we eliminate the absolute value bars. (x2−x1)2+(y2−y1)2=d2
Use the Square Root Property. d=±(x2−x1)2+(y2−y1)2
Distance is positive, so eliminate the negative value. d=(x2−x1)2+(y2−y1)2

This is the Distance Formula we use to find the distance d between the two points (x1,y1) and (x2,y2).

Distance Formula

The distance d between the two points (x1,y1) and (x2,y2) is

d=(x2−x1)2+(y2−y1)2

Use the Distance Formula to find the distance between the points (−5,−3) and (7,2).

Solution
This table illustrates the step-by-step application of the distance formula to find the distance between two given points.
Write the Distance Formula. d=(x2−x1)2+(y2−y1)2
Label the points, (−5,−3x1,y1),(7,2x2,y2) and substitute. d=(7−(−5))2+(2−(−3))2
Simplify. d=122+52
d=144+25
d=169
d=13

Use the Distance Formula to find the distance between the points (−4,−5) and (5,7).

Solution

d=15

Use the Distance Formula to find the distance between the points (−2,−5) and (−14,−10).

Solution

d=13

Use the Distance Formula to find the distance between the points (10,−4) and (−1,5). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Solution
Demonstrates the step-by-step application of the Distance Formula to calculate the distance between two points, providing both exact and approximate results.
Write the Distance Formula. d=(x2−x1)2+(y2−y1)2
Label the points, (10,−4x1,y1),(−1,5x2,y2) and substitute. d=(−1−10)2+(5−(−4))2
Simplify. d=(−11)2+92
d=121+81
d=202
Since 202 is not a perfect square, we can leave the answer in exact form or find a decimal approximation. d=202ord≈14.2

Use the Distance Formula to find the distance between the points (−4,−5) and (3,4). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Solution

d=130,d≈11.4

Use the Distance Formula to find the distance between the points (−2,−5) and (−3,−4). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Solution

d=2,d≈1.4

Use the Midpoint Formula

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points (x1,y1) and (x2,y2) is

(x1+x22,y1+y22)

To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.

Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are (−5,−4) and (7,2). Plot the endpoints and the midpoint on a rectangular coordinate system.

Solution
Write the Midpoint Formula. (x1+x22,y1+y22)
Label the points, (−5,−4x1,y1),(7,2x2,y2)
and substitute.
(−5+72,−4+22)
Simplify. (22,−22)
(1,−1)
The midpoint of the segment is the point
(1,−1).
Plot the endpoints and midpoint. A line graph plotted on a coordinate plane, showing a straight line passing through three points: (-5, -4), (1, -1), and (7, 2). The x-axis ranges from -8 to 8, and the y-axis from -8 to 8.

Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are (−3,−5) and (5,7). Plot the endpoints and the midpoint on a rectangular coordinate system.

Solution

This graph shows a line segment with endpoints (negative 3, negative 5) and (5, 7) and midpoint (1, negative 1).

Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are (−2,−5) and (6,−1). Plot the endpoints and the midpoint on a rectangular coordinate system.

Solution

This graph shows a line segment with endpoints (negative 2, negative 5) and (6, negative 1) and midpoint (2, negative 3).

Both the Distance Formula and the Midpoint Formula depend on two points, (x1,y1) and (x2,y2). It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, it may be easier to remember the formulas.

The distance formula is d equals square root of open parentheses x2 minus x1 close parentheses squared plus open parentheses y2 minus y1 close parentheses squared end of root. This is labeled subtract the coordinates. The midpoint formula is open parentheses open parentheses x1 plus x2 close parentheses upon 2 comma open parentheses y1 plus y2 close parentheses upon 2 close parentheses. This is labeled add the coordinates.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

This figure shows a double cone and an intersecting plane, which form a circle.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, (h,k), and the fixed distance is called the radius, r, of the circle.

Circle

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, (h,k), and the fixed distance is called the radius, r, of the circle.

We look at a circle in the rectangular coordinate system.
The radius is the distance from the center, (h,k), to a
point on the circle, (x,y).
A circle is plotted on a Cartesian coordinate system with its center at (h, k), a point on the circle at (x, y), and its radius labeled r, connecting the center to the point on the circle.
To derive the equation of a circle, we can use the
distance formula with the points (h,k), (x,y) and the
distance, r.
d=(x2−x1)2+(y2−y1)2
Substitute the values. r=(x−h)2+(y−k)2
Square both sides. r2=(x−h)2+(y−k)2

This is the standard form of the equation of a circle with center, (h,k), and radius, r.

Standard Form of the Equation a Circle

The standard form of the equation of a circle with center, (h,k), and radius, r, is

Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.

Write the standard form of the equation of the circle with radius 3 and center (0,0).

Solution
Use the standard form of the equation of a circle (x−h)2+(y−k)2=r2
Substitute in the values r=3,h=0, and k=0. (x−0)2+(y−0)2=32
The image displays the word 'Center:' followed by a stacked coordinate system, showing (h, k) in red above (0, 0) in black, enclosed within a single set of large parentheses. It illustrates the concept of a center point in two forms.
Simplify. x2+y2=9

Write the standard form of the equation of the circle with a radius of 6 and center (0,0).

Solution

x2+y2=36

Write the standard form of the equation of the circle with a radius of 8 and center (0,0).

Solution

x2+y2=64

In the last example, the center was (0,0). Notice what happened to the equation. Whenever the center is (0,0), the standard form becomes x2+y2=r2.

Write the standard form of the equation of the circle with radius 2 and center (−1,3).

Solution
Use the standard form of the equation of a
circle.
(x−h)2+(y−k)2=r2
Substitute in the values. (x−(−1))2+(y−3)2=22
The image displays the word 'Center:' followed by a parenthesis containing two lines of text: the top line reads 'h, k' in red, and the bottom line reads '-1, 3' in black, indicating the center coordinates.
Simplify. (x+1)2+(y−3)2=4

Write the standard form of the equation of the circle with a radius of 7 and center (2,−4).

Solution

(x−2)2+(y+4)2=49

Write the standard form of the equation of the circle with a radius of 9 and center (−3,−5).

Solution

(x+3)2+(y+5)2=81

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Write the standard form of the equation of the circle with center (2,4) that also contains the point (−2,1).

This graph shows circle with center at (2, 4, radius 5 and a point on the circle minus 2, 1.
Solution

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center (2,4) and point (−2,1)

This table demonstrates the step-by-step calculation of a radius using the distance formula with given coordinate values.
Use the Distance Formula to find the radius. r=(x2−x1)2+(y2−y1)2
Substitute the values. (2,4x1,y1),(−2,1x2,y2) r=(−2−2)2+(1−4)2
Simplify. r=(−4)2+(−3)2
r=16+9
r=25
r=5

Now that we know the radius, r=5, and the center, (2,4), we can use the standard form of the equation of a circle to find the equation.

Steps to derive the standard equation of a circle by substituting values and simplifying.
Use the standard form of the equation of a circle. (x−h)2+(y−k)2=r2
Substitute in the values. (x−2)2+(y−4)2=52
Simplify. (x−2)2+(y−4)2=25

Write the standard form of the equation of the circle with center (2,1) that also contains the point (−2,−2).

Solution

(x−2)2+(y−1)2=25

Write the standard form of the equation of the circle with center (7,1) that also contains the point (−1,−5).

Solution

(x−7)2+(y−1)2=100

Graph a Circle

Any equation of the form (x−h)2+(y−k)2=r2 is the standard form of the equation of a circle with center, (h,k), and radius, r. We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from x and y. In the next example, the equation has x+2, so we need to rewrite the addition as subtraction of a negative.

Find the center and radius, then graph the circle: (x+2)2+(y−1)2=9.

Solution
The equation of a circle is displayed as (x + 2)^2 + (y - 1)^2 = 9.
Use the standard form of the equation of a circle.
Identify the center, (h,k) and radius, r.
The standard form of a circle's equation, (x-h)^2 + (y-k)^2 = r^2, is shown with blue arrows indicating how to identify the values for h, k, and r in an example equation: (x - (-2))^2 + (y - 1)^2 = 3^2.
Center: (−2,1) radius: 3
Graph the circle. A circle with center at (-2, 1) and radius r=3 plotted on a Cartesian plane.

ⓐ Find the center and radius, then ⓑ graph the circle: (x−3)2+(y+4)2=4.

Solution

ⓐ The circle is centered at (3,−4) with a radius of 2.
ⓑ
This graph shows a circle with center at (3, negative 4) and a radius of 2.

ⓐ Find the center and radius, then ⓑ graph the circle: (x−3)2+(y−1)2=16.

Solution

ⓐ The circle is centered at (3,1) with a radius of 4.
ⓑ
This graph shows circle with center at (3, 1) and a radius of 4.

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of x2,y2 to be one.

Find the center and radius and then graph the circle, 4x2+4y2=64.

Solution
A mathematical equation is displayed reading '4x^2 + 4y^2 = 64' in black text against a white background.
Divide each side by 4. The equation x^2 + y^2 = 16 is displayed on a white background, representing a circle centered at the origin with a radius of 4.
Use the standard form of the equation of a circle.
Identify the center, (h,k) and radius, r.
Substitution of h=0, k=0, and r=4 into the standard equation of a circle, (x-h)^2 + (y-k)^2 = r^2, resulting in (x-0)^2 + (y-0)^2 = 4^2.
Center: (0,0) radius: 4
Graph the circle. A circle centered at the origin (0,0) with a radius of 4, displayed on a Cartesian coordinate system.

ⓐ Find the center and radius, then ⓑ graph the circle: 3x2+3y2=27

Solution

ⓐ The circle is centered at (0,0) with a radius of 3.
ⓑ
This graph shows circle with center at (0, 0) and a radius of 3.

ⓐ Find the center and radius, then ⓑ graph the circle: 5x2+5y2=125

Solution

ⓐ The circle is centered at (0,0) with a radius of 5.
ⓑ
This graph shows circle with center at (0, 0) and a radius of 5.

If we expand the equation from Example 8, (x+2)2+(y−1)2=9, the equation of the circle looks very different.

Steps to convert a circle's equation from standard form to general form, demonstrating squaring binomials and rearranging terms to achieve zero on one side.
(x+2)2+(y−1)2=9
Square the binomials. x2+4x+4+y2−2y+1=9
Arrange the terms in descending degree order, and get zero on the right x2+y2+4x−2y−4=0

This form of the equation is called the general form of the equation of the circle.

General Form of the Equation of a Circle

The general form of the equation of a circle is

x2+y2+ax+by+c=0

If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y. Then we can graph the circle using its center and radius.

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−4x−6y+4=0.

Solution
We need to rewrite this general form into standard form in order to find the center and radius.
The image shows the equation of a circle in general form: x^2 + y^2 - 4x - 6y + 4 = 0.
Group the x-terms and y-terms.
Collect the constants on the right side.
The image displays the algebraic equation x^2 - 4x + y^2 - 6y = -4 on a white background, representing a circle in its general form.
Complete the squares. The equation x² - 4x + 4 + y² - 6y + 9 = -4 + 4 + 9, a step in completing the square for a circle's equation.
Rewrite as binomial squares. The equation of a circle, (x-2)^2 + (y-3)^2 = 9, is displayed against a white background.
Identify the center and radius. Center: (2,3) radius: 3
Graph the circle. Graphic representation of a circle in a coordinate plane with center at (2, 3) and radius of 3 units.

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−6x−8y+9=0.

Solution

ⓐ The circle is centered at (3,4) with a radius of 4.
ⓑ
This graph shows circle with center at (3, 4) and a radius of 4.

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2+6x−2y+1=0.

Solution

ⓐ The circle is centered at (−3,1) with a radius of 3.
ⓑ
This graph shows circle with center at (negative 3, 1) and a radius of 3.

In the next example, there is a y-term and a y2-term. But notice that there is no x-term, only an x2-term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2+8y=0.

Solution
We need to rewrite this general form into standard form in order to find the center and radius.
The equation x^2 + y^2 + 8y = 0 is shown in black text against a white background, representing a circle in coordinate geometry.
Group the x-terms and y-terms. A mathematical equation is displayed on a white background, reading x^2 + y^2 + 8y = 0. The equation represents a circle in standard form.
There are no constants to collect on the
right side.
Complete the square for y2+8y. The equation x^2 + y^2 + 8y + 16 = 0 + 16 is displayed, with the number 16 highlighted in red on both sides, indicating a step in an algebraic manipulation, likely completing the square.
Rewrite as binomial squares. A mathematical equation representing a circle is shown, specifically (x-0)^2 + (y+4)^2 = 16. This is the standard form equation for a circle centered at (0, -4) with a radius of 4.
Identify the center and radius. Center: (0,−4) radius: 4
Graph the circle. A circle is plotted on a coordinate plane, centered at (0, -4) with a radius of 4. The x-axis extends from -6 to 6, and the y-axis from -10 to 2, with grid lines every unit.

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−2x−3=0.

Solution

ⓐ The circle is centered at (1,0) with a radius of 2.
ⓑ
This graph shows circle with center at (1, 0) and a radius of 2.

ⓐ Find the center and radius, then ⓑ graph the circle: x2+y2−12y+11=0.

Solution

ⓐ The circle is centered at (0,6) with a radius of 5.
ⓑ
This graph shows circle with center at (0, 6) and a radius of 5.

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

  • Distance-Midpoint Formulas and Circles
  • Finding the Distance and Midpoint Between Two Points
  • Completing the Square to Write Equation in Standard Form of a Circle

Key Concepts

  • Distance Formula: The distance d between the two points (x1,y1) and (x2,y2) is
    d=(x2−x1)2+(y2−y1)2
  • Midpoint Formula: The midpoint of the line segment whose endpoints are the two points (x1,y1) and (x2,y2) is
    (x1+x22,y1+y22)

    To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.
  • Circle: A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the center, (h,k), and the fixed distance is called the radius, r, of the circle.
  • Standard Form of the Equation a Circle: The standard form of the equation of a circle with center, (h,k), and radius, r, is
    Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.
  • General Form of the Equation of a Circle: The general form of the equation of a circle is
    x2+y2+ax+by+c=0

Practice Makes Perfect

Use the Distance Formula

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

(2,0) and (5,4)

Solution

d=5

(−4,−3) and (2,5)

(−4,−3) and (8,2)

Solution

13

(−7,−3) and (8,5)

(−1,4) and (2,0)

Solution

5

(−1,3) and (5,−5)

(1,−4) and (6,8)

Solution

13

(−8,−2) and (7,6)

(−3,−5) and (0,1)

Solution

d=35,d≈6.7

(−1,−2) and (−3,4)

(3,−1) and (1,7)

Solution

d=217,d≈8.2

(−4,−5) and (7,4)

Use the Midpoint Formula

In the following exercises, ⓐ find the midpoint of the line segment whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

(0,−5) and (4,−3)

Solution

ⓐ Midpoint: (2,−4)
ⓑ
This graph shows line segment with endpoints (0, negative 5) and (4, negative 3) and midpoint (2, negative 4).

(−2,−6) and (6,−2)

(3,−1) and (4,−2)

Solution

ⓐ Midpoint: (312,−112)
ⓑ
This graph shows line segment with endpoints (3, negative 1) and (4, negative 2) and midpoint (3 and a half, negative 1 and a half).

(−3,−3) and (6,−1)

Write the Equation of a Circle in Standard Form

In the following exercises, write the standard form of the equation of the circle with the given radius and center (0,0).

Radius: 7

Solution

x2+y2=49

Radius: 9

Radius: 2

Solution

x2+y2=2

Radius: 5

In the following exercises, write the standard form of the equation of the circle with the given radius and center

Radius: 1, center: (3,5)

Solution

(x−3)2+(y−5)2=1

Radius: 10, center: (−2,6)

Radius: 2.5, center: (1.5,−3.5)

Solution

(x−1.5)2+(y+3.5)2=6.25

Radius: 1.5, center: (−5.5,−6.5)

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

Center (3,−2) with point (3,6)

Solution

(x−3)2+(y+2)2=64

Center (6,−6) with point (2,−3)

Center (4,4) with point (2,2)

Solution

(x−4)2+(y−4)2=8

Center (−5,6) with point (−2,3)

Graph a Circle

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

(x+5)2+(y+3)2=1

Solution

ⓐ The circle is centered at (−5,−3) with a radius of 1.
ⓑ
This graph shows a circle with center at (negative 5, negative 3) and a radius of 1.

(x−2)2+(y−3)2=9

(x−4)2+(y+2)2=16

Solution

ⓐ The circle is centered at (4,−2) with a radius of 4.
ⓑ
This graph shows circle with center at (4, negative 2) and a radius of 4.

(x+2)2+(y−5)2=4

x2+(y+2)2=25

Solution

ⓐ The circle is centered at (0,−2) with a radius of 5.
ⓑ
This graph shows circle with center at (negative 2, 5) and a radius of 5.

(x−1)2+y2=36

(x−1.5)2+(y+2.5)2=0.25

Solution

ⓐ The circle is centered at (1.5,−2.5) with a radius of 0.5.
ⓑ
This graph shows circle with center at (1.5, 2.5) and a radius of 0.5

(x−1)2+(y−3)2=94

x2+y2=64

Solution

ⓐ The circle is centered at (0,0) with a radius of 8.
ⓑ
This graph shows circle with center at (0, 0) and a radius of 8.

x2+y2=49

2x2+2y2=8

Solution

ⓐ The circle is centered at (0,0) with a radius of 2.
ⓑ
This graph shows circle with center at (0, 0) and a radius of 2.

6x2+6y2=216

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

x2+y2+2x+6y+9=0

Solution

ⓐ Center: (−1,−3), radius: 1
ⓑ
This graph shows circle with center at (negative 1, negative 3) and a radius of 1.

x2+y2−6x−8y=0

x2+y2−4x+10y−7=0

Solution

ⓐ Center: (2,−5), radius: 6
ⓑ
This graph shows circle with center at (2, negative 5) and a radius of 6.

x2+y2+12x−14y+21=0

x2+y2+6y+5=0

Solution

ⓐ Center: (0,−3), radius: 2
ⓑ
This graph shows circle with center at (0, negative 3) and a radius of 2.

x2+y2−10y=0

x2+y2+4x=0

Solution

ⓐ Center: (−2,0), radius: 2
ⓑ
This graph shows circle with center at (negative 2, 0) and a radius of 2.

x2+y2−14x+13=0

Writing Exercises

Explain the relationship between the distance formula and the equation of a circle.

Solution

Answers will vary.

Is a circle a function? Explain why or why not.

In your own words, state the definition of a circle.

Solution

Answers will vary.

In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A self-assessment table for math skills, where students rate their confidence in using distance and midpoint formulas, and writing or graphing circle equations. Options: Confidently, With some help, No-I don't get it!

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

circle
A circle is all points in a plane that are a fixed distance from a fixed point in the plane.

Parabolas

Learning Objectives

By the end of this section, you will be able to:

  • Graph vertical parabolas
  • Graph horizontal parabolas
  • Solve applications with parabolas

Before you get started, take this readiness quiz.

Graph: y=−3x2+12x−12.
If you missed this problem, review Example 6 in Graph Quadratic Functions Using Properties.

Solution

A downward-opening parabola is graphed on a Cartesian coordinate system. Its vertex is located at (2, 0), and it passes through points such as (0, -4) and (4, -4).

Solve by completing the square: x2−6x+6=0.
If you missed this problem, review Example 2 in Solve Quadratic Equations by Completing the Square.

Solution

x=3±3

Write in standard form: y=3x2−6x+5.
If you missed this problem, review Example 7 in Graph Quadratic Functions Using Transformations.

Solution

y=3x−12+2

Graph Vertical Parabolas

The next conic section we will look at is a parabola. We define a parabola as all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

This figure shows a double cone. The bottom nappe is intersected by a plane in such a way that the intersection forms a parabola.

Parabola

A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

This figure shows a parabola opening upwards. Below the parabola is a horizontal line labeled directrix. A vertical dashed line through the center of the parabola is labeled axis of symmetry. The point where the axis intersects the parabola is labeled vertex. A point on the axis, within the parabola is labeled focus. A line perpendicular to the directrix connects the directrix to a point on the parabola and another line connects this point to the focus. Both these lines are of the same length.

Previously, we learned to graph vertical parabolas from the general form or the standard form using properties. Those methods will also work here. We will summarize the properties here.

Vertical Parabolas
General form
y=ax2+bx+c
Standard form
y=a(x−h)2+k
Orientation a>0 up; a<0 down a>0 up; a<0 down
Axis of symmetry x=−b2a x=h
Vertex Substitute x=−b2a and
solve for y.
(h,k)
y-intercept Let x=0 Let x=0
x-intercepts Let y=0 Let y=0

The graphs show what the parabolas look like when they open up or down. Their position in relation to the x- or y-axis is merely an example.

This figure shows two parabolas with axis x equals h and vertex h, k. The one on the left opens up and A is greater than 0. The one on the right opens down. Here A is less than 0.

To graph a parabola from these forms, we used the following steps.

Graph vertical parabolas (y=ax2+bx+corf(x)=a(x−h)2+k) using properties.

  1. Determine whether the parabola opens upward or downward.
  2. Find the axis of symmetry.
  3. Find the vertex.
  4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
  5. Find the x-intercepts.
  6. Graph the parabola.

The next example reviews the method of graphing a parabola from the general form of its equation.

Graph y=−x2+6x−8 by using properties.

Solution
The general form of a quadratic equation and a particular example with y = -x^2 + 6x - 8, showing the relationship between them.
Since a is −1, the parabola opens downward.
A red, inverted U-shaped arrow with both ends pointing downwards, suggesting a flow, connection, or dual downward direction.
To find the axis of symmetry, find x=−b2a. Formula for the symmetry axis of a parabola: x = -b/2a.
The image shows the mathematical equation: x = -6 / 2(-1).
The equation 'x = 3' is clearly displayed on a white background, representing a fundamental and concise mathematical expression.
The axis of symmetry is x=3.
Coordinate graph with a vertical dashed line at x=3.
The vertex is on the line x=3. The image displays the quadratic equation y = -x^2 + 6x - 8.
Let x=3. Equation showing how to calculate y using the formula y equals negative three squared plus six times three minus eight. The numbers three and eight are highlighted.
The equation y equals negative nine plus eighteen minus eight.
The equation 'y = 1' written on the image.
The vertex is (3,1).
A graph shows a dashed vertical line at x=3 and a point at (3,1).
The y-intercept occurs when x=0. The equation of a quadratic function is shown as y = -x^2 + 6x - 8.
Substitute x=0. An algebraic equation showing the calculation of 'y' by substituting the value of 0 into the expression: y = -0^2 + 6*0 - 8.
Simplify. Ecuación: y = -8
The y-intercept is (0,−8).
The point (0,−8) is three units to the left of the
line of symmetry. The point three units to the
right of the line of symmetry is (6,−8).
Point symmetric to the y-intercept is (6,−8).
Cartesian graph with three points: (3,1), (0,-8) and (6,-8). A dashed vertical line passes through x=3.
The x-intercept occurs when y=0. The quadratic equation is y equals negative x squared plus six x minus eight.
Let y=0. The quadratic equation negative x squared plus six x minus eight equals zero.
Factor the GCF. Quadratic equation showing zero equals negative of x squared minus six x plus eight.
Factor the trinomial. The equation 0 = -(x-4)(x-2), an example of a factored algebraic expression.
Solve for x. The mathematical expressions X=4 and X=2 appear in the image.
The x-intercepts are (4,0),(2,0).
Graph the parabola. This graph displays a downward-opening parabola on a coordinate plane. The vertex is at (3, 1), and the dashed line x=3 is the axis of symmetry.

Graph y=−x2+5x−6 by using properties.

Solution

This graph shows a parabola opening downward, with x intercepts (2, 0) and (3, 0) and y intercept (0, negative 6).

Graph y=−x2+8x−12 by using properties.

Solution

This graph shows a parabola opening downward, with vertex (4, 4) and x intercepts (2, 0) and (6, 0).

The next example reviews the method of graphing a parabola from the standard form of its equation, y=a(x−h)2+k.

Writey=3x2−6x+5 in standard form and then use properties of standard form to graph the equation.

Solution
Rewrite the function in y=a(x−h)2+k form
by completing the square.
y=3x2−6x+5
y=3(x2−2x)+5
y=3(x2−2x+1)+5−3
y=3(x−1)2+2
Identify the constants a, h, k. a=3, h=1, k=2
Since a=3, the parabola opens upward.
Two red arrows point upward, connected by a curve, symbolizing an upward trend or a cyclical motion with an upward force.
The axis of symmetry is x=h. The axis of symmetry is x=1.
The vertex is (h,k). The vertex is (1,2).
Find the y-intercept by substituting x=0. y=3(x−1)2+2
y=3·02−6·0+5
y=5
y-intercept (0,5)
Find the point symmetric to (0,5) across the axis of symmetry. (2,5)
Find the x-intercepts. y=3(x−1)2+2 0=3(x−1)2+2−2=3(x−1)2−23=(x−1)2±−23=x−1
The square root of a negative number
tells us the solutions are complex
numbers. So there are no x-intercepts.
Graph the parabola. A graph of a parabola opening upwards, with its vertex at (1, 2). The axis of symmetry is a dashed vertical line at x=1. Two points (0, 5) and (2, 5) are marked on the curve.

ⓐ Write y=2x2+4x+5 in standard form and ⓑ use properties of standard form to graph the equation.

Solution

ⓐ y=2(x+1)2+3
ⓑ
This graph shows a parabola opening upwards, with vertex (negative 1, 3) and y intercept (0, 5). It has the point minus (2, 5) on it.

ⓐ Write y=−2x2+8x−7 in standard form and ⓑ use properties of standard form to graph the equation.

Solution

ⓐ y=−2(x−2)2+1
ⓑ
This graph shows a parabola opening downwards, with vertex (2, 1) and axis of symmetry x equals 2. Its y intercept is (0, negative 7).

Graph Horizontal Parabolas

Our work so far has only dealt with parabolas that open up or down. We are now going to look at horizontal parabolas. These parabolas open either to the left or to the right. If we interchange the x and y in our previous equations for parabolas, we get the equations for the parabolas that open to the left or to the right.

Horizontal Parabolas
General form
x=ay2+by+c
Standard form
x=a(y−k)2+h
Orientation a>0 right; a<0 left a>0 right; a<0 left
Axis of symmetry y=−b2a y=k
Vertex Substitute y=−b2a and
solve for x.
(h,k)
y-intercepts Let x=0 Let x=0
x-intercept Let y=0 Let y=0

The graphs show what the parabolas look like when they to the left or to the right. Their position in relation to the x- or y-axis is merely an example.

This figure shows two parabolas with axis of symmetry y equals k,) and vertex (h, k. The one on the left is labeled a greater than 0 and opens to the right. The other parabola opens to the left.

Looking at these parabolas, do their graphs represent a function? Since both graphs would fail the vertical line test, they do not represent a function.

To graph a parabola that opens to the left or to the right is basically the same as what we did for parabolas that open up or down, with the reversal of the x and y variables.

Graph horizontal parabolas (x=ay2+by+corx=a(y−k)2+h) using properties.

  1. Determine whether the parabola opens to the left or to the right.
  2. Find the axis of symmetry.
  3. Find the vertex.
  4. Find the x-intercept. Find the point symmetric to the x-intercept across the axis of symmetry.
  5. Find the y-intercepts.
  6. Graph the parabola.

Graph x=2y2 by using properties.

Solution
Two mathematical equations are displayed: x = ay^2 + by + c, shown in red, and x = 2y^2, shown in black. Both equations define x in terms of y, with the first being a general quadratic in y and the second a specific instance.
Since a=2, the parabola opens to the right.
Two red arrows form a curved, cyclical path, suggesting connection, flow, or a continuous process.
To find the axis of symmetry, find y=−b2a. A mathematical equation is displayed, showing y = -b/2a, typically used to find the x-coordinate of the vertex of a parabola given the quadratic equation y = ax^2 + bx + c.
A mathematical equation showing y equals negative zero divided by the product of two and two, written as y = - 0 / 2(2).
The mathematical equation 'y = 0' is displayed on a white background.
The axis of symmetry is y=0.
The vertex is on the liney=0. A mathematical equation is displayed on a white background, which reads 'x = 2y²'.
Let y=0. The mathematical equation x = 2 * 0^2 is shown, where the '0' is highlighted with a red outline, indicating a step in a calculation or problem.
The mathematical notation X = Ø, indicating that X is equal to the empty set.
The vertex is (0,0).

Since the vertex is (0,0), both the x- and y-intercepts are the point (0,0). To graph the parabola we need more points. In this case it is easiest to choose values of y.
In the equation x equals 2 y squared, when y is 1, x is 2 and when y is 2, x is 8. The points are (2, 1) and (8, 2).
We also plot the points symmetric to (2,1) and (8,2) across the y-axis, the points (2,−1),(8,−2).

Graph the parabola.

This graph shows right opening parabola with vertex (0, 0). Four points are marked on it: point (2, 1), point (2, negative 1), point (8, 2) and point (8 minus 2).

Graph x=y2 by using properties.

Solution

This graph shows right opening parabola with vertex at origin. Two points on it are (4, 2) and (4, negative 2).

Graph x=−y2 by using properties.

Solution

This graph shows left opening parabola with vertex at origin. Two points on it are (negative 4, 2) and (negative 4, negative 2).

In the next example, the vertex is not the origin.

Graph x=−y2+2y+8 by using properties.

Solution
Two equations representing parabolas opening horizontally are presented: the general form x = ay^2 + by + c, and the specific equation x = -y^2 + 2y + 8.
Since a=−1, the parabola opens to the left.
A red, curved arrow with arrowheads pointing both upwards and downwards, illustrating a continuous loop, cyclical process, or interconnected relationship.
To find the axis of symmetry, find y=−b2a. A mathematical equation shows 'y equals negative b over two a'.
A mathematical equation shows 'y = -2 / (2(-1))' centered on a white background.
The mathematical equation 'y = 1' is shown, indicating a horizontal line on a coordinate plane where the y-value is constant at 1.
The axis of symmetry is y=1.
The vertex is on the liney=1. The image displays the quadratic equation x = -y^2 + 2y + 8, written in a clear, legible font against a plain white background.
Let y=1. A mathematical expression on a white background reads X = -1^2 + 2 * 1 + 8, with the numbers 1 highlighted in red.
The equation 'X=9' is visible against a white background.
The vertex is (9,1).
The x-intercept occurs when y=0. The image displays a mathematical equation written in black text on a white background. The equation is 'x = -y^2 + 2y + 8', representing a quadratic function where x is expressed in terms of y.
A mathematical equation, X = -0^2 + 2*0 + 8, with the number 0 highlighted in red outlines, is shown against a plain white background.
The image displays a mathematical equation 'X = 8' in a simple black font centered against a plain white background.
The x-intercept is (8,0).
The point (8,0) is one unit below the line of
symmetry. The symmetric point one unit
above the line of symmetry is (8,2)
Symmetric point is (8,2).
The y-intercept occurs when x=0. A mathematical equation is displayed, reading x = -y^2 + 2y + 8, shown against a plain white background.
Substitute x=0. A quadratic equation is displayed on a white background: 0 = -y^2 + 2y + 8. The equation is presented clearly in black text, ready for solving or analysis.
Solve. A quadratic equation displayed on a white background: y^2 - 2y - 8 = 0.
A mathematical equation is displayed, showing the expression (y-4)(y+2) = 0. The equation appears to be a factored form of a quadratic equation, set equal to zero to find the roots for 'y'.
The image displays the mathematical equations 'y=4' and 'y=-2' in a simple, clear font against a white background.
The y-intercepts are (0,4) and (0,−2).
Connect the points to graph the parabola. A graph displays a parabola opening to the left, with its vertex marked at (9, 1). Additional points (0, 4), (0, -2), (8, 2), and (8, 0) are highlighted on the curve.

Graph x=−y2−4y+12 by using properties.

Solution

This graph shows left opening parabola with vertex (16, negative 2) and x intercept (12, 0).

Graph x=−y2+2y−3 by using properties.

Solution

This graph shows left opening parabola with vertex (negative 2, 1) and x intercept minus (3, 0).

In Table 4, we see the relationship between the equation in standard form and the properties of the parabola. The How To box lists the steps for graphing a parabola in the standard form x=a(y−k)2+h. We will use this procedure in the next example.

Graph x=2(y−2)2+1 using properties.

Solution
Representation of the standard equation of a parabola x = a(y-k)^2 + h, with a numerical example x = 2(y-2)^2 + 1.
Identify the constants a, h, k. a=2,h=1,k=2
Since a=2, the parabola opens to the right.
Two red arrows in a loop, indicating a cycle or repetition.
The axis of symmetry is y=k. The axis of symmetry is y=2.
The vertex is (h,k). The vertex is (1,2).
Find the x-intercept by substituting y=0. x=2(y−2)2+1 x=2(0−2)2+1 x=9
The x-intercept is (9,0).
Find the point symmetric to (9,0) across the
axis of symmetry.
(9,4)
Find the y-intercepts. Let x=0. x=2(y−2)2+1 0=2(y−2)2+1 −1=2(y−2)2
A square cannot be negative, so there is no real
solution. So there are no y-intercepts.
Graph the parabola. Graph of a horizontal parabola with vertex at (1, 2). Shows the points (9, 4) and (9, 0) and their symmetry axis at y=2.

Graph x=3(y−1)2+2 using properties.

Solution

This graph shows a parabola opening right with vertex (2, 1) and x intercept (5, 0).

Graph x=2(y−3)2+2 using properties.

Solution

This graph shows a parabola opening right with vertex (2, 3) and symmetric points (4, 2) and (4, 4).

In the next example, we notice the a is negative and so the parabola opens to the left.

Graph x=−4(y+1)2+4 using properties.

Solution
Two mathematical equations are displayed on a white background. The top equation, in red, is x = a(y - k)^2 + h. Below it, in black, is a specific example: x = -4(y + 1)^2 + 4.
Identify the constants a, h, k. a=−4,h=4,k=−1
Since a=−4, the parabola opens to the left.
Two red curved arrows, one arcing upwards and the other downwards, both pointing to the left from a common origin on the right, suggesting a continuous loop or a back-and-forth interaction.
The axis of symmetry is y=k. The axis of symmetry is y=−1.
The vertex is (h,k). The vertex is (4,−1).
Find the x-intercept by substituting y=0. x=−4(y+1)2+4 x=−4(0+1)2+4 x=0
The x-intercept is (0,0).
Find the point symmetric to (0,0) across the
axis of symmetry.
(0,−2)
Find the y-intercepts. x=−4(y+1)2+4
Let x=0. 0=−4(y+1)2+4 −4=−4(y+1)2 1=(y+1)2 y+1=±1
y=−1+1y=−1−1
y=0y=−2
The y-intercepts are (0,0) and (0,−2).
Graph the parabola. A parabola opening left on a Cartesian plane with vertex (4, -1), passing through (0, 0) and (0, -2). A dashed line at y=-1 indicates the axis of symmetry.

Graph x=−4(y+2)2+4 using properties.

Solution

This figure shows a parabola opening to the left with vertex (4, negative 2) and y intercepts (0, negative 1) and (0, negative 3).

Graph x=−2(y+3)2+2 using properties.

Solution

This figure shows a parabola opening to the left with vertex (2, negative 3) and y intercepts (0, negative 2) and (0, negative 4).

The next example requires that we first put the equation in standard form and then use the properties.

Write x=2y2+12y+17 in standard form and then use the properties of the standard form to graph the equation.

Solution
A mathematical equation is displayed: x = 2y^2 + 12y + 17. It shows a quadratic relationship between x and y, where x is expressed as a function of y.
Rewrite the function in
x=a(y−k)2+h form by completing
the square.
A mathematical equation is displayed on a white background: x = 2(y^2 + 6y) + 17.
A mathematical equation is displayed on a white background: x = 2(y^2 + 6y + 9) + 17 - 18. The numbers '2', '9', and '18' are highlighted in red, indicating a step in a solution or a focus point.
A mathematical equation is presented against a white background. The equation reads: x = 2(y + 3)^2 - 1.
Two math equations are displayed: the general vertex form of a horizontal parabola, x = a(y-k)^2 + h, in red, and a specific example, x = 2(y+3)^2 - 1, in black.
Identify the constants a, h, k. a=2,h=−1,k=−3
Since a=2, the parabola opens to
the right.
Two red arrows create a semi-circular path, one pointing right and up, the other left and down, suggesting a cycle, continuous flow, or return motion within a process or system.
The axis of symmetry is y=k. The axis of symmetry is y=−3.
The vertex is (h,k). The vertex is (−1,−3).
Find the x-intercept by substituting
y=0.
x=2(y+3)2−1 x=2(0+3)2−1 x=17
The x-intercept is (17,0).
Find the point symmetric to (17,0)
across the axis of symmetry.
(17,−6)
Find the y-intercepts.

Let x=0.
x=2(y+3)2−10=2(y+3)2−1 1=2(y+3)2 12=(y+3)2 y+3=±12 y=−3±22
y=−3+22y=−3−22
y≈−2.3y≈−3.7
The y-intercepts are (0,−3+22),(0,−3−22).
Graph the parabola. A parabola on a coordinate plane with vertex (-1, -3), opening to the right. The parabola passes through points (17, 0) and (17, -6). A dashed line marks y=-3.

ⓐ Write x=3y2+6y+7 in standard form and ⓑ use properties of the standard form to graph the equation.

Solution

ⓐ x=3(y+1)2+4
ⓑ
This graph shows a parabola opening to the right with vertex (4, negative 1) and x intercept (7, 0).

ⓐ Write x=−4y2−16y−12 in standard form and ⓑ use properties of the standard form to graph the equation.

Solution

ⓐ x=−4(y+2)2+4
ⓑ
This graph shows a parabola opening to the left with vertex (4, negative 2) and x intercept minus (12, 0).

Solve Applications with Parabolas

Many architectural designs incorporate parabolas. It is not uncommon for bridges to be constructed using parabolas as we will see in the next example.

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 10 feet high and 20 feet wide at the base.
Solution
We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard formy=a(x−h)2+k.
Let the lower left side of the bridge be the
origin of the coordinate grid at the point (0,0).
Since the base is 20 feet wide the point
(20,0) represents the lower right side.
The bridge is 10 feet high at the highest
point. The highest point is the vertex of
the parabola so the y-coordinate of the
vertex will be 10.
Since the bridge is symmetric, the vertex
must fall halfway between the left most
point, (0,0), and the rightmost point
(20,0). From this we know that the
x-coordinate of the vertex will also be 10.
A parabolic curve is shown on a coordinate plane. The curve starts at (0,0), peaks at (10,10), and returns to the x-axis at (20,0). The x-axis is labeled from 0 to 20, and the y-axis from 0 to 10.
Identify the vertex, (h,k). (h,k)=(10,10)
h=10, k=10
Substitute the values into the standard form.

The value of a is still unknown. To find
the value of a use one of the other points
on the parabola.
y=a(x−h)2+k y=a(x−10)2+10 (x,y)=(0,0)
Substitute the values of the other point
into the equation.
y=a(x−10)2+10 0=a(0−10)2+10
Solve for a. 0=a(0−10)2+10 −10=a(−10)2−10=100a−10100=aa=−110
y=a(x−10)2+10
Substitute the value for a into the
equation.
y=−110(x−10)2+10

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 20 feet high and 40 feet wide at the base.
Solution

y=−120(x−20)2+20

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 5 feet high and 10 feet wide at the base.
Solution

y=−15(x−5)2+5

Access these online resources for additional instructions and practice with quadratic functions and parabolas.

  • Quadratic Functions
  • Introduction to Conics and Graphing Horizontal Parabolas

Key Concepts

  • Parabola: A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.
    Vertical Parabolas
    General form
    y=ax2+bx+c
    Standard form
    y=a(x−h)2+k
    Orientation a>0 up; a<0 down a>0 up; a<0 down
    Axis of symmetry x=−b2a x=h
    Vertex Substitute x=−b2a and
    solve for y.
    (h,k)
    y- intercept Let x=0 Let x=0
    x-intercepts Let y=0 Let y=0

    This figure shows two parabolas with axis x equals h and vertex (h, k). The one on the left opens up and a is greater than 0. The one on the right opens down. Here a is less than 0.
  • How to graph vertical parabolas (y=ax2+bx+c or f(x)=a(x−h)2+k) using properties.
    1. Determine whether the parabola opens upward or downward.
    2. Find the axis of symmetry.
    3. Find the vertex.
    4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
    5. Find the x-intercepts.
    6. Graph the parabola.

    Horizontal Parabolas
    General form
    x=ay2+by+c
    Standard form
    x=a(y−k)2+h
    Orientation a>0 right; a<0 left a>0 right; a<0 left
    Axis of symmetry y=−b2a y=k
    Vertex Substitute y=−b2a and
    solve for x.
    (h,k)
    y-intercepts Let x=0 Let x=0
    x-intercept Let y=0 Let y=0

    This figure shows two parabolas with axis of symmetry y equals k, and vertex (h, k). The one on the left is labeled a greater than 0 and opens to the right. The other parabola opens to the left.
  • How to graph horizontal parabolas (x=ay2+by+c or x=a(y−k)2+h) using properties.
    1. Determine whether the parabola opens to the left or to the right.
    2. Find the axis of symmetry.
    3. Find the vertex.
    4. Find the x-intercept. Find the point symmetric to the x-intercept across the axis of symmetry.
    5. Find the y-intercepts.
    6. Graph the parabola.

Practice Makes Perfect

Graph Vertical Parabolas

In the following exercises, graph each equation by using properties.

y=−x2+4x−3

Solution

This graph shows a parabola opening downward with vertex (2, 1) and x intercepts (1, 0) and (3, 0).

y=−x2+8x−15

y=6x2+2x−1

Solution

This graph shows a parabola opening upward. The vertex is (negative 0.167, negative 1.167), the x intercepts are (negative 0.608) and (negative 0.274, 0), and the y-intercept is (0, negative 1).

y=8x2−10x+3

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

y=−x2+2x−4

Solution

ⓐ y=−(x−1)2−3
ⓑ
This graph shows a parabola opening downward with vertex (1, negative 3) and y intercept (0, 4).

y=2x2+4x+6

y=−2x2−4x−5

Solution

ⓐ y=−2(x+1)2−3
ⓑ
This graph shows a parabola opening downward with vertex (negative 1, negative 3) and x intercepts (negative 5, 0).

y=3x2−12x+7

Graph Horizontal Parabolas

In the following exercises, graph each equation by using properties.

x=−2y2

Solution

This graph shows a parabola opening to the left with vertex (0, 0). Two points on it are (negative 2, 1) and (negative 2, negative 1).

x=3y2

x=4y2

Solution

This graph shows a parabola opening to the right with vertex (0, 0). Two points on it are (4, 1) and (4, negative 1).

x=−4y2

x=−y2−2y+3

Solution

This graph shows a parabola opening to the left with vertex (4, negative 1) and y intercepts (0, 1) and (0, negative 3).

x=−y2−4y+5

x=y2+6y+8

Solution

This graph shows a parabola opening to the right with vertex (negative 1, negative 3) and y intercepts (0, negative 2) and (0, negative 4).

x=y2−4y−12

x=(y−2)2+3

Solution

This graph shows a parabola opening to the right with vertex (3, 2) and x intercept (7, 0).

x=(y−1)2+4

x=−(y−1)2+2

Solution

This graph shows a parabola opening to the left with vertex (2, 1) and x intercept (1, 0).

x=−(y−4)2+3

x=(y+2)2+1

Solution

This graph shows a parabola opening to the right with vertex (1, negative 2) and x intercept (5, 0).

x=(y+1)2+2

x=−(y+3)2+2

Solution

This graph shows a parabola opening to the left with vertex (2, negative 3). Two points on it are (negative 2, negative 1) and (negative 2, 5).

x=−(y+4)2+3

x=−3(y−2)2+3

Solution

This graph shows a parabola opening to the left with vertex (3, 2) and y intercepts (0, 1) and (0, 3).

x=−2(y−1)2+2

x=4(y+1)2−4

Solution

This graph shows a parabola opening to the right with vertex (negative 4, negative 1) and y intercepts (0, 0) and (0, negative 2).

x=2(y+4)2−2

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

x=y2+4y−5

Solution

ⓐ x=(y+2)2−9
ⓑ
This graph shows a parabola opening to the right with vertex (negative 9, negative 2) and y intercepts (0, 1) and (0, negative 5).

x=y2+2y−3

x=−2y2−12y−16

Solution

ⓐ x=−2(y+3)2+2
ⓑ
This graph shows a parabola opening to the left with vertex (2, negative 3) and y intercepts (0, negative 2) and (0, negative 4).

x=−3y2−6y−5

Mixed Practice

In the following exercises, match each graph to one of the following equations: ⓐ x2 + y2 = 64 ⓑ x2 + y2 = 49
ⓒ (x + 5)2 + (y + 2)2 = 4 ⓓ (x − 2)2 + (y − 3)2 = 9 ⓔ y = −x2 + 8x − 15 ⓕ y = 6x2 + 2x − 1

This graph shows circle with center (0, 0) and radius 8 units.
Solution

ⓐ

This graph shows a parabola opening upwards. Its vertex has an x value of slightly less than 0 and a y value of slightly less than negative 1. A point on it is close to (negative 1, 3).
This graph shows circle with center (0, 0) and radius 7 units.
Solution

ⓑ

This graph shows a parabola opening downwards with vertex (4, 1) and x intercepts (3, 0) and (5, 0).
This graph shows circle with center (2, 3) and radius 3 units.
Solution

ⓓ

This graph shows circle with center (negative 5, negative 2) and radius 2 units.

Solve Applications with Parabolas

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This graph shows circle with center (negative 5, negative 2) and radius 2 units.
Solution

y=−115(x−15)2+15

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 50 feet high and 100 feet wide at the base.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 90 feet high and 60 feet wide at the base.
Solution

y=−110(x−30)2+90

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 45 feet high and 30 feet wide at the base.

Writing Exercises

In your own words, define a parabola.

Solution

Answers will vary.

Is the parabola y=x2 a function? Is the parabola x=y2 a function? Explain why or why not.

Write the equation of a parabola that opens up or down in standard form and the equation of a parabola that opens left or right in standard form. Provide a sketch of the parabola for each one, label the vertex and axis of symmetry.

Solution

Answers will vary.

Explain in your own words, how you can tell from its equation whether a parabola opens up, down, left or right.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns, 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: graph vertical parabolas, graph horizontal parabolas, solve applications with parabolas. The remaining columns are blank.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

parabola
A parabola is all points in a plane that are the same distance from a fixed point and a fixed line.

Ellipses

Learning Objectives

By the end of this section, you will be able to:

  • Graph an ellipse with center at the origin
  • Find the equation of an ellipse with center at the origin
  • Graph an ellipse with center not at the origin
  • Solve application with ellipses

Before you get started, take this readiness quiz.

Graph y=(x−1)2−2 using transformations.
If you missed this problem, review Example 5 in Graph Quadratic Functions Using Transformations.

Solution

A graph displays a blue upward-opening parabola on a Cartesian coordinate plane. The curve has its vertex at approximately (1, -2) and crosses the x-axis at roughly x=0 and x=2.5. Grid lines are visible.

Complete the square: x2−8x=8.
If you missed this problem, review Example 2 in Solve Quadratic Equations by Completing the Square.

Solution

x−42=8

Write in standard form. y=2x2−12x+14
If you missed this problem, review Example 7 in Graph Quadratic Functions Using Transformations.

Solution

y=2x−32−4

Graph an Ellipse with Center at the Origin

The next conic section we will look at is an ellipse. We define an ellipse as all points in a plane where the sum of the distances from two fixed points is constant. Each of the given points is called a focus of the ellipse.

Ellipse

An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a focus of the ellipse.

This figure shows a double cone intersected by a plane to form an ellipse.

We can draw an ellipse by taking some fixed length of flexible string and attaching the ends to two thumbtacks. We use a pen to pull the string taut and rotate it around the two thumbtacks. The figure that results is an ellipse.

This figure shows a pen attached to two strings, the other ends of which are attached to two thumbtacks. The strings are pulled taut and the pen is rotated to draw an ellipse. The thumbtacks are labeled F subscript 1 and F subscript 2.

A line drawn through the foci intersect the ellipse in two points. Each point is called a vertex of the ellipse. The segment connecting the vertices is called the major axis. The midpoint of the segment is called the center of the ellipse. A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.

This figure shows two ellipses. In each, two points within the ellipse are labeled foci. A line drawn through the foci intersects the ellipse in two points. Each point is labeled a vertex. In the figure on the left, the segment connecting the vertices is called the major axis. A segment perpendicular to the major axis that passes through its midpoint and intersects the ellipse in two points is labeled minor axis. The minor axis is shorter than the minor axis. In the figure on the right, the segment through the foci, connecting the vertices is longer and is labeled major axis. Its midpoint is labeled center.

We mentioned earlier that our goal is to connect the geometry of a conic with algebra. Placing the ellipse on a rectangular coordinate system gives us that opportunity. In the figure, we placed the ellipse so the foci ((−c,0),(c,0)) are on the x-axis and the center is the origin.

The figure on the left shows an ellipse with its center at the origin of the coordinate axes and its foci at points minus (c, 0) and (c, 0). A segment connects (negative c, 0) to a point (x, y) on the ellipse. The segment is labeled d subscript 1. Another segment, labeled d subscript 2 connects (c, 0) to (x, y). The figure on the right shows an ellipse with center at the origin, foci (negative c, 0) and (c, 0) and vertices (negative a, 0) and (a, 0). The point where the ellipse intersects the y axis is labeled (0, b). The segments connecting (0, 0) to (c, 0), (c, 0) to (0, b) and (0, b) to (0, 0) form a tight angled triangle with sides c, a and b respectively. The equation is a squared equals b squared plus c squared.

The definition states the sum of the distance from the foci to a point (x,y) is constant. So d1+d2 is a constant that we will call 2a so, d1+d2=2a. We will use the distance formula to lead us to an algebraic formula for an ellipse.

Use the distance formula to findd1,d2.d1+d2=2a(x−(−c))2+(y−0)2+(x−c)2+(y−0)2=2a After eliminating radicals and simplifying,we get:x2a2+y2a2−c2=1 To simplify the equation of the ellipse, weleta2−c2=b2. So, the equation of an ellipse centered at theorigin in standard form is:x2a2+y2b2=1

To graph the ellipse, it will be helpful to know the intercepts. We will find the x-intercepts and y-intercepts using the formula.

y-intercepts Letx=0.x2a2+y2b2=102a2+y2b2=1y2b2=1y2=b2y=±b x-interceptsLety=0.x2a2+y2b2=1x2a2+02b2=1x2a2=1x2=a2x=±a They-intercepts are(0,b)and(0,−b).Thex-intercepts are(a,0)and(−a,0).

Standard Form of the Equation an Ellipse with Center (0,​​0)

The standard form of the equation of an ellipse with center (0,​​0), is

x2a2+y2b2=1

The x-intercepts are (a,0) and (−a,0).

The y-intercepts are (0,b) and (0,−b).

Two figures show ellipses with their centers on the origin of the coordinate axes. They intersect the x axis at points (negative a, 0) and (a, 0) and the y axis at points (0, b) and (0, negative b). In the figure on the left the major axis of the ellipse is along the x axis and in the figure on the right, it is along the y axis.

Notice that when the major axis is horizontal, the value of a will be greater than the value of b and when the major axis is vertical, the value of b will be greater than the value of a. We will use this information to graph an ellipse that is centered at the origin.

Ellipse with Center (0,0)
x2a2+y2b2=1 a>b b>a
Major axis on the x- axis. on the y-axis.
x-intercepts (−a,0),(a,0)
y-intercepts (0,−b),(0,b)

How to Graph an Ellipse with Center (0, 0)

Graph: x24+y29=1.

Solution

Step 1. Write the equation in standard form. It is in standard form x squared upon 6 plus y squared upon 9 equals 1. Step 2. Determine whether the major axis is horizontal or vertical. Since 9 is greater than 4 and 9 is in the y squared term, the major axis is vertical. Step 3. Find the endpoints of the major axis. The endpoints will be the y-intercepts. Since b squared is 9, b is plus or minus 3. The endpoints of the major axis are (0, 3) and (0, negative 3). Step 4. Find the endpoints of the minor axis. The endpoints will be the x-intercepts. Since a squared is 4, a is plus or minus 2. The endpoints of the minor axis are (2, 0) and (negative 2, 0). Step 5. Sketch the ellipse using the x and y intercepts. The graph shows an ellipse with center at (0, 0) and foci at (0, 3), (0, negative 3), (negative 2, 0), and (2, 0).

Graph: x24+y216=1.

Solution

This graph shows an ellipse with x intercepts (negative 2, 0) and (2, 0) and y intercepts (0, 4) and (0, negative 4).

Graph: x29+y216=1.

Solution

This graph shows an ellipse with x intercepts (negative 3, 0) and (3, 0) and y intercepts (0, 4) and (0, negative 4).

We summarize the steps for reference.

How to Graph an Ellipse with Center (0,0).

  1. Write the equation in standard form.
  2. Determine whether the major axis is horizontal or vertical.
  3. Find the endpoints of the major axis.
  4. Find the endpoints of the minor axis
  5. Sketch the ellipse.

Sometimes our equation will first need to be put in standard form.

Graph x2+4y2=16.

Solution
We recognize this as the equation of an
ellipse since both the x and y terms are
squared and have different coefficients.
x2+4y2=16
To get the equation in standard form, divide
both sides by 16 so that the equation is equal
to 1.
x216+4y216=1616
Simplify. x216+y24=1
The equation is in standard form.
The ellipse is centered at the origin.
The center is (0,0).
Since 16>4 and 16 is in the x2 term,
the major axis is horizontal.
  a2=16,a=±4
  b2=4,b=±2
The vertices are (4,0),(−4,0).
The endpoints of the minor axis are
(0,2),(0,−2).
Sketch the ellipse. A graph displays an ellipse centered at the origin (0,0) on a Cartesian coordinate system. It passes through points (-4,0), (4,0), (0,2), and (0,-2).

Graph 9x2+16y2=144.

Solution

This graph shows an ellipse with x intercepts (negative 4, 0) and (4, 0) and y intercepts (0, 3) and (0, negative 3).

Graph 16x2+25y2=400.

Solution

This graph shows an ellipse with x intercepts (negative 5, 0) and (5, 0) and y intercepts (0, 4) and (0, negative 4).

Find the Equation of an Ellipse with Center at the Origin

If we are given the graph of an ellipse, we can find the equation of the ellipse.

Find the equation of the ellipse shown.

This graph shows an ellipse with x intercepts (negative 4, 0) and (4, 0) and y intercepts (0, 3) and (0, negative 3).
Solution
Step-by-step derivation of an ellipse's standard equation by substituting its major and minor axis values.
We recognize this as an ellipse that is centered at the origin. x2a2+y2b2=1
Since the major axis is horizontal and the distance from the center to the vertex is 4, we know a=4 and so a2=16. x216+y2b2=1
The minor axis is vertical and the distance from the center to the ellipse is 3, we know b=3 and so b2=9. x216+y29=1

Find the equation of the ellipse shown.

This graph shows an ellipse with x intercepts (negative 2, 0) and (2, 0) and y intercepts (0, 5) and (0, negative 5).
Solution

x24+y225=1

Find the equation of the ellipse shown.

This graph shows an ellipse with x intercepts (negative 3, 0) and (3, 0) and y intercepts (0, 2) and (0, negative 2).
Solution

x29+y24=1

Graph an Ellipse with Center Not at the Origin

The ellipses we have looked at so far have all been centered at the origin. We will now look at ellipses whose center is (h,k).

The equation is (x−h)2a2+(y−k)2b2=1 and when a>b, the major axis is horizontal so the distance from the center to the vertex is a. When b>a, the major axis is vertical so the distance from the center to the vertex is b.

Standard Form of the Equation an Ellipse with Center (h,k)

The standard form of the equation of an ellipse with center (h,k), is

(x−h)2a2+(y−k)2b2=1

When a>b, the major axis is horizontal so the distance from the center to the vertex is a.

When b>a, the major axis is vertical so the distance from the center to the vertex is b.

Graph: (x−3)29+(y−1)24=1.

Solution
The equation is in standard form,
(x−h)2a2+(y−k)2b2=1.
(x−3)29+(y−1)24=1
The ellipse is centered at (h,k). The center is (3,1).
Since 9>4 and 9 is in the x2 term,
the major axis is horizontal.
  a2=9,a=±3
  b2=4,b=±2
The distance from the center to the vertices is 3.
The distance from the center to the endpoints of the
minor axis is 2.
Sketch the ellipse. An ellipse centered at (3,1) is plotted on a Cartesian coordinate plane, showing its center, vertices at (3,3) and (3,-1), and co-vertices at (0,1) and (6,1).

Graph: (x+3)24+(y−5)216=1.

Solution

This graph shows an ellipse with center at (negative 3, 5), vertices at (negative 3, 9) and (negative 3, 1) and endpoints of minor axis at (negative 5, 5) and (negative 1, 5).

Graph: (x−1)225+(y+3)216=1.

Solution

This graph shows an ellipse with center at 1, negative 3, vertices at (negative 4, negative 3) and (6, negative 3) and endpoints of minor axis at 1, 1) and (negative 1, negative 7).

If we look at the equations of x29+y24=1 and (x−3)29+(y−1)24=1, we see that they are both ellipses with a=3 and b=2. So they will have the same size and shape. They are different in that they do not have the same center.

The equation in the first figure is x squared upon 9 plus y squared upon 4 equals 1. Here, a is 3 and b is 2. The ellipse is graphed with center at (0, 0). The equation on the right is open parentheses x minus 3 close parentheses squared upon 9 plus open parentheses y minus 1 close parentheses squared upon 4 equals 1. Here, too, a is 3 and b is 2, but the center is (3, 1). The ellipse is shown on the same graph along with the first ellipse. The center is shown to have moved 3 units right and 1 unit up.

Notice in the graph above that we could have graphed (x−3)29+(y−1)24=1 by translations. We moved the original ellipse to the right 3 units and then up 1 unit.

This graph shows an ellipse translated from center (0, 0) to center (3, 1). The center has moved 3 units right and 1 unit up. The original ellipse has vertices at (negative 3, 0) and (3, 0) and endpoint of minor axis at (negative 2, 0) and (2, 0). The translated ellipse has vertices at (0, 1) and (6, 1) and endpoints of minor axis at (3, negative 1) and (3, 3).

In the next example we will use the translation method to graph the ellipse.

Graph (x+4)216+(y−6)29=1 by translation.

Solution
This ellipse will have the same size and shape as x216+y29=1 whose center is (0,0). We graph this ellipse first.
The center is (0,0). Center (0,0)
Since 16>9, the major axis is horizontal.
  a2=16,a=±4
  b2=9,b=±3
The vertices are (4,0),(−4,0).
The endpoints of the minor axis are
(0,3),(0,−3).
Sketch the ellipse. A graph shows an ellipse centered at (0, 0) with x-intercepts at (-4, 0) and (4, 0), and y-intercepts at (0, 3) and (0, -3). Key points are marked.
The original equation is in standard form,
(x−h)2a2+(y−k)2b2=1.
(x−(−4))216+(y−6)29=1
The ellipse is centered at (h,k). The center is (−4,6).
We translate the graph of x216+y29=1 four
units to the left and then up 6 units.
Verify that the center is (−4,6).
The new ellipse is the ellipse whose equation
is
(x+4)216+(y−6)29=1.
A coordinate plane displays two ellipses with labeled points. The upper ellipse has a center at (-4, 6), and the lower ellipse is centered at the origin, with dimensions indicated by arrows.

Graph (x−5)29+(y+4)24=1 by translation.

Solution

This graph shows an ellipse with center (5, negative 4), vertices (2, negative 4) and (8, negative 4) and endpoints of minor axis (5, negative 2) and (5, negative 6).

Graph (x+6)216+(y+2)225=1 by translation.

Solution

This graph shows an ellipse with center (negative 6, negative 2), vertices (negative 6, 3) and (negative 6, negative 7) and endpoints of minor axis (negative 10, negative 2), and (negative 2, negative 2).

When an equation has both an x2 and a y2 with different coefficients, we verify that it is an ellipsis by putting it in standard form. We will then be able to graph the equation.

Write the equation x2+4y2−4x+24y+24=0 in standard form and graph.

Solution
We put the equation in standard form by completing the squares in both x and y.
x2+4y2−4x+24y+24=0
Rewrite grouping the x terms and y terms. A mathematical equation showing the initial steps to complete the square: (x^2 - 4x + _) + (4y^2 + 24y + _) = -24.
Make the coefficients of x2 and y2 equal 1. An algebraic equation showing a step in completing the square: (x^2 - 4x + ) + 4(y^2 + 6y + ) = -24. This form is used to identify properties of conic sections.
Complete the squares. Mathematical equation: (x^2 - 4x + 4) + 4(y^2 + 6y + 9) = -24 + 4 + 36. This illustrates completing the square. Constants like the blue 4 and red 4, 9, 36 are highlighted, showing terms added to both sides.
Write as binomial squares. The mathematical equation (x-2)^2 + 4(y+3)^2 = 16, representing an ellipse, is shown on a white background.
Divide both sides by 16 to get 1 on the right. A mathematical equation is displayed, showing an ellipse in a non-standard form: (x-2)^2 / 16 + 4(y+3)^2 / 16 = 16 / 16. It simplifies to (x-2)^2 / 16 + (y+3)^2 / 4 = 1.
Simplify. The image displays the equation of an ellipse in standard form: (x-2)^2/16 + (y+3)^2/4 = 1.
The equation is in standard form,
(x−h)2a2+(y−k)2b2=1
An equation for an ellipse is displayed: the quantity x minus 2 squared over 16, plus the quantity y plus 3 squared over 4, equals 1.
The ellipse is centered at (h,k). The center is (2,−3).
Since 16>4 and 16 is in the x2 term,
the major axis is horizontal.
  a2=16,a=±4
  b2=4,b=±2
The distance from the center to the vertices is 4.
The distance from the center to the endpoints of
the minor axis is 2.
Sketch the ellipse. A graph showing an ellipse centered at (2, -3) with vertices at (-2, -3) and (6, -3), and co-vertices at (2, -1) and (2, -5).

ⓐ Write the equation 6x2+4y2+12x−32y+34=0 in standard form and ⓑ graph.

Solution

ⓐ (x+1)26+(y−4)29=1
ⓑ
This graph shows an ellipse with center (negative 1, 4), vertices minus (1, 1) and (negative 1, 7) and endpoints of minor axis approximately (negative 3.5, 4) and (approximately 1.5, 4).

ⓐ Write the equation 4x2+y2−16x−6y+9=0 in standard form and ⓑ graph.

Solution

ⓐ (x−2)24+(y−3)216=1
ⓑ
This graph shows an ellipse with center (2, 3), vertices (2, negative 1) and (2, 7) and endpoints of minor axis (0, 3) and (4, 3).

Solve Application with Ellipses

The orbits of the planets around the sun follow elliptical paths.

Pluto (a dwarf planet) moves in an elliptical orbit around the Sun. The closest Pluto gets to the Sun is approximately 30 astronomical units (AU) and the furthest is approximately 50 AU. The Sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of Pluto.

This graph shows an ellipse with center (0, 0) and vertices (negative 40, 0) and (40, 0). The sun is shown at point (10, 0). This is 30 units from the right vertex and 50 units from the left vertex.
Solution
Steps to determine the standard form equation of an ellipse by identifying parameters like axes and foci, and performing relevant calculations.
We recognize this as an ellipse that is centered at the origin. x2a2+y2b2=1
Since the major axis is horizontal and the distance from the center to the vertex is 40, we know a=40 and so a2=1600. x21600+y2b2=1
The minor axis is vertical but the end points aren’t given. To find b we will use the location of the Sun. Since the Sun is a focus of the ellipse at the point (10,0), we know c=10. Use this to solve for b2. b2=a2−c2b2=402−102b2=1600−100b2=1500
Substitute a2 and b2 into the standard form of the ellipse. x21600+y21500=1

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 20 AU and the furthest is approximately 30 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

This graph shows an ellipse with center (0, 0) and vertices (negative 25, 0) and (25, 0). The sun is shown at point (5, 0). This is 20 units from the right vertex and 30 units from the left vertex.
Solution

x2625+y2600=1

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 20 AU and the furthest is approximately 50 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

This graph shows an ellipse with center (0, 0) and vertices (negative 35, 0) and (35, 0). The sun is shown at point (15, 0). This is 20 units from the right vertex and 50 units from the left vertex.
Solution

x21225+y21000=1

Access these online resources for additional instructions and practice with ellipses.

  • Conic Sections: Graphing Ellipses Part 1
  • Conic Sections: Graphing Ellipses Part 2
  • Equation for Ellipse From Graph

Key Concepts

  • Ellipse: An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a focus of the ellipse.
    This figure shows two ellipses. In each, two points within the ellipse are labeled foci. A line drawn through the foci intersects the ellipse in two points. Each point is labeled a vertex. In The figure on the left, the segment connecting the vertices is called the major axis. A segment perpendicular to the major axis that passes through its midpoint and intersects the ellipse in two points is labeled minor axis. The major axis is longer than the minor axis. In The figure on the right, the segment through the foci, connecting the vertices is shorter and is labeled minor axis. Its midpoint is labeled center.
    If we draw a line through the foci intersects the ellipse in two points—each is called a vertex of the ellipse.
    The segment connecting the vertices is called the major axis.
    The midpoint of the segment is called the center of the ellipse.
    A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.
  • Standard Form of the Equation an Ellipse with Center (0,0): The standard form of the equation of an ellipse with center (0,0), is
    x2a2+y2b2=1

    The x-intercepts are (a,0) and (−a,0).
    The y-intercepts are (0,b) and (0,−b).
  • How to an Ellipse with Center (0,0)
    1. Write the equation in standard form.
    2. Determine whether the major axis is horizontal or vertical.
    3. Find the endpoints of the major axis.
    4. Find the endpoints of the minor axis
    5. Sketch the ellipse.
  • Standard Form of the Equation an Ellipse with Center (h,k): The standard form of the equation of an ellipse with center (h,k), is
    (x−h)2a2+(y−k)2b2=1

    When a>b, the major axis is horizontal so the distance from the center to the vertex is a.
    When b>a, the major axis is vertical so the distance from the center to the vertex is b.

Practice Makes Perfect

Graph an Ellipse with Center at the Origin

In the following exercises, graph each ellipse.

x24+y225=1

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (2, 0) and (negative 2, 0).

x29+y225=1

x225+y236=1

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (5, 0) and (negative 5, 0).

x216+y236=1

x236+y216=1

Solution

This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).

x225+y29=1

x2+y24=1

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 2) and (0, negative 2) and endpoints of minor axis (1, 0) and (negative 1, 0).

x29+y2=1

4x2+25y2=100

Solution

This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).

16x2+9y2=144

16x2+36y2=576

Solution

This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).

9x2+25y2=225

Find the Equation of an Ellipse with Center at the Origin

In the following exercises, find the equation of the ellipse shown in the graph.


This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (negative 3, 0) and (3, 0).

Solution

x29+y225=1


This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).


This graph shows an ellipse with center (0, 0), vertices (0, 4) and (0, negative 4) and endpoints of minor axis (negative 3, 0) and (3, 0).

Solution

x29+y216=1


This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 4, 0) and (4, 0).

Graph an Ellipse with Center Not at the Origin

In the following exercises, graph each ellipse.

(x+1)24+(y+6)225=1

Solution

This graph shows an ellipse with center (negative 1, negative 6, vertices (negative 1, negative 1) and (negative 1, negative 11) and endpoints of minor axis (negative 3, negative 6) and (1, negative 6).

(x−3)225+(y+2)29=1

(x+4)24+(y−2)29=1

Solution

This graph shows an ellipse with center (negative 4, 2, vertices (negative 4, 5) and (negative 4, negative 1) and endpoints of minor axis (3, 1) and (negative 6, 2) and (negative 2, 2).

(x−4)29+(y−1)216=1

In the following exercises, graph each equation by translation.

(x−3)24+(y−7)225=1

Solution

This graph shows an ellipse with center (3, 7), vertices (3, 2) and (3, 12), and endpoints of minor axis (1, 7) and (5, 7).

(x+6)216+(y+5)24=1

(x−5)29+(y+4)225=1

Solution

This graph shows an ellipse with center (5, negative 4), vertices (5, 1) and (5, negative 9) and endpoints of minor axis (2, negative 4) and (8, negative 4).

(x+5)236+(y−3)216=1

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

25x2+9y2−100x−54y−44=0

Solution

ⓐ (x−2)29+(y−3)225=1
ⓑ
This graph shows an ellipse with center (2, 3), vertices (2, negative 2) and (2, 8) and endpoints of minor axis (negative 1, 3) and (5, 3).

4x2+25y2+8x+100y+4=0

4x2+25y2−24x−64=0

Solution

ⓐ y24+(x−3)225=1
ⓑ
This graph shows an ellipse with center (3, 0), vertices (negative 2, 0) and (8, 0) and endpoints of minor axis (3, 2) and (3, negative 2).

9x2+4y2+56y+160=0

In the following exercises, graph the equation.

x=−2(y−1)2+2

Solution

This graph shows a parabola with vertex (2, 1) and y intercepts (0, 0) and (2, 0).

x2+y2=49

(x+5)2+(y+2)2=4

Solution

This graph shows a circle with center (negative 5, negative 2) and a radius of 2 units.

y=−x2+8x−15

(x+3)216+(y+1)24=1

Solution

This graph shows an ellipse with center (negative 3, negative 1), vertices (1, negative 1) and (negative 7, negative 1) and endpoints of minor axis (negative 3, 1) and (negative 3, negative 3).

(x−2)2+(y−3)2=9

x225+y236=1

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 5, 0) and (5, 0).

x=4(y+1)2−4

x2+y2=64

Solution

This graph shows circle with center (0, 0) and with radius 8 units.

x29+y225=1

y=6x2+2x−1

Solution

This graph shows upward opening parabola. Its vertex has an x value of slightly less than 0 and a y value of slightly less than minus 1. A point on it is approximately at (negative 1, 3).

(x−2)29+(y+3)225=1

Solve Application with Ellipses

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 30 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

This graph shows an ellipse with center (0, 0), vertices (negative 20, 0) and (20, 0). The sun is shown at point (10, 0), which is 30 units from the left vertex and 10 units from the right vertex.
Solution

x2400+y2300=1

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 70 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

This graph shows an ellipse with center (0, 0), vertices (negative 40, 0) and (40, 0). The sun is shown at point (30, 0), which is 70 units from the left vertex and 10 units from the right vertex.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 85 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

This graph shows an ellipse with center (0, 0), vertices (negative 50, 0) and (50, 0). The sun is shown at point (35, 0), which is 85 units from the left vertex and 15 units from the right vertex.
Solution

x22500+y21275=1

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 95 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

This graph shows an ellipse with center (0, 0), vertices (negative 55, 0) and (55, 0). The sun is shown at point (40, 0), which is 95 units from the left vertex and 15 units from the right vertex.

Writing Exercises

In your own words, define an ellipse and write the equation of an ellipse centered at the origin in standard form. Draw a sketch of the ellipse labeling the center, vertices and major and minor axes.

Solution

Answers will vary.

Explain in your own words how to get the axes from the equation in standard form.

Compare and contrast the graphs of the equations x24+y29=1 and x29+y24=1.

Solution

Answers will vary.

Explain in your own words, the difference between a vertex and a focus of the ellipse.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns 4 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first columns has the following statements: graph an ellipse with center at the origin, find the equation of an ellipse with center at the origin, graph an ellipse with center not at the origin, solve applications with ellipses. The remaining columns are blank.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

ellipse
An ellipse is all points in a plane where the sum of the distances from two fixed points is constant.

Hyperbolas

Learning Objectives

By the end of this section, you will be able to:

  • Graph a hyperbola with center at (0,0)
  • Graph a hyperbola with center at (h,k)
  • Identify conic sections by their equations

Before you get started, take this readiness quiz.

Solve: x2=12.
If you missed this problem, review Example 1 in Solve Quadratic Equations Using the Square Root Property.

Solution

x=±23

Expand: (x−4)2.
If you missed this problem, review Example 8 in Multiply Polynomials.

Solution

x2−8x+16

Graph y=−23x.
If you missed this problem, review Example 4 in Graph Linear Equations in Two Variables.

Solution

A coordinate plane with a line graph passing through the origin (0,0), showing a negative slope, extending infinitely in both directions.

Graph a Hyperbola with Center at (0, 0)

The last conic section we will look at is called a hyperbola. We will see that the equation of a hyperbola looks the same as the equation of an ellipse, except it is a difference rather than a sum. While the equations of an ellipse and a hyperbola are very similar, their graphs are very different.

We define a hyperbola as all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

Hyperbola

A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.

The line through the foci, is called the transverse axis. The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola. The midpoint of the segment joining the foci is called the center of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the conjugate axis. Each piece of the graph is called a branch of the hyperbola.

The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.

Again our goal is to connect the geometry of a conic with algebra. Placing the hyperbola on a rectangular coordinate system gives us that opportunity. In the figure, we placed the hyperbola so the foci ((−c,0),(c,0)) are on the x-axis and the center is the origin.

The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The foci (negative c, 0) and (c, 0) are marked with a point and lie on the x-axis. The vertices are marked with a point and lie on the x-axis. The branches pass through the vertices and open left and right. The distance from (negative c, 0) to a point on the branch (x, y) is marked d sub 1. The distance from (x, y) on the branch to (c, 0) is marked d sub 2.

The definition states the difference of the distance from the foci to a point (x,y) is constant. So |d1−d2| is a constant that we will call 2a so |d1−d2|=2a. We will use the distance formula to lead us to an algebraic formula for an ellipse.

|d1−d2|=2aUse the distance formula to findd1,d2|(x−(−c))2+(y−0)2−(x−c)2+(y−0)2|=2aEliminate the radicals.To simplify the equation of the ellipse, weletc2−a2=b2.x2a2+y2c2−a2=1So, the equation of a hyperbola centered atthe origin in standard form is:x2a2−y2b2=1

To graph the hyperbola, it will be helpful to know about the intercepts. We will find the x-intercepts and y-intercepts using the formula.

x-interceptsy-intercepts x2a2−y2b2=1x2a2−y2b2=1 Lety=0.x2a2−02b2=1Letx=0.02a2−y2b2=1 x2a2=1−y2b2=1 x2=a2y2=−b2 x=±ay=±−b2 Thex-intercepts are(a,0)and(−a,0).There are noy-intercepts.

The a, b values in the equation also help us find the asymptotes of the hyperbola. The asymptotes are intersecting straight lines that the branches of the graph approach but never intersect as the x, y values get larger and larger.

To find the asymptotes, we sketch a rectangle whose sides intersect the x-axis at the vertices (−a,0), (a,0) and intersect the y-axis at (0,−b), (0,b). The lines containing the diagonals of this rectangle are the asymptotes of the hyperbola. The rectangle and asymptotes are not part of the hyperbola, but they help us graph the hyperbola.

The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes.

The asymptotes pass through the origin and we can evaluate their slope using the rectangle we sketched. They have equations y=bax and y=−bax.

There are two equations for hyperbolas, depending whether the transverse axis is vertical or horizontal. We can tell whether the transverse axis is horizontal by looking at the equation. When the equation is in standard form, if the x2-term is positive, the transverse axis is horizontal. When the equation is in standard form, if the y2-term is positive, the transverse axis is vertical.

The second equations could be derived similarly to what we have done. We will summarize the results here.

Standard Form of the Equation a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center (0,0), is

x2a2−y2b2=1ory2a2−x2b2=1
The figure shows the graph of two hyperbolas. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (0, a) and (0, negative a) and are marked with a point and lie on the y-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the y-axis at the vertices (0, a) and (0, negative a) and intersect the y-axis at (negative b, 0) and (b, 0). The branches of the hyperbola pass through the vertices, open up and down, and approach the asymptotes.

Notice that, unlike the equation of an ellipse, the denominator of x2 is not always a2 and the denominator of y2 is not always b2.

Notice that when the x2-term is positive, the transverse axis is on the x-axis. When the y2-term is positive, the transverse axis is on the y-axis.

Standard Forms of the Equation a Hyperbola with Center (0,0)
x2a2−y2b2=1 y2a2−x2b2=1
Orientation Transverse axis on the x-axis.
Opens left and right
Transverse axis on the y-axis.
Opens up and down
Vertices (−a,0), (a,0) (0,−a), (0,a)
x-intercepts (−a,0), (a,0) none
y-intercepts none (0,−a), (0,a)
Rectangle Use (±a,0) (0,±b) Use (0,±a) (±b,0)
asymptotes y=bax, y=−bax y=abx, y=−abx

We will use these properties to graph hyperbolas.

How to Graph a Hyperbola with Center (0,0)

Graph x225−y24=1.

Solution
Step 1 is to write the equation in standard form. The the quantity x squared divided by 25 end quantity minus the quantity y squared divided by 4 end quantity is equal to 1 is already in standard form. Step 2 is to determine whether the transverse axis is horizontal or vertical. Since the x squared term is positive, the transverse axis is horizontal. Step 3 is to find the vertices. Since a squared is equal to 25, then a is equal to plus or minus 5. The vertices lie on the x-axis and are (negative 5, 0) and (5, 0). Step 4 is to sketch the rectangle centered at the origin, intersecting one axis at plus or minus a and the other at plus or minus b. Since a is equal to plus or minus 5, the rectangle will intersect the x-axis at the vertices. Since b is equal to plus or minus 2, the rectangle will intersect the y-axis at (0, negative 2) and (0, 2). The rectangle is shown on a coordinate plane with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled. Step 5 is to sketch the asymptotes, the lines through the diagonals of the rectangle. The asymptotes have the equations y is equal to five-halves times x and y is equal to negative five-halves x. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled and the lines that represent the asymptotes. Step 6 is to draw the two branches of the hyperbola. Start at each vertex and use the asymptotes as a guide. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled, the lines that represent the asymptotes, y is equal to plus or minus five-halves times x, and the branches that pass through (plus or minus 5, 0) and open left and right.

Graph x216−y24=1.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus one-half times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.

Graph x29−y216=1.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus four-thirds times x, and branches that pass through the vertices (plus or minus 3, 0) and open left and right.

We summarize the steps for reference.

Graph a hyperbola centered at (0,0).

  1. Write the equation in standard form.
  2. Determine whether the transverse axis is horizontal or vertical.
  3. Find the vertices.
  4. Sketch the rectangle centered at the origin intersecting one axis at ±a and the other at ±b.
  5. Sketch the asymptotes—the lines through the diagonals of the rectangle.
  6. Draw the two branches of the hyperbola.

Sometimes the equation for a hyperbola needs to be first placed in standard form before we graph it.

Graph 4y2−16x2=64.

Solution
4y2−16x2=64
To write the equation in standard form, divide
each term by 64 to make the equation equal to 1.
4y264−16x264=6464
Simplify. y216−x24=1
Since the y2-term is positive, the transverse axis is vertical.
Since a2=16 then a=±4.
The vertices are on the y-axis, (0,−a), (0,a).
Since b2=4 then b=±2.
(0,−4), (0,4)
Sketch the rectangle intersecting the x-axis at (−2,0), (2,0) and the y-axis at the vertices.
Sketch the asymptotes through the diagonals of the rectangle.
Draw the two branches of the hyperbola.
Hyperbola centered at the origin with vertices at (0,4) and (0,-4), showing asymptotes and the reference rectangle.

Graph 4y2−25x2=100.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus five-halves times x, and branches that pass through the vertices (0, plus or minus 5) and open up and down.

Graph 25y2−9x2=225.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus three-fifths times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.

Graph a Hyperbola with Center at (h,k)

Hyperbolas are not always centered at the origin. When a hyperbola is centered at (h,k) the equations changes a bit as reflected in the table.

Standard Forms of the Equation a Hyperbola with Center (h,k)
(x−h)2a2−(y−k)2b2=1 (y−k)2a2−(x−h)2b2=1
Orientation Transverse axis is horizontal.
Opens left and right
Transverse axis is vertical.
Opens up and down
Center (h,k) (h,k)
Vertices a units to the left and right of the center a units above and below the center
Rectangle Use a units left/right of center
b units above/ below the center
Use a units above/below the center
b units left/right of center

How to Graph a Hyperbola with Center (h,k)

Graph (x−1)29−(y−2)216=1

Solution
Step 1 is to write the equation in standard form. Notice that the equation the quantity x minus 1 squared all divided by 9 end quantity minus the quantity y minus 2 squared all divided by 16 end quantity is equal to 1 is already in standard form. Step 2 is to deteremine whether the transverse axis is horizonal or vertical. Since the x squared term is positive, the hyperbola opens left and right. The transverse axis is horizontal. The hyperbola opens left and right. Step 3 is to find the center and a and b. h is equal to 1 and k is equal 2. a squared is equal to 9 and b squared is equal to 16. You can see tha x minus h is x minus 1, and that y minus k is y minus 2. So, the center is (1, 2) and a is equal to 3 and b is equal to 4. Step 4 is to sketch the rectangle centered at (h, k) using a and b. Mark the center (1, 2) on a coordinate plane. Sketch the rectangle that goes through the points 3 units to the left and right of the center and 4 units above and below the center. Step 5 is to sketch the asymptotes on the coordinate plane. They are the lines through the diagonals of the retcangle. Mark the vertices which lie on the rectangle 3 units to the left and right of the center. The vertices are (negative 2, 2) and (4, 2). Step 6 is to draw the branches of the hyperbola. Start the vertices, (negative 2, 2) and (4, 2) and use the asymptotes as a guide. The branches should open left and right.

Graph (x−3)225−(y−1)29=1.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with an asymptote that passes through (negative 2, negative 2) and (8, 4) and an asymptote that passes through (negative 2, 4) and (8, negative 2), and branches that pass through the vertices (negative 2, 2) and (8, 2) and opens left and right.

Graph (x−2)24−(y−2)29=1.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (2, 2), an asymptote that passes through (0, negative 1) and (4, 5) and an asymptote that passes through (0, 5) and (4, negative 1), and branches that pass through the vertices (0, 2) and (4, 2) and opens left and right.

We summarize the steps for easy reference.

Graph a hyperbola centered at (h,k).

  1. Write the equation in standard form.
  2. Determine whether the transverse axis is horizontal or vertical.
  3. Find the center and a, b.
  4. Sketch the rectangle centered at (h,k) using a, b.
  5. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
  6. Draw the two branches of the hyperbola.

Be careful as you identify the center. The standard equation has x−h and y−k with the center as (h,k).

Graph (y+2)29−(x+1)24=1.

Solution
A mathematical equation for a hyperbola is displayed, reading as (y+2)^2 / 9 - (x+1)^2 / 4 = 1. The equation shows two squared binomials over numerical denominators, set equal to one.
Since the y2-term is positive, the hyperbola
opens up and down.
A mathematical equation for a hyperbola is displayed, with \( (y - (-2))^2 \) over 9 minus \( (x - (-1))^2 \) over 4, equaling 1. The variables 'k' and 'h' are shown in red above -(-2) and -(-1), respectively.
Find the center, (h,k). Center: (−1,−2)
Find a, b. a=3 b=2
Sketch the rectangle that goes through the
points 3 units above and below the center and
2 units to the left/right of the center.
Sketch the asymptotes—the lines through the
diagonals of the rectangle.
Mark the vertices.
Graph the branches.
A graph displays a hyperbola centered at (-1,-2) on a coordinate plane. The hyperbola opens vertically, with its two branches extending upwards and downwards. Dashed blue lines represent the asymptotes, and a dashed red rectangle indicates the central box.

Graph (y+3)216−(x+2)29=1.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 3), an asymptote that passes through (negative 5, negative 7) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, 7), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 7) and opens up and down.

Graph (y+2)29−(x+2)29=1.

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 2), an asymptote that passes through (negative 5, negative 5) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, negative 5), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 5) and opens up and down.

Again, sometimes we have to put the equation in standard form as our first step.

Write the equation in standard form and graph 4x2−9y2−24x−36y−36=0.

Solution
A mathematical equation is displayed on a white background: 4x^2 - 9y^2 - 24x - 36y - 36 = 0.
To get to standard form, complete the squares. The image shows the mathematical equation 4(x^2 - 6x) - 9(y^2 + 4y) = 36, which is an equation of a hyperbola.
An algebraic equation: 4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36, illustrating a step in completing the square for a conic section, with specific constants highlighted.
The image displays the mathematical equation 4(x-3)^2 - 9(y+2)^2 = 36, which is the standard form of a hyperbola.
Divide each term by 36 to get the constant to be 1. An equation for a hyperbola is displayed, showing the difference of two fractions: 4(x-3)^2/36 minus 9(y+2)^2/36, which equals 36/36.
A mathematical equation for a hyperbola is displayed, reading: (x-3)^2 / 9 - (y+2)^2 / 4 = 1.
Since the x2-term is positive, the hyperbola
opens left and right.
Find the center, (h,k). Center: (3,−2)
Find a, b. a=3b=4
Sketch the rectangle that goes through the
points 3 units to the left/right of the center
and 2 units above and below the center.
Sketch the asymptotes—the lines through the
diagonals of the rectangle.
Mark the vertices.
Graph the branches.
This image depicts a hyperbola on a coordinate grid, with its center clearly marked at (3, -2). The hyperbola's branches extend outward, guided by its dashed asymptotes.

ⓐ Write the equation in standard form and ⓑ graph 9x2−16y2+18x+64y−199=0.

Solution

ⓐ (x+1)216−(y−2)29=1
ⓑ
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 1, 2), an asymptote that passes through (negative 5, 5) and (3, negative 1) and an asymptote that passes through (3, 5) and (negative 5, negative 1), and branches that pass through the vertices (negative 5, 2) and (3, 2) and opens left and right.

ⓐ Write the equation in standard form and ⓑ graph 16x2−25y2+96x−50y−281=0.

Solution

ⓐ (x+3)225−(y+1)216=1
ⓑ
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 3, negative 1), an asymptote that passes through (negative 8, negative 5) and (2, 3) and an asymptote that passes through (negative 8, 3) and (2, negative 5), and branches that pass through the vertices (negative 8, negative 1) and (2, negative 1) and opens left and right.

Identify Conic Sections by their Equations

Now that we have completed our study of the conic sections, we will take a look at the different equations and recognize some ways to identify a conic by its equation. When we are given an equation to graph, it is helpful to identify the conic so we know what next steps to take.

To identify a conic from its equation, it is easier if we put the variable terms on one side of the equation and the constants on the other.

Conic Characteristics of x2- and y2- terms Example
Parabola Either x2 OR y2. Only one variable is squared. x=3y2−2y+1
Circle x2- and y2- terms have the same coefficients x2+y2=49
Ellipse x2- and y2- terms have the same sign, different coefficients 4x2+25y2=100
Hyperbola x2- and y2- terms have different signs, different coefficients 25y2−4x2=100

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ 9x2+4y2+56y+160=0 ⓑ 9x2−16y2+18x+64y−199=0 ⓒ x2+y2−6x−8y=0 ⓓ y=−2x2−4x−5

Solution
ⓐ
This table illustrates an equation example and the identifying characteristic of its terms that classify it as a specific conic section, an ellipse.
9x2+4y2+56y+160=0
The x2- and y2-terms have the same sign and different coefficients. Ellipse


ⓑ
This table illustrates an example equation and its identifying characteristic that classify it as a hyperbola.
9x2−16y2+18x+64y−199=0
The x2- and y2-terms have different signs and different coefficients. Hyperbola


ⓒ
Characteristics of conic section equations and their corresponding types or examples.
x2+y2−6x−8y=0
The x2- and y2-terms have the same coefficients. Circle


ⓓ
This table describes how a quadratic equation, identified by only one variable being squared, results in a parabolic graph.
y=−2x2−4x−5
Only one variable, x, is squared. Parabola

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ x2+y2−8x−6y=0 ⓑ 4x2+25y2=100 ⓒ y=6x2+2x−1 ⓓ 16y2−9x2=144

Solution

ⓐ circle ⓑ ellipse ⓒ parabola ⓓ hyperbola

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ 16x2+9y2=144 ⓑ y=2x2+4x+6 ⓒ x2+y2+2x+6y+9=0 ⓓ 4x2−16y2=64

Solution

ⓐ ellipse ⓑ parabola ⓒ circle ⓓ hyperbola

Access these online resources for additional instructions and practice with hyperbolas.

  • Graph a Hyperbola with Center at the Origin
  • Graph a Hyperbola with Center not at the Origin
  • Graph a Hyperbola in General Form
  • Identifying Conic Sections in General Form

Key Concepts

  • Hyperbola: A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant.
    The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.
    Each of the fixed points is called a focus of the hyperbola.
    The line through the foci, is called the transverse axis.
    The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola.
    The midpoint of the segment joining the foci is called the center of the hyperbola.
    The line perpendicular to the transverse axis that passes through the center is called the conjugate axis.
    Each piece of the graph is called a branch of the hyperbola.
    The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.
    Standard Forms of the Equation a Hyperbola with Center (0,0)
    x2a2−y2b2=1 y2a2−x2b2=1
    Orientation Transverse axis on the x-axis.
    Opens left and right
    Transverse axis on the y-axis.
    Opens up and down
    Vertices (−a,0), (a,0) (0,−a), (0,a)
    x-intercepts (−a,0), (a,0) none
    y-intercepts none (0,−a), (0,a)
    Rectangle Use (±a,0) (0,±b) Use (0,±a) (±b,0)
    asymptotes y=bax, y=−bax y=abx, y=−abx
  • How to graph a hyperbola centered at (0,0).
    1. Write the equation in standard form.
    2. Determine whether the transverse axis is horizontal or vertical.
    3. Find the vertices.
    4. Sketch the rectangle centered at the origin intersecting one axis at ±a and the other at ±b.
    5. Sketch the asymptotes—the lines through the diagonals of the rectangle.
    6. Draw the two branches of the hyperbola.

    Standard Forms of the Equation a Hyperbola with Center (h,k)
    (x−h)2a2−(y−k)2b2=1 (y−k)2a2−(x−h)2b2=1
    Orientation Transverse axis is horizontal.
    Opens left and right
    Transverse axis is vertical.
    Opens up and down
    Center (h,k) (h,k)
    Vertices a units to the left and right of the center a units above and below the center
    Rectangle Use a units left/right of center
    b units above/below the center
    Use a units above/below the center
    b units left/right of center
  • How to graph a hyperbola centered at (h,k).
    1. Write the equation in standard form.
    2. Determine whether the transverse axis is horizontal or vertical.
    3. Find the center and a,b.
    4. Sketch the rectangle centered at (h,k) using a,b.
    5. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
    6. Draw the two branches of the hyperbola.

    Conic Characteristics of x2- and y2- terms Example
    Parabola Either x2 OR y2. Only one variable is squared. x=3y2−2y+1
    Circle x2- and y2- terms must have the same coefficients and they must be the same sign as the constant after the = sign x2+y2=49
    Ellipse x2- and y2- terms have the same sign, different coefficients 4x2+25y2=100
    Hyperbola x2- and y2- terms have different signs 25y2−4x2=100

Practice Makes Perfect

Graph a Hyperbola with Center at (0,0)

In the following exercises, graph.

x29−y24=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus two-thirds times x, and branches that pass through the vertices (plus or minus 3, 0) and open left and right.

x225−y29=1

x216−y225=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus five-fourths times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.

x29−y236=1

y225−x24=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus five-halves times x, and branches that pass through the vertices (0, plus or minus 5) and open up and down.

y236−x216=1

16y2−9x2=144

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus three-fourths times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.

25y2−9x2=225

4y2−9x2=36

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus three-halves times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.

16y2−25x2=400

4x2−16y2=64

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with asymptotes y is equal to plus or minus one-half times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.

9x2−4y2=36

Graph a Hyperbola with Center at (h,k)

In the following exercises, graph.

(x−1)216−(y−3)24=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 3) an asymptote that passes through (negative 3, 1) and (5, 5) and an asymptote that passes through (5, 1) and (negative 3, 5), and branches that pass through the vertices (negative 3, 3) and (5, 3) and opens left and right.

(x−2)24−(y−3)216=1

(y−4)29−(x−2)225=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 3) an asymptote that passes through (negative 3, 1) and (5, 5) and an asymptote that passes through (5, 1) and (negative 3, 5), and branches that pass through the vertices (negative 3, 3) and (5, 3) and opens left and right.

(y−1)225−(x−4)216=1

(y+4)225−(x+1)236=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, negative 4) an asymptote that passes through (negative 7, 1) and (5, negative 9) and an asymptote that passes through (5, 1) and (negative 7, negative 9), and branches that pass through the vertices (1, 1) and (1, negative 9) and open up and down.

(y+1)216−(x+1)24=1

(y−4)216−(x+1)225=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (negative 1, 4) an asymptote that passes through (4, 8) and (negative 6, 0) and an asymptote that passes through (negative 6, 8) and (4, 0), and branches that pass through the vertices (negative 1, 0) and (negative 1, 8) and open up and down.

(y+3)216−(x−3)236=1

(x−3)225−(y+2)29=1

Solution

The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (3, negative 2) an asymptote that passes through (8, 1) and (negative 2, negative 5) and an asymptote that passes through (negative 2, negative 1) and (8, negative 5), and branches that pass through the vertices (negative 2, negative 2) and (8, negative 2) and opens left and right.

(x+2)24−(y−1)29=1

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

9x2−4y2−18x+8y−31=0

Solution

ⓐ (x−1)24−(y−1)29=1
ⓑ
The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 1) an asymptote that passes through (3, 4) and (negative 1, negative 2) and an asymptote that passes through (negative 1, 4) and (3, negative 2), and branches that pass through the vertices (negative 1, 1) and (3, 1) and opens left and right.

16x2−4y2+64x−24y−36=0

y2−x2−4y+2x−6=0

Solution

ⓐ (y−2)29−(x−1)29=1
ⓑ
The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (1, 2) an asymptote that passes through (4, 5) and (negative 2, negative 1) and an asymptote that passes through (negative 2, 5) and (4, negative 1), and branches that pass through the vertices (1, 5) and (1, negative 1) and open up and down.

4y2−16x2−24y+96x−172=0

9y2−x2+18y−4x−4=0

Solution

ⓐ (y+1)21−(x+2)29=1
ⓑ
The graph shows the x-axis and y-axis that both run in the negative and positive directions with the center (negative 2, negative 1) an asymptote that passes through (1, 0) and (negative 5, negative 2) and an asymptote that passes through (3, 0) and (1, negative 2), and branches that pass through the vertices (negative 2, 0) and (negative 2, negative 2) and open up and down.

Identify the Graph of each Equation as a Circle, Parabola, Ellipse, or Hyperbola

In the following exercises, identify the type of graph.

ⓐ x=−y2−2y+3 ⓑ 9y2−x2+18y−4x−4=0 ⓒ 9x2+25y2=225 ⓓ x2+y2−4x+10y−7=0

ⓐ x=−2y2−12y−16 ⓑ x2+y2=9 ⓒ 16x2−4y2+64x−24y−36=0 ⓓ 16x2+36y2=576

Solution

ⓐ parabola ⓑ circle ⓒ hyperbola ⓓ ellipse

Mixed Practice

In the following exercises, graph each equation.

(y−3)29−(x+2)216=1

x2+y2−4x+10y−7=0

Solution

The graph shows the x y coordinate plane with a circle whose center is (2, negative 5) and whose radius is 6 units.

y=(x−1)2+2

x29+y225=1

Solution

The graph shows the x y coordinate plane with an ellipse whose major axis is vertical, vertices are (0, plus or minus 5) and co-vertices are (plus or minus 3, 0).

(x+2)2+(y−5)2=4

y2−x2−4y+2x−6=0

Solution

The graph shows the x y coordinate plane with the center (1, 2) an asymptote that passes through (negative 2, 5) and (5, negative 1) and an asymptote that passes through (4, 5) and (2, 0), and branches that pass through the vertices (1, 5) and (negative 2, negative 1) and open up and down.

x=−y2−2y+3

16x2+9y2=144

Solution

The graph shows the x y coordinate plane with an ellipse whose major axis is vertical, vertices are (0, plus or minus 4) and co-vertices are (plus or minus 3, 0).

Writing Exercises

In your own words, define a hyperbola and write the equation of a hyperbola centered at the origin in standard form. Draw a sketch of the hyperbola labeling the center, vertices, and asymptotes.

Explain in your own words how to create and use the rectangle that helps graph a hyperbola.

Solution

Answers will vary.

Compare and contrast the graphs of the equations x24−y29=1 and y29−x24=1.

Explain in your own words, how to distinguish the equation of an ellipse with the equation of a hyperbola.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was graph a hyperbola with center at (0, 0). In row 3, the I can was graph a hyperbola with a center at (h, k). In row 4, the I can was identify conic sections by their equations.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

hyperbola
A hyperbola is defined as all points in a plane where the difference of their distances from two fixed points is constant.

Solve Systems of Nonlinear Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve a system of nonlinear equations using graphing
  • Solve a system of nonlinear equations using substitution
  • Solve a system of nonlinear equations using elimination
  • Use a system of nonlinear equations to solve applications

Solve the system by graphing: {x−3y=−3x+y=5.
If you missed this problem, review Example 2 in Solve Systems of Linear Equations with Two Variables.

Solution

3,2

Solve the system by substitution: {x−4y=−4−3x+4y=0.
If you missed this problem, review Example 7 in Solve Systems of Linear Equations with Two Variables.

Solution

2,32

Solve the system by elimination: {3x−4y=−95x+3y=14.
If you missed this problem, review Example 9 in Solve Systems of Linear Equations with Two Variables.

Solution

1,3

Solve a System of Nonlinear Equations Using Graphing

We learned how to solve systems of linear equations with two variables by graphing, substitution and elimination. We will be using these same methods as we look at nonlinear systems of equations with two equations and two variables. A system of nonlinear equations is a system where at least one of the equations is not linear.

For example each of the following systems is a system of nonlinear equations.

{x2+y2=9x2−y=9{9x2+y2=9y=3x−3{x+y=4y=x2+2

System of Nonlinear Equations

A system of nonlinear equations is a system where at least one of the equations is not linear.

Just as with systems of linear equations, a solution of a nonlinear system is an ordered pair that makes both equations true. In a nonlinear system, there may be more than one solution. We will see this as we solve a system of nonlinear equations by graphing.

When we solved systems of linear equations, the solution of the system was the point of intersection of the two lines. With systems of nonlinear equations, the graphs may be circles, parabolas or hyperbolas and there may be several points of intersection, and so several solutions. Once you identify the graphs, visualize the different ways the graphs could intersect and so how many solutions there might be.

To solve systems of nonlinear equations by graphing, we use basically the same steps as with systems of linear equations modified slightly for nonlinear equations. The steps are listed below for reference.

Solve a system of nonlinear equations by graphing.

  1. Identify the graph of each equation. Sketch the possible options for intersection.
  2. Graph the first equation.
  3. Graph the second equation on the same rectangular coordinate system.
  4. Determine whether the graphs intersect.
  5. Identify the points of intersection.
  6. Check that each ordered pair is a solution to both original equations.

Solve the system by graphing: {x−y=−2y=x2.

Solution
Identify each graph. {x−y=−2liney=x2parabola
Sketch the possible options for
intersection of a parabola and a line.
Diagram showing how a parabola and a line can have 0, 1, or 2 points of intersection, representing the number of solutions for a system.
Graph the line, x−y=−2.
Slope-intercept form y=x+2.
Graph the parabola, y=x2.
A Cartesian graph featuring a dark blue parabola, y = x^2, and a light blue line, y = x+2. The two functions intersect at (-1,1) and (2,4) on a grid with axes ranging from -5 to 5.
Identify the points of intersection. The points of intersection appear to be (2,4) and (−1,1).
Check to make sure each solution makes
both equations true.
(2,4)
x−y=−2y=x22−4=?−24=?22−2=−2✓4=4✓

(−1,1)
x−y=−2y=x2−1−1=?−21=?(−1)2−2=−2✓1=1✓
The solutions are (2,4) and (−1,1).

Solve the system by graphing: {x+y=4y=x2+2.

Solution

This graph shows the equations of a system, x plus y is equal to 4 and y is equal x squared plus 2, and the x y-coordinate plane. The line has a slope of negative 1 and a y-intercept at 4. The vertex of the parabola is (0, 2) and opens upward. The line and parabola intersect at the points (negative 2, 6) and (1, 3), which are labeled.

Solve the system by graphing: {x−y=−1y=−x2+3.

Solution

This graph shows the equations of a system, x minus y is equal to negative 1 and y is equal to negative x squared plus three, and the x y-coordinate plane. The line has a slope of 1 and a y-intercept at 1. The vertex of the parabola is (0, negative 3) and opens upward. The line and parabola intersect at the points (negative 2, negative 1) and (1, 2), which are labeled.

To identify the graph of each equation, keep in mind the characteristics of the x2 and y2 terms of each conic.

Solve the system by graphing: {y=−1(x−2)2+(y+3)2=4.

Solution
Identify each graph. {y=−1line(x−2)2+(y+3)2=4circle
Sketch the possible options for the
intersection of a circle and a line.
Visual representation of a line intersecting a circle. It shows 0 solutions (line misses circle), 1 solution (line tangent), and 2 solutions (line crosses through).
Graph the circle, (x−2)2+(y+3)2=4
Center: (2,−3) radius: 2
Graph the line, y=−1.
It is a horizontal line.
A circle centered at (2, -3) with a radius of 2 is plotted on a Cartesian coordinate system. The point (2, -1) is marked on the circle.
Identify the points of intersection. The point of intersection appears to be (2,−1).
Check to make sure the solution makes
both equations true.
(2,−1)
(x−2)2+(y+3)2=4y=−1(2−2)2+(−1+3)2=?4−1=−1✓(0)2+(2)2=?44=4✓
The solution is (2,−1).

Solve the system by graphing: {x=−6(x+3)2+(y−1)2=9.

Solution

This graph shows the equations of a system, x is equal to negative 6 and the quantity x plus 3 squared plus the quantity y minus 1 squared is equal to 9, which is a circle, on the x y-coordinate plane. The line is a vertical line. The center of the circle is (negative 3, 1) and it has a radius of 3 units. The point of intersection between the line and circle is (negative 6, 1).

Solve the system by graphing: {y=4(x−2)2+(y+3)2=4.

Solution

This graph shows the equations of a system, y is equal to negative 4 and the quantity x minus 2 squared plus the quantity y plus 3 squared is equal to 4, which is a circle, on the x y-coordinate plane. The line is a horizontal line. The center of the circle is (2, negative 3) and it has a radius of 2 units. There is no point of intersection between the line and circle, so the system has no solution.

Solve a System of Nonlinear Equations Using Substitution

The graphing method works well when the points of intersection are integers and so easy to read off the graph. But more often it is difficult to read the coordinates of the points of intersection. The substitution method is an algebraic method that will work well in many situations. It works especially well when it is easy to solve one of the equations for one of the variables.

The substitution method is very similar to the substitution method that we used for systems of linear equations. The steps are listed below for reference.

Solve a system of nonlinear equations by substitution.

  1. Identify the graph of each equation. Sketch the possible options for intersection.
  2. Solve one of the equations for either variable.
  3. Substitute the expression from Step 2 into the other equation.
  4. Solve the resulting equation.
  5. Substitute each solution in Step 4 into one of the original equations to find the other variable.
  6. Write each solution as an ordered pair.
  7. Check that each ordered pair is a solution to both original equations.

Solve the system by using substitution: {9x2+y2=9y=3x−3.

Solution
Identify each graph. {9x2+y2=9ellipsey=3x−3line
Sketch the possible options for intersection of an
ellipse and a line.
The diagram shows how a line can intersect an oval at 0, 1, or 2 points, representing the number of solutions for their system of equations.
The equation y=3x−3 is solved for y. A mathematical equation, 'y = 3x - 3,' is displayed on a white background, suggesting a linear function or algebraic expression.
A mathematical equation is displayed on a white background, which reads 9x squared plus y squared equals 9.
Substitute 3x−3 for y in the first equation. A mathematical equation is displayed, showing '9x^2 + (3x - 3)^2 = 9'. The '3x - 3' part is highlighted in red, indicating a potential focus or area of interest in the equation.
Solve the equation for x. A mathematical equation is displayed on a white background: 9x^2 + 9x^2 - 18x + 9 = 9.
A step-by-step solution to the quadratic equation 18x^2 - 18x = 0 is displayed, showing factorization into 18x(x - 1) = 0, which yields the solutions x = 0 and x = 1.
Substitute x=0 and x=1 into y=3x−3 to find y. Two identical mathematical equations, y = 3x - 3, are displayed side-by-side on a white background.
Two sets of calculations for a linear equation, showing y = 3 * 0 - 3 resulting in y = -3, and y = 3 * 1 - 3 resulting in y = 0.
The ordered pairs are (0,−3), (1,0).
Check both ordered pairs in both equations.
(0,−3)
9x2+y2=9y=3x−39·02+(−3)2=?9−3=?3·0−30+9=?9−3=?0−39=9✓−3=−3✓
(1,0)
9x2+y2=9y=3x−39·12+02=?90=?3·1−39+0=?90=?3−39=9✓0=0✓
The solutions are (0,−3),(1,0).

Solve the system by using substitution: {x2+9y2=9y=13x−3.

Solution

No solution

Solve the system by using substitution: {4x2+y2=4y=x+2.

Solution

(−45,65),(0,2)

So far, each system of nonlinear equations has had at least one solution. The next example will show another option.

Solve the system by using substitution: {x2−y=0y=x−2.

Solution
Identify each graph. {x2−y=0parabolay=x−2line
Sketch the possible options for
intersection of a parabola and a line
Three diagrams illustrate the number of solutions when a parabola intersects a line: 0 solutions (no intersection), 1 solution (tangent point), and 2 solutions (two intersection points).
The equation y=x−2 is solved for y. The equation y = x - 2 is displayed in black text on a white background, with the number 2 in red.
A mathematical equation is displayed on a white background, reading 'x^2 - y = 0' in a black serif font.
Substitute x−2 for y in the first equation. A mathematical equation on a white background showing x squared minus (x minus 2) equals 0, with the (x-2) term highlighted in red.
Solve the equation for x. A mathematical equation, x squared minus x plus two equals zero, is displayed in black text on a white background.
This doesn’t factor easily, so we can
check the discriminant.
b2−4ac(−1)2−4·1·2−7 The discriminant is negative, so there is no real solution.
The system has no solution.

Solve the system by using substitution: {x2−y=0y=2x−3.

Solution

No solution

Solve the system by using substitution: {y2−x=0y=3x−2.

Solution

(49,−23),(1,1)

Solve a System of Nonlinear Equations Using Elimination

When we studied systems of linear equations, we used the method of elimination to solve the system. We can also use elimination to solve systems of nonlinear equations. It works well when the equations have both variables squared. When using elimination, we try to make the coefficients of one variable to be opposites, so when we add the equations together, that variable is eliminated.

The elimination method is very similar to the elimination method that we used for systems of linear equations. The steps are listed for reference.

Solve a system of equations by elimination.

  1. Identify the graph of each equation. Sketch the possible options for intersection.
  2. Write both equations in standard form.
  3. Make the coefficients of one variable opposites.
    Decide which variable you will eliminate.
    Multiply one or both equations so that the coefficients of that variable are opposites.
  4. Add the equations resulting from Step 3 to eliminate one variable.
  5. Solve for the remaining variable.
  6. Substitute each solution from Step 5 into one of the original equations. Then solve for the other variable.
  7. Write each solution as an ordered pair.
  8. Check that each ordered pair is a solution to both original equations.

Solve the system by elimination: {x2+y2=4x2−y=4.

Solution
Identify each graph. A system of two equations is shown with their corresponding conic sections: a circle (x^2 + y^2 = 4) and a parabola (x^2 - y = 4).
Sketch the possible options for
intersection of a circle and a parabola.
Illustrations demonstrating the possible number of solutions (intersection points) when a parabola and a circle intersect, ranging from 0 to 4.
Both equations are in standard form. A system of two equations is shown, featuring a circle equation x^2 + y^2 = 4 and a parabola-like equation x^2 - y = 4. This setup is typical for finding intersection points of these two graphs.
To get opposite coefficients of x2,
we will multiply the second equation by −1.
Two mathematical equations are displayed: x^2 + y^2 = 4, and -1(x^2 - y) = -1(4). The second equation has '-1' highlighted in red on both sides of the equal sign.
Simplify. A system of two equations is shown, consisting of x squared plus y squared equals 4, and negative x squared plus y equals 4, enclosed by a blue brace.
Add the two equations to eliminate x2. Solving a system of nonlinear equations by elimination. Adding x^2+y^2=4 and -x^2+y=4 eliminates x^2, yielding the simplified equation y^2+y=0.
Solve for y. The equation y(y+1) = 0 is displayed in black text on a white background.
Mathematical expressions showing y=0 and the step-by-step solution of y+1=0 leading to y=-1.
Substitute y=0 and y=−1 into one of
the original equations. Then solve for x.
Two mathematical equations are displayed on a white background. On the left, 'y = 0' is visible. To its right, the equation 'y = -1' is shown.
Two examples show solutions to the equation x squared minus y equals 4. When y equals zero, x equals plus or minus two. When y equals negative one, x equals plus or minus square root of three.
Write each solution as an ordered pair. The ordered pairs are
(−2,0) (2,0).
(3,−1)(−3,−1)
Check that each ordered pair is a
solution to both original equations.
We will leave the checks for each of
the four solutions to you.
The solutions are (−2,0), (2,0), (3,−1), and
(−3,−1).

Solve the system by elimination: {x2+y2=9x2−y=9.

Solution

(−3,0),(3,0),(−22,−1),(22,−1)

Solve the system by elimination: {x2+y2=1−x+y2=1.

Solution

(−1,0),(0,1),(0,−1)

There are also four options when we consider a circle and a hyperbola.

Solve the system by elimination: {x2+y2=7x2−y2=1.

Solution
Identify each graph. {x2+y2=7circlex2−y2=1hyperbola
Sketch the possible options for intersection
of a circle and hyperbola.
Diagrams demonstrating how the number of intersections (solutions) between a circle and a curve can range from zero to four, depending on their relative positions.
Both equations are in standard form. {x2+y2=7x2−y2=1
The coefficients of y2 are opposite, so we
will add the equations.
{x2+y2=7x2−y2=1__________2x2=8
Simplify. x2=4x=±2
x=2x=−2
Substitute x=2 and x=−2 into one of the
original equations. Then solve for y.
x2+y2=7x2+y2=722+y2=7(−2)2+y2=74+y2=74+y2=7y2=3y2=3y=±3y=±3
Write each solution as an ordered pair. The ordered pairs are (−2,3), (−2,−3),
(2,3), and (2,−3).
Check that the ordered pair is a solution to
both original equations.
We will leave the checks for each of the four
solutions to you.
The solutions are (−2,3), (−2,−3), (2,3),
and (2,−3).

Solve the system by elimination: {x2+y2=25y2−x2=7.

Solution

(−3,−4),(−3,4),(3,−4),(3,4)

Solve the system by elimination: {x2+y2=4x2−y2=4.

Solution

(−2,0),(2,0)

Use a System of Nonlinear Equations to Solve Applications

Systems of nonlinear equations can be used to model and solve many applications. We will look at an everyday geometric situation as our example.

The difference of the squares of two numbers is 15. The sum of the numbers is 5. Find the numbers.

Solution
Identify what we are looking for. Two different numbers.
Define the variables. x= first number
y= second number
Translate the information into a system of
equations.
First sentence. The difference of the squares of two numbers is 15.
A mathematical equation displays 'x² - y² = 15' on a white background, representing a hyperbola or a difference of squares problem.
Second sentence. The sum of the numbers is 5.
The mathematical equation 'x + y = 5' is displayed in a minimalist presentation on a white background.
Solve the system by substitution A system of two algebraic equations is shown. The first equation is x² - y² = 15, and the second equation is x + y = 5. Both equations are vertically aligned and enclosed by a curly brace on the left.
Solve the second equation for x. A mathematical equation on a white background reads 'x = 5 - y'. The 'x' and '=' are in black, while '5', '-', and 'y' are in varying shades of red, with 'y' being the reddest.
Substitute x into the first equation. A mathematical equation reads x squared minus y squared equals 15, displayed in black text against a plain white background.
A mathematical equation is displayed on a white background: (5-y)^2 - y^2 = 15. The 'y' in the first term is highlighted in red, indicating a variable.
Expand and simplify. A mathematical equation is displayed: (25 - 10y + y^2) - y^2 = 15. This equation involves variables and constants, representing a quadratic expression that simplifies to a linear one.
An algebraic equation showing the simplification of 25 - 10y + y^2 - y^2 = 15 to 25 - 10y = 15 by canceling out the y squared terms.
Solve for y. The image shows the mathematical equation -10y = -10 in a black font against a white background.
The image shows a mathematical equation 'y=1' in black text on a plain white background. The equation is centrally placed within the frame.
Substitute back into the second equation. The image displays the algebraic equation 'x + y = 5' centered on a plain white background.
An image displaying an algebraic equation and its solution. The equation is x + (1) = 5, and the solution shown directly below it is x = 4. The text is black on a white background.
The numbers are 1 and 4.

The difference of the squares of two numbers is −20. The sum of the numbers is 10. Find the numbers.

Solution

4 and 6

The difference of the squares of two numbers is 35. The sum of the numbers is −1. Find the numbers.

Solution

−18 and 17

Myra purchased a small 25” TV for her kitchen. The size of a TV is measured on the diagonal of the screen. The screen also has an area of 300 square inches. What are the length and width of the TV screen?

Solution
Identify what we are looking for. The length and width of the rectangle
Define the variables. Let x= width of the rectangle
y= length of the rectangle
Draw a diagram to help visualize the situation. A light blue rectangle with sides labeled 'x' and 'y', and a diagonal line across it labeled '25''.
Area is 300 square inches.
Translate the information into a system of
equations.
The diagonal of the right triangle is 25 inches.
Two mathematical equations are displayed: x squared plus y squared equals 25 squared, which simplifies to x squared plus y squared equals 625. These represent the equation of a circle.
The area of the rectangle is 300 square inches.
A system of three algebraic equations is shown: x * y = 300, x^2 + y^2 = 625, and x * y = 300. The equations are grouped by a curly brace on the left.
Solve the system using substitution. A simple mathematical equation 'x * y = 300' is displayed on a plain white background.
Solve the second equation for x. The image displays the algebraic equation x = 300 / y, where 'x' is equal to 300 divided by 'y'.
Substitute x into the first equation. The mathematical equation x^2 + y^2 = 625 is displayed on a white background, representing a circle centered at the origin with a radius of 25 units.
A mathematical equation shows (300/y)^2 + y^2 = 625, presented in black and orange text on a white background, representing a problem to solve for the variable y.
Simplify. A mathematical equation is displayed on a white background: 90000/y^2 + y^2 = 625. The numbers and symbols are rendered in black.
Multiply by y2 to clear the fractions. A mathematical equation is displayed, reading '90000 + y^4 = 625y^2' against a white background.
Put in standard form. An algebraic equation: y to the fourth power minus six hundred twenty-five times y squared plus nine thousand equals zero.
Solve by factoring. An algebraic equation is shown, displaying the product of two binomials, (y^2 - 225) and (y^2 - 400), set equal to zero.
Two quadratic equations, y^2 - 225 = 0 and y^2 - 400 = 0, are displayed in black text on a white background, representing calculations involving squared variables and constants.
Two mathematical equations are displayed: y squared equals 225, solved as y equals plus or minus 15; and y squared equals 400, solved as y equals plus or minus 20.
Since y is a side of the rectangle, we discard
the negative values.
Two mathematical equations are displayed horizontally, showing y = 15 on the left and y = 20 on the right, both rendered in a clear, dark sans-serif font against a white background.
Substitute back into the second equation. Equation showing x times y equals three hundred.
Two math problems are shown with their solutions. The first equation is x * 15 = 300, with x = 20. The second equation is x * 20 = 300, with x = 15.
If the length is 15 inches, the width is 20 inches.
If the length is 20 inches, the width is 15 inches.

Edgar purchased a small 20” TV for his garage. The size of a TV is measured on the diagonal of the screen. The screen also has an area of 192 square inches. What are the length and width of the TV screen?

Solution

If the length is 12 inches, the width is 16 inches. If the length is 16 inches, the width is 12 inches.

The Harper family purchased a small microwave for their family room. The diagonal of the door measures 15 inches. The door also has an area of 108 square inches. What are the length and width of the microwave door?

Solution

If the length is 12 inches, the width is 9 inches. If the length is 9 inches, the width is 12 inches.

Access these online resources for additional instructions and practice with solving nonlinear equations.

  • Nonlinear Systems of Equations
  • Solve a System of Nonlinear Equations
  • Solve a System of Nonlinear Equations by Elimination
  • System of Nonlinear Equations – Area and Perimeter Application

Key Concepts

  • How to solve a system of nonlinear equations by graphing.
    1. Identify the graph of each equation. Sketch the possible options for intersection.
    2. Graph the first equation.
    3. Graph the second equation on the same rectangular coordinate system.
    4. Determine whether the graphs intersect.
    5. Identify the points of intersection.
    6. Check that each ordered pair is a solution to both original equations.
  • How to solve a system of nonlinear equations by substitution.
    1. Identify the graph of each equation. Sketch the possible options for intersection.


    2. Solve one of the equations for either variable.
    3. Substitute the expression from Step 2 into the other equation.
    4. Solve the resulting equation.
    5. Substitute each solution in Step 4 into one of the original equations to find the other variable.
    6. Write each solution as an ordered pair.
    7. Check that each ordered pair is a solution to both original equations.
  • How to solve a system of equations by elimination.
    1. Identify the graph of each equation. Sketch the possible options for intersection.
    2. Write both equations in standard form.
    3. Make the coefficients of one variable opposites.
      Decide which variable you will eliminate.
      Multiply one or both equations so that the coefficients of that variable are opposites.
    4. Add the equations resulting from Step 3 to eliminate one variable.
    5. Solve for the remaining variable.
    6. Substitute each solution from Step 5 into one of the original equations. Then solve for the other variable.
    7. Write each solution as an ordered pair.
    8. Check that each ordered pair is a solution to both original equations.

Section Exercises

Practice Makes Perfect

Solve a System of Nonlinear Equations Using Graphing

In the following exercises, solve the system of equations by using graphing.

{y=2x+2y=−x2+2

{y=6x−4y=2x2

Solution

This graph shows the equations of a system, y is equal to 6 x minus 4 which is a line and y is equal to 2 x squared which is a parabola, on the x y-coordinate plane. The vertex of the parabola is (0, 0) and the parabola opens upward. The line has a slope of 6. The line and parabola intersect at the points (1, 2) and (2, 8), which are labeled. The solutions are (1, 2) and (2, 8).

{x+y=2x=y2

{x−y=−2x=y2

Solution

This graph shows the equations of a system, x minus y is equal to negative 2 which is a line and x is equal to y squared which is a rightward-opening parabola, on the x y-coordinate plane. The vertex of the parabola is (0, 0) and it passes through the points (1, 1) and (1, negative 1). The line has a slope of 1 and a y-intercept at 2. The line and parabola do not intersect, so the system has no solution.

{y=32x+3y=−x2+2

{y=x−1y=x2+1

Solution

This graph shows the equations of a system, y is x minus 1 which is a line and y is equal to x squared plus 1 which is an upward-opening parabola, on the x y-coordinate plane. The vertex of the parabola is (0, 1) and it passes through the points (negative 1, 2) and (1, 2). The line has a slope of 1 and a y-intercept at negative 1. The line and parabola do not intersect, so the system has no solution.

{x=−2x2+y2=4

{y=−4x2+y2=16

Solution

This graph shows the equations of a system, x is equal to negative 2 which is a line and x squared plus y squared is equal to 16 which is a circle, on the x y-coordinate plane. The line is horizontal. The center of the circle is (0, 0) and the radius of the circle is 4. The line and circle intersect at (negative 2, 0), so the solution of the system is (negative 2, 0).

{x=2(x+2)2+(y+3)2=16

{y=−1(x−2)2+(y−4)2=25

Solution

This graph shows the equations of a system, x is equal to 2 which is a line and the quantity x minus 2 end quantity squared plus the quantity y minus 4 end quantity squared is equal to 25 which is a circle, on the x y-coordinate plane. The line is horizontal. The center of the circle is (2, 4) and the radius of the circle is 5. The line and circle intersect at (2, negative 1), so the solution of the system is (2, negative 1).

{y=−2x+4y=x+1

{y=−12x+2y=x−2

Solution

This graph shows the equations of a system, y is equal to negative one-half x plus 2 which is a line and the y is equal to the square root of x minus 2, on the x y-coordinate plane. The curve for y is equal to the square root of x minus 2 The curve for y is equal to the square root of x plus 1 where x is greater than or equal to 0 and y is greater than or equal to negative 2. The line and square root curve intersect at (4, 0), so the solution is (4, 0).

Solve a System of Nonlinear Equations Using Substitution

In the following exercises, solve the system of equations by using substitution.

{x2+4y2=4y=12x−1

{9x2+y2=9y=3x+3

Solution

(−1,0),(0,3)

{9x2+y2=9y=x+3

{9x2+4y2=36x=2

Solution

(2,0)

{4x2+y2=4y=4

{x2+y2=169x=12

Solution

(12,−5),(12,5)

{3x2−y=0y=2x−1

{2y2−x=0y=x+1

Solution

No solution

{y=x2+3y=x+3

{y=x2−4y=x−4

Solution

(0,−4),(1,−3)

{x2+y2=25x−y=1

{x2+y2=252x+y=10

Solution

(3,4),(5,0)

Solve a System of Nonlinear Equations Using Elimination

In the following exercises, solve the system of equations by using elimination.

{x2+y2=16x2−2y=8

{x2+y2=16x2−y=4

Solution

(0,−4),(−7,3),(7,3)

{x2+y2=4x2+2y=1

{x2+y2=4x2−y=2

Solution

(0,−2),(−3,1),(3,1)

{x2+y2=9x2−y=3

{x2+y2=4y2−x=2

Solution

(−2,0),(1,−3),(1,3)

{x2+y2=252x2−3y2=5

{x2+y2=20x2−y2=−12

Solution

(−2,−4),(−2,4),(2,−4),(2,4)

{x2+y2=13x2−y2=5

{x2+y2=16x2−y2=16

Solution

(−4,0),(4,0)

{4x2+9y2=362x2−9y2=18

{x2−y2=32x2+y2=6

Solution

(−3,0),(3,0)

{4x2−y2=44x2+y2=4

{x2−y2=−53x2+2y2=30

Solution

(−2,−3),(−2,3),(2,−3),(2,3)

{x2−y2=1x2−2y=4

{2x2+y2=11x2+3y2=28

Solution

(−1,−3),(−1,3),(1,−3),(1,3)

Use a System of Nonlinear Equations to Solve Applications

In the following exercises, solve the problem using a system of equations.

The sum of two numbers is −6 and the product is 8. Find the numbers.

The sum of two numbers is 11 and the product is −42. Find the numbers.

Solution

−3 and 14

The sum of the squares of two numbers is 65. The difference of the numbers is 3. Find the numbers.

The sum of the squares of two numbers is 113. The difference of the numbers is 1. Find the numbers.

Solution

−7 and −8 or 8 and 7

The difference of the squares of two numbers is 15. The difference of twice the square of the first number and the square of the second number is 30. Find the numbers.

The difference of the squares of two numbers is 20. The difference of the square of the first number and twice the square of the second number is 4. Find the numbers.

Solution

−6 and −4 or −6 and 4 or 6 and −4 or 6 and 4

The perimeter of a rectangle is 32 inches and its area is 63 square inches. Find the length and width of the rectangle.

The perimeter of a rectangle is 52 cm and its area is 165 cm2. Find the length and width of the rectangle.

Solution

If the length is 11 cm, the width is 15 cm. If the length is 15 cm, the width is 11 cm.

Dion purchased a new microwave. The diagonal of the door measures 17 inches. The door also has an area of 120 square inches. What are the length and width of the microwave door?

Jules purchased a microwave for his kitchen. The diagonal of the front of the microwave measures 26 inches. The front also has an area of 240 square inches. What are the length and width of the microwave?

Solution

If the length is 10 inches, the width is 24 inches. If the length is 24 inches, the width is 10 inches.

Roman found a widescreen TV on sale, but isn’t sure if it will fit his entertainment center. The TV is 60”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1728 square inches. His entertainment center has an insert for the TV with a length of 50 inches and width of 40 inches. What are the length and width of the TV screen and will it fit Roman’s entertainment center?

Donnette found a widescreen TV at a garage sale, but isn’t sure if it will fit her entertainment center. The TV is 50”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1200 square inches. Her entertainment center has an insert for the TV with a length of 38 inches and width of 27 inches. What are the length and width of the TV screen and will it fit Donnette’s entertainment center?

Solution

The length is 40 inches and the width is 30 inches. The TV will not fit Donnette’s entertainment center.

Writing Exercises

In your own words, explain the advantages and disadvantages of solving a system of equations by graphing.

Explain in your own words how to solve a system of equations using substitution.

Solution

Answers will vary.

Explain in your own words how to solve a system of equations using elimination.

A circle and a parabola can intersect in ways that would result in 0, 1, 2, 3, or 4 solutions. Draw a sketch of each of the possibilities.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve a system of nonlinear equations using graphing. In row 3, the I can solve a system of nonlinear equations using substitution. In row 4, the I can was solve a system of a nonlinear equations using the elimination. In row 5, the I can was use a system of nonlinear equations to solve applications.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Distance and Midpoint Formulas; Circles

Use the Distance Formula

In the following exercises, find the distance between the points. Round to the nearest tenth if needed.

(−5,1) and (−1,4)

(−2,5) and (1,5)

Solution

d=3

(8,2) and (−7,−3)

(1,−4) and (5,−5)

Solution

d=17,d≈4.1

Use the Midpoint Formula

In the following exercises, find the midpoint of the line segment whose endpoints are given.

(−2,−6) and (−4,−2)

(3,7) and (5,1)

Solution

(4,4)

(−8,−10) and (9,5)

(−3,2) and (6,−9)

Solution

(32,−72)

Write the Equation of a Circle in Standard Form

In the following exercises, write the standard form of the equation of the circle with the given information.

radius is 15 and center is (0,0)

radius is 7 and center is (0,0)

Solution

x2+y2=7

radius is 9 and center is (−3,5)

radius is 7 and center is (−2,−5)

Solution

(x+2)2+(y+5)2=49

center is (3,6) and a point on the circle is (3,−2)

center is (2,2) and a point on the circle is (4,4)

Solution

(x−2)2+(y−2)2=8

Graph a Circle

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

2x2+2y2=450

3x2+3y2=432

Solution

ⓐ radius: 12, center: (0,0)
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 20 to 20. The y-axis of the plane runs from negative 15 to 15. The center of the circle is (0, 0) and the radius of the circle is 12.

(x+3)2+(y−5)2=81

(x+2)2+(y+5)2=49

Solution

ⓐ radius: 7, center: (−2,−5)
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 20 to 20. The y-axis of the plane runs from negative 15 to 15. The center of the circle is (negative 2, negative 5) and the radius of the circle is 7.

x2+y2−6x−12y−19=0

x2+y2−4y−60=0

Solution

ⓐ radius: 8, center: (0,2)
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 20 to 20. The y-axis of the plane runs from negative 15 to 15. The center of the circle is (0, 2) and the radius of the circle is 8.

Parabolas

Graph Vertical Parabolas

In the following exercises, graph each equation by using its properties.

y=x2+4x−3

y=2x2+10x+7

Solution

The figure shows an upward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 7 to 7. The vertex is (negative five-halves, negative eleven-halves) and the parabola passes through the points (negative 4, negative 1) and (negative 1, negative 1).

y=−6x2+12x−1

y=−x2+10x

Solution

The figure shows a downward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 36 to 36. The y-axis of the plane runs from negative 26 to 26. The vertex is (5, 25) and the parabola passes through the points (2, 16) and (8, 16).

In the following exercises, ⓐ write the equation in standard form, then ⓑ use properties of the standard form to graph the equation.

y=x2+4x+7

y=2x2−4x−2

Solution

ⓐ y=2(x−1)2−4
ⓑ
The figure shows an upward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 22 to 22. The y-axis of the plane runs from negative 16 to 16. The vertex is (1, negative 4) and the parabola passes through the points (0, negative 2) and (2, negative 2).

y=−3x2−18x−29

y=−x2+12x−35

Solution

ⓐ y=−(x−6)2+1
ⓑ
The figure shows a downward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 60 to 60. The y-axis of the plane runs from negative 46 to 46. The vertex is (6, 1) and the parabola passes through the points (5, 0) and (7, 0).

Graph Horizontal Parabolas

In the following exercises, graph each equation by using its properties.

x=2y2

x=2y2+4y+6

Solution

The figure shows a rightward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (4, negative 1) and the parabola passes through the points (6, 0) and (6, negative 2).

x=−y2+2y−4

x=−3y2

Solution

The figure shows a leftward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (0, 0) and the parabola passes through the points (negative 3, 1) and (negative 3, negative 1).

In the following exercises, ⓐ write the equation in standard form, then ⓑ use properties of the standard form to graph the equation.

x=4y2+8y

x=y2+4y+5

Solution

ⓐ x=(y+2)2+1
ⓑ
The figure shows a rightward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (1, negative 2) and the parabola passes through the points (5, 0) and (5, negative 4).

x=−y2−6y−7

x=−2y2+4y

Solution

ⓐ x=−2(y−1)2+2
ⓑ
The figure shows a leftward-opening parabola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The vertex is (2, negative 3) and the parabola passes through the points (0, 2) and (0, 0).

Solve Applications with Parabolas

In the following exercises, create the equation of the parabolic arch formed in the foundation of the bridge shown. Give the answer in standard form.

The figure shows a parabolic arch formed in the foundation of the bridge. The arch is 5 feet high and 20 feet wide.
The figure shows a parabolic arch formed in the foundation of the bridge. The arch is 25 feet high and 30 feet wide.
Solution

y=−19x2+103x

Ellipses

Graph an Ellipse with Center at the Origin

In the following exercises, graph each ellipse.

x236+y225=1

x24+y281=1

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 9), and co-vertices at (plus or minus 2, 0).

49x2+64y2=3136

9x2+y2=9

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 9 to 9. The y-axis of the plane runs from negative 7 to 7. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 3), and co-vertices at (plus or minus 1, 0).

Find the Equation of an Ellipse with Center at the Origin

In the following exercises, find the equation of the ellipse shown in the graph.

The figure shows an ellipse graphed on the x y coordinate plane. The ellipse has a center at (0, 0), a horizontal major axis, vertices at (plus or minus 10, 0), and co-vertices at (0, plus or minus 4).
The figure shows an ellipse graphed on the x y coordinate plane. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 8), and co-vertices at (plus or minus 6, 0).
Solution

x236+y264=1

Graph an Ellipse with Center Not at the Origin

In the following exercises, graph each ellipse.

(x−1)225+(y−6)24=1

(x+4)216+(y+1)29=1

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (negative 4, negative 1), a horizontal major axis, vertices at (negative 8, negative 1) and (0, negative 1) and co-vertices at (negative 4, 2) and (negative 4, negative 4).

(x−5)216+(y+3)236=1

(x+3)29+(y−2)225=1

Solution

The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (negative 3, 2), a vertical major axis, vertices at (negative 3, 7) and (negative 3, negative 3) and co-vertices at (negative 6, 2) and (0, 2).

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

4x2+16y2+48x+160y+480=0

25x2+4y2−150x−56y+321=0

Solution

ⓐ (x−3)24+(y−7)225=1
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 18 to 18. The y-axis of the plane runs from negative 14 to 14. The ellipse has a center at (3, 7), a vertical major axis, vertices at (3, 2) and (3, 12) and co-vertices at (negative 1, 7) and (5, 7).

25x2+4y2+150x+125=0

4x2+9y2−126y+405=0

Solution

ⓐ x29+(y−7)24=1
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 15 to 15. The y-axis of the plane runs from negative 11 to 11. The ellipse has a center at (0, 7), a horizontal major axis, vertices at (3, 7) and (negative 3, 7) and co-vertices at (0, 5) and (0, 9).

Solve Applications with Ellipses

In the following exercises, write the equation of the ellipse described.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 10 AU and the furthest is approximately 90 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

The figure shows a model of an elliptical orbit around the sun on the x y coordinate plane. The ellipse has a center at (0, 0), a horizontal major axis, vertices marked at (plus or minus 50, 0), the sun marked as a foci and labeled (50, 0), the closest distance the comet is from the sun marked as 10 A U, and the farthest a comet is from the sun marked as 90 A U.

Hyperbolas

Graph a Hyperbola with Center at (0,0)

In the following exercises, graph.

x225−y29=1

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 9 to 9. The hyperbola has a center at (0, 0) and branches that pass through the vertices (plus or minus 5, 0), and that open left and right.

y249−x216=1

9y2−16x2=144

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 19 to 19. The y-axis of the plane runs from negative 15 to 15. The hyperbola has a center at (0, 0) and branches that pass through the vertices (0, plus or minus 4), and that open up and down.

16x2−4y2=64

Graph a Hyperbola with Center at (h,k)

In the following exercises, graph.

(x+1)24−(y+1)29=1

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (negative 1, negative 1) and branches that pass through the vertices (negative 3, negative 1) and (1, negative 1), and that open left and right.

(x−2)24−(y−3)216=1

(y+2)29−(x+1)29=1

Solution

The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (negative 1, negative 2) and branches that pass through the vertices (negative 1, 1) and (negative 1, negative 5), and that open up and down.

(y−1)225−(x−2)29=1

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

4x2−16y2+8x+96y−204=0

Solution

ⓐ (x+1)216−(y−3)24=1
ⓑ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (negative 1, 3) and branches that pass through the vertices (negative 5, 3) and (3, 3), and that open left and right.

16x2−4y2−64x−24y−36=0

4y2−16x2+32x−8y−76=0

Solution

ⓐ (y−1)216−(x−1)24=1
ⓑ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (1, 1) and branches that pass through the vertices (1, negative 3) and (1, 5), and that open up and down.

36y2−16x2−96x+216y−396=0

Identify the Graph of each Equation as a Circle, Parabola, Ellipse, or Hyperbola

In the following exercises, identify the type of graph.


ⓐ 16y2−9x2−36x−96y−36=0
ⓑ x2+y2−4x+10y−7=0
ⓒ y=x2−2x+3
ⓓ 25x2+9y2=225

Solution

ⓐ hyperbola ⓑ circle ⓒ parabola ⓓ ellipse


ⓐ x2+y2+4x−10y+25=0
ⓑ y2−x2−4y+2x−6=0
ⓒ x=−y2−2y+3
ⓓ 16x2+9y2=144

Solve Systems of Nonlinear Equations

Solve a System of Nonlinear Equations Using Graphing

In the following exercises, solve the system of equations by using graphing.

{3x2−y=0y=2x−1

Solution

The figure shows a parabola and line graphed on the x y coordinate plane. The x-axis of the plane runs from negative 5 to 5. The y-axis of the plane runs from negative 4 to 4. The parabola has a vertex at (0, 0) and opens upward. The line has a slope of 2 with a y-intercept at negative 1. The parabola and line do not intersect, so the system has no solution.

{y=x2−4y=x−4

{x2+y2=169x=12

Solution

The figure shows a circle and line graphed on the x y coordinate plane. The x-axis of the plane runs from negative 20 to 20. The y-axis of the plane runs from negative 15 to 15. The circle has a center at (0, 0) and a radius of 13. The line is vertical. The circle and line intersect at the points (12, 5) and (12, negative 5), which are labeled. The solution of the system is (12, 5) and (12, negative 5)

{x2+y2=25y=−5

Solve a System of Nonlinear Equations Using Substitution

In the following exercises, solve the system of equations by using substitution.

{y=x2+3y=−2x+2

Solution

(−1,4)

{x2+y2=4x−y=4

{9x2+4y2=36y−x=5

Solution

No solution

{x2+4y2=42x−y=1

Solve a System of Nonlinear Equations Using Elimination

In the following exercises, solve the system of equations by using elimination.

{x2+y2=16x2−2y−1=0

Solution

(−7,3),(7,3)

{x2−y2=5−2x2−3y2=−30

{4x2+9y2=363y2−4x=12

Solution

(−3,0),(0,−2),(0,2)

{x2+y2=14x2−y2=16

Use a System of Nonlinear Equations to Solve Applications

In the following exercises, solve the problem using a system of equations.

The sum of the squares of two numbers is 25. The difference of the numbers is 1. Find the numbers.

Solution

−3 and −4 or 4 and 3

The difference of the squares of two numbers is 45. The difference of the square of the first number and twice the square of the second number is 9. Find the numbers.

The perimeter of a rectangle is 58 meters and its area is 210 square meters. Find the length and width of the rectangle.

Solution

If the length is 14 inches, the width is 15 inches. If the length is 15 inches, the width is 14 inches.

Colton purchased a larger microwave for his kitchen. The diagonal of the front of the microwave measures 34 inches. The front also has an area of 480 square inches. What are the length and width of the microwave?

Practice Test

In the following exercises, find the distance between the points and the midpoint of the line segment with the given endpoints. Round to the nearest tenth as needed.

(−4,−3) and (−10,−11)

Solution

distance: 10, midpoint: (−7,−7)

(6,8) and (−5,−3)

In the following exercises, write the standard form of the equation of the circle with the given information.

radius is 11 and center is (0,0)

Solution

x2+y2=121

radius is 12 and center is (10,−2)

center is (−2,3) and a point on the circle is (2,−3)

Solution

(x+2)2+(y−3)2=52

Find the equation of the ellipse shown in the graph.

The figure shows an ellipse graphed on the x y coordinate plane. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 10), and co-vertices at (plus or minus 6, 0).

In the following exercises, ⓐ identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola, and ⓑ graph the equation.

4x2+49y2=196

Solution

ⓐ ellipse
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The ellipse has a center at (0, 0), a horizontal major axis, vertices at (plus or minus 7, 0) and co-vertices at (0, plus or minus 2).

y=3(x−2)2−2

3x2+3y2=27

Solution

ⓐ circle
ⓑ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The parabola circle has a center at (0, 0) and a radius of 3.

y2100−x236=1

x216+y281=1

Solution

ⓐ ellipse
ⓑ
The figure shows an ellipse graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The ellipse has a center at (0, 0), a vertical major axis, vertices at (0, plus or minus 9) and co-vertices at (plus or minus 4, 0).

x=2y2+10y+7

64x2−9y2=576

Solution

ⓐ hyperbola
ⓑ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 8 to 8. The hyperbola has a center at (0, 0) and branches that pass through the vertices (plus or minus 3, 0) and that open left and right.

In the following exercises, ⓐ identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola, ⓑ write the equation in standard form, and ⓒ graph the equation.

25x2+64y2+200x−256y−944=0

x2+y2+10x+6y+30=0

Solution

ⓐ circle
ⓑ (x+5)2+(y+3)2=4
ⓒ
The figure shows a circle graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The circle has a center at (negative 5, negative 3) and a radius 2.

x=−y2+2y−4

9x2−25y2−36x−50y−214=0

Solution

ⓐ hyperbola
ⓑ (x−2)225−(y+1)29=1
ⓒ
The figure shows a hyperbola graphed on the x y coordinate plane. The x-axis of the plane runs from negative 14 to 14. The y-axis of the plane runs from negative 10 to 10. The hyperbola has a center at (2, negative 1) and branches that pass through the vertices (negative 3, negative 1) and (7, negative 1) that open left and right.

y=x2+6x+8

Solve the nonlinear system of equations by graphing:
{3y2−x=0y=−2x−1.

Solution

No solution
A graph displays a downward-sloping line and a parabola opening to the right, both intersecting at the origin (0,0) on a Cartesian coordinate system.

Solve the nonlinear system of equations using substitution:
{x2+y2=8y=−x−4.

Solve the nonlinear system of equations using elimination:
{x2+9y2=92x2−9y2=18.

Solution

(3,0),(−3,0)

Create the equation of the parabolic arch formed in the foundation of the bridge shown. Give the answer in y=ax2+bx+c form.

The figure shows a parabolic arch formed in the foundation of the bridge. The arch is 10 feet high and 30 feet wide.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 20 AU and the furthest is approximately 70 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

The figure shows a model of an elliptical orbit around the sun on the x y coordinate plane. The ellipse has a center at (0, 0), a horizontal major axis, vertices marked at (plus or minus 45, 0), the sun marked as a foci and labeled (25, 0), the closest distance the comet is from the sun marked as 20 A U, and the farthest a comet is from the sun marked as 70 A U.
Solution

x22025+y21400=1

The sum of two numbers is 22 and the product is −240. Find the numbers.

For her birthday, Olive’s grandparents bought her a new widescreen TV. Before opening it she wants to make sure it will fit her entertainment center. The TV is 55”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1452 square inches. Her entertainment center has an insert for the TV with a length of 50 inches and width of 40 inches. What are the length and width of the TV screen and will it fit Olive’s entertainment center?

Solution

The length is 44 inches and the width is 33 inches. The TV will fit Olive’s entertainment center.

system of nonlinear equations
A system of nonlinear equations is a system where at least one of the equations is not linear.

Introduction

A photo of lines of code on a computer terminal screen.
Cryptographers protect private data by encrypting it; this means they convert the data into a code that hackers and thieves cannot easily break. (credit: “joffi”/pixabay)

A strange charge suddenly appears on your credit card. But your card is in your wallet—it’s not even lost or stolen. Sadly, you may have been a victim of cyber crime. In this day and age, most transactions take advantage of the benefit of computers in some way. Cyber crime is any type of crime that uses a computer or computer network. Thankfully, many people are working to prevent cyber crime. Sometimes known as cryptographers, these people develop complex patterns in computer codes that block access to would-be thieves as well as write codes to intercept and decode information from them so that they may be identified. In this chapter, you will explore basic sequences and series related to those used by computer programmers to prevent cyber crime.

Sequences

Learning Objectives

By the end of this section, you will be able to:

  • Write the first few terms of a sequence
  • Find a formula for the general term (nth term) of a sequence
  • Use factorial notation
  • Find the partial sum
  • Use summation notation to write a sum

Before you get started, take this readiness quiz.

Evaluate 2n+3 for the integers 1, 2, 3, and 4.
If you missed this problem, review Example 6 in Use the Language of Algebra.

Solution

5, 7, 9, 11

Evaluate (−1)n for the integers 1, 2, 3, and 4.
If you missed this problem, review Example 8 in Integers.

Solution

−1,1,−1,1

If f(n)=n2+2, find f(1)+f(2)+f(3).
If you missed this problem, review Example 8 in Relations and Functions.

Solution

20

Write the First Few Terms of a Sequence

Let’s look at the function f(x)=2x and evaluate it for just the counting numbers.

f(x)=2x
x 2x
1 2
2 4
3 6
4 8
5 10
… …

If we list the function values in order as 2, 4, 6, 8, and 10, … we have a sequence. A sequence is a function whose domain is the counting numbers.

Sequences

A sequence is a function whose domain is the counting numbers.

A sequence can also be seen as an ordered list of numbers and each number in the list is a term. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an infinite sequence. Its domain is all counting numbers and there is an infinite number of counting numbers.

2,4,6,8,10,…,

If we limit the domain to a finite number of counting numbers, then the sequence is a finite sequence. If we use only the first four counting numbers, 1, 2, 3, 4 our sequence would be the finite sequence,

2,4,6,8

Often when working with sequences we do not want to write out all the terms. We want more compact way to show how each term is defined. When we worked with functions, we wrote f(x)=2x and we said the expression 2x was the rule that defined values in the range. While a sequence is a function, we do not use the usual function notation. Instead of writing the function as f(x)=2x, we would write it as an=2n. The an is the nth term of the sequence, the term in the nth position where n is a value in the domain. The formula for writing the nth term of the sequence is called the general term or formula of the sequence.

General Term of a Sequence

The general term of the sequence is found from the formula for writing the nth term of the sequence. The nth term of the sequence, an, is the term in the nth position where n is a value in the domain.

When we are given the general term of the sequence, we can find the terms by replacing n with the counting numbers in order. For an=2n,

n 1 2 3 4 5 an
an 2·12 2·24 2·36 2·48 2·510 2n
a1,a2,a3,a4,a5,…,an,…2,4,6,8,10,…

To find the values of a sequence, we substitute in the counting numbers in order into the general term of the sequence.

Write the first five terms of the sequence whose general term is an=4n−3.

Solution

We substitute the values 1, 2, 3, 4, and 5 into the formula, an=4n−3, in order.

This figure shows three rows and five columns. The first row reads nth term equals 4 times n minus 3 written five times. The second row reads a sub 1 equals 4 times g times 1 minus 3, a sub 2 equals 4 times g times 2 minus 3, a sub 3 equals 4 times g times 3 minus 3, a sub 4 equals 4 times g times 4 minus 3, a sub 5 equals 4 times g times 5 minus 3. The third row reads, a sub 1 equals 1, a sub 2 equals 5, a sub 3 equals 9, a sub 4 equals 13, a sub 5 equals 17.

The first five terms of the sequence are 1, 5, 9, 13, and 17.

Write the first five terms of the sequence whose general term is an=3n−4.

Solution

−1,2,5,8,11

Write the first five terms of the sequence whose general term is an=2n−5.

Solution

−3,−1,1,3,5

For some sequences, the variable is an exponent.

Write the first five terms of the sequence whose general term is an=2n+1.

Solution

We substitute the values 1, 2, 3, 4, and 5 into the formula, an=2n+1, in order.

This figure shows three rows and five columns. The first row reads “nth term equals 2 to the nth power plus 1” written five times. The second row reads, “a sub 1 equals 2 times 1 plus 1, a sub 2 equals 2 to the power of 2 plus 1, a sub 3 equals 2 to the power 3 plus 1, a sub 4 equals 2 to the power of 4 plus 1, a sub 5 equals 2 to the power 5 plus 1”. The last row reads “a sub 1 equals 3, a sub 2 equals 5, a sub 3 equals 9, a sub 4 equals 17, a sub 5 equals 33”.

The first five terms of the sequence are 3, 5, 9, 17, and 33.

Write the first five terms of the sequence whose general term is an=3n+4.

Solution

7,13,31,85,247

Write the first five terms of the sequence whose general term is an=2n−5.

Solution

−3,−1,3,11,27

It is not uncommon to see the expressions (−1)n or (−1)n+1 in the general term for a sequence. If we evaluate each of these expressions for a few values, we see that this expression alternates the sign for the terms.

n 1 2 3 4 5
(−1)n (−1)1−1 (−1)21 (−1)3−1 (−1)41 (−1)5−1
(−1)n+1 (−1)1+11 (−1)2+1−1 (−1)3+11 (−1)4+1−1 (−1)5+11
a1,a2,a3,a4,a5,…,an,…−1,1,−1,1,−1…1,−1,1,−1,1…

The terms in the next example will alternate signs as a result of the powers of −1.

Write the first five terms of the sequence whose general term is an=(−1)nn3.

Solution

We substitute the values 1, 2, 3, 4, and 5 into the formula, an=(−1)nn3, in order.

This figure shows three rows and five columns. The first row reads “nth term equals negative 1 to the nth power times n cubed” written five times. The second row reads a sub 1 equals negative 1 to the power of 1 times g times 1 cubed, a sub 2 equals negative 1 squared time g times 2 cubed, a sub 3 equals negative 1 cubed times g times 23 cubed, a sub 4 equals negative 1 to the power of 4 times g times 4 cubed, a sub 5 equals negative 1 to the power of 5 times g times 5 cubed. The last row reads, “a sub 1 equals negative 1, a sub 2 equals 8, a sub 3 equals negative 27, a sub 4 equals 64, and a sub 5 equals negative 125.

The first five terms of the sequence are −1,8,−27,64, and −125.

Write the first five terms of the sequence whose general term is an=(−1)nn2.

Solution

−1,4,−9,16,−25

Write the first five terms of the sequence whose general term is an=(−1)n+1n3.

Solution

1,−8,27,−64,125

Find a Formula for the General Term (nth Term) of a Sequence

Sometimes we have a few terms of a sequence and it would be helpful to know the general term or nth term. To find the general term, we look for patterns in the terms. Often the patterns involve multiples or powers. We also look for a pattern in the signs of the terms.

Find a general term for the sequence whose first five terms are shown.

4,8,12,16,20,…
Solution
A numerical sequence showing multiples of 4: 4, 8, 12, 16, 20, ...
Numerical sequence ranging from 1 to n, indicated by n.
We look for a pattern in the terms. A mathematical sequence titled 'Terms:' showing the numbers 4, 8, 12, 16, 20, indicating an arithmetic progression where each term increases by 4.
The numbers are all multiples of 4. Number pattern showing the result of multiplying four by consecutive whole numbers: 1, 2, 3, 4, 5, and so on to n. Demonstrates an arithmetic sequence.
The general term of the sequence is an=4n.

Find a general term for the sequence whose first five terms are shown.

3,6,9,12,15,…

Solution

an=3n

Find a general term for the sequence whose first five terms are shown.

5,10,15,20,25,…

Solution

an=5n

Find a general term for the sequence whose first five terms are shown.

2,−4,8,−16,32,…
Solution
A mathematical sequence displaying numbers 2, -4, 8, -16, 32, ..., indicating a pattern of multiplying by -2 for each subsequent term.
A numerical sequence or index labeled 'n:' followed by the numbers 1, 2, 3, 4, 5, and then an ellipsis leading to '...n', indicating a sequence of integers from 1 to n.
We look for a pattern in the terms. A sequence of terms is shown: 2, -4, 8, -16, 32, ... This is a geometric progression where each term is multiplied by -2 to get the next term.
The numbers are powers of 2. The signs are
alternating, with even n negative.
A mathematical pattern illustrating an alternating series, with each term defined by the formula (-1)^(n+1) * 2^n, where n represents the term number starting from 1.
The general term of the sequence is an=(−1)n+12n.

Find a general term for the sequence whose first five terms are shown.

−3,9,−27,81,−243,…

Solution

an=(−1)n3n

Find a general term for the sequence whose first five terms are shown

1,−4,9,−16,25,…

Solution

an=(−1)n+1n2

Find a general term for the sequence whose first five terms are shown.

13,19,127,181,1243,…
Solution
A mathematical sequence showing fractions 1/3, 1/9, 1/27, 1/81, 1/243, indicating a geometric progression where each term is 1 divided by increasing powers of three.
A series of numbers for n: 1, 2, 3, 4, 5, as n continues.
We look for a pattern in the terms. Terms: 1/3, 1/9, 1/27, 1/81, 1/243... a geometric progression where each term is one third of the previous one.
The numerators are all 1. Mathematical pattern: a sequence of fractions where the numerator is 1 and the denominator is 3 raised to a positive integer power (1/3^n).
The denominators are powers of 3. The general term of the sequence is an=13n.

Find a general term for the sequence whose first five terms are shown.

12,14,18,116,132,…

Solution

an=12n

Find a general term for the sequence whose first five terms are shown.

11,14,19,116,125,…

Solution

an=1n2

Use Factorial Notation

Sequences often have terms that are products of consecutive integers. We indicate these products with a special notation called factorial notation. For example,5!, read 5 factorial, means 5·4·3·2·1. The exclamation point is not punctuation here; it indicates the factorial notation.

Factorial Notation

If n is a positive integer, then n! is

n!=n(n−1)(n−2)…

We define 0! as 1, so 0!=1.

The values of n! for the first 5 positive integers are shown.

1!2!3!4!5!12⋅13⋅2⋅14⋅3⋅2⋅15⋅4⋅3⋅2⋅112624120

Write the first five terms of the sequence whose general term is an=1n!.

Solution

We substitute the values 1, 2, 3, 4, 5 into the formula, an=1n!, in order.

This figure shows four rows and five columns. The first row reads, “nth term equals one divided by n factorial” written five times. The second row reads “a sub 1 equals one divided by 1 factorial, a sub 2 equals 1 divided by 2 factorial, a sub 3 equals 1 divided by 3 factorial, a sub 4 equals 1 divided by 4 factorial, a sub 5 equals 1 divided by 5 factorial”. The third row reads “a sub 1 equals 1 divided 1”, “a sub 2 equals 1 divided by 2 times g times 1”, “a sub 3 equals 1 divided by 3 times g times 2 g times 1”, “a sub 4 equals 1 divided 4 times g times 3 times g times 2 times g times 1”, “a sub 5 equals 1 divided by 5 g times 4 times g times 3 times g times 2 times g times 1”, “a sub 1 equals 1, a sub 2 equals one-half”, “a sub 3 equals one-sixth”, “a sub 4 equals 1 divided by 24”, “a sub 5 equals 1 divided by 120”.

The first five terms of the sequence are 1,12,16,124,1120.

Write the first five terms of the sequence whose general term is an=2n!.

Solution

2,1,13,112,160

Write the first five terms of the sequence whose general term is an=3n!.

Solution

3,32,12,18,140

When there is a fraction with factorials in the numerator and denominator, we line up the factors vertically to make our calculations easier.

Write the first five terms of the sequence whose general term is an=(n+1)!(n−1)!.

Solution

We substitute the values 1, 2, 3, 4, 5 into the formula, an=(n+1)!(n−1)!, in order.

This figure shows five columns and five rows. The first row shows the sequence “nth term equals n plus 1 times factorial divided by n minus 1 times factorial” written five times. The second row is “a sub 1 equals 1 plus 1 times factorial divided by 1 minus 1 times factorial”, “a sub 2 equals 2 plus 1 times factorial divided by 2 minus 1 times factorial”, “a sub 3 equals 3 plus 1 times factorial divided by 3 minus 1 times factorial”, “a sub 4 equals 4 plus 1 times factorial divided by 4 minus 1 times factorial”, “a sub 5 equals 5 plus 1 times factorial divided by 5 minus 1 times factorial”. The third row reads “a sub 1 equals 2 times factorial divided by 0 times factorial”, “a sub 2 equals 3 times factorial divided by 1 times factorial”, “a sub 3 equals 4 times factorial divided by 2 times factorial”, “a sub 3 equals 4 times factorial divided by 2 times factorial”, “a sub 4 equals 5 times factorial divided by 3 times factorial”, “a sub 5 equals 6 times factorial divided by 4 times factorial”. The fourth row reads, “a sub 1 equals 2 times g time 1 divided by 1”, “a sub 2 equals 3 times g times 2 times g times 1 divided by 1”, “a sub 3 equals 4 times g times 3 times g times 2 times g times 1 divided by 2 times g times 1”, “a sub 4 equals 5 times g times 4 times g times 3 times g times 2 times g times 1 divided by 3 g times 2 times g times 1”, and “a sub 5 equals 6 times g times 5 times g times 4 times g times 3 times g times 2 times g times 1 divided by 4 times g times 3 times g times 2 times g times 1”. The fifth row reads “a sub 1 equals 2”, “a sub 2 equals 6”, “a sub 3 equals 12”, “a sub 4 equals 20”, “a sub 5 equals 30”.

The first five terms of the sequence are 2, 6, 12, 20, and 30.

Write the first five terms of the sequence whose general term is an=(n−1)!(n+1)!.

Solution

12,16,112,120,130

Write the first five terms of the sequence whose general term is an=n!(n+1)!.

Solution

12,13,14,15,16

Find the Partial Sum

Sometimes in applications, rather than just list the terms, it is important for us to add the terms of a sequence. Rather than just connect the terms with plus signs, we can use summation notation.

For example, a1+a2+a3+a4+a5 can be written as ∑i=15ai. We read this as “the sum of a sub i from i equals one to five.” The symbol ∑ means to add and the i is the index of summation. The 1 tells us where to start (initial value) and the 5 tells us where to end (terminal value).

Summation Notation

The sum of the first n terms of a sequence whose nth term is an is written in summation notation as:

∑i=1nai=a1+a2+a3+a4+a5+…+an

The i is the index of summation and the 1 tells us where to start and the n tells us where to end.

When we add a finite number of terms, we call the sum a partial sum.

Expand the partial sum and find its value: ∑i=152i.

Solution
Step-by-step evaluation of the summation of 2i from i=1 to 5, detailing the expansion and calculation.
∑i=152i
We substitute the values 1, 2, 3, 4, 5 in order. 2·1+2·2+2·3+2·4+2·5
Simplify. 2+4+6+8+10
Add. 30
∑i=152i=30

Expand the partial sum and find its value: ∑i=153i.

Solution

45

Expand the partial sum and find its value: ∑i=154i.

Solution

60

The index does not always have to be i we can use any letter, but i and k are commonly used. The index does not have to start with 1 either—it can start and end with any positive integer.

Expand the partial sum and find its value: ∑k=031k!.

Solution
Step-by-step evaluation of the summation of 1/k! from k=0 to 3, demonstrating algebraic simplification to the final result of 8/3.
∑k=031k!
We substitute the values 0, 1, 2, 3, in order. 10!+11!+12!+13!
Evaluate the factorials. 11+11+12+16
Simplify. 1+1+36+16
Simplify. 166
Simplify. 83
∑k=031k!=83

Expand the partial sum and find its value: ∑k=032k!.

Solution

163

Expand the partial sum and find its value: ∑k=033k!.

Solution

8

Use Summation Notation to Write a Sum

In the last two examples, we went from summation notation to writing out the sum. Now we will start with a sum and change it to summation notation. This is very similar to finding the general term of a sequence. We will need to look at the terms and find a pattern. Often the patterns involve multiples or powers.

Write the sum using summation notation: 1+12+13+14+15.

Solution
Demonstrating the process of converting a series of fractions into summation notation.
1+12+13+14+15
n:1,2,3,4,5
We look for a pattern in the terms. Terms: 1,12,13,14,15
The numerators are all one. Pattern: 11,12,13,14,15,…1n
The denominators are the counting numbers
from one to five.
The sum written in summation notation is
1+12+13+14+15=∑n=151n.

Write the sum using summation notation: 12+14+18+116+132.

Solution

∑n=1512n

Write the sum using summation notation: 1+14+19+116+125.

Solution

∑n=151n2

When the terms of a sum have negative coefficients, we must carefully analyze the pattern of the signs.

Write the sum using summation notation: −1+8−27+64−125.

Solution
A mathematical expression showing the alternating sum of the first five cubes: -1 + 8 - 27 + 64 - 125.
The image displays the variable 'n:' followed by the numbers 1, 2, 3, 4, 5, arranged horizontally in a sequence.
We look for a pattern in the terms. A mathematical sequence labeled 'Terms:' shows the numbers -1, 8, -27, 64, -125, representing alternating positive and negative perfect cubes (1^3, 2^3, 3^3, 4^3, 5^3).
The signs of the terms alternate,
and the odd terms are negative.
The image shows a mathematical sequence: (-1)^n multiplied by n^3 for n = 1, 2, 3, 4, 5. This creates an alternating series of cubic numbers.
The numbers are the cubes of the
counting numbers from one to five.
A mathematical sequence displaying the pattern (-1)^n ו n^3 for n ranging from 1 to 5, with the cubed number shown in red.
The image displays a mathematical pattern: (-1)^n * n^3.
The sum written in summation notation is
−1+8−27+64−125=∑n=15(−1)n⋅n3

Write each sum using summation notation: 1−4+9−16+25.

Solution

∑n=15(−1)n+1n2

Write each sum using summation notation: −2+4−6+8−10.

Solution

∑n=15(−1)n2n

Access this online resource for additional instruction and practice with sequences.

  • Series and Sequences-Finding Patterns

Key Concepts

  • Factorial Notation
    If n is a positive integer, then n! is
    n!=n(n−1)(n−2)…(3)(2)(1).

    We define 0! as 1, so 0!=1
  • Summation Notation
    The sum of the first n terms of a sequence whose nth term an is written in summation notation as:
    ∑i=1nai=a1+a2+a3+a4+a5+…+an

    The i is the index of summation and the 1 tells us where to start and the n tells us where to end.

Practice Makes Perfect

Write the First Few Terms of a Sequence

In the following exercises, write the first five terms of the sequence whose general term is given.

an=2n−7

Solution

−5,−3,−1,1,3

an=5n−1

an=3n+1

Solution

4,7,10,13,16

an=4n+2

an=2n+3

Solution

5,7,11,19,35

an=3n−1

an=3n−2n

Solution

1,5,21,73,233

an=2n−3n

an=2nn2

Solution

2,1,89,1,3225

an=3nn3

an=4n−22n

Solution

1,32,54,78,916

an=3n+33n

an=(−1)n·2n

Solution

−2,4,−6,8,−10

an=(−1)n·3n

an=(−1)n+1n2

Solution

1,−4,9,−16,25

an=(−1)n+1n4

an=(−1)n+1n2

Solution

1,−14,19,−116,125

an=(−1)n+12n

Find a Formula for the General Term (nth Term) of a Sequence

In the following exercises, find a general term for the sequence whose first five terms are shown.

8,16,24,32,40,…

Solution

an=8n

7,14,21,28,35,…

6,7,8,9,10,…

Solution

an=n+5

−3,−2,−1,0,1,…

e3,e4,e5,e6,e7,…

Solution

an=en+2

1e2,1e,1,e,e2,…

−5,10,−15,20,−25,…

Solution

an=(−1)n5n

−6,11,−16,21,−26,…

−1,8,−27,64,−125,…

Solution

an=(−1)nn3

2,−5,10,−17,26,…

−2,4,−6,8,−10,…

Solution

an=(−1)n2n

1,−3,5,−7,9,…

14,116,164,1256,11,024,…

Solution

an=14n

11,18,127,164,1125,…

−12,−23,−34,−45,−56,…

Solution

an=−nn+1

−2,−32,−43,−54,−65,…

−52,−54,−58,−516,−532,…

Solution

an=−52n

4,12,427,464,4125,…

Use Factorial Notation

In the following exercises, using factorial notation, write the first five terms of the sequence whose general term is given.

an=4n!

Solution

4,2,23,16,130

an=5n!

an=3n!

Solution

3,6,18,72,360

an=2n!

an=(2n)!

Solution

2,24,720,40320,3628800

an=(3n)!

an=(n−1)!(n)!

Solution

1,12,13,14,15

an=n!(n+1)!

an=n!n2

Solution

1,12,23,32,245

an=n2n!

an=(n+1)!n2

Solution

2,32,83,152,1445

an=(n+1)!2n

Find the Partial Sum

In the following exercises, expand the partial sum and find its value.

∑i=15i2

Solution

1+4+9+16+25=55

∑i=15i3

∑i=16(2i+3)

Solution

5+7+9+11+13+15=60

∑i=16(3i−2)

∑i=142i

Solution

2+4+8+16=30

∑i=143i

∑k=034k!

Solution

41+41+42+46=323=1023

∑k=04−1k!

∑k=15k(k+1)

Solution

2+6+12+20+30=70

∑k=15k(2k−3)

∑n=15nn+1

Solution

12+23+34+45+56=7120

∑n=14nn+2

Use Summation Notation to write a Sum

In the following exercises, write each sum using summation notation.

13+19+127+181+1243

Solution

∑n=1513n

14+116+164+1256

1+18+127+164+1125

Solution

∑n=151n3

15+125+1125+1625

2+1+23+12+25

Solution

∑n=152n

3+32+1+34+35+12

3−6+9−12+15

Solution

∑n=15(−1)n+13n

−5+10−15+20−25

−2+4−6+8−10+…+20

Solution

∑n=110(−1)n2n

1−3+5−7+9+…+21

14+16+18+20+22+24+26

Solution

∑n=17(2n+12)

9+11+13+15+17+19+21

Writing Exercises

In your own words, explain how to write the terms of a sequence when you know the formula. Show an example to illustrate your explanation.

Solution

Answers will vary.

Which terms of the sequence are negative when the nth term of the sequence is an=(−1)n(n+2)?

In your own words, explain what is meant by n! Show some examples to illustrate your explanation.

Solution

Answers will vary.

Explain what each part of the notation ∑k=1122k means.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This figure shows a table with four columns and six rows. The first row is the header row and labels each column, “I can”, “Confidently”, “With some help”, and “No I don’t get it!”. The first row in the second column reads, “Write the first few terms of a sequence”, the third row, first column reads, “Find a Formula for the nth Term of a Sequence”, the fourth row first column reads “Use Factorial Notation, the fifth row, first column reads, Find the partial sum”, and the last row, first column reads, “Use Summation Notation to write a Sum”. The remaining three columns and rows are blank.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

finite sequence
A sequence with a domain that is limited to a finite number of counting numbers.
general term of a sequence
The general term of the sequence is the formula for writing the nth term of the sequence. The nth term of the sequence, an, is the term in the nth position where n is a value in the domain.
infinite sequence
A sequence whose domain is all counting numbers and there is an infinite number of counting numbers.
partial sum
When we add a finite number of terms of a sequence, we call the sum a partial sum.
sequence
A sequence is a function whose domain is the counting numbers.

Arithmetic Sequences

Learning Objectives

By the end of this section, you will be able to:

  • Determine if a sequence is arithmetic
  • Find the general term (nth term) of an arithmetic sequence
  • Find the sum of the first n terms of an arithmetic sequence

Before you get started, take this readiness quiz.

Evaluate 4n−1 for the integers 1, 2, 3, and 4.
If you missed this problem, review Example 6 in Use the Language of Algebra.

Solution

3, 7, 11, 15

Solve the system of equations: {x+y=73x+4y=23.
If you missed this problem, review Example 9 in Solve Systems of Linear Equations with Two Variables.

Solution

(5,2)

If f(n)=n2(3n+5), find f(1)+f(20).
If you missed this problem, review Example 8 in Relations and Functions.

Solution

654

Determine if a Sequence is Arithmetic

The last section introduced sequences and now we will look at two specific types of sequences that each have special properties. In this section we will look at arithmetic sequences and in the next section, geometric sequences.

An arithmetic sequence is a sequence where the difference between consecutive terms is constant. The difference between consecutive terms in an arithmetic sequence, an−an−1, is d, the common difference, for n greater than or equal to two.

Arithmetic Sequence

An arithmetic sequence is a sequence where the difference between consecutive terms is always the same.

The difference between consecutive terms, an−an−1, is d, the common difference, for n greater than or equal to two.

This figure has two rows and three columns. The first row reads “7”, “10”,”13”, “16”, “19”, “22”, and an ellipsis, “10 minus 7, divided by 3”, “13 minus 10, divided by 3”, “16 minus 13, divided by 3”, nth term equals nth term minus 1 divided by d”

In each of these sequences, the difference between consecutive terms is constant, and so the sequence is arithmetic.

Determine if each sequence is arithmetic. If so, indicate the common difference.

ⓐ 5,9,13,17,21,25,…

ⓑ 4,9,12,17,20,25,…

ⓒ 10,3,−4,−11,−18,−25,…

Solution

To determine if the sequence is arithmetic, we find the difference of the consecutive terms shown.

ⓐ
Example of calculating the common difference for a given arithmetic sequence.



Find the difference of
the consecutive terms.
5,9,13,1721,25,… 9−513−917−1321−1725−21 44444
The sequence is arithmetic. The common difference is d=4.

ⓑ
This table demonstrates how to determine if a sequence is arithmetic by calculating consecutive term differences, concluding that the example sequence is not arithmetic.



Find the difference of
the consecutive terms.
4,9,12,1720,25,… 9−412−917−1220−1725−20 53535
The sequence is not arithmetic as all the differences between
the consecutive terms are not the same.
There is no common difference.

ⓒ
Demonstration of finding the common difference for an arithmetic sequence by subtracting consecutive terms, using a specific numerical example.



Find the difference of
the consecutive terms.
10,3,−4,−11−18,−25,… 3−10−4−3−11−(−4)−18−(−11)−25−(−18) −7−7−7−7−7
The sequence is arithmetic. The common difference is d=−7.

Determine if each sequence is arithmetic. If so, indicate the common difference.

ⓐ 9,20,31,42,53,64,… ⓑ 12,6,0,−6,−12,−18,… ⓒ 7,1,10,4,13,7,…

Solution

ⓐ The sequence is arithmetic with common difference d=11. ⓑ The sequence is arithmetic with common difference d=−6.
ⓒ The sequence is not arithmetic as all the differences between the consecutive terms are not the same.

Determine if each sequence is arithmetic. If so, indicate the common difference.

ⓐ −4,4,2,10,8,16,… ⓑ −3,−1,1,3,5,7,… ⓒ 7,2,−3,−8,−13,−18,…

Solution

ⓐ The sequence is not arithmetic as all the differences between the consecutive terms are not the same. ⓑ The sequence is arithmetic with common difference d=2.
ⓒ The sequence is arithmetic with common difference d=−5.

If we know the first term, a1, and the common difference, d, we can list a finite number of terms of the sequence.

Write the first five terms of the sequence where the first term is 5 and the common difference is d=−6.

Solution

We start with the first term and add the common difference. Then we add the common difference to that result to get the next term, and so on.

a1a2a3a4a555+(−6)−1+(−6)−7+(−6)−13+(−6)−1−7−13−19

The sequence is 5,−1,−7,−13,−19,…

Write the first five terms of the sequence where the first term is 7 and the common difference is d=−4.

Solution

7,3,−1,−5,−9,…

Write the first five terms of the sequence where the first term is 11 and the common difference is d=−8.

Solution

11,3,−5,−13,−21,…

Find the General Term (nth Term) of an Arithmetic Sequence

Just as we found a formula for the general term of a sequence, we can also find a formula for the general term of an arithmetic sequence.

Let’s write the first few terms of a sequence where the first term is a1 and the common difference is d. We will then look for a pattern.

As we look for a pattern we see that each term starts with a1.

This figures shows an image of a sequence.

The first term adds 0d to the a1, the second term adds 1d, the third term adds 2d, the fourth term adds 3d, and the fifth term adds 4d. The number of ds that were added to a1 is one less than the number of the term. This leads us to the following

an=a1+(n−1)d

General Term (nth term) of an Arithmetic Sequence

The general term of an arithmetic sequence with first term a1 and the common difference d is

an=a1+(n−1)d

We will use this formula in the next example to find the 15th term of a sequence.

Find the fifteenth term of a sequence where the first term is 3 and the common difference is 6.

Solution
Calculation of the 15th term of an arithmetic sequence using the formula a_n = a_1 + (n-1)d.
To find the fifteenth term, a15, use the
formula with a1=3andd=6.
an=a1+(n−1)d
Substitute in the values. a15=3+(15−1)6
Simplify. a15=3+(14)6
a15=87

Find the twenty-seventh term of a sequence where the first term is 7 and the common difference is 9.

Solution

241

Find the eighteenth term of a sequence where the first term is 13 and the common difference is −7.

Solution

−106

Sometimes we do not know the first term and we must use other given information to find it before we find the requested term.

Find the twelfth term of a sequence where the seventh term is 10 and the common difference is −2. Give the formula for the general term.

Solution
This table illustrates the step-by-step calculation for finding the first, 12th, and general terms of an arithmetic sequence.
To first find the first term, a1, use the
formula with a7=10,n=7,andd=−2.
an=a1+(n−1)d
Substitute in the values. 10=a1+(7−1)(−2)
Simplify. 10=a1+(6)(−2)
10=a1−12
a1=22
Find the twelfth term, a12, using the
formula with a1=22,n=12,andd=−2.
an=a1+(n−1)d
Substitute in the values. a12=22+(12−1)(−2)
Simplify. a12=22+(11)(−2)
a12=0
The twelfth term of the sequence is 0, a12=0.
To find the general term, substitute
the values into the formula.
an=a1+(n−1)d
an=22+(n−1)(−2)
an=22−2n+2
The general term is an=−2n+24.

Find the eleventh term of a sequence where the ninth term is 8 and the common difference is −3. Give the formula for the general term.

Solution

a11=2. The general term is an=−3n+35.

Find the nineteenth term of a sequence where the fifth term is 1 and the common difference is −4. Give the formula for the general term.

Solution

a19=−55. The general term is an=−4n+21.

Sometimes the information given leads us to two equations in two unknowns. We then use our methods for solving systems of equations to find the values needed.

Find the first term and common difference of a sequence where the fifth term is 19 and the eleventh term is 37. Give the formula for the general term.

Solution

Since we know two terms, we can make a system of equations using the formula for the general term.

The formula for the nth term of an arithmetic sequence, expressed as a_n = a_1 + (n-1)d, where a_n is the nth term, a_1 is the first term, n is the term number, and d is the common difference.
We know the value of a5 and a11, so we will use n=5 and n=11. This image displays two equations representing specific terms of an arithmetic sequence. The first equation, a_5 = a_1 + (5 - 1)d, calculates the 5th term (a_5). The second equation, a_11 = a_1 + (11 - 1)d, calculates the 11th term (a_11). In both formulas, a_1 represents the first term of the sequence and d represents the common difference.
Substitute in the values, a5=19 and a11=37. A system of two linear equations, 19 = a_1 + (5 - 1)d and 37 = a_1 + (11 - 1)d, used to find the first term (a_1) and common difference (d) of an arithmetic sequence. Key numbers are highlighted in red.
Simplify. A system of two linear equations is presented, with the first equation being 19 = a1 + 4d and the second equation being 37 = a1 + 10d, enclosed by a brace on the left.
Prepare to eliminate the a1 term by multiplying the top equation by −1.
Add the equations.
This image illustrates the elimination method for solving a system of linear equations. By adding the two equations, 'a1' is eliminated, leading to 18 = 6d, and subsequently d = 3.

Substituting d=3 back into the first equation.

The equation 19 = a₁ + 4 • 3 is displayed, where the dot multiplication sign and the number 3 are colored red.
Solve for a1. A mathematical equation shown in two lines. The first line is 19 = a sub 1 + 12, and the second line is the solution, 7 = a sub 1.

Use the formula with a1=7 and d=3.

The image shows the formula for the nth term of an arithmetic progression, which is a_n = a_1 + (n-1)d.
Substitute in the values. The image shows the mathematical formula a_n = 7 + (n - 1)3, which represents an arithmetic sequence where a_n is the nth term, 7 is the first term, and 3 is the common difference.
Simplify. Two lines of mathematical equations showing the simplification of an algebraic expression: a_n = 7 + 3n - 3 is simplified to a_n = 3n + 4.
The first term is a1=7.
The common difference is d=3.

The general term of the sequence is an=3n+4.

Find the first term and common difference of a sequence where the fourth term is 17 and the thirteenth term is 53. Give the formula for the general term.

Solution

a1=5,d=4. The general term is an=4n+1.

Find the first term and common difference of a sequence where the third term is 2 and the twelfth term is −25. Give the formula for the general term.

Solution

a1=8,d=−3. The general term is an=−3n+11.

Find the Sum of the First n Terms of an Arithmetic Sequence

As with the general sequences, it is often useful to find the sum of an arithmetic sequence. The sum, Sn, of the first n terms of any arithmetic sequence is written as Sn=a1+a2+a3+...+an. To find the sum by merely adding all the terms can be tedious. So we can also develop a formula to find the sum of a sequence using the first and last term of the sequence.

We can develop this new formula by first writing the sum by starting with the first term, a1, and keep adding a d to get the next term as:

Sn=a1+(a1+d)+(a1+2d)+…+an.

We can also reverse the order of the terms and write the sum by starting with an and keep subtracting d to get the next term as

Sn=an+(an–d)+(an–2d)+…+a1.

If we add these two expressions for the sum of the first n terms of an arithmetic sequence, we can derive a formula for the sum of the first n terms of any arithmetic series.

Sn=a1+(a1+d)+(a1+2d)+…+an+Sn=an+(an−d)+(an−2d)+…+a1_________________________________________________________2Sn=(a1+an)+(a1+an)+(a1+an)+…+(a1+an)

Because there are n sums of (a1+an) on the right side of the equation, we rewrite the right side as n(a1+an).

2Sn=n(a1+an)

We divide by two to solve for Sn.

Sn=n2(a1+an)

This gives us a general formula for the sum of the first n terms of an arithmetic sequence.

Sum of the First n Terms of an Arithmetic Sequence

The sum, Sn, of the first n terms of an arithmetic sequence is

Sn=n2(a1+an)

where a1 is the first term and an is the nth term.

We apply this formula in the next example where the first few terms of the sequence are given.

Find the sum of the first 30 terms of the arithmetic sequence: 8, 13, 18, 23, 28, …

Solution

To find the sum, we will use the formula Sn=n2(a1+an). We know a1=8, d=5 and n=30, but we need to find an in order to use the sum formula.

Illustrates the step-by-step calculation of the nth term and the sum of the first n terms for an arithmetic sequence.
an=a1+(n−1)d
Findanwherea1=8,d=5andn=30. a30=8+(30−1)5
Simplify. a30=8+(29)5
a30=153
Knowinga1=8,n=30,anda30=153,
use the sum formula.
Sn=n2(a1+an)
Substitute in the values. S30=302(8+153)
Simplify. S30=15(161)
Simplify. S30=2,415

Find the sum of the first 30 terms of the arithmetic sequence: 5, 9, 13, 17, 21, …

Solution

1,890

Find the sum of the first 30 terms of the arithmetic sequence: 7, 10, 13, 16, 19, …

Solution

1,515

In the next example, we are given the general term for the sequence and are asked to find the sum of the first 50 terms.

Find the sum of the first 50 terms of the arithmetic sequence whose general term is an=3n−4.

Solution

To find the sum, we will use the formula Sn=n2(a1+an). We know n=50, but we need to find a1 and an in order to use the sum formula.

A mathematical equation is displayed on a white background: a subscript n equals 3n minus 4.
Find a1, by substituting n=1. The calculation of the first term, a_1, from the expression 3 * 1 - 4, resulting in a_1 = -1. The value '1' is highlighted in red, indicating its substitution.
Find an by substituting n=50. Mathematical expressions show the nth term of a sequence as a_n = 3n - 4, and its 50th term calculated as a_50 = 3 * 50 - 4, highlighting the substituted value 50.
Simplify. The mathematical equation a_50 = 146 is displayed on a white background, representing the 50th term of a sequence equaling 146.
Knowing n=50,a1=−1, and a50=146 use the sum formula. The formula for the sum of the first 'n' terms of an arithmetic sequence is shown: S_n = (n/2)(a_1 + a_n).
Substitute in the values. A mathematical equation for the sum of an arithmetic sequence, S_n = (50/2)(-1 + 146), which is S_n = 25 * 145.
Simplify. A mathematical equation shows S with a subscript 50, an equals sign, 25, and (145), indicating S_50 = 25(145).
Simplify. A mathematical equation S with a subscript 50, followed by an equals sign and the number 3,625, is displayed against a white background.

Find the sum of the first 50 terms of the arithmetic sequence whose general term is an=2n−5.

Solution

2,300

Find the sum of the first 50 terms of the arithmetic sequence whose general term is an=4n+3.

Solution

5,250

In the next example we are given the sum in summation notation. To add all the terms would be tedious, so we extract the information needed to use the formula to find the sum of the first n terms.

Find the sum: ∑i=125(4i+7).

Solution

To find the sum, we will use the formula Sn=n2(a1+an). We know n=25, but we need to find a1 and an in order to use the sum formula.

Expand the summation notation. A mathematical equation shows the expansion of a summation expression from i=1 to 25 for (4i + 7), illustrating the first few terms and the last term of the series.
Simplify. A mathematical equation showing the expansion of the summation of 4i + 7 from i=1 to 25, resulting in the series 11 + 15 + 19 + ... + 107.
Identify a1. A mathematical equation is displayed on a white background, reading 'a_1 = 11' in black text.
Identify a25. A mathematical expression shows 'a sub 25 equals 107' on a plain white background.
Knowing n=25,a1=11, and a25=107
use the sum formula.
A white background displays the mathematical formula S_n = n/2 (a_1 + a_n), which is used to calculate the sum of an arithmetic sequence, representing S sub n equals n over 2 times the quantity of a sub 1 plus a sub n.
Substitute in the values. A mathematical equation shows S_25 equals 25 divided by 2, multiplied by the sum of 11 and 107, representing the sum of an arithmetic series.
Simplify. A mathematical equation shows S subscript 25 equals 25 over 2 multiplied by 118, displayed in black text on a white background.
Simplify. The mathematical equation S₂₅ = 1,475 is displayed in black text against a plain white background, indicating a sum or a value for an indexed variable.

Find the sum: ∑i=130(6i−4).

Solution

2,670

Find the sum: ∑i=135(5i−3).

Solution

3,045

Access these online resources for additional instruction and practice with arithmetic sequences

  • Arithmetic Sequences
  • Arithmetic Sequences: A Formula for the ‘n-th’ Term
  • Arithmetic Series

Key Concepts

  • General Term (nth term) of an Arithmetic Sequence
    The general term of an arithmetic sequence with first term a1 and the common difference d is
    an=a1+(n−1)d
  • Sum of the First n Terms of an Arithmetic Sequence
    The sum, Sn, of the first n terms of an arithmetic sequence, where a1 is the first term and an is the nth term is
    Sn=n2(a1+an)

Practice Makes Perfect

Determine if a Sequence is Arithmetic

In the following exercises, determine if each sequence is arithmetic, and if so, indicate the common difference.

4,12,20,28,36,44,…

Solution

The sequence is arithmetic with common difference d=8.

−7,−2,3,8,13,18,…

−15,−16,3,12,21,30,…

Solution

The sequence is not arithmetic.

11,5,−1,−7−13,−19,…

8,5,2,−1,−4,−7,…

Solution

The sequence is arithmetic with common difference d=−3.

15,5,−5,−15,−25,−35,…

In the following exercises, write the first five terms of each sequence with the given first term and common difference.

a1=11 and d=7

Solution

11,18,25,32,39

a1=18 and d=9

a1=−7 and d=4

Solution

−7,−3,1,5,9

a1=−8 and d=5

a1=14 and d=−9

Solution

14,5,−4,−13,−22

a1=−3 and d=−3

Find the General Term (nth Term) of an Arithmetic Sequence

In the following exercises, find the term described using the information provided.

Find the twenty-first term of a sequence where the first term is three and the common difference is eight.

Solution

163

Find the twenty-third term of a sequence where the first term is six and the common difference is four.

Find the thirtieth term of a sequence where the first term is −14 and the common difference is five.

Solution

131

Find the fortieth term of a sequence where the first term is −19 and the common difference is seven.

Find the sixteenth term of a sequence where the first term is 11 and the common difference is −6.

Solution

−79

Find the fourteenth term of a sequence where the first term is eight and the common difference is −3.

Find the twentieth term of a sequence where the fifth term is −4 and the common difference is −2. Give the formula for the general term.

Solution

a20=−34. The general term is an=−2n+6.

Find the thirteenth term of a sequence where the sixth term is −1 and the common difference is −4. Give the formula for the general term.

Find the eleventh term of a sequence where the third term is 19 and the common difference is five. Give the formula for the general term.

Solution

a11=59. The general term is an=5n+4.

Find the fifteenth term of a sequence where the tenth term is 17 and the common difference is seven. Give the formula for the general term.

Find the eighth term of a sequence where the seventh term is −8 and the common difference is −5. Give the formula for the general term.

Solution

a8=−13. The general term is an=−5n+27.

Find the fifteenth term of a sequence where the tenth term is −11 and the common difference is −3. Give the formula for the general term.

In the following exercises, find the first term and common difference of the sequence with the given terms. Give the formula for the general term.

The second term is 14 and the thirteenth term is 47.

Solution

a1=11,d=3. The general term is an=3n+8.

The third term is 18 and the fourteenth term is 73.

The second term is 13 and the tenth term is −51.

Solution

a1=21,d=−8. The general term is an=−8n+29.

The third term is four and the tenth term is −38.

The fourth term is −6 and the fifteenth term is 27.

Solution

a1=−15,d=3. The general term is an=3n−18.

The third term is −13 and the seventeenth term is 15.

Find the Sum of the First n Terms of an Arithmetic Sequence

In the following exercises, find the sum of the first 30 terms of each arithmetic sequence.

11,14,17,20,23,…

Solution

1,635

12,18,24,30,36,…

8,5,2,−1,−4,…

Solution

−1,065

16,10,4,−2,−8,…

−17,−15,−13,−11,−9,…

Solution

360

−15,−12,−9,−6,−3,…

In the following exercises, find the sum of the first 50 terms of the arithmetic sequence whose general term is given.

an=5n−1

Solution

6,325

an=2n+7

an=−3n+5

Solution

–3,575

an=−4n+3

In the following exercises, find each sum.

∑i=140(8i−7)

Solution

6,280

∑i=145(7i−5)

∑i=150(3i+6)

Solution

4,125

∑i=125(4i+3)

∑i=135(−6i−2)

Solution

−3,850

∑i=130(−5i+1)

Writing Exercises

In your own words, explain how to determine whether a sequence is arithmetic.

Solution

Answers will vary.

In your own words, explain how the first two terms are used to find the tenth term. Show an example to illustrate your explanation.

In your own words, explain how to find the general term of an arithmetic sequence.

Solution

Answers will vary.

In your own words, explain how to find the sum of the first n terms of an arithmetic sequence without adding all the terms.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This figure shows a chart with four rows and four columns. The first row is the header row and reads, “I can”, “Confidently”, “With some help”, and “No, I don’t get it!” The column, beginning with second row reads 1. Determine if a Sequence is Arithmetic, 2. Find the General Term (nth term) of Arithmetic Sequence, and 3. Find the sum of the first Terms of an Arithmetic Sequence”. The remaining columns are blank.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

arithmetic sequence
An arithmetic sequence is a sequence where the difference between consecutive terms is constant.
common difference
The difference between consecutive terms in an arithmetic sequence, an−an−1, is d, the common difference, for n greater than or equal to two.

Geometric Sequences and Series

Learning Objectives

By the end of this section, you will be able to:

  • Determine if a sequence is geometric
  • Find the general term (nth term) of a geometric sequence
  • Find the sum of the first n terms of a geometric sequence
  • Find the sum of an infinite geometric series
  • Apply geometric sequences and series in the real world

Before you get started, take this readiness quiz.

Simplify: 2432.
If you missed this problem, review Example 1 in Fractions.

Solution

34

Evaluate: ⓐ 34 ⓑ (12)4.
If you missed this problem, review Example 8 in Integers.

Solution

ⓐ 81; ⓑ 116

If f(x)=4·3x, find ⓐ f(1) ⓑ f(2) ⓒ f(3).
If you missed this problem, review Example 8 in Relations and Functions.

Solution

ⓐ 12; ⓑ 36; ⓒ 108

Determine if a Sequence is Geometric

We are now ready to look at the second special type of sequence, the geometric sequence.

A sequence is called a geometric sequence if the ratio between consecutive terms is always the same. The ratio between consecutive terms in a geometric sequence is r, the common ratio, where n is greater than or equal to two.

Geometric Sequence

A geometric sequence is a sequence where the ratio between consecutive terms is always the same.

The ratio between consecutive terms, anan−1, is r, the common ratio. n is greater than or equal to two.

Consider these sequences.

This figure shows two sets of sequences where r is the common ratio.

Determine if each sequence is geometric. If so, indicate the common ratio.

ⓐ 4,8,16,32,64,128,…

ⓑ −2,6,−12,36,−72,216,…

ⓒ 27,9,3,1,13,19,…

Solution

To determine if the sequence is geometric, we find the ratio of the consecutive terms shown.

ⓐ
This table demonstrates finding the common ratio of a geometric sequence with a worked example.
Find the ratio of the consecutive terms. 4,8,16,32,64,128,…84168321664321286422222
The sequence is geometric. The common ratio is r=2.
ⓑ
This table demonstrates finding ratios of consecutive terms in a sequence to determine if it is geometric, concluding it lacks a common ratio.
Find the ratio of the consecutive terms. −2,6,−12,36,−72,216,… 6−2−12636−12−7236216−72 −3−2−3−2−3
The sequence is not geometric. There is no common ratio.
ⓒ
Analysis of a numerical sequence (27, 9, 3, 1, ...) to determine its common ratio (1/3) and confirm it as a geometric sequence.
27,9,3,1,13,19,…
Find the ratio of the consecutive terms. 927391313119131313131313
The sequence is geometric. The common ratio is r=13.

Determine if each sequence is geometric. If so indicate the common ratio.

ⓐ 7,21,63,189,567,1,701,…

ⓑ 64,16,4,1,14,116,…

ⓒ 2,4,12,48,240,1,440,…

Solution

ⓐ The sequence is geometric with common ratio r=3. ⓑ The sequence is geometric with common ratio r=14. ⓒ The sequence is not geometric. There is no common ratio.

Determine if each sequence is geometric. If so indicate the common ratio.

ⓐ −150,−30,−15,−5,−52,0,…

ⓑ 5,10,20,40,80,160,…

ⓒ 8,4,2,1,12,14,…

Solution

ⓐ The sequence is not geometric. There is no common ratio. ⓑ The sequence is geometric with common ratio r=2. ⓒ The sequence is geometric with common ratio r=12.

If we know the first term, a1, and the common ratio, r, we can list a finite number of terms of the sequence.

Write the first five terms of the sequence where the first term is 3 and the common ratio is r=−2.

Solution

We start with the first term and multiply it by the common ratio. Then we multiply that result by the common ratio to get the next term, and so on.

a1a2a3a4a5 33·(−2)−6·(−2)12·(−2)−24·(−2) −612−2448

The sequence is 3,−6,12,−24,48,…

Write the first five terms of the sequence where the first term is 7 and the common ratio is r=−3.

Solution

7,−21,63,−189,567

Write the first five terms of the sequence where the first term is 6 and the common ratio is r=−4.

Solution

6,−24,96,−384,1536

Find the General Term (nth Term) of a Geometric Sequence

Just as we found a formula for the general term of a sequence and an arithmetic sequence, we can also find a formula for the general term of a geometric sequence.

Let’s write the first few terms of the sequence where the first term is a1 and the common ratio is r. We will then look for a pattern.

This figure shows an image of a geometric sequence.

As we look for a pattern in the five terms above, we see that each of the terms starts with a1.

The first term, a1, is not multiplied by any r. In the second term, the a1 is multiplied by r. In the third term, the a1 is multiplied by r two times (r·r or r2). In the fourth term, the a1 is multiplied by r three times (r·r·r or r3) and in the fifth term, the a1 is multiplied by r four times. In each term, the number of times a1 is multiplied by r is one less than the number of the term. This leads us to the following

an=a1rn−1

General Term (nth term) of a Geometric Sequence

The general term of a geometric sequence with first term a1 and the common ratio r is

an=a1rn−1

We will use this formula in the next example to find the fourteenth term of a sequence.

Find the fourteenth term of a sequence where the first term is 64 and the common ratio is r=12.

Solution
Steps to find the 14th term (a14) of a geometric sequence where a1=64 and r=1/2.
To find the fourteenth term, a14,
use the formula with a1=64 and r=12.
an=a1rn−1
Substitute in the values. a14=64(12)14−1
Simplify. a14=64(12)13
a14=1128

Find the thirteenth term of a sequence where the first term is 81 and the common ratio is r=13.

Solution

16,561

Find the twelfth term of a sequence where the first term is 256 and the common ratio is r=14.

Solution

116,384

Sometimes we do not know the common ratio and we must use the given information to find it before we find the requested term.

Find the twelfth term of the sequence 3, 6, 12, 24, 48, 96, … Find the general term for the sequence.

Solution

To find the twelfth term, we use the formula, an=a1rn−1, and so we need to first determine a1 and the common ratio r.

Illustrates finding terms and the general formula for the geometric sequence 3, 6, 12... with first term 3 and common ratio 2.
3,6,12,24,48,96,…
The first term is three. a1=3
Find the common ratio. 6312624124824964822222
The common ratio is r=2.
To find the twelfth term, a12, use the
formula with a1=3andr=2.
an=a1rn−1
Substitute in the values. a12=3·212−1
Simplify. a12=3·211
a12=6,144
Find the general term. an=a1rn−1
We use the formula with a1=3andr=2. an=3(2)n−1

Find the ninth term of the sequence 6, 18, 54, 162, 486, 1,458, … Then find the general term for the sequence.

Solution

a9=39,366. The general term is an=6(3)n−1.

Find the eleventh term of the sequence 7, 14, 28, 56, 112, 224, … Then find the general term for the sequence.

Solution

a11=7,168. The general term is an=7(2)n−1.

Find the Sum of the First n Terms of a Geometric Sequence

We found the sum of both general sequences and arithmetic sequence. We will now do the same for geometric sequences. The sum,Sn, of the first n terms of a geometric sequence is written as Sn=a1+a2+a3+...+an. We can write this sum by starting with the first term, a1, and keep multiplying by r to get the next term as:

Sn=a1+a1r+a1r2+...+a1rn−1

Let’s also multiply both sides of the equation by r.

rSn=a1r+a1r2+a1r3+...+a1rn

Next, we subtract these equations. We will see that when we subtract, all but the first term of the top equation and the last term of the bottom equation subtract to zero.

This table illustrates the step-by-step derivation of the formula for the sum of the first n terms of a geometric series (Sn).
Sn=a1+a1r+a1r2+a1r3+…+a1rn−1rSn=a1r+a1r2+a1r3+…+a1rn−1+a1rn____________________________________________________Sn−rSn=a1−a1rn
We factor both sides. Sn(1−r)=a1(1−rn)
To obtain the formula for Sn,
divide both sides by (1−r).
Sn=a1(1−rn)1−r

Sum of the First n Terms of a Geometric Series

The sum, Sn, of the first n terms of a geometric sequence is

Sn=a1(1−rn)1−r

where a1 is the first term and r is the common ratio, and r is not equal to one.

We apply this formula in the next example where the first few terms of the sequence are given. Notice the sum of a geometric sequence typically gets very large when the common ratio is greater than one.

Find the sum of the first 20 terms of the geometric sequence 7, 14, 28, 56, 112, 224, …

Solution

To find the sum, we will use the formula Sn=a1(1−rn)1−r. We know a1=7,r=2, and n=20.

Steps and calculations for finding the sum of a geometric sequence given its first term, common ratio, and number of terms.
Knowing a1=7,r=2, and n=20,
use the sum formula.
Sn=a1(1−rn)1−r
Substitute in the values. S20=7(1−220)1−2
Simplify. S20=7,340,025

Find the sum of the first 20 terms of the geometric sequence 3, 6, 12, 24, 48, 96, …

Solution

3,145,725

Find the sum of the first 20 terms of the geometric sequence 6, 18, 54, 162, 486, 1,458, …

Solution

10,460,353,200

In the next example, we are given the sum in summation notation. While adding all the terms might be possible, most often it is easiest to use the formula to find the sum of the first n terms.

To use the formula, we need r. We can find it by writing out the first few terms of the sequence and find their ratio. Another option is to realize that in summation notation, a sequence is written in the form ∑i=1ka(r)i, where r is the common ratio.

Find the sum: ∑i=1152(3)i.

Solution

To find the sum, we will use the formula Sn=a1(1−rn)1−r, which requires a1 and r. We will write out a few of the terms, so we can get the needed information.

A mathematical summation expression: Sum from i equals 1 to 15 of 2 multiplied by 3 to the power of i.
Write out the first few terms. A mathematical sequence showing the terms 2*3^1, 2*3^2, 2*3^3 and their numerical values 6, 18, 54. This illustrates a geometric progression where each term is three times the previous one.
Identify a1. The mathematical equation 'a1 = 6' is displayed on a plain white background.

Find the common ratio.

Mathematical equations calculating the common ratio as 3, shown through fractions (18/6, 54/18) and represented in a summation formula, confirming 'The common ratio is r = 3.'

Knowing a1=6,r=3, and n=15,
use the sum formula.

The formula for the sum of the first n terms of a geometric sequence is displayed, S_n = a_1(1-r^n) / (1-r).
Substitute in the values. A mathematical formula for S subscript 15 equals 6 multiplied by the quantity 1 minus 3 to the power of 15, all divided by the quantity 1 minus 3.
Simplify. S_15 = 43,046,718

Find the sum: ∑i=1156(2)i.

Solution

393,204

Find the sum: ∑i=1105(2)i.

Solution

10,230

Find the Sum of an Infinite Geometric Series

If we take a geometric sequence and add the terms, we have a sum that is called a geometric series. An infinite geometric series is an infinite sum whose first term is a1 and common ratio is r and is written

a1+a1r+a1r2+…+a1rn−1+…

Infinite Geometric Series

An infinite geometric series is an infinite sum whose first term is a1 and common ratio is r and is written

a1+a1r+a1r2+…+a1rn−1+…

We know how to find the sum of the first n terms of a geometric series using the formula, Sn=a1(1−rn)1−r. But how do we find the sum of an infinite sum?

Let’s look at the infinite geometric series 3+6+12+24+48+96+…. Each term gets larger and larger so it makes sense that the sum of the infinite number of terms gets larger. Let’s look at a few partial sums for this series. We see a1=3 and r=2

Sn=a1(1−rn)1−rSn=a1(1−rn)1−rSn=a1(1−rn)1−r S10=3(1−210)1−2S30=3(1−230)1−2S50=3(1−250)1−2 S10=3,069S30=3,221,225,469S50≈3.38×1015

As n gets larger and larger, the sum gets larger and larger. This is true when |r|≥1 and we call the series divergent. We cannot find a sum of an infinite geometric series when |r|≥1.

Let’s look at an infinite geometric series whose common ratio is a fraction less than one,
12+14+18+116+132+164+…. Here the terms get smaller and smaller as n gets larger. Let’s look at a few finite sums for this series. We see a1=12 and r=12.

Sn=a1(1−rn)1−rSn=a1(1−rn)1−rSn=a1(1−rn)1−r S10=12(1−(12)10)1−12S20=12(1−(12)20)1−12S30=12(1−(12)30)1−12 S10≈.9990234375S20≈0.9999990463S30≈0.9999999991

Notice the sum gets larger and larger but also gets closer and closer to one. When |r|<1, the expression rn gets smaller and smaller. In this case, we call the series convergent. As n approaches infinity, (gets infinitely large), rn gets closer and closer to zero. In our sum formula, we can replace the rn with zero and then we get a formula for the sum, S, for an infinite geometric series when |r|<1.

Sn=a1(1−rn)1−rS=a1(1−0)1−rS=a11−r

This formula gives us the sum of the infinite geometric sequence. Notice the S does not have the subscript n as in Sn as we are not adding a finite number of terms.

Sum of an Infinite Geometric Series

For an infinite geometric series whose first term is a1 and common ratio r,

If|r|<1,the sum is

S=a11−r

If|r|≥1,the infinite geometric series does not have a sum. We say the series diverges.

Find the sum of the infinite geometric series 54+18+6+2+23+29+…

Solution

To find the sum, we first have to verify that the common ratio |r|<1 and then we can use the sum formula S=a11−r.

Steps for calculating the sum of an infinite geometric series with a1=54 and r=1/3.
Find the common ratio. r=1854r=618…
r=13r=13|r|<1
Identify a1. a1=54
Knowing a1=54,r=13,
use the sum formula.
S=a11−r
Substitute in the values. S=541−13
Simplify. S=81

Find the sum of the infinite geometric series 48+24+12+6+3+32+…

Solution

96

Find the sum of the infinite geometric series 64+16+4+1+14+116+…

Solution

2563

An interesting use of infinite geometric series is to write a repeating decimal as a fraction.

Write the repeating decimal 0.5– as a fraction.

Solution
Steps to convert a repeating decimal (0.5 recurring) to a fraction using the infinite geometric series formula.
Rewrite the 0.5– showing the repeating five. 0.5555555555555…
Use place value to rewrite this as a sum. 0.5+0.05+0.005+0.0005+...
This is an infinite geometric series.
Find the common ratio. r=0.050.5r=0.0050.05…
r=0.1r=0.1|r|<1
Identify a1. a1=0.5
Knowing a1=0.5,r=0.1,
use the sum formula.
S=a11−r
Substitute in the values. S=0.51−0.1
Simplify. S=0.50.9
Multiply numerator and denominator by 10. S=59
We are asked to find the fraction form. 0.5–=59

Write the repeating decimal 0.4– as a fraction.

Solution

49

Write the repeating decimal 0.8– as a fraction.

Solution

89

Apply Geometric Sequences and Series in the Real World

One application of geometric sequences has to do with consumer spending. If a tax rebate is given to each household, the effect on the economy is many times the amount of the individual rebate.

The government has decided to give a $1,000 tax rebate to each household in order to stimulate the economy. The government statistics say that each household will spend 80% of the rebate in goods and services. The businesses and individuals who benefitted from that 80% will then spend 80% of what they received and so on. The result is called the multiplier effect. What is the total effect of the rebate on the economy?

Solution

Every time money goes into the economy, 80% of it is spent and is then in the economy to be spent. Again, 80% of this money is spent in the economy again. This situation continues and so leads us to an infinite geometric series.

1000+1000(0.8)+1000(0.8)2+…

Here the first term is 1,000, a1=1000. The common ratio is 0.8,r=0.8. We can evaluate this sum since 0.8<1. We use the formula for the sum on an infinite geometric series.

Step-by-step calculation of the sum of an infinite geometric series.
S=a11−r
Substitute in the values, a1=1,000 and r=0.8. S=1,0001−0.8
Evaluate. S=5,000

The total effect of the $1,000 received by each household will be a $5,000 growth in the economy.

What is the total effect on the economy of a government tax rebate of $1,000 to each household in order to stimulate the economy if each household will spend 90% of the rebate in goods and services?

Solution

$10,000

What is the total effect on the economy of a government tax rebate of $500 to each household in order to stimulate the economy if each household will spend 85% of the rebate in goods and services?

Solution

$3,333.33

We have looked at a compound interest formula where a principal, P, is invested at an interest rate, r, for t years. The new balance, A, is A=P(1+rn)nt when interest is compounded n times a year. This formula applies when a lump sum was invested upfront and tells us the value after a certain time period.

An annuity is an investment that is a sequence of equal periodic deposits. We will be looking at annuities that pay the interest at the time of the deposits. As we develop the formula for the value of an annuity, we are going to let n=1. That means there is one deposit per year.

Simplification of the compound interest formula when the compounding frequency (n) is set to 1.
A=P(1+rn)nt
Let n=1. A=P(1+r1)1t
Simplify. A=P(1+r)t

Suppose P dollars is invested at the end of each year. One year later that deposit is worth P(1+r)1 dollars, and another year later it is worth P(1+r)2 dollars. After t years, it will be worth A=P(1+r)t dollars.

End of year 1 End of year 2 End of year 3
First Deposit P
@ end of year 1
P Amount 1 year later
P(1+r)1
Amount 2 years later
P(1+r)2
2nd Deposit P
@ end of year 2
P Amount 1 year later
P(1+r)1
3rd Deposit P
@ end of year 3
P

After three years, the value of the annuity is

P plus P times the quantity 1 plus r in parentheses, to the first power, plus P times the quantity 1 plus r, in parentheses, squared. This equals the money deposited at the end of year three, plus the money deposited at the end of year two, plus the money deposited at the end of year 1.

This a sum of the terms of a geometric sequence where the first term is P and the common ratio is 1+r. We substitute these values into the sum formula. Be careful, we have two different uses of r. The r in the sum formula is the common ratio of the sequence. In this case, that is 1+r where r is the interest rate.

Steps showing the derivation and simplification of a financial formula for S_t.
St=a1(1−rt)1−r
Substitute in the values. St=P(1−(1+r)t)1−(1+r)
Simplify. St=P(1−(1+r)t)−r
St=P((1+r)t−1)r

Remember our premise was that one deposit was made at the end of each year.

We can adapt this formula for n deposits made per year and the interest is compounded n times a year.

Value of an Annuity with Interest Compounded n Times a Year

For a principal, P, invested at the end of a compounding period, with an interest rate, r, which is compounded n times a year, the new balance, A, after t years, is

At=P((1+rn)nt−1)rn

New parents decide to invest $100 per month in an annuity for their baby daughter. The account will pay 5% interest per year which is compounded monthly. How much will be in the child’s account at her eighteenth birthday?

Solution

To find the Annuity formula, At=P((1+rn)nt−1)rn, we need to identify P, r, n, and t.

Step-by-step calculation of the future value of an annuity, showing variable definitions, formula application, and final investment outcome.
Identify P, the amount invested each month. P=100
Identify r, the annual interest rate, in decimal form. r=0.05
Identify n,
the number of times the deposit
will be made and the interest compounded
each year.
n=12
Identify t, the number of years. t=18
Knowing P=100,r=0.05,n=12and
t=18, use the sum formula.
At=P((1+rn)nt−1)rn
Substitute in the values. At=100((1+0.0512)12·18−1)0.0512
Use the calculator to evaluate. Be sure to
use parentheses as needed.
At=34,920.20
The child will have $34,920.20 when she turns
18.

New grandparents decide to invest $200 per month in an annuity for their grandson. The account will pay 5% interest per year which is compounded monthly. How much will be in the child’s account at his twenty-first birthday?

Solution

$88,868.36

Arturo just got his first full-time job after graduating from college at age 27. He decided to invest $200 per month in an IRA (an annuity). The interest on the annuity is 8%, which is compounded monthly. How much will be in the Arturo’s account when he retires at his sixty-seventh birthday?

Solution

$698,201.57

Access these online resources for additional instruction and practice with sequences.

  • Geometric Sequences
  • Geometric Series
  • Future Value Annuities and Geometric Series
  • Application of a Geometric Series: Tax Rebate

Key Concepts

  • General Term (nth term) of a Geometric Sequence: The general term of a geometric sequence with first term a1 and the common ratio r is
    an=a1rn−1
  • Sum of the First n Terms of a Geometric Series: The sum, Sn, of the n terms of a geometric sequence is
    Sn=a1(1−rn)1−r

    where a1 is the first term and r is the common ratio.
  • Infinite Geometric Series: An infinite geometric series is an infinite sum whose first term is a1 and common ratio is r and is written
    a1+a1r+a1r2+…+a1rn−1+…
  • Sum of an Infinite Geometric Series: For an infinite geometric series whose first term is a1 and common ratio r,
    If|r|<1,the sum is S=a11−r We say the series converges. If|r|≥1,the infinite geometric series does not have a sum. We say the series diverges.
  • Value of an Annuity with Interest Compounded n Times a Year: For a principal, P, invested at the end of a compounding period, with an interest rate, r, which is compounded n times a year, the new balance, A, after t years, is
    At=P((1+rn)nt−1)rn

Practice Makes Perfect

Determine if a Sequence is Geometric

In the following exercises, determine if the sequence is geometric, and if so, indicate the common ratio.

3,12,48,192,768,3072,…

Solution

The sequence is geometric with common ratio r=4.

2,10,50,250,1250,6250,…

72,36,18,9,92,94,…

Solution

The sequence is geometric with common ratio r=12.

54,18,6,2,23,29,…

−3,6,−12,24,−48,96,…

Solution

The sequence is geometric with a common ratio r=−2.

2,−6,18,−54,162,−486,…

In the following exercises, determine if each sequence is arithmetic, geometric or neither. If arithmetic, indicate the common difference. If geometric, indicate the common ratio.

48,24,12,6,3,32,…

Solution

The sequence is geometric with common ratio r=12.

12,6,0,−6,−12,−18,…

−7,−2,3,8,13,18,…

Solution

The sequence is arithmetic with common difference d=5.

5,9,13,17,21,25,…

12,14,18,116,132,164,…

Solution

The sequence is geometric with common ratio r=12.

4,8,12,24,48,96,…

In the following exercises, write the first five terms of each geometric sequence with the given first term and common ratio.

a1=4 and r=3

Solution

4,12,36,108,324

a1=9 and r=2

a1=−4 and r=−2

Solution

−4,8,−16,32,−64

a1=−5 and r=−3

a1=27 and r=13

Solution

27,9,3,1,13

a1=64 and r=14

Find the General Term (nth Term) of a Geometric Sequence

In the following exercises, find the indicated term of a sequence where the first term and the common ratio is given.

Find a11 given a1=8 and r=3.

Solution

472,392

Find a13 given a1=7 and r=2.

Find a10 given a1=−6 and r=−2.

Solution

3,072

Find a15 given a1=−4 and r=−3.

Find a10 given a1=100,000 and r=0.1.

Solution

0.0001

Find a8 given a1=1,000,000 and r=0.01.

In the following exercises, find the indicated term of the given sequence. Find the general term for the sequence.

Find a9 of the sequence, 9,18,36,72,144,288,…

Solution

a9=2,304. The general term is an=9(2)n−1.

Find a12 of the sequence, 5,15,45,135,405,1215,…

Find a15 of the sequence, −486,162,−54,18,−6,2,…

Solution

a15=−219,683. The general term is an=−486(−13)n−1.

Find a16 of the sequence, 224,−112,56,−28,14,−7,…

Find a10 of the sequence, 1,0.1,0.01,0.001,0.0001,0.00001,…

Solution

a10=0.000000001. The general term is an=(0.1)n−1.

Find a9 of the sequence, 1000,100,10,1,0.1,0.01,…

Find the Sum of the First n terms of a Geometric Sequence

In the following exercises, find the sum of the first fifteen terms of each geometric sequence.

8,24,72,216,648,1944,…

Solution

57,395,624

7,14,28,56,112,224,…

−6,12,−24,48,−96,192,…

Solution

−65,538

−4,12,−36,108,−324,972,…

81,27,9,3,1,13,…

Solution

7,174,45359,049≈121.5

256,64,16,4,1,14,116,…

In the following exercises, find the sum of the geometric sequence.

∑i=115(2)i

Solution

65,534

∑i=110(3)i

∑i=194(2)i

Solution

4088

∑i=185(3)i

∑i=1109(13)i

Solution

29,5246561≈4.5

∑i=1154(12)i

Find the Sum of an Infinite Geometric Series

In the following exercises, find the sum of each infinite geometric series.

1+13+19+127+181+1243+1729+…

Solution

32

1+12+14+18+116+132+164+…

6−2+23−29+227−281+…

Solution

92

−4+2−1+12−14+18−…

6+12+24+48+96+192+…

Solution

no sum as r≥1

5+15+45+135+405+1215+…

1,024+512+256+128+64+32+…

Solution

2,048

6,561+2187+729+243+81+27+…

In the following exercises, write each repeating decimal as a fraction.

0.3–

Solution

13

0.6–

0.7–

Solution

79

0.2–

0.45—

Solution

511

0.27—

Apply Geometric Sequences and Series in the Real World

In the following exercises, solve the problem.

Find the total effect on the economy of each government tax rebate to each household in order to stimulate the economy if each household will spend the indicated percent of the rebate in goods and services.

Tax rebate to each household Percent spent on goods and services Total Effect on the economy
ⓐ $1,000 85%
ⓑ $1,000 75%
ⓒ $1,500 90%
ⓓ $1,500 80%
Solution

ⓐ $6666.67 ⓑ $4000 ⓒ $15,000 ⓒ $7500

New grandparents decide to invest $100 per month in an annuity for their grandchild. The account will pay 6% interest per year which is compounded monthly (12 times a year). How much will be in the child’s account at their twenty-first birthday?

Berenice just got her first full-time job after graduating from college at age 30. She decided to invest $500 per quarter in an IRA (an annuity). The interest on the annuity is 7% which is compounded quarterly (4 times a year). How much will be in the Berenice’s account when she retires at age 65?

Solution

$295,581.88

Alice wants to purchase a home in about five years. She is depositing $500 a month into an annuity that earns 5% per year that is compounded monthly (12 times a year). How much will Alice have for her down payment in five years?

Myra just got her first full-time job after graduating from college. She plans to get a master’s degree, and so is depositing $2,500 a year from her year-end bonus into an annuity. The annuity pays 6.5% per year and is compounded yearly. How much will she have saved in five years to pursue her master’s degree?

Solution

$14,234.10

Writing Exercises

In your own words, explain how to determine whether a sequence is geometric.

In your own words, explain how to find the general term of a geometric sequence.

Solution

Answers will vary.

In your own words, explain the difference between a geometric sequence and a geometric series.

In your own words, explain how to determine if an infinite geometric series has a sum and how to find it.

Solution

Answers will vary.

Self Check

This figure shows seven rows and four columns. The first row is the header row and reads, “I can”, “Confidently”, “With some help”, and “No, I don’t get it. The first column reads, “Determine if a sequence is geometric”, “Find the general term (nth term) of a”, “Geometric sequence”, “Find the sum of an Infinite geometric series”, Use geometric sequences to solve applications”. The remaining columns are blank.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

annuity
An annuity is an investment that is a sequence of equal periodic deposits.
common ratio
The ratio between consecutive terms in a geometric sequence, anan−1, is r, the common ratio, where n is greater than or equal to two.
geometric sequence
A geometric sequence is a sequence where the ratio between consecutive terms is always the same
infinite geometric series
An infinite geometric series is an infinite sum infinite geometric sequence.

Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Use Pascal’s Triangle to expand a binomial
  • Evaluate a binomial coefficient
  • Use the Binomial Theorem to expand a binomial

Before you get started, take this readiness quiz.

Simplify: 7·6·5·44·3·2·1.
If you missed this problem, review Example 2 in Fractions.

Solution

35

Expand: (3x+5)2.
If you missed this problem, review Example 8 in Multiply Polynomials.

Solution

9x2+30x+25

Expand: (x−y)2.
If you missed this problem, review Example 8 in Multiply Polynomials.

Solution

x2−2xy+y2

Use Pascal’s Triangle to Expand a Binomial

In our previous work, we have squared binomials either by using FOIL or by using the Binomial Squares Pattern. We can also say that we expanded (a+b)2.

(a+b)2=a2+2ab+b2

To expand (a+b)3, we recognize that this is (a+b)2(a+b) and multiply.

(a+b)3 (a+b)2(a+b) (a2+2ab+b2)(a+b) a3+2a2b+ab2+a2b+2ab2+b3 a3+3a2b+3ab2+b3 (a+b)3=a3+3a2b+3ab2+b3

To find a method that is less tedious that will work for higher expansions like (a+b)7, we again look for patterns in some expansions.

Number of terms First term Last term
(a+b)1=a+b 2 a1 b1
(a+b)2=a2+2ab+b2 3 a2 b2
(a+b)3=a3+3a2b+3ab2+b3 4 a3 b3
(a+b)4=a4+4a3b+6a2b2+4ab3+b4 5 a4 b4
(a+b)5=a5+5a4b+10a3b2+10a2b3+5ab4+b5 6 a5 b5
(a+b)n +1 an bn

Notice the first and last terms show only one variable. Recall that a0=1, so we could rewrite the first and last terms to include both variables. For example, we could expand (a+b)3 to show each term with both variables.

This figure shows the pattern a plus b to the power of 3 equals a to a power of 3 times b to a power of 0 plus 3 times a to a power of 2 times b to a power of 1 plus 3 a to a power of 0 times b to a power of 3.

Generally, we don’t show the zero exponents, just as we usually write x rather than 1x.

Patterns in the expansion of (a+b)n

  • The number of terms is n+1.
  • The first term is an and the last term is bn.
  • The exponents on a decrease by one on each term going left to right.
  • The exponents on b increase by one on each term going left to right.
  • The sum of the exponents on any term is n.

Let’s look at an example to highlight the last three patterns.

This figure shows the pattern a plus b to the power of 5 equals a plus 5 times a times b plus 10 times a times b plus 5 times a times b plus b.

From the patterns we identified, we see the variables in the expansion of (a+b)n, would be

(a+b)n=an+___an−1b1+___an−2b2+...+___a1bn−1+bn.

To find the coefficients of the terms, we write our expansions again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients.

A plus b to the power of 0 equals 1. The top level of Pascal’s Triangle is 1. A plus b to the power of 1 equals 1 a plus 1 b. The second level of Pascal’s Triangle is 1, 1. A plus b to the power of 2 equals 1 a to the power of 2 plus 2 a b plus 1 b to the power of 2. The third level of Pascal’s Triangle is 1, 2, 1. A plus b to the power of 3 equals 1 a to the power of 3 plus 3 a to the power of 2 b plus 3 a b to the power of 2 plus 1 b to the power of 3. The fourth level of Pascal’s Triangle is 1,3,3,1. A plus b to the power of 4 equals 1 a to the power of 4 plus 4 a to the power of 3 b plus 6 a to the power of 2 b to the power of 2 plus 4 a b to the power of 3 plus 1 b to the power of 4. The fifth level of Pascal’s Triangle is 1, 4, 6, 4, 1. A plus b to the power of 5 equals 1 a to the power of 5 plus 5 a to the power of 4 b plus 10 a to the power of 3 b to the power of 2 plus 10 a to the power of 2 b to the power of 3. The sixth row of the Pascal’s Triangle is 1, 5, 10, 10, 5, 1.

The array to the right is called Pascal’s Triangle. Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers.

This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.

This triangle gives the coefficients of the terms when we expand binomials.

Pascal’s Triangle

This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.

In the next example, we will use this triangle and the patterns we recognized to expand the binomial.

Use Pascal’s Triangle to expand (x+y)6.

Solution

We know the variables for this expansion will follow the pattern we identified. The nonzero exponents of x will start at six and decrease to one. The nonzero exponents of y will start at one and increase to six. The sum of the exponents in each term will be six. In our pattern, a=x and b=y.

(a+b)n=an+___an−1b1+___an−2b2+...+___a1bn−1+bn(x+y)6=x6+___x5y1+___x4y2+___x3y3+___x2y4+___x1y5+y6


This figure shows a plus b to the power of n equals a to the power of n plus a to the power if n minus 1 b to the power of 1 plus a to the power of n minus 2 b to the power if 2 plus ellipsis plus a to the power of 1 b to the power of n minus 1 plus b to the power of n. The next figure shows x plus y to the power of 6 equals x to the power of 6 plus x to the power of 5 y to the power of 1 plus x to the power of 4 y to the power of 2 plus x to the power of 3 y to the power of 3 plus x to the power of 2 y to the power of 4 plus x to the power of 1 y to the power of 5 plus y to the power of 6.

Use Pascal’s Triangle to expand (x+y)5.

Solution

x5+5x4y+10x3y2+10x2y3
+5xy4+y5

Use Pascal’s Triangle to expand (p+q)7.

Solution

p7+7p6q+21p5q2+35p4q3
+35p3q4+21p2q5+7pq6+q7

In the next example we want to expand a binomial with one variable and one constant. We need to identify the a and b to carefully apply the pattern.

Use Pascal’s Triangle to expand (x+3)5.

Solution

We identify the a and b of the pattern.

This figure shows how we identify a plus b to the power of n, in the pattern x plus 3 to the power of 5.

In our pattern, a=x and b=3.

We know the variables for this expansion will follow the pattern we identified. The sum of the exponents in each term will be five.

(a+b)n=an+___an−1b1+___an−2b2+...+___a1bn−1+bn(x+3)5=x5+___x4·31+___x3·32+___x2·33+___x1·34+35


This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1. This figure shows X plus 3 to the power of 5 equals 1 x to the power of 5 g 3 x to the power of 4 plus 10 g 9 x to the power of 3 plus 10 g 27 x to the power of 2 plus 5 g 81 x to the power of 1 plus 1 g 243. Then, x plus 3 to the power of 5 equals x to the power of 5 plus 15 x to the power of 4 plus 90 x to the power of 3 plus 270 x to the power of 2 plus 405 plus 243.

Use Pascal’s Triangle to expand (x+2)4.

Solution

x4+8x3+24x2+32x+16

Use Pascal’s Triangle to expand (x+1)6.

Solution

x6+6x5+15x4+20x3+15x2
+6x+1

In the next example, the binomial is a difference and the first term has a constant times the variable. Once we identify the a and b of the pattern, we must once again carefully apply the pattern.

Use Pascal’s Triangle to expand (3x−2)4.

Solution

We identify the a and b of the pattern.

This figure shows how we identify a plus b to the power of n, in the pattern 3 x minus 2 to the power of 4.

In our pattern, a=3x and b=−2.

This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.


(a+b)n=an+___an−1b1+___an−2b2+...+___a1bn−1+bn(3x−2)4=1·(3x)4+4(3x)3(−2)1+6(3x)2(−2)2+4 (3x)1 (−2)3+1·(−2)4 (3x−2) 4 =81 x4+4(27 x3)(−2)+6(9 x2)(4)+4(3x)(−8)+1·16 (3x−2) 4 =81 x4−216 x3+216 x2−96x+16

Use Pascal’s Triangle to expand (2x−3)4.

Solution

16x4−96x3+216x2−216x+81

Use Pascal’s Triangle to expand (2x−1)6.

Solution

64x6−192x5+240x4−160x3
+60x2−12x+1

Evaluate a Binomial Coefficient

While Pascal’s Triangle is one method to expand a binomial, we will also look at another method. Before we get to that, we need to introduce some more factorial notation. This notation is not only used to expand binomials, but also in the study and use of probability.

To find the coefficients of the terms of expanded binomials, we will need to be able to evaluate the notation (nr) which is called a binomial coefficient. We read (nr) as “n choose r” or “n taken r at a time”.

Binomial Coefficient (nr)

A binomial coefficient (nr), where r and n are integers with 0≤r≤n, is defined as

(nr)=n!r!(n−r)!

We read (nr) as “n choose r” or “n taken r at a time”.

Evaluate: ⓐ (51) ⓑ (77) ⓒ (40) ⓓ (85).

Solution

ⓐ We will use the definition of a binomial coefficient, (nr)=n!r!(n−r)!.

This table demonstrates the step-by-step calculation of the binomial coefficient (5 choose 1) using its factorial definition.
(51)
Use the definition, (nr)=n!r!(n−r)!, where
n=5,r=1.
5!1!(5−1)!
Simplify. 5!1!(4)!
Rewrite 5!as5·4! 5·4!1!·4!
Simplify, by removing common factors. 5·4!1·4!
Simplify. 5
(51)=5

ⓑ

This table demonstrates the step-by-step calculation of the binomial coefficient (7 choose 7) using its combinatorial definition.
(77)
Use the definition, (nr)=n!r!(n−r)!, where
n=7,r=7.
7!7!(7−7)!
Simplify. 7!7!(0)!
Simplify. Remember 0!=1. 1
(77)=1

ⓒ

Step-by-step calculation of the binomial coefficient (4 choose 0) using the factorial definition.
(40)
Use the definition, (nr)=n!r!(n−r)!, where
n=4,r=0.
4!0!(4−0)!
Simplify. 4!0!(4)!
Simplify. 1
(40)=1

ⓓ

Step-by-step calculation of the binomial coefficient (8 choose 5) using the factorial formula.
(85)
Use the definition, (nr)=n!r!(n−r)!, where
n=8,r=5.
8!5!(8−5)!
Simplify. 8!5!(3)!
Rewrite 8!as8·7·6·5! and remove common factors. 8·7·6·5!5!·3·2·1
Simplify. 56
(85)=56

Evaluate each binomial coefficient:

ⓐ (61) ⓑ (88) ⓒ (50) ⓓ (73).

Solution

ⓐ 6 ⓑ 1 ⓒ 1 ⓓ 35

Evaluate each binomial coefficient:

ⓐ (21) ⓑ (1111) ⓒ (90) ⓓ (65).

Solution

ⓐ 2 ⓑ 1 ⓒ 1 ⓓ 6

In the previous example, parts (a), (b), (c) demonstrate some special properties of binomial coefficients.

Properties of Binomial Coefficients

(n1)=n(nn)=1(n0)=1

Use the Binomial Theorem to Expand a Binomial

We are now ready to use the alternate method of expanding binomials. The Binomial Theorem uses the same pattern for the variables, but uses the binomial coefficient for the coefficient of each term.

Binomial Theorem

For any real numbers a and b, and positive integer n,

(a+b)n=(n0)an+(n1)an−1b1+(n2)an−2b2+...+(nr)an−rbr+...+(nn)bn

Use the Binomial Theorem to expand (p+q)4.

Solution

We identify the a and b of the pattern.

This figure shows how we identify a plus b to the power of n, in the pattern p plus q to the power of 4.

In our pattern, a=p and b=q.

We use the Binomial Theorem.

(a+b)n=(n0)an+(n1)an−1b1+(n2)an−2b2+...+(nr)an−rbr+...+(nn)bn

Substitute in the values a=p,b=q and n=4.

(p+q)4=(40)p4+(41)p4−1q1+(42)p4−2q2+(43)p4−3q3+(44)q4

Simplify the exponents.

(p+q)4=(40)p4+(41)p3q+(42)p2q2+(43)pq3+(44)q4

Evaluate the coefficients. Remember, (n1)=n,(nn)=1,(n0)=1.

(p+q)4=1p4+4p3q1+4!2!(2)!p2q2+4!3!(4−3)!p1q3+1q4(p+q)4=p4+4p3q+6p2q2+4pq3+q4

Use the Binomial Theorem to expand (x+y)5.

Solution

x5+5x4y+10x3y2+10x2y3
+5xy4+y5

Use the Binomial Theorem to expand (m+n)6.

Solution

m6+6m5n+15m4n2+20m3n3
+15m2n4+6mn5+n6

Notice that when we expanded (p+q)4 in the last example, using the Binomial Theorem, we got the same coefficients we would get from using Pascal’s Triangle.

The figure above is P plus q to the power of 4 equals 4 choose 0 times p to the power of 4 plus 4 choose 1 times p to the power of 3 q plus 4 choose 2 times p to the power of 2 q to the power of 2 plus 4 choose 3 times p q to the power of 3 plus 4 choose 4 times q to the power of 4. P plus q to the power of 4 equals p to the power of 4 p to the power of 3 q plus 6 p to the power of 2 q to the power of 2 plus 4 p q to the power of 3 plus q to the power of 4. This figure on the right shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.

The next example, the binomial is a difference. When the binomial is a difference, we must be careful in identifying the values we will use in the pattern.

Use the Binomial Theorem to expand (x−2)5.

Solution

We identify the a and b of the pattern.

This figure shows x minus 2 to the power of 5.

In our pattern, a=x and b=−2.

We use the Binomial Theorem.

(a+b)n=(n0)an+(n1)an−1b1+(n2)an−2b2+...+(nr)an−rbr+...+(nn)bn

Substitute in the values a=x,b=−2, and n=5.

(x−2)5=(50)x5+(51)x5−1(−2)1+(52)x5−2(−2)2+(53)x5−3(−2)3+(54)x5−4(−2)4+(55)(−2)5

Simplify the exponents and evaluate the coefficients. Remember,(n1)=n,(nn)=1,(n0)=1.

(x−2)5=(50)x5+(51)x4(−2)+(52)x3(−2)2+(53)x2(−2)3+(54)x(−2)4+(55)(−2)5(x−2)5=1x5+5(−2)x4+5!2!·3!(−2)2x3+5!3!·2!(−2)3x2+5!4!·1!(−2)4x+1(−2)5(x−2)5=x5+5(−2)x4+10·4·x3+10(−8)x2+5·16·x+1(−32)(x−2)5=x5−10x4+40x3−80x2+80x−32

Use the Binomial Theorem to expand (x−3)5.

Solution

x5−15x4+90x3−270x2
+405x−243

Use the Binomial Theorem to expand (y−1)6.

Solution

y6−6y5+15y4−20y3+15y2
−6y+1

Things can get messy when both terms have a coefficient and a variable.

Use the Binomial Theorem to expand (2x−3y)4.

Solution

We identify the a and b of the pattern.

This figure shows how we identify a plus b to the power of n, in the pattern 2 x minus 3 y times the power of 4.

In our pattern, a=2x and b=−3y.

We use the Binomial Theorem.

(a+b)n=(n0)an+(n1)an−1b1+(n2)an−2b2+...+(nr)an−rbr+...+(nn)bn

Substitute in the values a=2x,b=−3y and n=4.

(2x−3y)4=(40)(2x)4+(41)(2x)4−1(−3y)1+(42)(2x)4−2(−3y)2+(43)(2x)4−3(−3y)3+(44)(−3y)4

Simplify the exponents.

(2x−3y)4=(40)(2x)4+(41)(2x)3(−3y)1+(42)(2x)2(−3y)2+(43)(2x)1(−3y)3+(44)(−3y)4

Evaluate the coefficients. Remember, (n1)=n,(nn)=1,(n0)=1.

(2x−3y)4=1(2x)4+4(2x)3(−3y)1+4!2!(2)!(2x)2(−3y)2+4!3!(4−3)!(2x)1(−3y)3+1(−3y)4

(2x−3y)4=16x4+4·8x3(−3y)+6(4x2)(9y2)+4(2x)(−27y3)+81y4

(2x−3y)4=16x4−96x3y+216x2y2−216xy3+81y4

Use the Binomial Theorem to expand (3x−2y)5.

Solution

243x5−810x4y+1080x3y2
−720x2y3+240xy4−32y5

Use the Binomial Theorem to expand (4x−3y)4.

Solution

256x4−768x3y+864x2y2
−432xy3+81y4

The real beauty of the Binomial Theorem is that it gives a formula for any particular term of the expansion without having to compute the whole sum. Let’s look for a pattern in the Binomial Theorem.

This figure shows a plus b to the power of n equals n choose 0 times a to the power of n b to the power of 0 plus n choose 1 times a to the power of n minus 1 b to the 1 plus n choose 2 times a to the power of n minus 2 b to the power of 2 plus ellipsis plus n choose r times a to the power of n minus r plus ellipsis plus n choose n times b to the power of n.

Notice, that in each case the exponent on the b is one less than the number of the term. The (r+1)st term is the term where the exponent of b is r. So we can use the format of the (r+1)st term to find the value of a specific term.

Find a Specific Term in a Binomial Expansion

The (r+1)st term in the expansion of (a+b)n is

(nr)an−rbr

Find the fourth term of (x+y)7.

Solution
In our pattern, n=7,a=x and b=y. Two binomial expressions are shown: (a + b) in red raised to the power of n, and (x + y) in black raised to the power of 7, illustrating algebraic notation.
We are looking for the fourth term.
Sincer+1=4,thenr=3.
Write the formula. The image shows a mathematical expression: the binomial coefficient 'n choose r' multiplied by 'a' raised to the power of 'n-r' and 'b' raised to the power of 'r', often seen in binomial expansion.
Substitute in the values, n=7,r=3,a=x, and b=y. A mathematical expression showing the binomial coefficient (7 choose 3) multiplied by a raised to the power of (7-3) and b cubed, representing a term from a binomial expansion.
A mathematical formula stating 'Use (n over r) = n! / (r!(n - r)!)', which defines the combination formula (n choose r). A mathematical expression featuring factorials, exponents, and variables is displayed against a white background. The expression reads: '7! / (3!4!) a^(7-3) b^3'.
Simplify. A mathematical expression: (7*6*5*4!)/(4!*3*2*1) * a^4b^3. Cancellation of 4! and (3*2*1) with 6 simplifies the fraction to 7*5, resulting in 35a^4b^3.
Simplify. The image displays the algebraic expression 35a^4b^3 on a white background, representing a monomial with a coefficient and two variables raised to positive integer exponents.

Find the third term of (x+y)6.

Solution

15x4y2

Find the fifth term of (a+b)8.

Solution

70a4b4

Find the coefficient of the x6 term of (x+3)9.

Solution
In our pattern, then n=9,a=x, and b=3. The image displays two binomial expressions: the general form (a+b)^n in red, and a specific example (x+3)^9 in black, illustrating the concept of binomial expansion.
We are looking for the coefficient of the x6 term.
    Since a=x, and x9−r=x6, we know r=3.
Write the formula. A mathematical expression featuring the binomial coefficient, represented as (n choose r), multiplied by a raised to the power of (n-r) and b raised to the power of r. This is a common term from binomial expansion.
Substitute in the values, n=9,r=3,a=x, and b=3. A mathematical expression featuring a binomial coefficient (9 choose 3), variable x raised to the power of (9-3), and the number 3 raised to the power of 3, all multiplied together.
The image displays the formula for combinations, written as 'Use (n over r) = n! / (r!(n-r)!)', which defines how to calculate the number of ways to choose r items from a set of n items without regard to the order of selection. A mathematical expression featuring a fraction with factorials (9!/(3!6!)), multiplied by X raised to the power of (theta-3), and then by 3 raised to the power of 3.
Simplify. The mathematical expression (9 * 8 * 7 * 6!) / (3! * 6!) multiplied by x^6 and 27, demonstrating algebraic and factorial notation.
Simplify. A mathematical expression featuring the numbers 84 and 27, multiplied by 'x' raised to the power of 6, on a white background.
Simplify. The mathematical expression 2268x^6 is displayed in a dark gray font against a plain white background, appearing as part of a larger equation or problem.
The coefficient of the x6 term is 2268.

Find the coefficient of the x5 term of (x+4)8.

Solution

3,584

Find the coefficient of the x4 term of (x+2)7.

Solution

280

Access these online resources for additional instruction and practice with sequences.

  • Binomial Expansion Using Pascal’s Triangle
  • Binomial Coefficients

Key Concepts

  • Patterns in the expansion of (a+b)n
    • The number of terms is n+1.
    • The first term is an and the last term is bn.
    • The exponents on a decrease by one on each term going left to right.
    • The exponents on b increase by one on each term going left to right.
    • The sum of the exponents on any term is n.
  • Pascal’s Triangle This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1
  • Binomial Coefficient (nr) : A binomial coefficient (nr), where r and n are integers with 0≤r≤n, is defined as
    (nr)=n!r!(n−r)!

    We read (nr) as “n choose r” or “n taken r at a time”.
  • Properties of Binomial Coefficients
    (n1)=n(nn)=1(n0)=1
  • Binomial Theorem: For any real numbers a, b, and positive integer n,
    (a+b)n=(n0)an+(n1)an−1b1+(n2)an−2b2+...+(nr)an−rbr+...+(nn)bn

Section Exercises

Practice Makes Perfect

Use Pascal’s Triangle to Expand a Binomial

In the following exercises, expand each binomial using Pascal’s Triangle.

(x+y)4

(a+b)8

Solution

a8+8a7b+28a6b2+56a5b3
+70a4b4+56a3b5+28a2b6
+8ab7+b8

(m+n)10

(p+q)9

Solution

p9+9p8q+36p7q2+84p6q3
+126p5q4+126p4q5+84p3q6
+36p2q7+9pq8+q9

(x−y)5

(a−b)6

Solution

a6−6a5b+15a4b2−20a3b3
+15a2b4−6ab5+b6

(x+4)4

(x+5)3

Solution

x3+15x2+75x+125

(y+2)5

(y+1)7

Solution

y7+7y6+21y5+35y4+35y3
+21y2+7y+1

(z−3)5

(z−2)6

Solution

z6−12z5+60z4−160z3+240z2
−192z+64

(4x−1)3

(3x−1)5

Solution

243x5−405x4+270x3−90x2
+15x−1

(3x−4)4

(3x−5)3

Solution

27x3−135x2+225x−125

(2x+3y)3

(3x+5y)3

Solution

27x3+135x2y+225xy2+125y3

Evaluate a Binomial Coefficient

In the following exercises, evaluate.

ⓐ (81) ⓑ (1010) ⓒ (60) ⓓ (93)

ⓐ (71) ⓑ (44) ⓒ (30) ⓓ (108)

Solution

ⓐ 7 ⓑ 1 ⓒ 1 ⓓ 45

ⓐ (31) ⓑ (99) ⓒ (70) ⓓ (53)

ⓐ (41) ⓑ (55) ⓒ (80) ⓓ (119)

Solution

ⓐ 4 ⓑ 1 ⓒ 1 ⓓ 55

Use the Binomial Theorem to Expand a Binomial

In the following exercises, expand each binomial.

(x+y)3

(m+n)5

Solution

m5+5m4n+10m3n2+10m2n3
+5mn4+n5

(a+b)6

(s+t)7

Solution

s7+7s6t+21s5t2+35s4t3
+35s3t4+21s2t5+7st6+t7

(x−2)4

(y−3)4

Solution

y4−12y3+54y2−108y+81

(p−1)5

(q−4)3

Solution

q3−12q2+48q−64

(3x−y)5

(5x−2y)4

Solution

625x4−1000x3y+600x2y2
−160xy3+16y4

(2x+5y)4

(3x+4y)5

Solution

243x5+1620x4y+4320x3y2
+5760x2y3+3840xy4+1024y5

In the following exercises, find the indicated term in the expansion of the binomial.

Sixth term of (x+y)10

Fifth term of (a+b)9

Solution

126a5b4

Fourth term of (x−y)8

Seventh term of (x−y)11

Solution

462x5y6

In the following exercises, find the coefficient of the indicated term in the expansion of the binomial.

y3 term of (y+5)4

x6 term of (x+2)8

Solution

112

x5 term of (x−4)6

x7 term of (x−3)9

Solution

324

a4b2 term of (2a+b)6

p5q4 term of (3p+q)9

Solution

30,618

Writing Exercises

In your own words explain how to find the rows of the Pascal’s Triangle. Write the first five rows of Pascal’s Triangle.

In your own words, explain the pattern of exponents for each variable in the expansion of.

Solution

Answers will vary.

In your own words, explain the difference between (a+b)n and (a−b)n.

In your own words, explain how to find a specific term in the expansion of a binomial without expanding the whole thing. Use an example to help explain.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This figure shows a table with four rows and four columns. The first row is the header row and reads. “I can”, “Confidently”, “With some help” and “No, I don’t get it”. The first column, beginning at the second row reads, “Use Pascal’s Triangle to Expand a Binomial”, “Evaluate a Binomial Coefficient” and “Use the Binomial Theorem to Expand a Binomial”. The remaining columns are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Chapter Review Exercises

Sequences

Write the First Few Terms of a Sequence

In the following exercises, write the first five terms of the sequence whose general term is given.

an=7n−5

an=3n+4

Solution

7,13,31,85,247

an=2n+n

an=2n+14n

Solution

34,516,764,9256,111024

an=(−1)nn2

Find a Formula for the General Term (nth Term) of a Sequence

In the following exercises, find a general term for the sequence whose first five terms are shown.

9,18,27,36,45,…

Solution

an=9n

−5,−4,−3,−2,−1,…

1e3,1e2,1e,1,e,…

Solution

an=en−4

1,−8,27,−64,125,…

−13,−12,−35,−23,−57,…

Solution

an=−nn+2

Use Factorial Notation

In the following exercises, using factorial notation, write the first five terms of the sequence whose general term is given.

an=4n!

an=n!(n+2)!

Solution

16,112,120,130,142

an=(n−1)!(n+1)2

Find the Partial Sum

In the following exercises, expand the partial sum and find its value.

∑i=17(2i−5)

Solution

−3+(−1)+1+3+5
+7+9=21

∑i=135i

∑k=044k!

Solution

4+4+2+23+16=656

∑k=14(k+1)(2k+1)

Use Summation Notation to write a Sum

In the following exercises, write each sum using summation notation.

−13+19−127+181−1243

Solution

∑n=15(−1)n13n

4−8+12−16+20−24

4+2+43+1+45

Solution

∑n=154n

Arithmetic Sequences

Determine if a Sequence is Arithmetic

In the following exercises, determine if each sequence is arithmetic, and if so, indicate the common difference.

1,2,4,8,16,32,…

−7,−1,5,11,17,23,…

Solution

The sequence is arithmetic with common difference d=6.

13,9,5,1,−3,−7,…

In the following exercises, write the first five terms of each arithmetic sequence with the given first term and common difference.

a1=5 and d=3

Solution

5,8,11,14,17

a1=8 and d=−2

a1=−13 and d=6

Solution

−13,−7,−1,5,11

Find the General Term (nth Term) of an Arithmetic Sequence

In the following exercises, find the term described using the information provided.

Find the twenty-fifth term of a sequence where the first term is five and the common difference is three.

Find the thirtieth term of a sequence where the first term is 16 and the common difference is −5.

Solution

−129

Find the seventeenth term of a sequence where the first term is −21 and the common difference is two.

In the following exercises, find the indicated term and give the formula for the general term.

Find the eighteenth term of a sequence where the fifth term is 12 and the common difference is seven.

Solution

a18=103. The general term is an=7n−23.

Find the twenty-first term of a sequence where the seventh term is 14 and the common difference is −3.

In the following exercises, find the first term and common difference of the sequence with the given terms. Give the formula for the general term.

The fifth term is 17 and the fourteenth term is 53.

Solution

a1=1,d=4. The general term is an=4n−3.

The third term is −26 and the sixteenth term is −91.

Find the Sum of the First n Terms of an Arithmetic Sequence

In the following exercises, find the sum of the first 30 terms of each arithmetic sequence.

7,4,1,−2,−5,…

Solution

−1,095

1,6,11,16,21,…

In the following exercises, find the sum of the first fifteen terms of the arithmetic sequence whose general term is given.

an=4n+7

Solution

585

an=−2n+19

In the following exercises, find each sum.

∑i=150(4i−5)

Solution

4,850

∑i=130(−3i−7)

∑i=135(i+10)

Solution

980

Geometric Sequences and Series

Determine if a Sequence is Geometric

In the following exercises, determine if the sequence is geometric, and if so, indicate the common ratio.

3,12,48,192,768,3072,…

5,10,15,20,25,30,…

Solution

The sequence is not geometric.

112,56,28,14,7,72,…

9,−18,36,−72,144,−288,…

Solution

The sequence is geometric with common ratio r=−2.

In the following exercises, write the first five terms of each geometric sequence with the given first term and common ratio.

a1=−3 and r=5

a1=128 and r=14

Solution

128,32,8,2,12

a1=5 and r=−3

Find the General Term (nth Term) of a Geometric Sequence

In the following exercises, find the indicated term of a sequence where the first term and the common ratio is given.

Find a9 given a1=6 and r=2.

Solution

1,536

Find a11 given a1=10,000,000 and r=0.1.

In the following exercises, find the indicated term of the given sequence. Find the general term of the sequence.

Find a12 of the sequence, 6,−24,96,−384,1536,−6144,…

Solution

a12=−25,165,824. The general term is an=6(−4)n−1.

Find a9 of the sequence, 4374,1458,486,162,54,18,…

Find the Sum of the First n terms of a Geometric Sequence

In the following exercises, find the sum of the first fifteen terms of each geometric sequence.

−4,8,−16,32,−64,128…

Solution

−43,692

3,12,48,192,768,3072…

3125,625,125,25,5,1…

Solution

3906.25

In the following exercises, find the sum

∑i=187(3)i

∑i=1624(12)i

Solution

1898=23.625

Find the Sum of an Infinite Geometric Series

In the following exercises, find the sum of each infinite geometric series.

1−13+19−127+181−1243+1729−…

49+7+1+17+149+1343+…

Solution

3436≈57.167

In the following exercises, write each repeating decimal as a fraction.

0.8–

0.36—

Solution

411

Apply Geometric Sequences and Series in the Real World

In the following exercises, solve the problem.

What is the total effect on the economy of a government tax rebate of $360 to each household in order to stimulate the economy if each household will spend 60% of the rebate in goods and services?

Adam just got his first full-time job after graduating from high school at age 17. He decided to invest $300 per month in an IRA (an annuity). The interest on the annuity is 7% which is compounded monthly. How much will be in Adam’s account when he retires at his sixty-seventh birthday?

Solution

$1,634,421.27

Binomial Theorem

Use Pascal’s Triangle to Expand a Binomial

In the following exercises, expand each binomial using Pascal’s Triangle.

(a+b)7

(x−y)4

Solution

x4−4x3y+6x2y2−4xy3+y4

(x+6)3

(2y−3)5

Solution

32y5−240y4+720y3−1080y2
+810y−243

(7x+2y)3

Evaluate a Binomial Coefficient

In the following exercises, evaluate.


ⓐ (111)
ⓑ (1212)
ⓒ (130)
ⓓ (83)

Solution

ⓐ 11 ⓑ 1 ⓒ 1 ⓓ 56


ⓐ (71)
ⓑ (55)
ⓒ (90)
ⓓ (95)


ⓐ (11)
ⓑ (1515)
ⓒ (40)
ⓓ (112)

Solution

ⓐ 1 ⓑ 1 ⓒ 1 ⓓ 55

Use the Binomial Theorem to Expand a Binomial

In the following exercises, expand each binomial, using the Binomial Theorem.

(p+q)6

(t−1)9

Solution

t9−9t8+36t7−84t6+126t5
−126t4+84t3−36t2+9t−1

(2x+1)4

(4x+3y)4

Solution

256x4+768x3y+864x2y2
+432xy3+81y4

(x−3y)5

In the following exercises, find the indicated term in the expansion of the binomial.

Seventh term of (a+b)9

Solution

84a3b6

Third term of (x−y)7

In the following exercises, find the coefficient of the indicated term in the expansion of the binomial.

y4 term of (y+3)6

Solution

135

x5 term of (x−2)8

a3b4 term of (2a+b)7

Solution

280

Practice Test

In the following exercises, write the first five terms of the sequence whose general term is given.

an=5n−33n

an=(n+2)!(n+3)!

Solution

14,15,16,17,18

Find a general term for the sequence, −23,−45,−67,−89,−1011,…

Expand the partial sum and find its value. ∑i=14(−4)i

Solution

−4+16−64+256=204

Write the following using summation notation. −1+14−19+116−125

Write the first five terms of the arithmetic sequence with the given first term and common difference. a1=−13 and d=3

Solution

−13,−10,−7,−4,−1

Find the twentieth term of an arithmetic sequence where the first term is two and the common difference is −7.

Find the twenty-third term of an arithmetic sequence whose seventh term is 11 and common difference is three. Then find a formula for the general term.

Solution

a23=59. The general term is an=3n−10.

Find the first term and common difference of an arithmetic sequence whose ninth term is −1 and the sixteenth term is −15. Then find a formula for the general term.

Find the sum of the first 25 terms of the arithmetic sequence, 5,9,13,17,21,…

Solution

1,325

Find the sum of the first 50 terms of the arithmetic sequence whose general term is an=−3n+100.

Find the sum. ∑i=140(5i−21)

Solution

3,260

In the following exercises, determine if the sequence is arithmetic, geometric, or neither. If arithmetic, then find the common difference. If geometric, then find the common ratio.

14,3,−8,−19,−30,−41,…

324,108,36,12,4,43,…

Solution

The sequence is geometric with common ratio r=13.

Write the first five terms of the geometric sequence with the given first term and common ratio. a1=6 and r=−2

In the geometric sequence whose first term and common ratio are a1=5 and r=4, find a11.

Solution

5,242,880

Find a10 of the geometric sequence, 1250,250,50,10,2,25,…. Then find a formula for the general term.

Find the sum of the first thirteen terms of the geometric sequence, 2,−6,18,−54,162,−486…

Solution

797,162

In the following exercises, find the sum.

∑i=195(2)i

1−15+125−1125+1625−13125+…

Solution

56

Write the repeating decimal as a fraction. 0.81—

Dave just got his first full-time job after graduating from high school at age 18. He decided to invest $450 per month in an IRA (an annuity). The interest on the annuity is 6% which is compounded monthly. How much will be in Adam’s account when he retires at his sixty-fifth birthday?

Solution

$1,409,344.19

Expand the binomial using Pascal’s Triangle. (m−2n)5

Evaluate each binomial coefficient. ⓐ (81)
ⓑ (1616) ⓒ (120) ⓓ (106)

Solution

ⓐ 8 ⓑ 1 ⓒ 1 ⓓ 210

Expand the binomial using the Binomial Theorem. (4x+5y)3