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Prealgebra 2e

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Contents

  1. Preface
  2. Whole Numbers
    1. Introduction
    2. Introduction to Whole Numbers
    3. Add Whole Numbers
    4. Subtract Whole Numbers
    5. Multiply Whole Numbers
    6. Divide Whole Numbers
  3. The Language of Algebra
    1. Introduction to the Language of Algebra
    2. Use the Language of Algebra
    3. Evaluate, Simplify, and Translate Expressions
    4. Solving Equations Using the Subtraction and Addition Properties of Equality
    5. Find Multiples and Factors
    6. Prime Factorization and the Least Common Multiple
  4. Integers
    1. Introduction to Integers
    2. Introduction to Integers
    3. Add Integers
    4. Subtract Integers
    5. Multiply and Divide Integers
    6. Solve Equations Using Integers; The Division Property of Equality
  5. Fractions
    1. Introduction to Fractions
    2. Visualize Fractions
    3. Multiply and Divide Fractions
    4. Multiply and Divide Mixed Numbers and Complex Fractions
    5. Add and Subtract Fractions with Common Denominators
    6. Add and Subtract Fractions with Different Denominators
    7. Add and Subtract Mixed Numbers
    8. Solve Equations with Fractions
  6. Decimals
    1. Introduction to Decimals
    2. Decimals
    3. Decimal Operations
    4. Decimals and Fractions
    5. Solve Equations with Decimals
    6. Averages and Probability
    7. Ratios and Rate
    8. Simplify and Use Square Roots
  7. Percents
    1. Introduction to Percents
    2. Understand Percent
    3. Solve General Applications of Percent
    4. Solve Sales Tax, Commission, and Discount Applications
    5. Solve Simple Interest Applications
    6. Solve Proportions and their Applications
  8. The Properties of Real Numbers
    1. Introduction to the Properties of Real Numbers
    2. Rational and Irrational Numbers
    3. Commutative and Associative Properties
    4. Distributive Property
    5. Properties of Identity, Inverses, and Zero
    6. Systems of Measurement
  9. Solving Linear Equations
    1. Introduction to Solving Linear Equations
    2. Solve Equations Using the Subtraction and Addition Properties of Equality
    3. Solve Equations Using the Division and Multiplication Properties of Equality
    4. Solve Equations with Variables and Constants on Both Sides
    5. Solve Equations with Fraction or Decimal Coefficients
  10. Math Models and Geometry
    1. Introduction
    2. Use a Problem Solving Strategy
    3. Solve Money Applications
    4. Use Properties of Angles, Triangles, and the Pythagorean Theorem
    5. Use Properties of Rectangles, Triangles, and Trapezoids
    6. Solve Geometry Applications: Circles and Irregular Figures
    7. Solve Geometry Applications: Volume and Surface Area
    8. Solve a Formula for a Specific Variable
  11. Polynomials
    1. Introduction to Polynomials
    2. Add and Subtract Polynomials
    3. Use Multiplication Properties of Exponents
    4. Multiply Polynomials
    5. Divide Monomials
    6. Integer Exponents and Scientific Notation
    7. Introduction to Factoring Polynomials
  12. Graphs
    1. Graphs
    2. Use the Rectangular Coordinate System
    3. Graphing Linear Equations
    4. Graphing with Intercepts
    5. Understand Slope of a Line
  13. Cumulative Review
  14. Powers and Roots Tables
  15. Geometric Formulas

Preface

Welcome to Prealgebra 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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About OpenStax Resources

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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 Prealgebra 2e

Prealgebra 2e is designed to meet scope and sequence requirements for a one-semester prealgebra course. The text introduces 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.

Students who are taking basic mathematics and prealgebra classes in college present a unique set of challenges. Many students in these classes have been unsuccessful in their prior math classes. They may think they know some math, but their core knowledge is full of holes. Furthermore, these students need to learn much more than the course content. They need to learn study skills, time management, and how to deal with math anxiety. Some students lack basic reading and arithmetic skills. The organization of Prealgebra makes it easy to adapt the book to suit a variety of course syllabi.

Coverage and Scope

Prealgebra 2e takes a student-support approach in its presentation of the content. The beginning, in particular, is presented as a sequence of small steps so that students gain confidence in their ability to succeed in the course. 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: Whole Numbers

    Each of the four basic operations with whole numbers—addition, subtraction, multiplication, and division—is modeled and explained. As each operation is covered, discussions of algebraic notation and operation signs, translation of algebraic expressions into word phrases, and the use of the operation in applications are included.

  • Chapter 2: The Language of Algebra

    Mathematical vocabulary as it applies to the whole numbers is presented. The use of variables, which distinguishes algebra from arithmetic, is introduced early in the chapter, and the development of and practice with arithmetic concepts use variables as well as numeric expressions. In addition, the difference between expressions and equations is discussed, word problems are introduced, and the process for solving one-step equations is modeled.

  • Chapter 3: Integers

    While introducing the basic operations with negative numbers, students continue to practice simplifying, evaluating, and translating algebraic expressions. The Division Property of Equality is introduced and used to solve one-step equations.

  • Chapter 4: Fractions

    Fraction circles and bars are used to help make fractions real and to develop operations on them. Students continue simplifying and evaluating algebraic expressions with fractions, and learn to use the Multiplication Property of Equality to solve equations involving fractions.

  • Chapter 5: Decimals

    Basic operations with decimals are presented, as well as methods for converting fractions to decimals and vice versa. Averages and probability, unit rates and unit prices, and square roots are included to provide opportunities to use and round decimals.

  • Chapter 6: Percents

    Conversions among percents, fractions, and decimals are explored. Applications of percent include calculating sales tax, commission, and simple interest. Proportions and solving percent equations as proportions are addressed as well.

  • Chapter 7: The Properties of Real Numbers

    The properties of real numbers are introduced and applied as a culmination of the work done thus far, and to prepare students for the upcoming chapters on equations, polynomials, and graphing.

  • Chapter 8: Solving Linear Equations

    A gradual build-up to solving multi-step equations is presented. Problems involve solving equations with constants on both sides, variables on both sides, variables and constants on both sides, and fraction and decimal coefficients.

  • Chapter 9: Math Models and Geometry

    The chapter begins with opportunities to solve “traditional” number, coin, and mixture problems. Geometry sections cover the properties of triangles, rectangles, trapezoids, circles, irregular figures, the Pythagorean Theorem, and volumes and surface areas of solids. Distance-rate-time problems and formulas are included as well.

  • Chapter 10: Polynomials

    Adding and subtracting polynomials is presented as an extension of prior work on combining like terms. Integer exponents are defined and then applied to scientific notation. The chapter concludes with a brief introduction to factoring polynomials.

  • Chapter 11: Graphs

    This chapter is placed last so that all of the algebra with one variable is completed before working with linear equations in two variables. Examples progress from plotting points to graphing lines by making a table of solutions to an equation. Properties of vertical and horizontal lines and intercepts are included. Graphing linear equations at the end of the course gives students a good opportunity to review evaluating expressions and solving equations.

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 Prealgebra 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

Learning Objectives

Each chapter is divided into multiple sections (or modules), each of which is organized around a set of learning objectives. The learning objectives are listed explicitly at the beginning of each section and are the focal point of every instructional element.

Narrative text

Narrative text is used to introduce key concepts, terms, and definitions, to provide real-world context, and to provide transitions between topics and examples. An informal voice was used to make the content accessible to students.

Throughout this book, we rely on a few basic conventions to highlight the most important ideas:
  • Key terms are boldfaced, typically when first introduced and/or when formally defined.
  • Key concepts and definitions are called out in a blue box for easy reference.

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 in order 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.

Figures

Prealgebra 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.

A diagram depicting the removal of selected (circled) groups of three yellow dots from two panels, leaving an envelope in one and five dots in the other, after the operation.

Supporting Features

Four small but important features serve to support Examples:

Be Prepared!

Each section, beginning with Section 1.2, 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.

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. A “How To” is a list of steps necessary to solve a certain type of problem. A "How To" typically precedes an Example.

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'. A “Try It” exercise immediately follows an Example, providing the student with an immediate opportunity to solve a similar problem. In the PDF and the Web View version of the text, answers to the Try It exercises are located in the Answer Key.

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. The “Media” icon appears at the conclusion of each section, just prior to the Section Exercises. This icon marks a list of links to online video tutorials that reinforce the concepts and skills introduced in the section.

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

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 math concepts from that particular section.
  • Writing Exercises are included in every Exercise Set to encourage conceptual understanding, critical thinking, and literacy.

Chapter Review Features

The end of each chapter includes a review of the most important takeaways, as well as additional practice problems that students can use to prepare for exams.

  • Key Terms provides a formal definition for each bold-faced term in the chapter.
  • Key Concepts summarizes the most important ideas introduced in each section, linking back to the relevant Example(s) in case students need to review.
  • Chapter Review Exercises includes 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

All answers to Try It questions are provided in the Answer Key. Answers to Examples are provided directly below the question. Students can find odd-numbered answers to Chapter Review Exercises, Practice Test, and Section Exercises in the Answer Key. Answers to all odd and even-numbered questions 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, Links to Literacy assignments, and an answer key. 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 and MaryAnne Anthony-Smith taught mathematics at Santa Ana College for many years and have worked together on several projects aimed at improving student learning in developmental math courses. They are the authors of Strategies for Success: Study Skills for the College Math Student, published by Pearson HigherEd.

Lynn Marecek, Santa Ana College

MaryAnne Anthony-Smith, Santa Ana College

Andrea Honeycutt Mathis, Northeast Mississippi Community College

Reviewers

Tony Ayers, Collin College Preston Ridge Campus
David Behrman, Somerset Community College
Brandie Biddy, Cecil College
Bryan Blount, Kentucky Wesleyan College
Steven Boettcher, Estrella Mountain Community College
Kimberlyn Brooks, Cuyahoga Community College
Pamela Burleson, Lone Star College University Park
Tamara Carter, Texas A&M University
Phil Clark, Scottsdale Community College
Christina Cornejo, Erie Community College
Denise Cutler, Bay de Noc Community College
Richard Darnell, Eastern Wyoming College
Robert Diaz, Fullerton College
Karen Dillon, Thomas Nelson Community College
Valeree Falduto, Palm Beach State
Bryan Faulkner, Ferrum College
David French, Tidewater Community College
Stephanie Gable, Columbus State University
Heather Gallacher, Cleveland State University
Rachel Gross, Towson University
Dianne Hendrickson, Becker College
Linda Hunt, Shawnee State University
Betty Ivory, Cuyahoga Community College
Joanne Kendall, Lone Star College System
Kevin Kennedy, Athens Technical College
Stephanie Krehl, Mid-South Community College
Allyn Leon, Imperial Valley College
Gerald LePage, Bristol Community College
Laurie Lindstrom, Bay de Noc Community College
Jonathan Lopez, Niagara University
Yixia Lu, South Suburban College
Mikal McDowell, Cedar Valley College
Kim McHale, Columbia College of Missouri
Allen Miller, Northeast Lakeview College
Michelle Moravec, Baylor University TX/McLennan Community College
Jennifer Nohai-Seaman, Housatonic Community College
Rick Norwood, East Tennessee State University
Linda Padilla, Joliet Junior College
Kelly Proffitt, Patrick Henry Community College
Teresa Richards, Butte-Glenn Community College
Christian Roldan-Johnson, College of Lake County Community College
Patricia C. Rome, Delgado Community College, City Park Campus
Kegan Samuel, Naugatuck Valley Community College
Bruny Santiago, Tarrant College Southeast Campus
Sutandra Sarkar, Georgia State University
Richard Sgarlotti, Bay Mills Community College
Chuang Shao, Rose State College
Carla VanDeSande, Arizona State University
Shannon Vinson, Wake Technical Community College
Maryam Vulis, Norwalk Community College
Toby Wagner, Chemeketa Community College
Libby Watts, Tidewater Community College
Becky Wheelock, San Diego City College

Introduction

A photograph of different types of fruit at a market.
Purchasing pounds of fruit at a fruit market requires a basic understanding of numbers. (credit: Dr. Karl-Heinz Hochhaus, Wikimedia Commons)

Even though counting is first taught at a young age, mastering mathematics, which is the study of numbers, requires constant attention. If it has been a while since you have studied math, it can be helpful to review basic topics. In this chapter, we will focus on numbers used for counting as well as four arithmetic operations—addition, subtraction, multiplication, and division. We will also discuss some vocabulary that we will use throughout this book.

Introduction to Whole Numbers

Learning Objectives

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

  • Identify counting numbers and whole numbers
  • Model whole numbers
  • Identify the place value of a digit
  • Use place value to name whole numbers
  • Use place value to write whole numbers
  • Round whole numbers

Identify Counting Numbers and Whole Numbers

Learning algebra is similar to learning a language. You start with a basic vocabulary and then add to it as you go along. You need to practice often until the vocabulary becomes easy to you. The more you use the vocabulary, the more familiar it becomes.

Algebra uses numbers and symbols to represent words and ideas. Let’s look at the numbers first. The most basic numbers used in algebra are those we use to count objects: 1,2,3,4,5,… and so on. These are called the counting numbers. The notation “…” is called an ellipsis, which is another way to show “and so on”, or that the pattern continues endlessly. Counting numbers are also called natural numbers.

Counting Numbers

The counting numbers start with 1 and continue.

1,2,3,4,5…

Counting numbers and whole numbers can be visualized on a number line as shown in Figure 1.

An image of a number line from 0 to 6 in increments of one. An arrow above the number line pointing to the right with the label “larger”. An arrow pointing to the left with the label “smaller”.
The numbers on the number line increase from left to right, and decrease from right to left.

The point labeled 0 is called the origin. The points are equally spaced to the right of 0 and labeled with the counting numbers. When a number is paired with a point, it is called the coordinate of the point.

The discovery of the number zero was a big step in the history of mathematics. Including zero with the counting numbers gives a new set of numbers called the whole numbers.

Whole Numbers

The whole numbers are the counting numbers and zero.

0,1,2,3,4,5…

We stopped at 5 when listing the first few counting numbers and whole numbers. We could have written more numbers if they were needed to make the patterns clear.

Which of the following are ⓐ counting numbers? ⓑ whole numbers?

0,14,3,5.2,15,105

Solution

Solution

  • ⓐ The counting numbers start at 1, so 0 is not a counting number. The numbers 3,15,and105 are all counting numbers.
  • ⓑ Whole numbers are counting numbers and 0. The numbers 0,3,15,and105 are whole numbers.

The numbers 14 and 5.2 are neither counting numbers nor whole numbers. We will discuss these numbers later.

Which of the following are ⓐ counting numbers ⓑ whole numbers?

0,23,2,9,11.8,241,376

Solution
  • ⓐ 2, 9, 241, 376
  • ⓑ 0, 2, 9, 241, 376

Which of the following are ⓐ counting numbers ⓑ whole numbers?

0,53,7,8.8,13,201

Solution
  • ⓐ 7, 13, 201
  • ⓑ 0, 7, 13, 201

Model Whole Numbers

Our number system is called a place value system because the value of a digit depends on its position, or place, in a number. The number 537 has a different value than the number 735. Even though they use the same digits, their value is different because of the different placement of the 7 and the 5.

Money gives us a familiar model of place value. Suppose a wallet contains three $100 bills, seven $10 bills, and four $1 bills. The amounts are summarized in Figure 2. How much money is in the wallet?

An image of three stacks of American currency. First stack from left to right is a stack of 3 $100 bills, with label “Three $100 bills, 3 times $100 equals $300”. Second stack from left to right is a stack of 7 $10 bills, with label “Seven $10 bills, 7 times $10 equals $70”. Third stack from left to right is a stack of 4 $1 bills, with label “Four $1 bills, 4 times $1 equals $4”.

Find the total value of each kind of bill, and then add to find the total. The wallet contains $374.

An image of “$300 + $70 +$4” where the “3” in “$300”, the “7” in “$70”, and the “4” in “$4” are all in red instead of black like the rest of the expression. Below this expression there is the value “$374”. An arrow points from the red “3” in the expression to the “3” in “$374”, an arrow points to the red “7” in the expression to the “7” in “$374”, and an arrow points from the red “4” in the expression to the “4” in “$374”.

Base-10 blocks provide another way to model place value, as shown in Figure 3. The blocks can be used to represent hundreds, tens, and ones. Notice that the tens rod is made up of 10 ones, and the hundreds square is made of 10 tens, or 100 ones.

An image with three items. The first item is a single block with the label “A single block represents 1”. The second item is a horizontal rod consisting of 10 blocks, with the label “A rod represents 10”. The third item is a square consisting of 100 blocks, with the label “A square represents 100”. The square is 10 blocks tall and 10 blocks wide.

Figure 4 shows the number 138 modeled with base-10 blocks.

An image consisting of three items. The first item is a square of 100 blocks, 10 blocks wide and 10 blocks tall, with the label “1 hundred”. The second item is 3 horizontal rods containing 10 blocks each, with the label “3 tens”. The third item is 8 individual blocks with the label “8 ones”.
We use place value notation to show the value of the number 138.
An image of “100 + 30 +8” where the “1” in “100”, the “3” in “30”, and the “8” are all in red instead of black like the rest of the expression. Below this expression there is the value “138”. An arrow points from the red “1” in the expression to the “1” in “138”, an arrow points to the red “3” in the expression to the “3” in “138”, and an arrow points from the red “8” in the expression to the “8” in 138.
Digit Place value Number Value Total value
1 hundreds 1 100 100
3 tens 3 10 30
8 ones 8 1 +8
Sum = 138

Use place value notation to find the value of the number modeled by the base-10 blocks shown.

An image consisting of three items. The first item is two squares of 100 blocks each, 10 blocks wide and 10 blocks tall. The second item is one horizontal rod containing 10 blocks. The third item is 5 individual blocks.
Solution

Solution

There are 2 hundreds squares, which is 200.

There is 1 tens rod, which is 10.

There are 5 ones blocks, which is 5.
An image of “200 + 10 + 5” where the “2” in “200”, the “1” in “10”, and the “5” are all in red instead of black like the rest of the expression. Below this expression there is the value “215”. An arrow points from the red “2” in the expression to the “2” in “215”, an arrow points to the red “1” in the expression to the “1” in “215”, and an arrow points from the red “5” in the expression to the “5” in 215.

Digit Place value Number Value Total value
2 hundreds 2 100 200
1 tens 1 10 10
5 ones 5 1 +5
215

The base-10 blocks model the number 215.

Use place value notation to find the value of the number modeled by the base-10 blocks shown.

An image consisting of three items. The first item is a square of 100, 10 blocks wide and 10 blocks tall. The second item is 7 horizontal rods containing 10 blocks each. The third item is 6 individual blocks.
Solution

176

Use place value notation to find the value of the number modeled by the base-10 blocks shown.

An image consisting of three items. The first item is two squares of 100 blocks each, 10 blocks wide and 10 blocks tall. The second item is three horizontal rods containing 10 blocks each. The third item is 7 individual blocks.
Solution

237

Doing the Manipulative Mathematics activity Number Line-Part 1 will help you develop a better understanding of the counting numbers and the whole numbers.

Identify the Place Value of a Digit

By looking at money and base-10 blocks, we saw that each place in a number has a different value. A place value chart is a useful way to summarize this information. The place values are separated into groups of three, called periods. The periods are ones, thousands, millions, billions, trillions, and so on. In a written number, commas separate the periods.

Just as with the base-10 blocks, where the value of the tens rod is ten times the value of the ones block and the value of the hundreds square is ten times the tens rod, the value of each place in the place-value chart is ten times the value of the place to the right of it.

Figure 5 shows how the number 5,278,194 is written in a place value chart.

A chart titled 'Place Value' with fifteen columns and 4 rows, with the columns broken down into five groups of three. The header row shows Trillions, Billions, Millions, Thousands, and Ones. The next row has the values 'Hundred trillions', 'Ten trillions', 'trillions', 'hundred billions', 'ten billions', 'billions', 'hundred millions', 'ten millions', 'millions', 'hundred thousands', 'ten thousands', 'thousands', 'hundreds', 'tens', and 'ones'. The first 8 values in the next row are blank. Starting with the ninth column, the values are '5', '2', '7', '8', '1', '9', and '4'.
  • The digit 5 is in the millions place. Its value is 5,000,000.
  • The digit 2 is in the hundred thousands place. Its value is 200,000.
  • The digit 7 is in the ten thousands place. Its value is 70,000.
  • The digit 8 is in the thousands place. Its value is 8,000.
  • The digit 1 is in the hundreds place. Its value is 100.
  • The digit 9 is in the tens place. Its value is 90.
  • The digit 4 is in the ones place. Its value is 4.

In the number 63,407,218; find the place value of each of the following digits:

  1. ⓐ 7
  2. ⓑ 0
  3. ⓒ 1
  4. ⓓ 6
  5. ⓔ 3
Solution

Solution

Write the number in a place value chart, starting at the right.
A figure titled “Place Values” with fifteen columns and 2 rows, with the colums broken down into five groups of three. The first row has the values “Hundred trillions”, “Ten trillions”, “trillions”, “hundred billions”, “ten billions”, “billions”, “hundred millions”, “ten millions”, “millions”, “hundred thuosands”, “ten thousands”, “thousands”, “hundreds”, “tens”, and “ones”. The first 7 values in the second row are blank. Starting with eighth column, the values are “6”, “3”, “4”, “0”, “7”, “2”, “1” and “8”. The first group is labeled “trillions” and contains the first row values of “Hundred trillions”, “ten trillions”, and “trillions”. The second group is labeled “billions” and contains the first row values of “Hundred billions”, “ten billions”, and “billions”. The third group is labeled “millions” and contains the first row values of “Hundred millions”, “ten millions”, and “millions”. The fourth group is labeled “thousands” and contains the first row values of “Hundred thousands”, “ten thousands”, and “thousands”. The fifth group is labeled “ones” and contains the first row values of “Hundreds”, “tens”, and “ones”.

  • ⓐ The 7 is in the thousands place.
  • ⓑ The 0 is in the ten thousands place.
  • ⓒ The 1 is in the tens place.
  • ⓓ The 6 is in the ten millions place.
  • ⓔ The 3 is in the millions place.

For each number, find the place value of digits listed: 27,493,615

  1. ⓐ 2
  2. ⓑ 1
  3. ⓒ 4
  4. ⓓ 7
  5. ⓔ 5
Solution
  • ⓐ ten millions
  • ⓑ tens
  • ⓒ hundred thousands
  • ⓓ millions
  • ⓔ ones

For each number, find the place value of digits listed: 519,711,641,328

  1. ⓐ 9
  2. ⓑ 4
  3. ⓒ 2
  4. ⓓ 6
  5. ⓔ 7
Solution
  • ⓐ billions
  • ⓑ ten thousands
  • ⓒ tens
  • ⓓ hundred thousands
  • ⓔ hundred millions

Use Place Value to Name Whole Numbers

When you write a check, you write out the number in words as well as in digits. To write a number in words, write the number in each period followed by the name of the period without the ‘s’ at the end. Start with the digit at the left, which has the largest place value. The commas separate the periods, so wherever there is a comma in the number, write a comma between the words. The ones period, which has the smallest place value, is not named.

An image with three values separated by commas. The first value is “37” and has the label “millions”. The second value is “519” and has the label thousands. The third value is “248” and has the label ones. Underneath, the value “37” has an arrow pointing to “Thirty-seven million”, the value “519” has an arrow pointing to “Five hundred nineteen thousand”, and the value “248” has an arrow pointing to “Two hundred forty-eight”.

So the number 37,519,248 is written thirty-seven million, five hundred nineteen thousand, two hundred forty-eight.

Notice that the word and is not used when naming a whole number.

Name a whole number in words.

  1. Starting at the digit on the left, name the number in each period, followed by the period name. Do not include the period name for the ones.
  2. Use commas in the number to separate the periods.

Name the number 8,165,432,098,710 in words.

Solution

Solution

Begin with the leftmost digit, which is 8. It is in the trillions place. eight trillion
The next period to the right is billions. one hundred sixty-five billion
The next period to the right is millions. four hundred thirty-two million
The next period to the right is thousands. ninety-eight thousand
The rightmost period shows the ones. seven hundred ten

An image with five values separated by commas. The first value is “8” and has the label “trillions”. The second value is “165” and has the label “bilions”. The third value is “432” and has the label “millions”. The fourth value is “098” and has the label “thousands”. The fifth value is “710” and has the label “ones”. Underneath, the value “8” has an arrow pointing to “Eight trillion”, the value “165” has an arrow pointing to “One hundred sixty-five billion”, the value “432” has an arrow pointing to “Four hundred thirty-two million”, the value “098” has an arrow pointing to “Ninety-eight thousand”, and the value “710” has an arrow pointing to “seven hundred ten”.

Putting all of the words together, we write 8,165,432,098,710 as eight trillion, one hundred sixty-five billion, four hundred thirty-two million, ninety-eight thousand, seven hundred ten.

Name each number in words: 9,258,137,904,061

Solution

nine trillion, two hundred fifty-eight billion, one hundred thirty-seven million, nine hundred four thousand, sixty-one

Name each number in words: 17,864,325,619,004

Solution

seventeen trillion, eight hundred sixty-four billion, three hundred twenty-five million, six hundred nineteen thousand, four

A student conducted research and found that the number of mobile phone users in the United States during one month in 2014 was 327,577,529. Name that number in words.

Solution

Solution

Identify the periods associated with the number.
An image with three values separated by commas. The first value is “327” and has the label “millions”. The second value is “577” and has the label “thousands”. The third value is “529” and has the label “ones”.

Name the number in each period, followed by the period name. Put the commas in to separate the periods.

Millions period: three hundred twenty-seven million

Thousands period: five hundred seventy-seven thousand

Ones period: five hundred twenty-nine

So the number of mobile phone users in the Unites States during the month of April was three hundred twenty-seven million, five hundred seventy-seven thousand, five hundred twenty-nine.

The population in a country is 316,128,839. Name that number.

Solution

three hundred sixteen million, one hundred twenty-eight thousand, eight hundred thirty-nine

One year is 31,536,000 seconds. Name that number.

Solution

thirty-one million, five hundred thirty-six thousand

Use Place Value to Write Whole Numbers

We will now reverse the process and write a number given in words as digits.

Use place value to write a whole number.

  1. Identify the words that indicate periods. (Remember the ones period is never named.)
  2. Draw three blanks to indicate the number of places needed in each period. Separate the periods by commas.
  3. Name the number in each period and place the digits in the correct place value position.

Write the following numbers using digits.

  • ⓐ fifty-three million, four hundred one thousand, seven hundred forty-two
  • ⓑ nine billion, two hundred forty-six million, seventy-three thousand, one hundred eighty-nine
Solution

Solution

ⓐ Identify the words that indicate periods.

Except for the first period, all other periods must have three places. Draw three blanks to indicate the number of places needed in each period. Separate the periods by commas.

Then write the digits in each period.
An image with three blocks of text pointing to numerical values. The first block of text is “fifty-three million”, has the label “millions”, and points to value 53. The second block of text is “four hundred one thousand”, has the label “thousands”, and points to value 401. The third block of text is “seven hundred forty-two”, has the label “ones”, and points to value 742.

Put the numbers together, including the commas. The number is 53,401,742.

ⓑ Identify the words that indicate periods.

Except for the first period, all other periods must have three places. Draw three blanks to indicate the number of places needed in each period. Separate the periods by commas.

Then write the digits in each period.
An image with four blocks of text pointing to numerical values. The first block of text is “nine billion”, has the label “billions”, and points to value 9. The second block of text is “two hundred forty-six million”, has the label “millions”, and points to value 246. The third block of text is “seventy-three thousand”, has the label “thousands”, and points to value 742. The fourth block of text is “one hundred eighty-nine”, has the label “ones”, and points to the value 189.

The number is 9,246,073,189.

Notice that in part ⓑ , a zero was needed as a place-holder in the hundred thousands place. Be sure to write zeros as needed to make sure that each period, except possibly the first, has three places.

Write each number in standard form:

fifty-three million, eight hundred nine thousand, fifty-one.

Solution

53,809,051

Write each number in standard form:

two billion, twenty-two million, seven hundred fourteen thousand, four hundred sixty-six.

Solution

2,022,714,466

A state budget was about $77 billion. Write the budget in standard form.

Solution

Solution

Identify the periods. In this case, only two digits are given and they are in the billions period. To write the entire number, write zeros for all of the other periods.
An image with four blocks of text pointing to numerical values. The first block of text is “77 billion”, has the label “billions”, and points to value “77”. The second block of text is null, has the label “millions”, and points to value “000”. The third block of text is null, has the label “thousands”, and points to value “000”. The fourth block of text is null, has the label “ones”, and points to the value “000”.

So the budget was about $77,000,000,000.

Write each number in standard form:

The closest distance from Earth to Mars is about 34 million miles.

Solution

34,000,000 miles

Write each number in standard form:

The total weight of an aircraft carrier is 204 million pounds.

Solution

204,000,000 pounds

Round Whole Numbers

In 2013, the U.S. Census Bureau reported the population of the state of New York as 19,651,127 people. It might be enough to say that the population is approximately 20 million. The word approximately means that 20 million is not the exact population, but is close to the exact value.

The process of approximating a number is called rounding. Numbers are rounded to a specific place value depending on how much accuracy is needed. 20 million was achieved by rounding to the millions place. Had we rounded to the one hundred thousands place, we would have 19,700,000 as a result. Had we rounded to the ten thousands place, we would have 19,650,000 as a result, and so on. The place value to which we round to depends on how we need to use the number.

Using the number line can help you visualize and understand the rounding process. Look at the number line in Figure 7. Suppose we want to round the number 76 to the nearest ten. Is 76 closer to 70 or 80 on the number line?

An image of a number line from 70 to 80 with increments of one. All the numbers on the number line are black except for 70 and 80 which are red. There is an orange dot at the value “76” on the number line.
We can see that 76 is closer to 80 than to 70. So 76 rounded to the nearest ten is 80.

Now consider the number 72. Find 72 in Figure 8.

An image of a number line from 70 to 80 with increments of one. All the numbers on the number line are black except for 70 and 80 which are red. There is an orange dot at the value “72” on the number line.
We can see that 72 is closer to 70, so 72 rounded to the nearest ten is 70.

How do we round 75 to the nearest ten. Find 75 in Figure 9.

An image of a number line from 70 to 80 with increments of one. All the numbers on the number line are black except for 70 and 80 which are red. There is an orange dot at the value “75” on the number line.
The number 75 is exactly midway between 70 and 80.

So that everyone rounds the same way in cases like this, mathematicians have agreed to round to the higher number, 80. So, 75 rounded to the nearest ten is 80.

Now that we have looked at this process on the number line, we can introduce a more general procedure. To round a number to a specific place, look at the number to the right of that place. If the number is less than 5, round down. If it is greater than or equal to 5, round up.

So, for example, to round 76 to the nearest ten, we look at the digit in the ones place.

An image of value “76”. The text “tens place” is in blue and points to number 7 in “76”. The text “is greater than 5” is in red and points to the number 6 in “76”.

The digit in the ones place is a 6. Because 6 is greater than or equal to 5, we increase the digit in the tens place by one. So the 7 in the tens place becomes an 8. Now, replace any digits to the right of the 8 with zeros. So, 76 rounds to 80.

An image of the value “76”. The “6” in “76” is crossed out and has an arrow pointing to it which says “replace with 0”. The “7” has an arrow pointing to it that says “add 1”. Under the value “76” is the value “80”.

Let’s look again at rounding 72 to the nearest 10. Again, we look to the ones place.

An image of value “72”. The text “tens place” is in blue and points to number 7 in “72”. The text “is less than 5” is in red and points to the number 2 in “72”.

The digit in the ones place is 2. Because 2 is less than 5, we keep the digit in the tens place the same and replace the digits to the right of it with zero. So 72 rounded to the nearest ten is 70.

An image of the value “72”. The “2” in “72” is crossed out and has an arrow pointing to it which says “replace with 0”. The “7” has an arrow pointing to it that says “do not add 1”. Under the value “72” is the value “70”.

Round a whole number to a specific place value.

  1. Locate the given place value. All digits to the left of that place value do not change unless the given place value is a 9, in which case it may. (See Step 3.)
  2. Underline the digit to the right of the given place value.
  3. Determine if this digit is greater than or equal to 5.
    • Yes—add 1 to the digit in the given place value. If that digit is 9, replace it with 0 and add 1 to the digit immediately to its left. If that digit is also a 9, repeat.
    • No—do not change the digit in the given place value.
  4. Replace all digits to the right of the given place value with zeros.

Round 843 to the nearest ten.

Solution

Solution

Locate the tens place. The number 843 with the label “tens place” pointed at the 4 in 843.
Underline the digit to the right of the tens place. The number 843 with the 3 underlined.
Since 3 is less than 5, do not change the digit in the tens place. The number 843 with the 3 underlined.
Replace all digits to the right of the tens place with zeros. The number 840 with the 0 underlined.
Rounding 843 to the nearest ten gives 840.

Round to the nearest ten: 157.

Solution

160

Round to the nearest ten: 884.

Solution

880

Round each number to the nearest hundred:

  1. ⓐ 23,658
  2. ⓑ 3,978
Solution

Solution

ⓐ
Locate the hundreds place. An arrow points from 'hundreds place' to the digit 6 in the number 23,658, illustrating the concept of place value.
The digit to the right of the hundreds place is 5. Underline the digit to the right of the hundreds place. The number 23,658 is written with a horizontal line drawn under the digit 5.
Since 5 is greater than or equal to 5, round up by adding 1 to the digit in the hundreds place. Then replace all digits to the right of the hundreds place with zeros. The number 23,658 is written with an arrow pointing to the digit 6 and the instruction to add 1, and an arrow pointing to the digits 58 with the instructions to replace with zeros. The final product is 23,700.
So 23,658 rounded to the nearest hundred is 23,700.
ⓑ
Locate the hundreds place. An arrow points to the '9' in the number 3,978, with the words 'hundreds place' written above, illustrating the hundreds digit in a four-digit number.
Underline the digit to the right of the hundreds place. The number 3,978 is written with a horizontal line drawn underneath the digit 7.
The digit to the right of the hundreds place is 7. Since 7 is greater than or equal to 5, round up by adding 1 to the 9. Then place all digits to the right of the hundreds place with zeros. An illustration of rounding 3,978 to the nearest thousand. The hundreds digit 9 is underlined, indicating that 1 should be added to the thousands place (3+1=4) and the remaining digits replaced with zeros, resulting in 4,000.
So 3,978 rounded to the nearest hundred is 4,000.

Round to the nearest hundred: 17,852.

Solution

17,900

Round to the nearest hundred: 4,951.

Solution

5,000

Round each number to the nearest thousand:

  1. ⓐ 147,032
  2. ⓑ 29,504
Solution

Solution

ⓐ
Locate the thousands place. Underline the digit to the right of the thousands place. The number 147,032 is written with an arrow pointing to the digit 7, with a label indicating that this is the 'thousands place.'
The digit to the right of the thousands place is 0. Since 0 is less than 5, we do not change the digit in the thousands place. The number 147,032 is displayed in bold black text on a white background, with the digit '0' in the hundreds place underlined.
We then replace all digits to the right of the thousands pace with zeros. The number 147,000 is written.
So 147,032 rounded to the nearest thousand is 147,000.
ⓑ
Locate the thousands place. The number 29,504 is written with an arrow pointing to the digit 9 and a label indicating that this number is in the thousands place.
Underline the digit to the right of the thousands place. The number 29,504 is written with a horizontal line drawn underneath the digit 5.
The digit to the right of the thousands place is 5. Since 5 is greater than or equal to 5, round up by adding 1 to the 9. Then replace all digits to the right of the thousands place with zeros. An image demonstrating how to round the number 29,504 up to the nearest ten thousand, resulting in 30,000, by adding 1 to the thousands place (9+1=10), writing 0, adding 1 to the ten thousands place, and replacing the last three digits with zeros.
So 29,504 rounded to the nearest thousand is 30,000.

Notice that in part ⓑ , when we add 1 thousand to the 9 thousands, the total is 10 thousands. We regroup this as 1 ten thousand and 0 thousands. We add the 1 ten thousand to the 2 ten thousands and put a 0 in the thousands place.

Round to the nearest thousand: 63,921.

Solution

64,000

Round to the nearest thousand: 156,437.

Solution

156,000

ACCESS ADDITIONAL ONLINE RESOURCES

  • Determine Place Value
  • Write a Whole Number in Digits from Words

Key Concepts

A chart titled 'Place Value' with fifteen columns and 4 rows, with the columns broken down into five groups of three. The header row shows Trillions, Billions, Millions, Thousands, and Ones. The next row has the values 'Hundred trillions', 'Ten trillions', 'trillions', 'hundred billions', 'ten billions', 'billions', 'hundred millions', 'ten millions', 'millions', 'hundred thousands', 'ten thousands', 'thousands', 'hundreds', 'tens', and 'ones'. The first 8 values in the next row are blank. Starting with the ninth column, the values are '5', '2', '7', '8', '1', '9', and '4'.
  • Name a whole number in words.
    1. Starting at the digit on the left, name the number in each period, followed by the period name. Do not include the period name for the ones.
    2. Use commas in the number to separate the periods.
  • Use place value to write a whole number.
    1. Identify the words that indicate periods. (Remember the ones period is never named.)
    2. Draw three blanks to indicate the number of places needed in each period.
    3. Name the number in each period and place the digits in the correct place value position.
  • Round a whole number to a specific place value.
    1. Locate the given place value. All digits to the left of that place value do not change unless the given place value is a 9, in which case it may. (See Step 3.).
    2. Underline the digit to the right of the given place value.
    3. Determine if this digit is greater than or equal to 5. If yes—add 1 to the digit in the given place value. If that digit is 9, replace it with 0 and add 1 to the digit immediately to its left. If that digit is also a 9, repeat. If no—do not change the digit in the given place value.
    4. Replace all digits to the right of the given place value with zeros.

Practice Makes Perfect

Identify Counting Numbers and Whole Numbers

In the following exercises, determine which of the following numbers are ⓐ counting numbers ⓑ whole numbers.

0,23,5,8.1,125

Solution
  1. ⓐ 5, 125
  2. ⓑ 0, 5, 125

0,710,3,20.5,300

0,49,3.9,50,221

Solution
  1. ⓐ 50, 221
  2. ⓑ 0, 50, 221

0,35,10,303,422.6

Model Whole Numbers

In the following exercises, use place value notation to find the value of the number modeled by the base-10 blocks.

An image consisting of three items. The first item is five squares of 100 blocks each, 10 blocks wide and 10 blocks tall. The second item is six horizontal rods containing 10 blocks each. The third item is 1 individual block.
Solution

561

An image consisting of three items. The first item is three squares of 100 blocks each, 10 blocks wide and 10 blocks tall. The second item is eight horizontal rods containing 10 blocks each. The third item is 4 individual blocks.
An image consisting of two items. The first item is four squares of 100 blocks each, 10 blocks wide and 10 blocks tall. The second item is 7 individual blocks.
Solution

407

An image consisting of two items. The first item is six squares of 100 blocks each, 10 blocks wide and 10 blocks tall. The second item is 2 horizontal rods with 10 blocks each.

Identify the Place Value of a Digit

In the following exercises, find the place value of the given digits.

579,601

  1. ⓐ 9
  2. ⓑ 6
  3. ⓒ 0
  4. ⓓ 7
  5. ⓔ 5
Solution
  1. ⓐ thousands
  2. ⓑ hundreds
  3. ⓒ tens
  4. ⓓ ten thousands
  5. ⓔ hundred thousands

398,127

  1. ⓐ 9
  2. ⓑ 3
  3. ⓒ 2
  4. ⓓ 8
  5. ⓔ 7

56,804,379

  1. ⓐ 8
  2. ⓑ 6
  3. ⓒ 4
  4. ⓓ 7
  5. ⓔ 0
Solution
  1. ⓐ hundred thousands
  2. ⓑ millions
  3. ⓒ thousands
  4. ⓓ tens
  5. ⓔ ten thousands

78,320,465

  1. ⓐ 8
  2. ⓑ 4
  3. ⓒ 2
  4. ⓓ 6
  5. ⓔ 7

Use Place Value to Name Whole Numbers

In the following exercises, name each number in words.

1,078

Solution

One thousand, seventy-eight

5,902

364,510

Solution

Three hundred sixty-four thousand, five hundred ten

146,023

5,846,103

Solution

Five million, eight hundred forty-six thousand, one hundred three

1,458,398

37,889,005

Solution

Thirty seven million, eight hundred eighty-nine thousand, five

62,008,465

The height of Mount Rainier is 14,410 feet.

Solution

Fourteen thousand, four hundred ten

The height of Mount Adams is 12,276 feet.

Seventy years is 613,200 hours.

Solution

Six hundred thirteen thousand, two hundred

One year is 525,600 minutes.

The U.S. Census estimate of the population of Miami-Dade county was 2,617,176.

Solution

Two million, six hundred seventeen thousand, one hundred seventy-six

The population of Chicago was 2,718,782.

There are projected to be 23,867,000 college and university students in the US in five years.

Solution

Twenty three million, eight hundred sixty-seven thousand

About twelve years ago there were 20,665,415 registered automobiles in California.

The population of China is expected to reach 1,377,583,156 in 2016.

Solution

One billion, three hundred seventy-seven million, five hundred eighty-three thousand, one hundred fifty-six

The population of India is estimated at 1,267,401,849 as of July 1,2014.

Use Place Value to Write Whole Numbers

In the following exercises, write each number as a whole number using digits.

four hundred twelve

Solution

412

two hundred fifty-three

thirty-five thousand, nine hundred seventy-five

Solution

35,975

sixty-one thousand, four hundred fifteen

eleven million, forty-four thousand, one hundred sixty-seven

Solution

11,044,167

eighteen million, one hundred two thousand, seven hundred eighty-three

three billion, two hundred twenty-six million, five hundred twelve thousand, seventeen

Solution

3,226,512,017

eleven billion, four hundred seventy-one million, thirty-six thousand, one hundred six

The population of the world was estimated to be seven billion, one hundred seventy-three million people.

Solution

7,173,000,000

The age of the solar system is estimated to be four billion, five hundred sixty-eight million years.

Lake Tahoe has a capacity of thirty-nine trillion gallons of water.

Solution

39,000,000,000,000

The federal government budget was three trillion, five hundred billion dollars.

Round Whole Numbers

In the following exercises, round to the indicated place value.

Round to the nearest ten:
  1. ⓐ 386
  2. ⓑ 2,931
Solution
  1. ⓐ 390
  2. ⓑ 2,930
Round to the nearest ten:
  1. ⓐ 792
  2. ⓑ 5,647
Round to the nearest hundred:
  1. ⓐ 13,748
  2. ⓑ 391,794
Solution
  1. ⓐ 13,700
  2. ⓑ 391,800
Round to the nearest hundred:
  1. ⓐ 28,166
  2. ⓑ 481,628

Round to the nearest ten:

  1. ⓐ 1,492
  2. ⓑ 1,497
Solution
  1. ⓐ 1,490
  2. ⓑ 1,500
Round to the nearest thousand:
  1. ⓐ 2,391
  2. ⓑ 2,795
Round to the nearest hundred:
  1. ⓐ 63,994
  2. ⓑ 63,949
Solution
  1. ⓐ 64,000
  2. ⓑ 63,900
Round to the nearest thousand:
  1. ⓐ 163,584
  2. ⓑ 163,246

Everyday Math

Writing a Check Jorge bought a car for $24,493. He paid for the car with a check. Write the purchase price in words.

Solution

Twenty four thousand, four hundred ninety-three dollars

Writing a Check Marissa’s kitchen remodeling cost $18,549. She wrote a check to the contractor. Write the amount paid in words.

Buying a Car Jorge bought a car for $24,493. Round the price to the nearest:

  1. ⓐ ten dollars
  2. ⓑ hundred dollars
  3. ⓒ thousand dollars
  4. ⓓ ten-thousand dollars
Solution
  1. ⓐ $24,490
  2. ⓑ $24,500
  3. ⓒ $24,000
  4. ⓓ $20,000

Remodeling a Kitchen Marissa’s kitchen remodeling cost $18,549. Round the cost to the nearest:

  1. ⓐ ten dollars
  2. ⓑ hundred dollars
  3. ⓒ thousand dollars
  4. ⓓ ten-thousand dollars

Population The population of China was 1,355,692,544 in 2014. Round the population to the nearest:

  1. ⓐ billion people
  2. ⓑ hundred-million people
  3. ⓒ million people
Solution
  1. ⓐ 1,000,000,000
  2. ⓑ 1,400,000,000
  3. ⓒ 1,356,000,000

Astronomy The average distance between Earth and the sun is 149,597,888 kilometers. Round the distance to the nearest:

  1. ⓐ hundred-million kilometers
  2. ⓑ ten-million kilometers
  3. ⓒ million kilometers

Writing Exercises

In your own words, explain the difference between the counting numbers and the whole numbers.

Solution

Answers may vary. The whole numbers are the counting numbers with the inclusion of zero.

Give an example from your everyday life where it helps to round numbers.

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 understanding of whole numbers, place value, and rounding skills with options: 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.

coordinate
A number paired with a point on a number line is called the coordinate of the point.
counting numbers
The counting numbers are the numbers 1, 2, 3, ….
number line
A number line is used to visualize numbers. The numbers on the number line get larger as they go from left to right, and smaller as they go from right to left.
origin
The origin is the point labeled 0 on a number line.
place value system
Our number system is called a place value system because the value of a digit depends on its position, or place, in a number.
rounding
The process of approximating a number is called rounding.
whole numbers
The whole numbers are the numbers 0, 1, 2, 3, ….

Add Whole Numbers

Learning Objectives

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

  • Use addition notation
  • Model addition of whole numbers
  • Add whole numbers without models
  • Translate word phrases to math notation
  • Add whole numbers in applications

Before you get started, take this readiness quiz.

What is the number modeled by the base-10 blocks?
An image consisting of three items. The first item is two squares of 100 blocks each, 10 blocks wide and 10 blocks tall. The second item is one horizontal rod containing 10 blocks. The third item is 5 individual blocks.
If you missed this problem, review Example 2 in Introduction to Whole Numbers.

Solution

215

Write the number three hundred forty-two thousand six using digits?
If you missed this problem, review Example 7 in Introduction to Whole Numbers.

Solution

342,006

Use Addition Notation

A college student has a part-time job. Last week he worked 3 hours on Monday and 4 hours on Friday. To find the total number of hours he worked last week, he added 3 and 4.

The operation of addition combines numbers to get a sum. The notation we use to find the sum of 3 and 4 is:

3+4

We read this as three plus four and the result is the sum of three and four. The numbers 3 and 4 are called the addends. A math statement that includes numbers and operations is called an expression.

Addition Notation

To describe addition, we can use symbols and words.

Operation Notation Expression Read as Result
Addition + 3+4 three plus four the sum of 3 and 4
Translate from math notation to words:
  1. ⓐ 7+1
  2. ⓑ 12+14
Solution

Solution

  • ⓐ The expression consists of a plus symbol connecting the addends 7 and 1. We read this as seven plus one. The result is the sum of seven and one.
  • ⓑ The expression consists of a plus symbol connecting the addends 12 and 14. We read this as twelve plus fourteen. The result is the sum of twelve and fourteen.

Translate from math notation to words:

  1. ⓐ 8+4
  2. ⓑ 18+11
Solution
  • ⓐ eight plus four; the sum of eight and four
  • ⓑ eighteen plus eleven; the sum of eighteen and eleven

Translate from math notation to words:

  1. ⓐ 21+16
  2. ⓑ 100+200
Solution
  1. ⓐ twenty-one plus sixteen; the sum of twenty-one and sixteen
  2. ⓑ one hundred plus two hundred; the sum of one hundred and two hundred

Model Addition of Whole Numbers

Addition is really just counting. We will model addition with base-10 blocks. Remember, a block represents 1 and a rod represents 10. Let’s start by modeling the addition expression we just considered, 3+4.

Each addend is less than 10, so we can use ones blocks.

We start by modeling the first number with 3 blocks. Three empty squares are displayed above the digit 3 on a white background.
Then we model the second number with 4 blocks. Two groups of empty squares are shown, with the first group containing three squares labeled '3' and the second group containing four squares labeled '4', illustrating basic counting or grouping.
Count the total number of blocks. The digit '7' displayed below a row of seven empty square placeholders.

There are 7 blocks in all. We use an equal sign (=) to show the sum. A math sentence that shows that two expressions are equal is called an equation. We have shown that. 3+4=7.

Doing the Manipulative Math Worksheets activity “Model Addition of Whole Numbers” will help you develop a better understanding of adding whole numbers.

Model the addition 2+6.

Solution

Solution

2+6 means the sum of 2 and 6

Each addend is less than 10, so we can use ones blocks.

Model the first number with 2 blocks. Two light blue outlined squares are shown above the number 2, indicating a count or selection of two items.
Model the second number with 6 blocks. The image shows two light blue squares above the number 2, and six light blue squares above the number 6.
Count the total number of blocks The image shows eight light blue squares above the number 8.
There are 8 blocks in all, so 2+6=8.

Model: 3+6.

Solution


The image shows three gray squares, followed by a gap, and then six more gray squares. Below this visual representation, the mathematical equation '3 + 6 = 9' is displayed, illustrating the sum of the squares.

Model: 5+1.

Solution


The image uses gray squares to model the addition problem 5 + 1.

When the result is 10 or more ones blocks, we will exchange the 10 blocks for one rod.

Model the addition 5+8.

Solution

Solution

5+8 means the sum of 5 and 8.

Each addend is less than 10, se we can use ones blocks.
Model the first number with 5 blocks. An image showing five blank, light gray squares with dark outlines, arranged horizontally. The number '5' is displayed directly below the center of the five squares on a white background, indicating a quantity.
Model the second number with 8 blocks. The image shows 5 outlined gray squares with the number 5 written beneath, and a group of eight outlined gray squares with the number 8 written beneath.
Count the result. There are more than 10 blocks so we exchange 10 ones blocks for 1 tens rod. In this image, there are two groups of 5 outlined gray squares and one group of 3 outlined gray squares. The two groups of 5 are circled and transformed into a single group of 10 red blocks below with the one group of three grey outlined squares next to it.
Now we have 1 ten and 3 ones, which is 13. 5 + 8 = 13

Notice that we can describe the models as ones blocks and tens rods, or we can simply say ones and tens. From now on, we will use the shorter version but keep in mind that they mean the same thing.

Model the addition: 5+7.

Solution


The image uses blocks to demonstrate the addition problem 5 + 7.

Model the addition: 6+8.

Solution


The image uses blocks to demonstrate the addition problem 6 + 8.

Next we will model adding two digit numbers.

Model the addition: 17+26.

Solution

Solution

17+26 means the sum of 17 and 26.

Model the 17. 1 ten and 7 ones In this image, a group of 10 blocks is on the left and a group of single 7 blocks is on the right.
Model the 26. 2 tens and 6 ones In this image, two groups of 10 outlined boxes are on the left, and 6 single outlined boxes are on the right.
Combine. 3 tens and 13 ones In this image, three groups of 10 outlined blocks is on the right and 13 single outlined blocks are on the right. 10 of the single outlined boxes on the right are colored red.
Exchange 10 ones for 1 ten. 4 tens and 3 ones
40+3=43
An image illustrating rows of squares, with three rows of ten light blue squares and one row of ten red squares aligned below them. To the right, three additional light blue squares are arranged horizontally.
We have shown that 17+26=43

Model the addition: 15+27.

Solution


Visual representation of 15 + 27 = 42 using base-ten blocks, showing four tens and two ones.

Model the addition: 16+29.

Solution


An image illustrating the addition 16 + 29 = 45 using base-ten blocks. Four rows of ten blocks represent 40, and five individual blocks represent 5, totaling 45. The sum is written below the blocks.

Add Whole Numbers Without Models

Now that we have used models to add numbers, we can move on to adding without models. Before we do that, make sure you know all the one digit addition facts. You will need to use these number facts when you add larger numbers.

Imagine filling in Table 6 by adding each row number along the left side to each column number across the top. Make sure that you get each sum shown. If you have trouble, model it. It is important that you memorize any number facts you do not already know so that you can quickly and reliably use the number facts when you add larger numbers.

+ 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
1 1 2 3 4 5 6 7 8 9 10
2 2 3 4 5 6 7 8 9 10 11
3 3 4 5 6 7 8 9 10 11 12
4 4 5 6 7 8 9 10 11 12 13
5 5 6 7 8 9 10 11 12 13 14
6 6 7 8 9 10 11 12 13 14 15
7 7 8 9 10 11 12 13 14 15 16
8 8 9 10 11 12 13 14 15 16 17
9 9 10 11 12 13 14 15 16 17 18

Did you notice what happens when you add zero to a number? The sum of any number and zero is the number itself. We call this the Identity Property of Addition. Zero is called the additive identity.

Identity Property of Addition

The sum of any number a and 0 is the number.

a+0=a0+a=a
Find each sum:
  1. ⓐ 0+11
  2. ⓑ 42+0
Solution

Solution

ⓐ The first addend is zero. The sum of any number and zero is the number. 0+11=11
ⓑ The second addend is zero. The sum of any number and zero is the number. 42+0=42
Find each sum:
  1. ⓐ 0+19
  2. ⓑ 39+0
Solution
  1. ⓐ 0+19=19
  2. ⓑ 39+0=39
Find each sum:
  1. ⓐ 0+24
  2. ⓑ 57+0
Solution
  1. ⓐ 0+24=24
  2. ⓑ 57+0=57

Look at the pairs of sums.

Demonstrates the commutative property of addition, where changing the order of operands does not change the sum.
2+3=5 3+2=5
4+7=11 7+4=11
8+9=17 9+8=17

Notice that when the order of the addends is reversed, the sum does not change. This property is called the Commutative Property of Addition, which states that changing the order of the addends does not change their sum.

Commutative Property of Addition

Changing the order of the addends a and b does not change their sum.

a+b=b+a

Add:

  1. ⓐ 8+7
  2. ⓑ 7+8
Solution

Solution

  • This table demonstrates an example of an addition operation, specifically 8 plus 7, yielding 15.
    ⓐ
    Add. 8+7
    15
  • Details for addition problem 'ⓑ', showing the expression '7+8' and its sum '15'.
    ⓑ
    Add. 7+8
    15

Did you notice that changing the order of the addends did not change their sum? We could have immediately known the sum from part ⓑ just by recognizing that the addends were the same as in part ⓐ , but in the reverse order. As a result, both sums are the same.

Add: 9+7 and 7+9.

Solution

9+7=16; 7+9=16

Add: 8+6 and 6+8.

Solution

8+6=14; 6+8=14

Add: 28+61.

Solution

Solution

To add numbers with more than one digit, it is often easier to write the numbers vertically in columns.

This table illustrates the step-by-step vertical addition of two-digit numbers, showing both the instructions and the corresponding mathematical representation.
Write the numbers so the ones and tens digits line up vertically. 28 +61____
Then add the digits in each place value.
Add the ones: 8+1=9
Add the tens: 2+6=8
28 +61____89

Add: 32+54.

Solution

32+54=86

Add: 25+74.

Solution

25+74=99

In the previous example, the sum of the ones and the sum of the tens were both less than 10. But what happens if the sum is 10 or more? Let’s use our base-10 model to find out. Figure 1 shows the addition of 17 and 26 again.

An image containing two groups of items. The left group includes 1 horizontal rod with 10 blocks and 7 individual blocks 2 horizontal rods with 10 blocks each and 6 individual blocks. The label to the left of this group of items is “17 + 26 =”. The right group contains two items. Four horizontal rods containing 10 blocks each. Then, 3 individual blocks. The label for this group is “17 + 26 = 43”.

When we add the ones, 7+6, we get 13 ones. Because we have more than 10 ones, we can exchange 10 of the ones for 1 ten. Now we have 4 tens and 3 ones. Without using the model, we show this as a small red 1 above the digits in the tens place.

When the sum in a place value column is greater than 9, we carry over to the next column to the left. Carrying is the same as regrouping by exchanging. For example, 10 ones for 1 ten or 10 tens for 1 hundred.

Add whole numbers.

  1. Write the numbers so each place value lines up vertically.
  2. Add the digits in each place value. Work from right to left starting with the ones place. If a sum in a place value is more than 9, carry to the next place value.
  3. Continue adding each place value from right to left, adding each place value and carrying if needed.

Add: 43+69.

Solution

Solution

Step-by-step example demonstrating how to add two-digit numbers with carrying, using 43 and 69 as an illustration.
Write the numbers so the digits line up vertically. 43 +69____
Add the digits in each place.
Add the ones: 3+9=12
Write the 2 in the ones place in the sum.
Add the 1 ten to the tens place.
413 +69____2
Now add the tens: 1+4+6=11
Write the 11 in the sum.
413 +69____112

Add: 35+98.

Solution

35+98=133

Add: 72+89.

Solution

72+89=161

Add: 324+586.

Solution

Solution

Step-by-step guide to adding multi-digit numbers with visual examples.
Write the numbers so the digits line up vertically. An addition problem in which 324 is added to 586.
Add the digits in each place value.
Add the ones: 4+6=10
Write the 0 in the ones place in the sum and carry the 1 ten to the tens place.
An addition problem in which 324 is added to 586. in this first step, the right digits of both numbers are added: 4 + 6 = 10. The 1 is written above the 2 and 0 is written below.
Add the tens: 1+2+8=11
Write the 1 in the tens place in the sum and carry the 1 hundred to the hundreds
A mathematical column addition problem features the number 586. Above the digits, the fractions 1/3 and 1/2 appear alongside the digit 4, placed as if they are carry-over numbers. Below the line, the partial sum '10' is shown.
Add the hundreds: 1+3+5=9
Write the 9 in the hundreds place.
A column addition problem showing 324 plus 586 equals 910, with the fractions 1/3 and 1/2 written above the hundreds and tens digits of the first operand.

Add: 456+376.

Solution

456+376=832

Add: 269+578.

Solution

269+578=847

Add: 1,683+479.

Solution

Solution

This table illustrates the step-by-step process of adding multi-digit numbers using the column addition method, including carrying.
Write the numbers so the digits line up vertically. 1,683 +479______
Add the digits in each place value.
Add the ones: 3+9=12.
Write the 2 in the ones place of the sum and carry the 1 ten to the tens place.
1,6813 +479______2
Add the tens: 1+7+8=16
Write the 6 in the tens place and carry the 1 hundred to the hundreds place.
1,61813 +479______62
Add the hundreds: 1+6+4=11
Write the 1 in the hundreds place and carry the 1 thousand to the thousands place.
1,61813 +479______162
Add the thousands 1+1=2.
Write the 2 in the thousands place of the sum.
1,161813 +479______2,162

When the addends have different numbers of digits, be careful to line up the corresponding place values starting with the ones and moving toward the left.

Add: 4,597+685.

Solution

4,597+685=5,282

Add: 5,837+695.

Solution

5,837+695=6,532

Add: 21,357+861+8,596.

Solution

Solution

Step-by-step illustration of adding multi-digit numbers with carrying, broken down by place value.
Write the numbers so the place values line up vertically. 21,357 861 +8,596_______
Add the digits in each place value.
Add the ones: 7+1+6=14
Write the 4 in the ones place of the sum and carry the 1 to the tens place.
21,3517 861 +8,596_______ 4
Add the tens: 1+5+6+9=21
Write the 1 in the tens place and carry the 2 to the hundreds place.
21,32517 861 +8,596_______ 14
Add the hundreds: 2+3+8+5=18
Write the 8 in the hundreds place and carry the 1 to the thousands place.
21,132517 861 +8,596_______ 814
Add the thousands 1+1+8=10.
Write the 0 in the thousands place and carry the 1 to the ten thousands place.
211,132517 861 +8,596_______ 0814
Add the ten-thousands 1+2=3.
Write the 3 in the ten thousands place in the sum.
211,132517 861 +8,596_______ 30,814

This example had three addends. We can add any number of addends using the same process as long as we are careful to line up the place values correctly.

Add: 46,195+397+6,281.

Solution

46,195+397+6,281=52,873

Add: 53,762+196+7,458.

Solution

53,762+196+7,458=61,416

Translate Word Phrases to Math Notation

Earlier in this section, we translated math notation into words. Now we’ll reverse the process. We’ll translate word phrases into math notation. Some of the word phrases that indicate addition are listed in Table 16.

Operation Words Example Expression
Addition plus
sum
increased by
more than
total of
added to
1 plus 2
the sum of 3 and 4
5 increased by 6
8 more than 7
the total of 9 and 5
6 added to 4
1+2
3+4
5+6
7+8
9+5
4+6

Translate and simplify: the sum of 19 and 23.

Solution

Solution

The word sum tells us to add. The words of 19 and 23 tell us the addends.

This table demonstrates the step-by-step process of translating a verbal addition problem into a mathematical expression and finding its sum.
The sum of 19 and 23
Translate. 19+23
Add. 42
The sum of 19 and 23 is 42.

Translate and simplify: the sum of 17 and 26.

Solution

Translate: 17+26; Simplify: 43

Translate and simplify: the sum of 28 and 14.

Solution

Translate: 28+14; Simplify: 42

Translate and simplify: 28 increased by 31.

Solution

Solution

The words increased by tell us to add. The numbers given are the addends.

Steps demonstrating how to translate a word problem into a mathematical expression and solve it.
28 increased by 31.
Translate. 28+31
Add. 59
So 28 increased by 31 is 59.

Translate and simplify: 29 increased by 76.

Solution

Translate: 29+76; Simplify 105

Translate and simplify: 37 increased by 69.

Solution

Translate 37 + 69; Simplify 106

Add Whole Numbers in Applications

Now that we have practiced adding whole numbers, let’s use what we’ve learned to solve real-world problems. We’ll start by outlining a plan. First, we need to read the problem to determine what we are looking for. Then we write a word phrase that gives the information to find it. Next we translate the word phrase into math notation and then simplify. Finally, we write a sentence to answer the question.

Hao earned grades of 87,93,68,95,and89 on the five tests of the semester. What is the total number of points he earned on the five tests?

Solution

Solution

We are asked to find the total number of points on the tests.

Step-by-step guide on translating a verbal phrase to a mathematical sum and solving for total points.
Write a phrase. the sum of points on the tests
Translate to math notation. 87+93+68+95+89
Then we simplify by adding.
Since there are several numbers, we will write them vertically. 837936895+89____432
Write a sentence to answer the question. Hao earned a total of 432 points.

Notice that we added points, so the sum is 432 points. It is important to include the appropriate units in all answers to applications problems.

Mark is training for a bicycle race. Last week he rode 18 miles on Monday, 15 miles on Wednesday, 26 miles on Friday, 49 miles on Saturday, and 32 miles on Sunday. What is the total number of miles he rode last week?

Solution

He rode 140 miles.

Lincoln Middle School has three grades. The number of students in each grade is 230,165,and325. What is the total number of students?

Solution

The total number is 720 students.

Some application problems involve shapes. For example, a person might need to know the distance around a garden to put up a fence or around a picture to frame it. The perimeter is the distance around a geometric figure. The perimeter of a figure is the sum of the lengths of its sides.

Find the perimeter of the patio shown.

This is an image of a perimeter of a patio. There are six sides. The far left side is labeled 4 feet, the top side is labeled 9 feet, the right side is short and labeled 2 feet, then extends across to the left and is labeled 3 feet. From here, the side extends down and is labeled 2 feet. Finally, the base is labeled 6 feet.
Solution

Solution

This table illustrates the step-by-step process of calculating the perimeter of an object, from defining the problem to providing the final solution.
We are asked to find the perimeter.
Write a phrase. the sum of the sides
Translate to math notation. 4+6+2+3+2+9
Simplify by adding. 26
Write a sentence to answer the question.
We added feet, so the sum is 26 feet. The perimeter of the patio is 26 feet.

Find the perimeter of the figure. All lengths are in inches.

This image includes 8 sides. Side one on the left is labeled 4 inches, side 2 on the top is labeled 9 inches, side 3 on the right is labeled 4 inches, side 4 is labeled 3 inches, side 5 is labeled 2 inches, side 6 is labeled 3 inches, side 7 is labeled 2 inches, and side 8 is labeled 3 inches.
Solution

The perimeter is 30 inches.

Find the perimeter of the figure. All lengths are in inches.

This image includes 8 sides. Moving in a clockwise direction, the first side is labeled 2 inches, side 2 is labeled 12 inches, side 3 is labeled 6 inches, side 4 is labeled 4 inches, side 5 is labeled 2 inches, side 6 is labeled 4 inches, side 7 is labeled 2 inches and side 8 is labeled 4 inches.
Solution

The perimeter is 36 inches.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Adding Two-Digit Numbers with base-10 blocks
  • Adding Three-Digit Numbers with base-10 blocks
  • Adding Whole Numbers

Key Concepts

  • Addition Notation To describe addition, we can use symbols and words.
    Operation Notation Expression Read as Result
    Addition + 3+4 three plus four the sum of 3 and 4
  • Identity Property of Addition
    • The sum of any number a and 0 is the number. a+0=a 0+a=a
  • Commutative Property of Addition
    • Changing the order of the addends a and b does not change their sum. a+b=b+a.
  • Add whole numbers.
    1. Write the numbers so each place value lines up vertically.
    2. Add the digits in each place value. Work from right to left starting with the ones place. If a sum in a place value is more than 9, carry to the next place value.
    3. Continue adding each place value from right to left, adding each place value and carrying if needed.

Practice Makes Perfect

Use Addition Notation

In the following exercises, translate the following from math expressions to words.

5+2

Solution

five plus two; the sum of 5 and 2.

6+3

13+18

Solution

thirteen plus eighteen; the sum of 13 and 18.

15+16

214+642

Solution

two hundred fourteen plus six hundred forty-two; the sum of 214 and 642

438+113

Model Addition of Whole Numbers

In the following exercises, model the addition.

2+4

Solution


The image uses blocks to demonstrate addition of whole numbers, in this case 2 + 4.

2+4=6

5+3

8+4

Solution


The image uses blocks to demonstrate addition of whole numbers, in this case 8 + 4.

8+4=12

5+9

14+75

Solution


The image uses blocks to demonstrate addition of whole numbers, in this case 14 + 75.

14+75=89

15+63

16+25

Solution


The image uses blocks to demonstrate addition of whole numbers, in this case 16 + 25.

16+25=41

14+27

Add Whole Numbers

In the following exercises, fill in the missing values in each chart.

An image of a table with 11 columns and 11 rows. The cells in the first row and first column are shaded darker than the other cells. The first column has the values “+; 0; 1; 2; 3; 4; 5; 6; 7; 8; 9”. The second column has the values “0; 0; 1; null; 3; 4; 5; 6; null; 8; 9”. The third column has the values “1; 1; 2; 3; null; 5; 6; 7; null; 9; 10”. The fourth column has the values “2; 2; 3; 4; 5; null; 7; 8; 9; null; 11”. The fifth column has the values “3; null; 4; 5; null; null; 8; null; 10; 11; null”. The sixth column has the values “4; 4; null; 6;7; 8; null; 10; null; null; 13”. The seventh column has the values “5; 5; null; null; 8; 9; null; null; 12; null; 14”. The eighth column has the values “6; 6; 7; 8; null; null; 11; null; null; 14; null”. The ninth column has the values “7; 7; 8; null; 10; 11; null; 13; null; null; null”. The tenth column has the values “8; null; 9; null; null; 12; 13; null; 15; 16; 17”. The eleventh column has the values “9; 9; null; 11; 12; null; null; 15; 16; null; null”.
Solution


An image of a table with 11 columns and 11 rows. The cells in the first row and first column are shaded darker than the other cells. The cells contain numbers and answers to the problem.

An image of a table with 11 columns and 11 rows. The cells in the first row and first column are shaded darker than the other cells. The first column has the values “+; 0; 1; 2; 3; 4; 5; 6; 7; 8; 9”. The second column has the values “0; 0; 1; 2; null; 4; 5; null; 7; 8; null”. The third column has the values “1; 1; 2; null; 4; 5; 6; null; 8; 9; null”. The fourth column has the values “2; 2; 3; 4; null; 6; null; 8; null; 10; 11”. The fifth column has the values “3; 3; null; null; 6; 7; 8; 9; 10; null; 12”. The sixth column has the values “4; 4; 5; 6; null; null; 9; null; null; 12; 13”. The seventh column has the values “5; null; 6; 7; null; null; null; null; 12; null; null”. The eighth column has the values “6; 6; null; null; 9; 10; 11; 12; null; 14; null”. The ninth column has the values “7; null; 8; 9; null; 11; 12; 13; null; null; 16”. The tenth column has the values “8; 8; null; 10; 11; null; 13; null; 15; 16; null”. The eleventh column has the values “9; 9; 10; null; null; 13; null; 15; 16; 17; null”.
An image of a table with 8 columns and 5 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null. The first column has the values “+; 6; 7; 8; 9”. The first row has the values “+; 3; 4; 5; 6; 7; 8; 9”.
Solution


An image of a table with 8 columns and 5 rows. The cells in the first row and first column are shaded darker than the other cells. The cells contain numbers and answers to the problem.

An image of a table with 8 columns and 5 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null. The first row has the values “+; 6; 7; 8; 9”. The first column has the values “+; 3; 4; 5; 6; 7; 8; 9”.
An image of a table with 6 columns and 6 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null. The first row has the values “+; 5; 6; 7; 8; 9”. The first column has the values “+; 5; 6; 7; 8; 9”.
Solution


An image of a table with 6 columns and 6 rows. The cells in the first row and first column are shaded darker than the other cells. The cells contain numbers and answers to the problem.

An image of a table with 5 columns and 5 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or first column are all null. The first row has the values “+; 6; 7; 8; 9”. The first column has the values “+; 6; 7; 8; 9”.

In the following exercises, add.

  1. ⓐ 0+13
  2. ⓑ 13+0
Solution
  1. ⓐ 13
  2. ⓑ 13
  1. ⓐ 0+5,280
  2. ⓑ 5,280+0
  1. ⓐ 8+3
  2. ⓑ 3+8
Solution
  1. ⓐ 11
  2. ⓑ 11
  1. ⓐ 7+5
  2. ⓑ 5+7

45+33

Solution

78

37+22

71+28

Solution

99

43+53

26+59

Solution

85

38+17

64+78

Solution

142

92+39

168+325

Solution

493

247+149

584+277

Solution

861

175+648

832+199

Solution

1,031

775+369

6,358+492

Solution

6,850

9,184+578

3,740+18,593

Solution

22,333

6,118+15,990

485,012+619,848

Solution

1,104,860

368,911+857,289

24,731+592+3,868

Solution

29,191

28,925+817+4,593

8,015+76,946+16,570

Solution

101,531

6,291+54,107+28,635

Translate Word Phrases to Math Notation

In the following exercises, translate each phrase into math notation and then simplify.

the sum of 13 and 18

Solution

13+18=31

the sum of 12 and 19

the sum of 90 and 65

Solution

90+65=155

the sum of 70 and 38

33 increased by 49

Solution

33+49=82

68 increased by 25

250 more than 599

Solution

599+250=849

115 more than 286

the total of 628 and 77

Solution

628+77=705

the total of 593 and 79

1,482 added to 915

Solution

915+1,482=2,397

2,719 added to 682

Add Whole Numbers in Applications

In the following exercises, solve the problem.

Home remodeling Sophia remodeled her kitchen and bought a new range, microwave, and dishwasher. The range cost $1,100, the microwave cost $250, and the dishwasher cost $525. What was the total cost of these three appliances?

Solution

The total cost was $1,875.

Sports equipment Aiden bought a baseball bat, helmet, and glove. The bat cost $299, the helmet cost $35, and the glove cost $68. What was the total cost of Aiden’s sports equipment?

Bike riding Ethan rode his bike 14 miles on Monday, 19 miles on Tuesday, 12 miles on Wednesday, 25 miles on Friday, and 68 miles on Saturday. What was the total number of miles Ethan rode?

Solution

Ethan rode 138 miles.

Business Chloe has a flower shop. Last week she made 19 floral arrangements on Monday, 12 on Tuesday, 23 on Wednesday, 29 on Thursday, and 44 on Friday. What was the total number of floral arrangements Chloe made?

Apartment size Jackson lives in a 7 room apartment. The number of square feet in each room is 238,120,156,196,100,132, and 225. What is the total number of square feet in all 7 rooms?

Solution

The total square footage in the rooms is 1,167 square feet.

Weight Seven men rented a fishing boat. The weights of the men were 175,192,148,169,205,181, and 225 pounds. What was the total weight of the seven men?

Salary Last year Natalie’s salary was $82,572. Two years ago, her salary was $79,316, and three years ago it was $75,298. What is the total amount of Natalie’s salary for the past three years?

Solution

Natalie’s total salary is $237,186.

Home sales Emma is a realtor. Last month, she sold three houses. The selling prices of the houses were $292,540,$505,875, and $423,699. What was the total of the three selling prices?

In the following exercises, find the perimeter of each figure.

An image of a triangle with side lengths of 14 inches, 12 inches, and 18 inches.
Solution

The perimeter of the figure is 44 inches.

An image of a right triangle with base of 12 centimeters, height of 5 centimeters, and diagonal hypotenuse of 13 centimeters.
A rectangle 21 meters wide and 7 meters tall.
Solution

The perimeter of the figure is 56 meters.

A rectangle 19 feet wide and 14 feet tall.
A trapezoid with horizontal top length of 19 yards, the side lengths are 18 yards and are diagonal, and the horizontal bottom length is 16 yards.
Solution

The perimeter of the figure is 71 yards.

A trapezoid with horizontal top length of 24 meters, the side lengths are 17 meters and are diagonal, and the horizontal bottom length is 29 meters.
This is a rectangle-like image with six sides. Starting from the top left of the figure, the first line runs right for 24 feet. From the end of this line, the second line runs down for 7 feet. Then the third line runs left from this point for 19 feet. The fourth line runs up 3 feet. The fifth line runs left for 5 feet. The sixth line runs up for 4 feet, connecting it at a corner with start of the first line.
Solution

The perimeter of the figure is 62 feet.

This is an image with 6 straight sides. Starting from the top left of the figure, the first line runs right for 25 inches. From the end of this line, the second line runs down for 10 inches. Then the third line runs left from this point for 14 inches. The fourth line runs up 7 inches. The fifth line runs left for 11 inches. The sixth line runs up, connecting it at a corner with start of the first line.

Everyday Math

Calories Paulette had a grilled chicken salad, ranch dressing, and a 16-ounce drink for lunch. On the restaurant’s nutrition chart, she saw that each item had the following number of calories:

Grilled chicken salad – 320 calories
Ranch dressing – 170 calories
16-ounce drink – 150 calories

What was the total number of calories of Paulette’s lunch?

Solution

The total number of calories was 640.

Calories Fred had a grilled chicken sandwich, a small order of fries, and a 12-oz chocolate shake for dinner. The restaurant’s nutrition chart lists the following calories for each item:

Grilled chicken sandwich – 420 calories
Small fries – 230 calories
12-oz chocolate shake – 580 calories

What was the total number of calories of Fred’s dinner?

Test scores A student needs a total of 400 points on five tests to pass a course. The student scored 82,91,75,88,and70. Did the student pass the course?

Solution

Yes, he scored 406 points.

Elevators The maximum weight capacity of an elevator is 1150 pounds. Six men are in the elevator. Their weights are 210,145,183,230,159,and164 pounds. Is the total weight below the elevator’s maximum capacity?

Solution

Yes, the total weight is 1091 pounds.

Writing Exercises

How confident do you feel about your knowledge of the addition facts? If you are not fully confident, what will you do to improve your skills?

Solution

Answers will vary.

How have you used models to help you learn the addition facts?

Self Check

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

A self-assessment chart helps students evaluate their addition skills, from using notation to applying whole numbers. Categories include Confidently, With some help, and No-I don't get it!

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

sum
The sum is the result of adding two or more numbers.

Subtract Whole Numbers

Learning Objectives

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

  • Use subtraction notation
  • Model subtraction of whole numbers
  • Subtract whole numbers
  • Translate word phrases to math notation
  • Subtract whole numbers in applications

Before you get started, take this readiness quiz.

Model 3+4 using base-ten blocks.
If you missed this problem, review Example 2 in Add Whole Numbers.

Solution
Pink squares demonstrate the addition problem 3 + 4 = 7, showing three squares plus four squares equals seven squares.

Add: 324+586.
If you missed this problem, review Example 10 in Add Whole Numbers.

Solution

910

Use Subtraction Notation

Suppose there are seven bananas in a bowl. Elana uses three of them to make a smoothie. How many bananas are left in the bowl? To answer the question, we subtract three from seven. When we subtract, we take one number away from another to find the difference. The notation we use to subtract 3 from 7 is

7−3

We read 7−3 as seven minus three and the result is the difference of seven and three.

Subtraction Notation

To describe subtraction, we can use symbols and words.

Operation Notation Expression Read as Result
Subtraction − 7−3 seven minus three the difference of 7 and 3

Translate from math notation to words: ⓐ 8−1 ⓑ 26−14.

Solution

Solution

  • ⓐ We read this as eight minus one. The result is the difference of eight and one.
  • ⓑ We read this as twenty-six minus fourteen. The result is the difference of twenty-six and fourteen.

Translate from math notation to words:

  1. ⓐ 12−4
  2. ⓑ 29−11
Solution
  1. ⓐ twelve minus four; the difference of twelve and four
  2. ⓑ twenty-nine minus eleven; the difference of twenty-nine and eleven

Translate from math notation to words:

  1. ⓐ 11−2
  2. ⓑ 29−12
Solution
  1. ⓐ eleven minus two; the difference of eleven and two
  2. ⓑ twenty-nine minus twelve; the difference of twenty-nine and twelve

Model Subtraction of Whole Numbers

A model can help us visualize the process of subtraction much as it did with addition. Again, we will use base-10 blocks. Remember a block represents 1 and a rod represents 10. Let’s start by modeling the subtraction expression we just considered, 7−3.

This table visually demonstrates the subtraction of 3 from 7 using blocks, illustrating each step from initial modeling to the final mathematical result.
We start by modeling the first number, 7. The image shows seven outlined gray boxes with the number 7 written underneath them.
Now take away the second number, 3. We'll circle 3 blocks to show that we are taking them away. The image shows three gray outlined boxes circled in red, a gap, and then four gray outlined boxes not circled.
Count the number of blocks remaining. Four light gray squares with dark outlines and subtle drop shadows are arranged horizontally on a clean white background.
There are 4 ones blocks left. We have shown that 7−3=4.

Model the subtraction: 8−2.

Solution

Solution

8−2 means the difference of 8 and 2.
Model the first, 8. A row of eight light blue squares with dark borders, with the number '8' centered directly below them on a white background.
Take away the second number, 2. A row of eight light blue squares with dark borders. The first two squares are circled in red.
Count the number of blocks remaining. A row of six light blue squares with dark borders.
There are 6 ones blocks left. We have shown that 8−2=6.

Model: 9−6.

Solution


The image shows the use of blocks to demonstrate the subtraction problem 9 – 6.

Model: 6−1.

Solution


The image shows the use of blocks to demonstrate the subtraction problem 9 – 6.

Model the subtraction: 13−8.

Solution

Solution

Model the first number, 13. We use 1 ten and 3 ones. Ten light blue blocks aligned horizontally, followed by a gap and then three more blocks. This simple graphic might illustrate quantities, data segments, or a step-by-step process.
Take away the second number, 8. However, there are not 8 ones, so we will exchange the 1 ten for 10 ones. Two groups of light blue squares with dark outlines are displayed on a white background, one group has 10 squares and the other has 3 squares, with the numbers 10 and 3 placed below them respectively.
Now we can take away 8 ones. A row of light blue squares. Ten squares are enclosed by a magenta oval, with an arrow pointing left underneath, indicating a loop or repetition. Four additional squares are to the right.
Count the blocks remaining. A row of five light blue squares with dark outlines. There is a small gap between the first two squares and the rest of the three squares.
There are five ones left. We have shown that 13−8=5.

As we did with addition, we can describe the models as ones blocks and tens rods, or we can simply say ones and tens.

Model the subtraction: 12−7.

Solution


The image shows blocks used to model the subtraction equation 12 – 7.

Model the subtraction: 14−8.

Solution


The image shows blocks used to model the subtraction equation 14 – 8.

Model the subtraction: 43−26.

Solution

Solution

Because 43−26 means 43 take away 26, we begin by modeling the 43.
An image containing two items. The first item is 4 horizontal rods containing 10 blocks each. The second item is 3 individual blocks.

Now, we need to take away 26, which is 2 tens and 6 ones. We cannot take away 6 ones from 3 ones. So, we exchange 1 ten for 10 ones.
This figure contains two groups. The first group on the left includes 3 rows of blue base 10 blocks and 1 red row of 10 blocks. This is labeled 4 tens. Alongside the first row of ten blocks are 3 individual blocks. This is labeled 3 ones. An arrow points to the right to the second group in which there are three rows of 10 base blocks labeled 3 tens. Next to this is a row of 3 blue individual blocks and two rows each with five individual blocks in red. This is labeled 13 ones.

Now we can take away 2 tens and 6 ones.
This image includes one row of base ten blocks at the top of the image; Next to it are seven individual blocks. Below this, is a group of two rows of base ten blocks, and two rows of 3 individual blocks with a circle around all. The arrow points to the right and shows one row of ten blocks and seven individual blocks underneath.

Count the number of blocks remaining. There is 1 ten and 7 ones, which is 17.

43−26=17

Model the subtraction: 42−27.

Solution


This image uses base ten blocks to visually represent the subtraction 42 - 27 = 15. It shows 4 tens and 2 ones as the starting number, and then partitions it into the number being subtracted (27) and the result (15).

Model the subtraction: 45−29.

Solution


An illustration of subtracting 29 from 45 using base-ten blocks. Initially showing 4 tens and 5 units, 2 tens and 9 units are circled and removed, leaving 1 ten and 6 units, demonstrating 45 - 29 = 16.

Subtract Whole Numbers

Addition and subtraction are inverse operations. Addition undoes subtraction, and subtraction undoes addition.

We know 7−3=4 because 4+3=7. Knowing all the addition number facts will help with subtraction. Then we can check subtraction by adding. In the examples above, our subtractions can be checked by addition.

7−3=4because4+3=713−8=5because5+8=1343−26=17because17+26=43
Subtract and then check by adding:
  1. ⓐ 9−7
  2. ⓑ 8−3.
Solution

Solution

Illustration of subtracting 7 from 9, with expression, result, and addition-based verification.
ⓐ
9−7
Subtract 7 from 9. 2
Check with addition.
2+7=9✓
This table demonstrates the subtraction of 3 from 8, including the problem, its result, and a verification step.
ⓑ
8−3
Subtract 3 from 8. 5
Check with addition.
5+3=8✓

Subtract and then check by adding:

7−0

Solution

7 − 0 = 7; 7 + 0 = 7

Subtract and then check by adding:

6−2

Solution

6 − 2 = 4; 2 + 4 = 6

To subtract numbers with more than one digit, it is usually easier to write the numbers vertically in columns just as we did for addition. Align the digits by place value, and then subtract each column starting with the ones and then working to the left.

Subtract and then check by adding: 89−61.

Solution

Solution

This table illustrates the step-by-step process of subtracting two-digit numbers, including lining up, subtracting by place value, and checking with addition.
Write the numbers so the ones and tens digits line up vertically. 89 −61____
Subtract the digits in each place value.

Subtract the ones: 9-1=8
Subtract the tens: 8-6=2
89 −61____ 28
Check using addition.
28 +61____ 89

Our answer is correct.

Subtract and then check by adding: 86−54.

Solution

86 − 54 = 32 because 54 + 32 = 86

Subtract and then check by adding: 99−74.

Solution

99 − 74 = 25 because 74 + 25 = 99

When we modeled subtracting 26 from 43, we exchanged 1 ten for 10 ones. When we do this without the model, we say we borrow 1 from the tens place and add 10 to the ones place.

Find the difference of whole numbers.

  1. Write the numbers so each place value lines up vertically.
  2. Subtract the digits in each place value. Work from right to left starting with the ones place. If the digit on top is less than the digit below, borrow as needed.
  3. Continue subtracting each place value from right to left, borrowing if needed.
  4. Check by adding.

Subtract: 43−26.

Solution

Solution

A step-by-step visual guide demonstrating the process of subtracting numbers with borrowing, including textual instructions and corresponding illustrations.
Write the numbers so each place value lines up vertically. A vertical subtraction problem is displayed, showing 43 minus 26, set up for manual calculation. A horizontal line underneath indicates where the answer would be placed.
Subtract the ones. We cannot subtract 6 from 3, so we borrow 1 ten. This makes 3 tens and 13 ones. We write these numbers above each place and cross out the original digits. A vertical subtraction problem setup, demonstrating the regrouping (borrowing) method for 43 minus 26. The 4 is shown as 3, and the 3 as 13, preparing for the subtraction calculation.
Now we can subtract the ones. 13−6=7. We write the 7 in the ones place in the difference. A vertical subtraction problem setup, demonstrating the regrouping (borrowing) method for 43 minus 26. The 4 is shown as 3, and the 3 as 13, preparing for the subtraction calculation to result in the answer 7.
Now we subtract the tens. 3−2=1. We write the 1 in the tens place in the difference. A handwritten example demonstrating subtraction with borrowing, showing 43 minus 26 equals 17.
Check by adding.

A vertical addition problem showing 17 plus 26 equals 43, with a black checkmark verifying the correct answer. The numbers are neatly aligned, demonstrating a basic arithmetic operation.
Our answer is correct.

Subtract and then check by adding: 93−58.

Solution

93 − 58 = 35 because 58 + 35 = 93

Subtract and then check by adding: 81−39.

Solution

81 − 39 = 42 because 42 + 39 = 81

Subtract and then check by adding: 207−64.

Solution

Solution

This table illustrates the step-by-step process of subtracting multi-digit numbers with borrowing, followed by a verification step.
Write the numbers so each place value lines up vertically. A vertical subtraction problem shows 207 minus 64, with a line beneath to indicate the answer should follow.
Subtract the ones. 7−4=3.
Write the 3 in the ones place in the difference.
A vertical subtraction problem of 207 - 64, demonstrating the initial borrowing from the hundreds to the tens place, and the calculation of the units digit as 3.
Subtract the tens. We cannot subtract 6 from 0 so we borrow 1 hundred and add 10 tens to the 0 tens we had. This makes a total of 10 tens. We write 10 above the tens place and cross out the 0. Then we cross out the 2 in the hundreds place and write 1 above it. A vertical subtraction problem of 207 - 64, demonstrating the initial borrowing from the hundreds to the tens place, and the calculation of the units digit as 3.
Now we subtract the tens. 10−6=4. We write the 4 in the tens place in the difference. A vertical subtraction problem demonstrating the borrowing method. The calculation shows 107 minus 64, resulting in 43, with regrouping indicated by the crossed-out digits and new values above them.
Finally, subtract the hundreds. There is no digit in the hundreds place in the bottom number so we can imagine a 0 in that place. Since 1−0=1, we write 1 in the hundreds place in the difference. A vertical subtraction problem showing 207 minus 64, with borrowing steps indicated, resulting in the answer 143.
Check by adding.
A vertical addition problem of 143 + 64, with the placement of 1 over the first digit 1 of 143. This results in the answer 207.
Our answer is correct.

Subtract and then check by adding: 439−52.

Solution

439 − 52 = 387 because 387 + 52 = 439

Subtract and then check by adding: 318−75.

Solution

318 − 75 = 243 because 243 + 75 = 318

Subtract and then check by adding: 910−586.

Solution

Solution

Step-by-step guide demonstrating multi-digit subtraction with borrowing, showing instructional text alongside visual examples and a final check.
Write the numbers so each place value lines up vertically. A vertical subtraction problem is presented, with 910 on top and 586 below it, separated by a horizontal line, and a minus sign to the left of 586, indicating the operation.
Subtract the ones. We cannot subtract 6 from 0, so we borrow 1 ten and add 10 ones to the 0 ones we had. This makes 10 ones. We write a 0 above the tens place and cross out the 1. We write the 10 above the ones place and cross out the 0. Now we can subtract the ones. 10−6=4. A vertical subtraction problem setup, showing 910 minus 586. The image illustrates the first step of borrowing for the ones column, where the '1' in the tens place becomes '0' and the '0' becomes '10'.
Write the 4 in the ones place of the difference. A vertical subtraction problem setup, showing 910 minus 586. The image illustrates the first step of borrowing for the ones column, where the '1' in the tens place becomes '0' and the '0' becomes '10', resulting in the answer 4.
Subtract the tens. We cannot subtract 8 from 0, so we borrow 1 hundred and add 10 tens to the 0 tens we had, which gives us 10 tens. Write 8 above the hundreds place and cross out the 9. Write 10 above the tens place. A vertical subtraction problem setup, showing 910 minus 586. The image illustrates the first step of borrowing for the ones column, where the '9' in the hundreds place becomes '8', the '1' in the tens place becomes '10' and the '0' becomes '10'. This results in an answer of 4.
Now we can subtract the tens. 10−8=2. An image illustrating long subtraction with borrowing: 910 minus 586. The borrowing steps for units and tens are shown, yielding 4 and 2 respectively. However, the final result of 24 is incorrect, as it should be 324.
Subtract the hundreds place. 8−5=3 Write the 3 in the hundreds place in the difference. Vertical subtraction problem 910 - 586 = 324, demonstrating the regrouping technique used to solve it.
Check by adding.

A vertical addition problem showing 324 plus 586 equals 910. The carries are indicated above the numbers, and a checkmark confirms the correctness of the sum.

Our answer is correct.

Subtract and then check by adding: 832−376.

Solution

832 − 376 = 456 because 456 + 376 = 832

Subtract and then check by adding: 847−578.

Solution

847 − 578 = 269 because 269 + 578 = 847

Subtract and then check by adding: 2,162−479.

Solution

Solution

Write the numbers so each place value lines up vertically. A vertical subtraction problem of 2162 minus 479.
Subtract the ones. Since we cannot subtract 9 from 2, borrow 1 ten and add 10 ones to the 2 ones to make 12 ones. Write 5 above the tens place and cross out the 6. Write 12 above the ones place and cross out the 2. A vertical subtraction problem of 2162 minus 479, illustrating the borrowing process. The '2' in the ones place becomes '12' by borrowing from the '6' in the tens place, which then becomes '5'.
Now we can subtract the ones. 12−9=3
Write 3 in the ones place in the difference. A vertical subtraction problem where 2162 is being subtracted by 479. The image shows the first step of borrowing, where the '2' in the ones place becomes '12' and the '6' in the tens place becomes '5', resulting in '3' for the ones column.
Subtract the tens. Since we cannot subtract 7 from 5, borrow 1 hundred and add 10 tens to the 5 tens to make 15 tens. Write 0 above the hundreds place and cross out the 1. Write 15 above the tens place. A vertical subtraction problem where 2162 is being subtracted by 479. The image shows the steps of borrowing, where the '2' in the ones place becomes '12' and the '6' in the tens place becomes '5' which then is crossed out to become '15', and the '1' in the hundreds place becomes '0'. This results in '3' written in the answer's ones place.
Now we can subtract the tens. 15−7=8
Write 8 in the tens place in the difference. A vertical subtraction problem where 2162 is being subtracted by 479. The image shows the steps of borrowing, where the '2' in the ones place becomes '12' and the '6' in the tens place becomes '15', and the '1' in the hundreds place becomes '0'. This results in '3' written in the answer's ones place and an '8' written in the tens place.
Now we can subtract the hundreds. A step-by-step vertical subtraction problem demonstrating the borrowing or regrouping method, with 2062 minus 479 resulting in 1583.
Write 6 in the hundreds place in the difference. A long subtraction problem showing 2162 minus 479, with borrowing indicated, resulting in 1683. The final answer, 1683, is partially highlighted in red (the '6').
Subtract the thousands. There is no digit in the thousands place of the bottom number, so we imagine a 0. 1−0=1. Write 1 in the thousands place of the difference. A long subtraction problem showing 2162 minus 479, with borrowing indicated, resulting in 1683. The final answer, 1683, is partially highlighted in red (the '1').
Check by adding.

11,61813+479______2,162✓

Our answer is correct.

Subtract and then check by adding: 4,585−697.

Solution

4,585 − 697 = 3,888 because 3,888 + 697 = 4,585

Subtract and then check by adding: 5,637−899.

Solution

5,637 − 899 = 4,738 because 4,738 + 899 = 5,637

Translate Word Phrases to Math Notation

As with addition, word phrases can tell us to operate on two numbers using subtraction. To translate from a word phrase to math notation, we look for key words that indicate subtraction. Some of the words that indicate subtraction are listed in Table 12.

Operation Word Phrase Example Expression
Subtraction minus 5 minus 1 5−1
difference the difference of 9 and 4 9−4
decreased by 7 decreased by 3 7−3
less than 5 less than 8 8−5
subtracted from 1 subtracted from 6 6−1

Translate and then simplify:

  1. ⓐ the difference of 13 and 8
  2. ⓑ subtract 24 from 43
Solution

Solution

  • ⓐ

    The word difference tells us to subtract the two numbers. The numbers stay in the same order as in the phrase.

    Steps to translate a verbal mathematical phrase into an expression and simplify it to its numerical value.
    the difference of 13 and 8
    Translate. 13−8
    Simplify. 5
  • ⓑ

    The words subtract from tells us to take the first number away from the second. We must be careful to get the order correct.

    This table illustrates the steps to translate a word problem into a mathematical expression and then simplify it to a numerical result.
    subtract 24 from 43
    Translate. 43−24
    Simplify. 19

Translate and simplify:

  1. ⓐ the difference of 14 and 9
  2. ⓑ subtract 21 from 37
Solution
  1. ⓐ 14 − 9 = 5
  2. ⓑ 37 − 21 = 16

Translate and simplify:

  1. ⓐ 11 decreased by 6
  2. ⓑ 18 less than 67
Solution
  1. ⓐ 11 − 6 = 5
  2. ⓑ 67 − 18 = 49

Subtract Whole Numbers in Applications

To solve applications with subtraction, we will use the same plan that we used with addition. First, we need to determine what we are asked to find. Then we write a phrase that gives the information to find it. We translate the phrase into math notation and then simplify to get the answer. Finally, we write a sentence to answer the question, using the appropriate units.

The temperature in Chicago one morning was 73 degrees Fahrenheit. A cold front arrived and by noon the temperature was 27 degrees Fahrenheit. What was the difference between the temperature in the morning and the temperature at noon?

Solution

Solution

We are asked to find the difference between the morning temperature and the noon temperature.

This table illustrates the step-by-step process of translating a word phrase into a subtraction problem and calculating its numerical difference.
Write a phrase. the difference of 73 and 27
Translate to math notation. Difference tells us to subtract. 73−27
Then we do the subtraction. The vertical subtraction of 73 by 27, resulting in 46, with the borrowing process from the tens column (7 to 6, 3 to 13) clearly shown.
Write a sentence to answer the question. The difference in temperatures was 46 degrees Fahrenheit.

The high temperature on June1st in Boston was 77 degrees Fahrenheit, and the low temperature was 58 degrees Fahrenheit. What was the difference between the high and low temperatures?

Solution

The difference is 19 degrees Fahrenheit.

The weather forecast for June 2 in St Louis predicts a high temperature of 90 degrees Fahrenheit and a low of 73 degrees Fahrenheit. What is the difference between the predicted high and low temperatures?

Solution

The difference is 17 degrees Fahrenheit.

A washing machine is on sale for $399. Its regular price is $588. What is the difference between the regular price and the sale price?

Solution

Solution

We are asked to find the difference between the regular price and the sale price.

This table illustrates the step-by-step process of solving a subtraction word problem, from phrasing to the final answer.
Write a phrase. the difference between 588 and 399
Translate to math notation. 588−399
Subtract. A three column subtraction problem is set up to show 588 minus 399. With borrowing indicated, the answer is 189.
Write a sentence to answer the question. The difference between the regular price and the sale price is $189.

A television set is on sale for $499. Its regular price is $648. What is the difference between the regular price and the sale price?

Solution

The difference is $149.

A patio set is on sale for $149. Its regular price is $285. What is the difference between the regular price and the sale price?

Solution

The difference is $136.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Model subtraction of two-digit whole numbers
  • Model subtraction of three-digit whole numbers
  • Subtract Whole Numbers

Key Concepts

Operation Notation Expression Read as Result
Subtraction − 7−3 seven minus three the difference of 7 and 3
  • Subtract whole numbers.
    1. Write the numbers so each place value lines up vertically.
    2. Subtract the digits in each place value. Work from right to left starting with the ones place. If the digit on top is less than the digit below, borrow as needed.
    3. Continue subtracting each place value from right to left, borrowing if needed.
    4. Check by adding.

Practice Makes Perfect

Use Subtraction Notation

In the following exercises, translate from math notation to words.

15−9

Solution

fifteen minus nine; the difference of fifteen and nine

18−16

42−35

Solution

forty-two minus thirty-five; the difference of forty-two and thirty-five

83−64

675−350

Solution

six hundred seventy-five minus three hundred fifty; the difference of six hundred seventy-five and three hundred fifty

790−525

Model Subtraction of Whole Numbers

In the following exercises, model the subtraction.

5−2

Solution


The image uses blocks to demonstrate the subtraction problem 5 - 2.

8−4

6−3

Solution


The image uses blocks to demonstrate the subtraction problem 6- 3.

7−5

18−5

Solution


The image uses blocks to demonstrate the subtraction problem 18 - 5.

19−8

17−8

Solution


The image uses blocks to demonstrate the subtraction problem 17 - 8.

17−9

35−13

Solution


The image uses blocks to demonstrate the subtraction problem 35 - 13.

32−11

61−47

Solution


The image uses blocks to demonstrate the subtraction problem 61 - 47.

55−36

Subtract Whole Numbers

In the following exercises, subtract and then check by adding.

9−4

Solution

5

9−3

8−0

Solution

8

2−0

38−16

Solution

22

45−21

85−52

Solution

33

99−47

493−370

Solution

123

268−106

5,946−4,625

Solution

1,321

7,775−3,251

75−47

Solution

28

63−59

461−239

Solution

222

486−257

525−179

Solution

346

542−288

6,318−2,799

Solution

3,519

8,153−3,978

2,150−964

Solution

1,186

4,245−899

43,650−8,982

Solution

34,668

35,162−7,885

Translate Word Phrases to Algebraic Expressions

In the following exercises, translate and simplify.

The difference of 10 and 3

Solution

10 − 3; 7

The difference of 12 and 8

The difference of 15 and 4

Solution

15 − 4; 11

The difference of 18 and 7

Subtract 6 from 9

Solution

9 − 6; 3

Subtract 8 from 9

Subtract 28 from 75

Solution

75 − 28; 47

Subtract 59 from 81

45 decreased by 20

Solution

45 − 20; 25

37 decreased by 24

92 decreased by 67

Solution

92 − 67; 25

75 decreased by 49

12 less than 16

Solution

16 − 12; 4

15 less than 19

38 less than 61

Solution

61 − 38; 23

47 less than 62

Mixed Practice

In the following exercises, simplify.

76−47

Solution

29

91−53

256−184

Solution

72

305−262

719+341

Solution

1,060

647+528

2,015−1,993

Solution

22

2,020−1,984

In the following exercises, translate and simplify.

Seventy-five more than thirty-five

Solution

75 + 35; 110

Sixty more than ninety-three

13 less than 41

Solution

41 − 13; 28

28 less than 36

The difference of 100 and 76

Solution

100 − 76; 24

The difference of 1,000 and 945

Subtract Whole Numbers in Applications

In the following exercises, solve.

Temperature The high temperature on June 2 in Las Vegas was 80 degrees and the low temperature was 63 degrees. What was the difference between the high and low temperatures?

Solution

The difference between the high and low temperature was 17 degrees

Temperature The high temperature on June 1 in Phoenix was 97 degrees and the low was 73 degrees. What was the difference between the high and low temperatures?

Class size Olivia’s third grade class has 35 children. Last year, her second grade class had 22 children. What is the difference between the number of children in Olivia’s third grade class and her second grade class?

Solution

The difference between the third grade and second grade was 13 children.

Class size There are 82 students in the school band and 46 in the school orchestra. What is the difference between the number of students in the band and the orchestra?

Shopping A mountain bike is on sale for $399. Its regular price is $650. What is the difference between the regular price and the sale price?

Solution

The difference between the regular price and sale price is $251.

Shopping A mattress set is on sale for $755. Its regular price is $1,600. What is the difference between the regular price and the sale price?

Savings John wants to buy a laptop that costs $840. He has $685 in his savings account. How much more does he need to save in order to buy the laptop?

Solution

John needs to save $155 more.

Banking Mason had $1,125 in his checking account. He spent $892. How much money does he have left?

Everyday Math

Road trip Noah was driving from Philadelphia to Cincinnati, a distance of 502 miles. He drove 115 miles, stopped for gas, and then drove another 230 miles before lunch. How many more miles did he have to travel?

Solution

157 miles

Test Scores Sara needs 350 points to pass her course. She scored 75,50,70,and80 on her first four tests. How many more points does Sara need to pass the course?

Writing Exercises

Explain how subtraction and addition are related.

Solution

Answers may vary.

How does knowing addition facts help you to subtract numbers?

Self Check

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

A self-assessment table for subtraction skills, allowing students to rate their proficiency as 'Confidently', 'With some help', or 'No-I don't get it!' across five different learning objectives.

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

difference
The difference is the result of subtracting two or more numbers.

Multiply Whole Numbers

Learning Objectives

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

  • Use multiplication notation
  • Model multiplication of whole numbers
  • Multiply whole numbers
  • Translate word phrases to math notation
  • Multiply whole numbers in applications

Before you get started, take this readiness quiz.

Add: 1,683+479.
If you missed this problem, review Example 11 in Add Whole Numbers.

Solution

2,162

Subtract: 605−321.
If you missed this problem, review Example 8 in Subtract Whole Numbers.

Solution

284

Use Multiplication Notation

Suppose you were asked to count all these pennies shown in Figure 1.

An image of 3 horizontal rows of pennies, each row containing 8 pennies.

Would you count the pennies individually? Or would you count the number of pennies in each row and add that number 3 times.

8+8+8

Multiplication is a way to represent repeated addition. So instead of adding 8 three times, we could write a multiplication expression.

3×8

We call each number being multiplied a factor and the result the product. We read 3×8 as three times eight, and the result as the product of three and eight.

There are several symbols that represent multiplication. These include the symbol × as well as the dot, ·, and parentheses ().

Operation Symbols for Multiplication

To describe multiplication, we can use symbols and words.

Operation Notation Expression Read as Result
Multiplication ×
·
()
3×8
3·8
3(8)
three times eight the product of 3 and 8

Translate from math notation to words:

  1. ⓐ 7×6
  2. ⓑ 12·14
  3. ⓒ 6(13)
Solution

Solution

  • ⓐ We read this as seven times six and the result is the product of seven and six.
  • ⓑ We read this as twelve times fourteen and the result is the product of twelve and fourteen.
  • ⓒ We read this as six times thirteen and the result is the product of six and thirteen.

Translate from math notation to words:

  1. ⓐ 8×7
  2. ⓑ 18·11
Solution
  1. ⓐ eight times seven ; the product of eight and seven
  2. ⓑ eighteen times eleven ; the product of eighteen and eleven

Translate from math notation to words:

  1. ⓐ (13)(7)
  2. ⓑ 5(16)
Solution
  1. ⓐ thirteen times seven ; the product of thirteen and seven
  2. ⓑ five times sixteen; the product of five and sixteen

Model Multiplication of Whole Numbers

There are many ways to model multiplication. Unlike in the previous sections where we used base-10 blocks, here we will use counters to help us understand the meaning of multiplication. A counter is any object that can be used for counting. We will use round blue counters.

Model: 3×8.

Solution

Solution

To model the product 3×8, we’ll start with a row of 8 counters.
An image of a horizontal row of 8 counters.

The other factor is 3, so we’ll make 3 rows of 8 counters.
An image of 3 horizontal rows of counters, each row containing 8 counters.

Now we can count the result. There are 24 counters in all.

3×8=24

If you look at the counters sideways, you’ll see that we could have also made 8 rows of 3 counters. The product would have been the same. We’ll get back to this idea later.

Model each multiplication: 4×6.

Solution


The image shows uses circles to demonstrate the multiplication problem of 4 × 6.

Model each multiplication: 5×7.

Solution


The image shows uses circles to demonstrate the multiplication problem of 5 × 7.

Multiply Whole Numbers

In order to multiply without using models, you need to know all the one digit multiplication facts. Make sure you know them fluently before proceeding in this section.

Table 2 shows the multiplication facts. Each box shows the product of the number down the left column and the number across the top row. If you are unsure about a product, model it. It is important that you memorize any number facts you do not already know so you will be ready to multiply larger numbers.

× 0 1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9
2 0 2 4 6 8 10 12 14 16 18
3 0 3 6 9 12 15 18 21 24 27
4 0 4 8 12 16 20 24 28 32 36
5 0 5 10 15 20 25 30 35 40 45
6 0 6 12 18 24 30 36 42 48 54
7 0 7 14 21 28 35 42 49 56 63
8 0 8 16 24 32 40 48 56 64 72
9 0 9 18 27 36 45 54 63 72 81

What happens when you multiply a number by zero? You can see that the product of any number and zero is zero. This is called the Multiplication Property of Zero.

Multiplication Property of Zero

The product of any number and 0 is 0.

a·0=00·a=0
Multiply:
  1. ⓐ 0·11
  2. ⓑ (42)0
Solution

Solution

This table demonstrates the Zero Property of Multiplication with examples and rules, illustrating that any number multiplied by zero equals zero.
ⓐ 0·11
The product of any number and zero is zero. 0
ⓑ (42)0
Multiplying by zero results in zero. 0
Find each product:
  1. ⓐ 0·19
  2. ⓑ (39)0
Solution
  1. ⓐ 0
  2. ⓑ 0
Find each product:
  1. ⓐ 0·24
  2. ⓑ (57)0
Solution
  1. ⓐ 0
  2. ⓑ 0

What happens when you multiply a number by one? Multiplying a number by one does not change its value. We call this fact the Identity Property of Multiplication, and 1 is called the multiplicative identity.

Identity Property of Multiplication

The product of any number and 1 is the number.

1·a=aa·1=a
Multiply:
  1. ⓐ (11)1
  2. ⓑ 1·42
Solution

Solution

Illustrates the identity property of multiplication with examples where any number multiplied by one retains its original value.
ⓐ (11)1
The product of any number and one is the number. 11
ⓑ 1·42
Multiplying by one does not change the value. 42

Find each product:

  1. ⓐ (19)1
  2. ⓑ 1·39
Solution
  1. ⓐ 19
  2. ⓑ 39

Find each product:

  1. ⓐ (24)(1)
  2. ⓑ 1×57
Solution
  1. ⓐ 24
  2. ⓑ 57

Earlier in this chapter, we learned that the Commutative Property of Addition states that changing the order of addition does not change the sum. We saw that 8+9=17 is the same as 9+8=17.

Is this also true for multiplication? Let’s look at a few pairs of factors.

4·7=287·4=28
9·7=637·9=63
8·9=729·8=72

When the order of the factors is reversed, the product does not change. This is called the Commutative Property of Multiplication.

Commutative Property of Multiplication

Changing the order of the factors does not change their product.

a·b=b·a
Multiply:
  1. ⓐ 8·7
  2. ⓑ 7·8
Solution

Solution

Multiplication problems 8x7 and 7x8 demonstrating their shared result of 56.
ⓐ 8·7
Multiply. 56
ⓑ 7·8
Multiply. 56

Changing the order of the factors does not change the product.

Multiply:
  1. ⓐ 9·6
  2. ⓑ 6·9
Solution

54 and 54; both are the same.

Multiply:
  1. ⓐ 8·6
  2. ⓑ 6·8
Solution

48 and 48; both are the same.

To multiply numbers with more than one digit, it is usually easier to write the numbers vertically in columns just as we did for addition and subtraction.

27×3___

We start by multiplying 3 by 7.

3×7=21

We write the 1 in the ones place of the product. We carry the 2 tens by writing 2 above the tens place.

The image shows a vertical multiplication problem of 27 times 3. An arrow indicates the 2 carried from 3 x 7 = 21 is in the tens place. Another arrow indicates the remaining 1 from 3 x 7 = 21 is written in the ones place below.

Then we multiply the 3 by the 2, and add the 2 above the tens place to the product. So 3×2=6, and 6+2=8. Write the 8 in the tens place of the product.

A vertical multiplication problem showing 27 multiplied by 3, resulting in 81. An arrow indicates that the '8' in 81 is derived from (3 x 2) + the '2' carried from 3 x 7 = 21.

The product is 81.

When we multiply two numbers with a different number of digits, it’s usually easier to write the smaller number on the bottom. You could write it the other way, too, but this way is easier to work with.

Multiply: 15·4.

Solution

Solution

Step-by-step vertical multiplication of 15 by 4, showing both explanatory text and mathematical calculation at each stage.
Write the numbers so the digits 5 and 4 line up vertically. 15 ×4_____
Multiply 4 by the digit in the ones place of 15. 4⋅5=20.
Write 0 in the ones place of the product and carry the 2 tens. 125 ×4_____ 0
Multiply 4 by the digit in the tens place of 15. 4⋅1=4.
Add the 2 tens we carried. 4+2=6.
Write the 6 in the tens place of the product. 125 ×4_____ 60

Multiply: 64·8.

Solution

512

Multiply: 57·6.

Solution

342

Multiply: 286·5.

Solution

Solution

Step-by-step vertical multiplication of a three-digit number by a single-digit number.
Write the numbers so the digits 5 and 6 line up vertically. 286 ×5_____
Multiply 5 by the digit in the ones place of 286. 5⋅6=30.
Write the 0 in the ones place of the product and carry the 3 to the tens place.Multiply 5 by the digit in the tens place of 286. 5⋅8=40. 2836 ×5_____ 0
Add the 3 tens we carried to get 40+3=43.
Write the 3 in the tens place of the product and carry the 4 to the hundreds place.
24836 ×5_____ 30
Multiply 5 by the digit in the hundreds place of 286. 5⋅2=10.
Add the 4 hundreds we carried to get 10+4=14.
Write the 4 in the hundreds place of the product and the 1 to the thousands place.
24836 ×5_____ 1,430

Multiply: 347·5.

Solution

1,735

Multiply: 462·7.

Solution

3,234

When we multiply by a number with two or more digits, we multiply by each of the digits separately, working from right to left. Each separate product of the digits is called a partial product. When we write partial products, we must make sure to line up the place values.

Multiply two whole numbers to find the product.

  1. Write the numbers so each place value lines up vertically.
  2. Multiply the digits in each place value.
    • Work from right to left, starting with the ones place in the bottom number.
      • Multiply the ones digit of the bottom number by the ones digit in the top number, then by the tens digit, and so on.
      • If a product in a place value is more than 9, carry to the next place value.
      • Write the partial products, lining up the digits in the place values with the numbers above.
    • Repeat for the tens place in the bottom number, the hundreds place, and so on.
    • Insert a zero as a placeholder with each additional partial product.
  3. Add the partial products.

Multiply: 62(87).

Solution

Solution

This table illustrates the step-by-step process of multiplying two-digit numbers (e.g., 87 x 62) with corresponding visual aids.
Write the numbers so each place lines up vertically. A vertical multiplication problem setup showing 62 multiplied by 87.
Start by multiplying 7 by 62. Multiply 7 by the digit in the ones place of 62. 7⋅2=14. Write the 4 in the ones place of the product and carry the 1 to the tens place. A vertical multiplication problem setup showing 62 multiplied by 87. Red digits indicate carried numbers ('1' above '6') and the unit digit of the first partial product ('4' below the line from 7x2=14).
Multiply 7 by the digit in the tens place of 62. 7⋅6=42. Add the 1 ten we carried. 42+1=43. Write the 3 in the tens place of the product and the 4 in the hundreds place. A vertical long multiplication problem showing 62 multiplied by 87. The first partial product, 62 x 7, is displayed as 434, along with red carry-over digits '1' and '6' above '62'.
The first partial product is 434.
Now, write a 0 under the 4 in the ones place of the next partial product as a placeholder since we now multiply the digit in the tens place of 87 by 62. Multiply 8 by the digit in the ones place of 62. 8⋅2=16. Write the 6 in the next place of the product, which is the tens place. Carry the 1 to the tens place. Vertical long multiplication problem: 62 multiplied by 87. The steps show 62 x 7 = 434 and 2 x 8 = 16 with the remainder 1 written above in the tens place and 6 written below in the tens place to create the number 60.
Multiply 8 by 6, the digit in the tens place of 62, then add the 1 ten we carried to get 49. Write the 9 in the hundreds place of the product and the 4 in the thousands place. Vertical long multiplication problem: 62 multiplied by 87. The steps show 62 x 7 = 434 and 62 x 80 = 4960, with red digits indicating carries and the second partial product.
The second partial product is 4960. Add the partial products. Vertical long multiplication problem: 62 multiplied by 87. This step shows the partial products 434 added to 4960 resulting in an answer of 5394.

The product is 5,394.

Multiply: 43(78).

Solution

3,354

Multiply: 64(59).

Solution

3,776

Multiply:
  1. ⓐ 47·10
  2. ⓑ 47·100.
Solution

Solution

This table demonstrates the long multiplication method for multiplying a two-digit number by powers of ten (10 and 100), showing the detailed steps for each calculation.
ⓐ 47·10. 47×10___00470___470
ⓑ 47·100 47×100_____000004700_____4,700

When we multiplied 47 times 10, the product was 470. Notice that 10 has one zero, and we put one zero after 47 to get the product. When we multiplied 47 times 100, the product was 4,700. Notice that 100 has two zeros and we put two zeros after 47 to get the product.

Do you see the pattern? If we multiplied 47 times 10,000, which has four zeros, we would put four zeros after 47 to get the product 470,000.

Multiply:

  1. ⓐ 54·10
  2. ⓑ 54·100.
Solution
  1. ⓐ 540
  2. ⓑ 5,400
Multiply:
  1. ⓐ 75·10
  2. ⓑ 75·100.
Solution
  1. ⓐ 750
  2. ⓑ 7,500

Multiply: (354)(438).

Solution

Solution

There are three digits in the factors so there will be 3 partial products. We do not have to write the 0 as a placeholder as long as we write each partial product in the correct place.
An image of the multiplication problem “354 times 438” worked out vertically. 354 is the top number, 438 is the second number. Below 438 is a multiplication bar. Below the bar is the number 2,832. 2832 has the label “Multiply 8 times 354”. Below 2832 is the number 1,062;  1062 has the label “Multiply 3 times 354”.  Below 1062 is the number 1,416; 1416 has the label “Multiply 4 times 354”.  Below this is a bar and below the bar is the number “155,052”, with the label “Add the partial products”.

Multiply: (265)(483).

Solution

127,995

Multiply: (823)(794).

Solution

653,462

Multiply: (896)201.

Solution

Solution

There should be 3 partial products. The second partial product will be the result of multiplying 896 by 0.
An image of the multiplication problem “896 times 201” worked out vertically. 896 is the top number, the 8 in the hundreds place, the 9 in the tens place, the 6 in the ones place. 201 is the second number,  the 2 in the hundreds place, the 0 in the tens place, the 1 in the ones place. Below 201 is a multiplcation bar. Below the bar is the number 896, the 8 in the hundreds place, the 9 in the tens place, the 6 in the ones place. 896 has the label “Multiply 1 times 896”. Below 896 is the number “000”, the 0 in the thousands place, the 0 in the hundreds place, and the 0 in the tens place. “000” has the label “Multiply 0 times 896”.  Below “000” is the number 1792, the 1 in the hundred thousands place, the 7 in the ten thousands place, the 9 in the thousands place, and the 2 in the hundreds place. 1792 has the label “Multiply 2 times 896”.  Below this is a bar and below the bar is the number “180,096”, with the label “Add the partial products”.

Notice that the second partial product of all zeros doesn’t really affect the result. We can place a zero as a placeholder in the tens place and then proceed directly to multiplying by the 2 in the hundreds place, as shown.

Multiply by 10, but insert only one zero as a placeholder in the tens place. Multiply by 200, putting the 2 from the 12. 2·6=12 in the hundreds place.

896×201_____89617920__________180,096

Multiply: (718)509.

Solution

365,462

Multiply: (627)804.

Solution

504,108

When there are three or more factors, we multiply the first two and then multiply their product by the next factor. For example:

Step-by-step demonstration of multiplying three numbers (8 × 3 × 2).
to multiply 8⋅3⋅2
first multiply 8⋅3 24⋅2
then multiply 24⋅2. 48

Translate Word Phrases to Math Notation

Earlier in this section, we translated math notation into words. Now we’ll reverse the process and translate word phrases into math notation. Some of the words that indicate multiplication are given in Table 11.

Operation Word Phrase Example Expression
Multiplication times
product
twice
3 times 8
the product of 3 and 8
twice 4
3×8,3·8,(3)(8),
(3)8,or3(8)
2·4

Translate and simplify: the product of 12 and 27.

Solution

Solution

The word product tells us to multiply. The words of 12 and 27 tell us the two factors.

Steps demonstrating the translation of a word problem ("the product of 12 and 27") into a mathematical expression and its calculated result.
the product of 12 and 27
Translate. 12⋅27
Multiply. 324

Translate and simplify: the product of 13 and 28.

Solution

13 · 28; 364

Translate and simplify: the product of 47 and 14.

Solution

47 · 14; 658

Translate and simplify: twice two hundred eleven.

Solution

Solution

The word twice tells us to multiply by 2.

This table illustrates the process of translating a verbal mathematical phrase into an algebraic expression and then solving for the numerical result.
twice two hundred eleven
Translate. 2(211)
Multiply. 422

Translate and simplify: twice one hundred sixty-seven.

Solution

2(167); 334

Translate and simplify: twice two hundred fifty-eight.

Solution

2(258); 516

Multiply Whole Numbers in Applications

We will use the same strategy we used previously to solve applications of multiplication. First, we need to determine what we are looking for. Then we write a phrase that gives the information to find it. We then translate the phrase into math notation and simplify to get the answer. Finally, we write a sentence to answer the question.

Humberto bought 4 sheets of stamps. Each sheet had 20 stamps. How many stamps did Humberto buy?

Solution

Solution

We are asked to find the total number of stamps.

This table demonstrates a step-by-step process for translating a word problem into a mathematical expression, performing the calculation, and providing the final answer.
Write a phrase for the total. the product of 4 and 20
Translate to math notation. 4⋅20
Multiply. A vertical multiplication problem showing 20 multiplied by 4, with the result 80, written in a clear, dark blue font on a white background.
Write a sentence to answer the question. Humberto bought 80 stamps.

Valia donated water for the snack bar at her son’s baseball game. She brought 6 cases of water bottles. Each case had 24 water bottles. How many water bottles did Valia donate?

Solution

Valia donated 144 water bottles.

Vanessa brought 8 packs of hot dogs to a family reunion. Each pack has 10 hot dogs. How many hot dogs did Vanessa bring?

Solution

Vanessa brought 80 hot dogs.

When Rena cooks rice, she uses twice as much water as rice. How much water does she need to cook 4 cups of rice?

Solution

Solution

We are asked to find how much water Rena needs.

This table illustrates the step-by-step process of translating a word problem into a mathematical expression and deriving the final numerical solution.
Write as a phrase. twice as much as 4 cups
Translate to math notation. 2⋅4
Multiply to simplify. 8
Write a sentence to answer the question. Rena needs 8 cups of water for 4 cups of rice.

Erin is planning her flower garden. She wants to plant twice as many dahlias as sunflowers. If she plants 14 sunflowers, how many dahlias does she need?

Solution

Erin needs 28 dahlias.

A college choir has twice as many women as men. There are 18 men in the choir. How many women are in the choir?

Solution

There are 36 women in the choir.

Van is planning to build a patio. He will have 8 rows of tiles, with 14 tiles in each row. How many tiles does he need for the patio?

Solution

Solution

We are asked to find the total number of tiles.

This table illustrates the step-by-step process of translating a word phrase into mathematical notation, solving it, and providing a final answer.
Write a phrase. the product of 8 and 14
Translate to math notation. 8⋅14
Multiply to simplify. 134×8___112
Write a sentence to answer the question. Van needs 112 tiles for his patio.

Jane is tiling her living room floor. She will need 16 rows of tile, with 20 tiles in each row. How many tiles does she need for the living room floor?

Solution

Jane needs 320 tiles.

Yousef is putting shingles on his garage roof. He will need 24 rows of shingles, with 45 shingles in each row. How many shingles does he need for the garage roof?

Solution

Yousef needs 1,080 tiles.

If we want to know the size of a wall that needs to be painted or a floor that needs to be carpeted, we will need to find its area. The area is a measure of the amount of surface that is covered by the shape. Area is measured in square units. We often use square inches, square feet, square centimeters, or square miles to measure area. A square centimeter is a square that is one centimeter (cm.) on a side. A square inch is a square that is one inch on each side, and so on.

An image of two squares, one larger than the other. The smaller square is 1 centimeter by 1 centimeter and has the label “1 square centimeter”. The larger square is 1 inch by 1 inch and has the label “1 square inch”.

For a rectangular figure, the area is the product of the length and the width. Figure 3 shows a rectangular rug with a length of 2 feet and a width of 3 feet. Each square is 1 foot wide by 1 foot long, or 1 square foot. The rug is made of 6 squares. The area of the rug is 6 square feet.

An image of a rectangle containing 6 blocks, 2 feet tall and 3 feet wide. This image has the label “2 times 3 = 6 feet squared”.
The area of a rectangle is the product of its length and its width, or 6 square feet.

Jen’s kitchen ceiling is a rectangle that measures 9 feet long by 12 feet wide. What is the area of Jen’s kitchen ceiling?

Solution

Solution

We are asked to find the area of the kitchen ceiling.

This table illustrates the step-by-step process of solving an area word problem, from phrasing and notation to calculation and the final answer.
Write a phrase for the area. the product of 9 and 12
Translate to math notation. 9⋅12
Multiply. 112×9___108
Answer with a sentence. The area of Jen's kitchen ceiling is 108 square feet.

Zoila bought a rectangular rug. The rug is 8 feet long by 5 feet wide. What is the area of the rug?

Solution

The area of the rug is 40 square feet.

Rene’s driveway is a rectangle 45 feet long by 20 feet wide. What is the area of the driveway?

Solution

The area of the driveway is 900 square feet

ACCESS ADDITIONAL ONLINE RESOURCES

  • Multiplying Whole Numbers
  • Multiplication with Partial Products
  • Example of Multiplying by Whole Numbers

Key Concepts

Operation Notation Expression Read as Result
Multiplication ×
·
()
3×8
3·8
3(8)
three times eight the product of 3 and 8
  • Multiplication Property of Zero
    • The product of any number and 0 is 0.
      a⋅0=0
      0⋅a=0
  • Identity Property of Multiplication
    • The product of any number and 1 is the number.
      1⋅a=a
      a⋅1=a
  • Commutative Property of Multiplication
    • Changing the order of the factors does not change their product.
      a⋅b=b⋅a
  • Multiply two whole numbers to find the product.
    1. Write the numbers so each place value lines up vertically.
    2. Multiply the digits in each place value.
    3. Work from right to left, starting with the ones place in the bottom number.
    4. Multiply the bottom number by the ones digit in the top number, then by the tens digit, and so on.
    5. If a product in a place value is more than 9, carry to the next place value.
    6. Write the partial products, lining up the digits in the place values with the numbers above. Repeat for the tens place in the bottom number, the hundreds place, and so on.
    7. Insert a zero as a placeholder with each additional partial product.
    8. Add the partial products.

Practice Makes Perfect

Use Multiplication Notation

In the following exercises, translate from math notation to words.

4×7

Solution

four times seven; the product of four and seven

8×6

5·12

Solution

five times twelve; the product of five and twelve

3·9

(10)(25)

Solution

ten times twenty-five; the product of ten and twenty-five

(20)(15)

42(33)

Solution

forty-two times thirty-three; the product of forty-two and thirty-three

39(64)

Model Multiplication of Whole Numbers

In the following exercises, model the multiplication.

3×6

Solution


The image shows how gray circles can be set up to demonstrate the multiplication problem of 3 × 6.

4×5

5×9

Solution


A rectangular array of 45 grey circles, arranged in 5 rows and 9 columns. Below the array, the multiplication equation '5 x 9 = 45' is displayed, illustrating the product of the rows and columns.

3×9

Multiply Whole Numbers

In the following exercises, fill in the missing values in each chart.

An image of a table with 11 columns and 11 rows. The cells in the first row and first column are shaded darker than the other cells. The first column has the values “x; 0; 1; 2; 3; 4; 5; 6; 7; 8; 9”. The second column has the values “0; 0; 0; null; 0; 0; 0; 0; null; 0; 0”. The third column has the values “1; 0; 1; 2; null; 4; 5; 6; null; 8; 9”. The fourth column has the values “2; 0; 2; 4; 6; null; 10; 12; 14; null; 18”. The fifth column has the values “3; null; 3; 6; null; null; 15; null; 21; 24; null”. The sixth column has the values “4; 0; null; 8; 12; 16; null; 24; null; null; 36”. The seventh column has the values “5; 0; null; null; 15; 20; null; null; 35; null; 45”. The eighth column has the values “6; 0; 6; 12; null; null; 30; null; null; 48; null”. The ninth column has the values “7; 0; 7; null; 21; 28; null; 42; null; null; null”. The tenth column has the values “8; null; 8; null; null; 32; 40; null; 56; 64; 72”. The eleventh column has the values “9; 0; null; 18; 27; null, null; 54; 63; null; null”.
Solution


An image of a table with 11 columns and 11 rows. The cells in the first row and first column are shaded darker than the other cells. The cells contain numbers and answers to the problem.

An image of a table with 11 columns and 11 rows. The cells in the first row and first column are shaded darker than the other cells. The first column has the values “x; 0; 1; 2; 3; 4; 5; 6; 7; 8; 9”. The second column has the values “0; 0; 0; 0 pink; 0; 0; 0; 0; 0; 0; 0”. The third column has the values “1; 0; 1; 2; 3; 4; 5; 6; 7; 8; 9”. The fourth column has the values “2; 0; 2; 4; 6; 8; 10; 12; 14; 16; 18”. The fifth column has the values “3; 0; 3; 6; 9; 12; 15; 18; 21; 24; 27”. The sixth column has the values “4; 0; 4; 8; 12; 16; 20; 24; 28; 32; 36”. The seventh column has the values “5; 0; 5; 10; 15; 20; 25; 30; 35; 40; 45”. The eighth column has the values “6; 0; 6; 12; 18; 24; 30; 36; 42; 48; 54”. The ninth column has the values “7; 0; 7; 14; 21; 28; 35; 42; 49; 56; 63”. The tenth column has the values “8; 0; 8; 16; 24; 32; 40; 48; 56; 64; 72”. The eleventh column has the values “9; 0; 9; 18; 27; 36, 45; 54; 63; 72; 81”.
An image of a table with 8 columns and 7 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null.  The first column has the values “x; 4; 5; 6; 7; 8; 9”. The first row has the values “x; 3; 4; 5; 6; 7; 8; 9”.
Solution


An image of a table with 8 columns and 7 rows. The cells in the first row and first column are shaded darker than the other cells. The cells contain numbers and answers to the problem.

PROD: An image of a table with 7 columns and 8 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null.  The first column has the values “x; 3; 4; 5; 6; 7; 8; 9”. The first row has the values “x; 4; 5; 6; 7; 8; 9”.
An image of a table with 8 columns and 5 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null.  The first row has the values “x; 3; 4; 5; 6; 7; 8; 9”. The first column has the values “x;  6; 7; 8; 9”.
Solution


An image of a table with 8 columns and 5 rows. The cells in the first row and first column are shaded darker than the other cells. The cells contain numbers and answers to the problem.

An image of a table with 5 columns and 8 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null. The first column has the values “x; 3; 4; 5; 6; 7; 8; 9”. The first row has the values “x; 6; 7; 8; 9”.
PROD: An image of a table with 6 columns and 6 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null.  The first column has the values “x; 5; 6; 7; 8; 9”. The first row has the values “x; 5; 6; 7; 8; 9”.
Solution


A dark teal and grey multiplication table showing the products of numbers 5 through 9. The rows and columns are labeled with these numbers, and the grid displays their respective multiplication results.

An image of a table with 6 columns and 6 rows. The cells in the first row and first column are shaded darker than the other cells. The cells not in the first row or column are all null.  The first column has the values “x; 5; 6; 7; 8; 9”. The first row has the values “x; 5; 6; 7; 8; 9”.

In the following exercises, multiply.

0·15

Solution

0

0·41

(99)0

Solution

0

(77)0

1·43

Solution

43

1·34

(28)1

Solution

28

(65)1

1(240,055)

Solution

240,055

1(189,206)

  1. ⓐ 7·6
  2. ⓑ 6·7
Solution
  1. ⓐ 42
  2. ⓑ 42
  1. ⓐ 8×9
  2. ⓑ 9×8

(79)(5)

Solution

395

(58)(4)

275·6

Solution

1,650

638·5

3,421×7

Solution

23,947

9,143×3

52(38)

Solution

1,976

37(45)

96·73

Solution

7,008

89·56

27×85

Solution

2,295

53×98

23·10

Solution

230

19·10

(100)(36)

Solution

3,600

(100)(25)

1,000(88)

Solution

88,000

1,000(46)

50×1,000,000

Solution

50,000,000

30×1,000,000

247×139

Solution

34,333

156×328

586(721)

Solution

422,506

472(855)

915·879

Solution

804,285

968·926

(104)(256)

Solution

26,624

(103)(497)

348(705)

Solution

245,340

485(602)

2,719×543

Solution

1,476,417

3,581×724

Translate Word Phrases to Math Notation

In the following exercises, translate and simplify.

the product of 18 and 33

Solution

18 · 33; 594

the product of 15 and 22

fifty-one times sixty-seven

Solution

51(67); 3,417

forty-eight times seventy-one

twice 249

Solution

2(249); 498

twice 589

ten times three hundred seventy-five

Solution

10(375); 3,750

ten times two hundred fifty-five

Mixed Practice

In the following exercises, simplify.

38×37

Solution

1,406

86×29

415−267

Solution

148

341−285

6,251+4,749

Solution

11,000

3,816+8,184

(56)(204)

Solution

11,424

(77)(801)

947·0

Solution

0

947+0

15,382+1

Solution

15,383

15,382·1

In the following exercises, translate and simplify.

the difference of 50 and 18

Solution

50 − 18; 32

the difference of 90 and 66

twice 35

Solution

2(35); 70

twice 140

20 more than 980

Solution

20 + 980; 1,000

65 more than 325

the product of 12 and 875

Solution

12(875); 10,500

the product of 15 and 905

subtract 74 from 89

Solution

89 − 74; 15

subtract 45 from 99

the sum of 3,075 and 95

Solution

3,075 + 95; 3,170

the sum of 6,308 and 724

366 less than 814

Solution

814 − 366; 448

388 less than 925

Multiply Whole Numbers in Applications

In the following exercises, solve.

Party supplies Tim brought 9 six-packs of soda to a club party. How many cans of soda did Tim bring?

Solution

Tim brought 54 cans of soda to the party.

Sewing Kanisha is making a quilt. She bought 6 cards of buttons. Each card had four buttons on it. How many buttons did Kanisha buy?

Field trip Seven school busses let off their students in front of a museum in Washington, DC. Each school bus had 44 students. How many students were there?

Solution

There were 308 students.

Gardening Kathryn bought 8 flats of impatiens for her flower bed. Each flat has 24 flowers. How many flowers did Kathryn buy?

Charity Rey donated 15 twelve-packs of t-shirts to a homeless shelter. How many t-shirts did he donate?

Solution

Rey donated 180 t-shirts.

School There are 28 classrooms at Anna C. Scott elementary school. Each classroom has 26 student desks. What is the total number of student desks?

Recipe Stephanie is making punch for a party. The recipe calls for twice as much fruit juice as club soda. If she uses 10 cups of club soda, how much fruit juice should she use?

Solution

Stephanie should use 20 cups of fruit juice.

Gardening Hiroko is putting in a vegetable garden. He wants to have twice as many lettuce plants as tomato plants. If he buys 12 tomato plants, how many lettuce plants should he get?

Government The United States Senate has twice as many senators as there are states in the United States. There are 50 states. How many senators are there in the United States Senate?

Solution

There are 100 senators in the U.S. senate.

Recipe Andrea is making potato salad for a buffet luncheon. The recipe says the number of servings of potato salad will be twice the number of pounds of potatoes. If she buys 30 pounds of potatoes, how many servings of potato salad will there be?

Painting Jane is painting one wall of her living room. The wall is rectangular, 13 feet wide by 9 feet high. What is the area of the wall?

Solution

The area of the wall is 117 square feet.

Home décor Shawnte bought a rug for the hall of her apartment. The rug is 3 feet wide by 18 feet long. What is the area of the rug?

Room size The meeting room in a senior center is rectangular, with length 42 feet and width 34 feet. What is the area of the meeting room?

Solution

The area of the room is 1,428 square feet.

Gardening June has a vegetable garden in her yard. The garden is rectangular, with length 23 feet and width 28 feet. What is the area of the garden?

NCAA basketball According to NCAA regulations, the dimensions of a rectangular basketball court must be 94 feet by 50 feet. What is the area of the basketball court?

Solution

The area of the court is 4,700 square feet.

NCAA football According to NCAA regulations, the dimensions of a rectangular football field must be 360 feet by 160 feet. What is the area of the football field?

Everyday Math

Stock market Javier owns 300 shares of stock in one company. On Tuesday, the stock price rose $12 per share. How much money did Javier’s portfolio gain?


Solution

Javier’s portfolio gained $3,600.

Salary Carlton got a $200 raise in each paycheck. He gets paid 24 times a year. How much higher is his new annual salary?

Writing Exercises

How confident do you feel about your knowledge of the multiplication facts? If you are not fully confident, what will you do to improve your skills?

Solution

Answers will vary.

How have you used models to help you learn the multiplication facts?

Self Check

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

A self-assessment chart for multiplication skills, including using notation, modeling, multiplying whole numbers, translating word phrases, and applying multiplication.

ⓑ 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?

product
The product is the result of multiplying two or more numbers.

Divide Whole Numbers

Learning Objectives

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

  • Use division notation
  • Model division of whole numbers
  • Divide whole numbers
  • Translate word phrases to math notation
  • Divide whole numbers in applications

Before you get started, take this readiness quiz.

Multiply: 27·3.
If you missed this problem, review Example 6 in Multiply Whole Numbers.

Solution

81

Subtract: 43−26.
If you missed this problem, review Example 7 in Subtract Whole Numbers

Solution

17

Multiply: 62(87).
If you missed this problem, review Example 7 in Multiply Whole Numbers.

Solution

5,394

Use Division Notation

So far we have explored addition, subtraction, and multiplication. Now let’s consider division. Suppose you have the 12 cookies in Figure 1 and want to package them in bags with 4 cookies in each bag. How many bags would we need?

An image of three rows of four cookies to show twelve cookies.

You might put 4 cookies in first bag, 4 in the second bag, and so on until you run out of cookies. Doing it this way, you would fill 3 bags.

An image of 3 bags of cookies, each bag containing 4 cookies.

In other words, starting with the 12 cookies, you would take away, or subtract, 4 cookies at a time. Division is a way to represent repeated subtraction just as multiplication represents repeated addition.

Instead of subtracting 4 repeatedly, we can write

12÷4

We read this as twelve divided by four and the result is the quotient of 12 and 4. The quotient is 3 because we can subtract 4 from 12 exactly 3 times. We call the number being divided the dividend and the number dividing it the divisor. In this case, the dividend is 12 and the divisor is 4.

In the past you may have used the notation 412, but this division also can be written as 12÷4,12/4,124. In each case the 12 is the dividend and the 4 is the divisor.

Operation Symbols for Division

To represent and describe division, we can use symbols and words.

Operation Notation Expression Read as Result
Division ÷
ab
ba
a/b
12÷4
124
412
12/4
Twelve divided by four the quotient of 12 and 4

Division is performed on two numbers at a time. When translating from math notation to English words, or English words to math notation, look for the words of and and to identify the numbers.

Translate from math notation to words.

ⓐ 64÷8 ⓑ 427 ⓒ 428

Solution

Solution

  • ⓐ We read this as sixty-four divided by eight and the result is the quotient of sixty-four and eight.
  • ⓑ We read this as forty-two divided by seven and the result is the quotient of forty-two and seven.
  • ⓒ We read this as twenty-eight divided by four and the result is the quotient of twenty-eight and four.

Translate from math notation to words:

ⓐ 84÷7 ⓑ 186 ⓒ 824

Solution
  • ⓐ eighty-four divided by seven; the quotient of eighty-four and seven
  • ⓑ eighteen divided by six; the quotient of eighteen and six.
  • ⓒ twenty-four divided by eight; the quotient of twenty-four and eight

Translate from math notation to words:

ⓐ 72÷9 ⓑ 213 ⓒ 654

Solution
  • ⓐ seventy-two divided by nine; the quotient of seventy-two and nine
  • ⓑ twenty-one divided by three; the quotient of twenty-one and three
  • ⓒ fifty-four divided by six; the quotient of fifty-four and six

Model Division of Whole Numbers

As we did with multiplication, we will model division using counters. The operation of division helps us organize items into equal groups as we start with the number of items in the dividend and subtract the number in the divisor repeatedly.

Doing the Manipulative Math Worksheets activity "Model Division of Whole Numbers" will help you develop a better understanding of dividing whole numbers.

Model the division: 24÷8.

Solution

Solution

To find the quotient 24÷8, we want to know how many groups of 8 are in 24.

Model the dividend. Start with 24 counters.
An image of 24 counters placed randomly.

The divisor tell us the number of counters we want in each group. Form groups of 8 counters.
An image of 24 counters, all contained in 3 bubbles, each bubble containing 8 counters.

Count the number of groups. There are 3 groups.

24÷8=3

Model: 24÷6.

Solution


Four irregular shapes, each containing seven gray circles, arranged in a grid pattern.

Model: 42÷7.

Solution


An illustration of six irregular shapes, each containing several smaller grey circles, arranged in a scattered pattern.

Divide Whole Numbers

We said that addition and subtraction are inverse operations because one undoes the other. Similarly, division is the inverse operation of multiplication. We know 12÷4=3 because 3·4=12. Knowing all the multiplication number facts is very important when doing division.

We check our answer to division by multiplying the quotient by the divisor to determine if it equals the dividend. In Example 2, we know 24÷8=3 is correct because 3·8=24.

Divide. Then check by multiplying. ⓐ 42÷6 ⓑ 729 ⓒ 763

Solution

Solution

  • Illustrates division of 42 by 6, with calculation and verification steps.
    ⓐ
    42÷6
    Divide 42 by 6. 7
    Check by multiplying.
    7·6
    42✓
  • Demonstration of dividing 72 by 9, including the calculation and verification steps.
    ⓑ
    729
    Divide 72 by 9. 8
    Check by multiplying.
    8·9
    72✓
  • Step-by-step demonstration of dividing 63 by 7 using long division, including a multiplication check.
    ⓒ
    763
    Divide 63 by 7. 9
    Check by multiplying.
    9·7
    63✓

Divide. Then check by multiplying:

ⓐ 54÷6 ⓑ 279

Solution

ⓐ 9 ⓑ 3

Divide. Then check by multiplying:

ⓐ 369 ⓑ 840

Solution

ⓐ 4 ⓑ 5

What is the quotient when you divide a number by itself?

1515=1because1·15=15

Dividing any number (except 0) by itself produces a quotient of 1. Also, any number divided by 1 produces a quotient of the number. These two ideas are stated in the Division Properties of One.

Division Properties of One

This table outlines basic division properties, presenting their descriptive rules and corresponding mathematical expressions.
Any number (except 0) divided by itself is one. a÷a=1
Any number divided by one is the same number. a÷1=a

Divide. Then check by multiplying:

  1. ⓐ 11÷11
  2. ⓑ 191
  3. ⓒ 17
Solution

Solution

  • ⓐ
    11÷11
    A number divided by itself is 1. 1
    Check by multiplying.
    1·11
    11✓
  • ⓑ
    191
    A number divided by 1 equals itself. 19
    Check by multiplying.
    19·1
    19✓
  • ⓒ
    17
    A number divided by 1 equals itself. 7
    Check by multiplying.
    7·1
    7✓

Divide. Then check by multiplying:

ⓐ 14÷14 ⓑ 271

Solution
  1. ⓐ 1
  2. ⓑ 27

Divide. Then check by multiplying:

ⓐ 161 ⓑ 14

Solution
  1. ⓐ 16
  2. ⓑ 4

Suppose we have $0, and want to divide it among 3 people. How much would each person get? Each person would get $0. Zero divided by any number is 0.

Now suppose that we want to divide $10 by 0. That means we would want to find a number that we multiply by 0 to get 10. This cannot happen because 0 times any number is 0. Division by zero is said to be undefined.

These two ideas make up the Division Properties of Zero.

Division Properties of Zero

Rules for division involving zero, showing verbal descriptions and corresponding mathematical expressions.
Zero divided by any number is 0. 0÷a=0
Dividing a number by zero is undefined. a÷0 undefined

Another way to explain why division by zero is undefined is to remember that division is really repeated subtraction. How many times can we take away 0 from 10? Because subtracting 0 will never change the total, we will never get an answer. So we cannot divide a number by 0.

Divide. Check by multiplying: ⓐ 0÷3 ⓑ 10/0.

Solution

Solution

  • Demonstration and verification of the property that zero divided by any non-zero number equals zero.
    ⓐ
    0÷3
    Zero divided by any number is zero. 0
    Check by multiplying.
    0·3
    0✓
  • Illustration of division by zero being undefined, showing a mathematical expression and its resulting state.
    ⓑ
    10/0
    Division by zero is undefined. undefined

Divide. Then check by multiplying:

ⓐ 0÷2 ⓑ 17/0

Solution

ⓐ 0 ⓑ undefined

Divide. Then check by multiplying:

ⓐ 0÷6 ⓑ 13/0

Solution

ⓐ 0 ⓑ undefined

When the divisor or the dividend has more than one digit, it is usually easier to use the 412 notation. This process is called long division. Let’s work through the process by dividing 78 by 3.

Divide the first digit of the dividend, 7, by the divisor, 3.
The divisor 3 can go into 7 two times since 2×3=6. Write the 2 above the 7 in the quotient. A long division problem showing 78 divided by 3. The initial step highlights dividing the 7 by 3, resulting in 2 as the first digit of the quotient.
Multiply the 2 in the quotient by 3 and write the product, 6, under the 7. A long division problem showing the first step of dividing 78 by 3. The digit 7 is divided by 3, resulting in a quotient of 2 and a remainder of 1 after subtracting 6.
Subtract that product from the first digit in the dividend. Subtract 7−6. Write the difference, 1, under the first digit in the dividend. A long division problem showing the first step of dividing 78 by 3. The digit 7 is divided by 3, resulting in a quotient of 2 and a remainder of 1 after subtracting 6.
Bring down the next digit of the dividend. Bring down the 8. A long division problem showing the next step of dividing 78 by 3. The digit 6 is subtracted from 7 resulting in 1. The digit 8 is brought down for a resulting number of 18.
Divide 18 by the divisor, 3. The divisor 3 goes into 18 six times. A long division problem showing the next step of dividing 78 by 3. In this step, the number 18 is divided by 3. The resulting number 6 is placed above the long division problem, revealing the overall answer to be 26.
Write 6 in the quotient above the 8.
Multiply the 6 in the quotient by the divisor and write the product, 18, under the dividend. Subtract 18 from 18. A long division problem showing the final step of dividing 78 by 3. The number 18 can be fully divisible by 3, leaving zero remainders and and overall answer of 26.

We would repeat the process until there are no more digits in the dividend to bring down. In this problem, there are no more digits to bring down, so the division is finished.

So78÷3=26.

Check by multiplying the quotient times the divisor to get the dividend. Multiply 26×3 to make sure that product equals the dividend, 78.

216×3___78✓

It does, so our answer is correct.

Divide whole numbers.

  1. Divide the first digit of the dividend by the divisor.
    If the divisor is larger than the first digit of the dividend, divide the first two digits of the dividend by the divisor, and so on.
  2. Write the quotient above the dividend.
  3. Multiply the quotient by the divisor and write the product under the dividend.
  4. Subtract that product from the dividend.
  5. Bring down the next digit of the dividend.
  6. Repeat from Step 1 until there are no more digits in the dividend to bring down.
  7. Check by multiplying the quotient times the divisor.

Divide 2,596÷4. Check by multiplying:

Solution

Solution

Let's rewrite the problem to set it up for long division. A long division problem set up to divide 2596 by 4.
Divide the first digit of the dividend, 2, by the divisor, 4. A long division problem set up to divide 2596 by 4. The first digit, 2, is colored red.
Since 4 does not go into 2, we use the first two digits of the dividend and divide 25 by 4. The divisor 4 goes into 25 six times.
We write the 6 in the quotient above the 5. A long division problem showing 2596 being divided by 4. The first digit of the quotient, 6, is shown above the 5, indicating that 25 divided by 4 is 6.
Multiply the 6 in the quotient by the divisor 4 and write the product, 24, under the first two digits in the dividend. A long division problem showing 2596 divided by 4. The first step of the division shows that 4 goes into 25 six times, with 24 (in red) written below 25, indicating the product of 4 and 6.
Subtract that product from the first two digits in the dividend. Subtract 25−24. Write the difference, 1, under the second digit in the dividend. A long division problem showing 2596 divided by 4. The first step of the division shows that 4 goes into 25 six times, with 24 (in red) written below 25, indicating the product of 4 and 6. After subtracting 24 from 25, the resulting remainder is 1.
Now bring down the 9 and repeat these steps. There are 4 fours in 19. Write the 4 over the 9. Multiply the 4 by 4 and subtract this product from 19. A long division problem showing 2596 divided by 4. The next step of the division brings down the third digit in the quotient, 9, resulting in 19, the next number to be divided by 4. 4 can go into 19 four times (16, written below 19). The number four is written above 2596. Finally 19 minus 16 equals a remainder of 3.
Bring down the 6 and repeat these steps. There are 9 fours in 36. Write the 9 over the 6. Multiply the 9 by 4 and subtract this product from 36. Long division of 2596 by 4, yielding a result of 649 with no remainder, demonstrating the step-by-step process of division.
So 2,596÷4=649.
Check by multiplying.
A multiplication problem in which 649 is multiplied by 4, resulting in 2596.

It equals the dividend, so our answer is correct.

Divide. Then check by multiplying: 2,636÷4

Solution

659

Divide. Then check by multiplying: 2,716÷4

Solution

679

Divide 4,506÷6. Check by multiplying:

Solution

Solution

Let's rewrite the problem to set it up for long division. A long division problem showing 4506 divided by 6.
First we try to divide 6 into 4. A long division problem showing 4506 divided by 6. The first digit, 4, is in red.
Since that won't work, we try 6 into 45.
There are 7 sixes in 45. We write the 7 over the 5.
A long division problem with 4506 being divided by 6. The number 7 is shown as the first digit of the quotient above the 5, indicating that 45 divided by 6 is 7 with a remainder.
Multiply the 7 by 6 and subtract this product from 45. An illustration of long division, specifically the initial step where 45 is divided by 6, yielding 7, and 42 is subtracted to find the remainder of 3.
Now bring down the 0 and repeat these steps. There are 5 sixes in 30.
Write the 5 over the 0. Multiply the 5 by 6 and subtract this product from 30.
An illustration of long division, specifically the step in which the digit 0 is brought down with the remainder 3 resulting in 30 which can then be further divided by 6. 30 can be fully divided by 6 five times. Thus, the number 5 is written above the division problem, resulting in a remainder of 0.
Now bring down the 6 and repeat these steps. There is 1 six in 6.
Write the 1 over the 6. Multiply 1 by 6 and subtract this product from 6.
An illustration of long division, showing the final step in which the digit 6 is brought down with the remainder 0 resulting in 6 which can then be divided by 6 one time. Thus, the number 1 is written above the division problem, resulting in a final remainder of 0 and a final answer of 751.
Check by multiplying.
A vertical multiplication problem showing 751 multiplied by 6, with the correct result of 4,506 marked by a checkmark.

It equals the dividend, so our answer is correct.

Divide. Then check by multiplying: 4,305÷5.

Solution

861

Divide. Then check by multiplying: 3,906÷6.

Solution

651

Divide 7,263÷9. Check by multiplying.

Solution

Solution

Let's rewrite the problem to set it up for long division. A long division problem showing 7263 divided by 9.
First we try to divide 9 into 7. A long division problem showing 7263 divided by 9. THe first digit, 7, is in red.
Since that won't work, we try 9 into 72. There are 8 nines in 72.
We write the 8 over the 2.
A long division problem showing 7263 divided by 9. The image shows the first step, dividing the first two digits, 72, by 9, resulting in 8 which is written above the division line.
Multiply the 8 by 9 and subtract this product from 72. A long division problem showing 7263 divided by 9. The image shows the next step, in which 9 is multiplied by 8, resulting in 72, which is then written below the intitial 72, creating a remainder of 0.
Now bring down the 6 and repeat these steps. There are 0 nines in 6.
Write the 0 over the 6. Multiply the 0 by 9 and subtract this product from 6.
A long division problem showing 7263 divided by 9. The image shows the next step in which the digit 6 is brought down. Six is not divisible by 9, so 0 is written above the division problem. Nine multiplied by 0 equals 0, which is written below, creating a remainder of 6.
Now bring down the 3 and repeat these steps. There are 7 nines in 63. Write the 7 over the 3.
Multiply the 7 by 9 and subtract this product from 63.
A long division problem showing 7263 divided by 9. The image shows the next step in which the final digit 3 is brought down next to the number 6, resulting in 63. 63 is divisible by 9 7 times. Thus, 7 is written above the division line, leaving zero remainder and a final answer of 807.
Check by multiplying.
A multiplication problem in which 807 is multipled by 9 to equal 7,263 with a checkmark next to the answer.

It equals the dividend, so our answer is correct.

Divide. Then check by multiplying: 4,928÷7.

Solution

704

Divide. Then check by multiplying: 5,663÷7.

Solution

809

So far all the division problems have worked out evenly. For example, if we had 24 cookies and wanted to make bags of 8 cookies, we would have 3 bags. But what if there were 28 cookies and we wanted to make bags of 8? Start with the 28 cookies as shown in Figure 2.

An image of 28 cookies placed at random.

Try to put the cookies in groups of eight as in Figure 3.

An image of 28 cookies. There are 3 circles, each containing 8 cookies, leaving 3 cookies outside the circles.

There are 3 groups of eight cookies, and 4 cookies left over. We call the 4 cookies that are left over the remainder and show it by writing R4 next to the 3. (The R stands for remainder.)

To check this division we multiply 3 times 8 to get 24, and then add the remainder of 4.

3×8___24+4___28

Divide 1,439÷4. Check by multiplying.

Solution

Solution

Let's rewrite the problem to set it up for long division. A mathematical expression showing the long division of 1439 by 4.
First we try to divide 4 into 1. Since that won't work, we try 4 into 14.
There are 3 fours in 14. We write the 3 over the 4.
A long division problem is shown, indicating the calculation of 1439 divided by 4, with the first digit of the quotient, 3, placed above the '14' of 1439.
Multiply the 3 by 4 and subtract this product from 14. A long division problem is shown where 1439 is divided by 4. In this step, multiplying 4 and 3 equals 12, which is written below the first two digits, 14. 14 minus 12 equals a remainder of 2.
Now bring down the 3 and repeat these steps. There are 5 fours in 23.
Write the 5 over the 3. Multiply the 5 by 4 and subtract this product from 23.
A long division problem is shown where 1439 is divided by 4. In this step, the digit 3 is brought down next to the remainder 2, resulting in the number 23. 23 is divisible by 4 5 times. 5 is written above and the number 20 is written below 23, resulting in a remainder of 3.
Now bring down the 9 and repeat these steps. There are 9 fours in 39.
Write the 9 over the 9. Multiply the 9 by 4 and subtract this product from 39.
There are no more numbers to bring down, so we are done.
The remainder is 3.
A long division problem is shown where 1439 is divided by 4. In this step, the final digit 9 is brought down next to the previous remainder of 3, creating the number 39. The number 4 can only go into 39 nine times, resulting in the number 36, creating a final remainder of 3. The answer written above is then 359R3.
Check by multiplying.
A multiplication problem showing 359 times 4, resulting in 1,436. This number is then added to 3, resulting in a final answer of 1,439.

So 1,439÷4 is 359 with a remainder of 3. Our answer is correct.

Divide. Then check by multiplying: 3,812÷8.

Solution

476 with a remainder of 4

Divide. Then check by multiplying: 4,319÷8.

Solution

539 with a remainder of 7

Divide and then check by multiplying: 1,461÷13.

Solution

Solution

Let's rewrite the problem to set it up for long division. 131,461
First we try to divide 13 into 1. Since that won't work, we try 13 into 14.
There is 1 thirteen in 14. We write the 1 over the 4.
A long division problem showing 1461 divided by 13 with the number 1 written above the long division line.
Multiply the 1 by 13 and subtract this product from 14. A long division problem is shown, with 1461 being divided by 13. The first step of the division is completed, showing 13 subtracted from 14, resulting in a remainder of 1. The quotient above is 1.
Now bring down the 6 and repeat these steps. There is 1 thirteen in 16.
Write the 1 over the 6. Multiply the 1 by 13 and subtract this product from 16.
A long division problem in progress, showing 1461 divided by 13. The current steps result in a partial quotient of 11 with a remainder of 3, indicating the full calculation is incomplete.
Now bring down the 1 and repeat these steps. There are 2 thirteens in 31.
Write the 2 over the 1. Multiply the 2 by 13 and subtract this product from 31. There are no more numbers to bring down, so we are done.
The remainder is 5. 1,462÷13 is 112 with a remainder of 5.
A long division problem in progress, showing 1461 divided by 13. This final step shows the answer is 112 with a remainder of 5.
Check by multiplying.
A handwritten arithmetic problem demonstrating the check for a division, where the quotient (112) is multiplied by the divisor (13), and the remainder (5) is added to reach the original number (1461).

Our answer is correct.

Divide. Then check by multiplying: 1,493÷13.

Solution

114 R11

Divide. Then check by multiplying: 1,461÷12.

Solution

121 R9

Divide and check by multiplying: 74,521÷241.

Solution

Solution

Let's rewrite the problem to set it up for long division. 24174,521
First we try to divide 241 into 7. Since that won’t work, we try 241 into 74. That still won’t work, so we try 241 into 745. Since 2 divides into 7 three times, we try 3.
Since 3×241=723, we write the 3 over the 5 in 745.
Note that 4 would be too large because 4×241=964, which is greater than 745.
Multiply the 3 by 241 and subtract this product from 745. A long division problem showing 74521 divided by 241. In this step, the first three digits 745 are divided by 241, which allows the student to write 4 above the division line, and subtract 723 from 745, creating a remainder of 22.
Now bring down the 2 and repeat these steps. 241 does not divide into 222.
We write a 0 over the 2 as a placeholder and then continue.
A long division problem showing 74521 divided by 241. In this step, the fourth digit, 2, is written next to the previous remainder of 22, creating the number 222, which is not divisible by 241. Thus, the number zero is written above the division line.
Now bring down the 1 and repeat these steps. Try 9. Since 9×241=2,169,
we write the 9 over the 1. Multiply the 9 by 241 and subtract this product from 2,221.
A handwritten long division problem showing 74521 divided by 241, resulting in a quotient of 309 with a remainder of 52.
There are no more numbers to bring down, so we are finished. The remainder is 52. So 74,521÷241
is 309 with a remainder of 52.
Check by multiplying.
A multiplication problem in which 309 is multiplied by 241. With the added remainder of 52, the final answer is 74,521.

Sometimes it might not be obvious how many times the divisor goes into digits of the dividend. We will have to guess and check numbers to find the greatest number that goes into the digits without exceeding them.

Divide. Then check by multiplying: 78,641÷256.

Solution

307 R49

Divide. Then check by multiplying: 76,461÷248.

Solution

308 R77

Translate Word Phrases to Math Notation

Earlier in this section, we translated math notation for division into words. Now we’ll translate word phrases into math notation. Some of the words that indicate division are given in Table 19.

Operation Word Phrase Example Expression
Division divided by
quotient of
divided into
12 divided by 4
the quotient of 12 and 4
4 divided into 12
12÷4
124
12/4
412

Translate and simplify: the quotient of 51 and 17.

Solution

Solution

The word quotient tells us to divide.

the quotient of 51 and 17Translate.51÷17Divide.3

We could just as correctly have translated the quotient of 51 and 17 using the notation

1751or5117.

Translate and simplify: the quotient of 91 and 13.

Solution

91 ÷ 13; 7

Translate and simplify: the quotient of 52 and 13.

Solution

52 ÷ 13; 4

Divide Whole Numbers in Applications

We will use the same strategy we used in previous sections to solve applications. First, we determine what we are looking for. Then we write a phrase that gives the information to find it. We then translate the phrase into math notation and simplify it to get the answer. Finally, we write a sentence to answer the question.

Cecelia bought a 160-ounce box of oatmeal at the big box store. She wants to divide the 160 ounces of oatmeal into 8-ounce servings. She will put each serving into a plastic bag so she can take one bag to work each day. How many servings will she get from the big box?

Solution

Solution

We are asked to find the how many servings she will get from the big box.

Demonstrates the process of solving a division word problem: phrasing, math notation, simplification, and final answer.
Write a phrase. 160 ounces divided by 8 ounces
Translate to math notation. 160÷8
Simplify by dividing. 20
Write a sentence to answer the question. Cecelia will get 20 servings from the big box.

Marcus is setting out animal crackers for snacks at the preschool. He wants to put 9 crackers in each cup. One box of animal crackers contains 135 crackers. How many cups can he fill from one box of crackers?

Solution

Marcus can fill 15 cups.

Andrea is making bows for the girls in her dance class to wear at the recital. Each bow takes 4 feet of ribbon, and 36 feet of ribbon are on one spool. How many bows can Andrea make from one spool of ribbon?

Solution

Andrea can make 9 bows.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Dividing Whole Numbers
  • Dividing Whole Numbers No Remainder
  • Dividing Whole Numbers With Remainder

Key Concepts

Operation Notation Expression Read as Result
Division ÷
ab
ba
a/b
12÷4
124
412
12/4
Twelve divided by four the quotient of 12 and 4
  • Division Properties of One
    • Any number (except 0) divided by itself is one. a÷a=1
    • Any number divided by one is the same number. a÷1=a
  • Division Properties of Zero
    • Zero divided by any number is 0. 0÷a=0
    • Dividing a number by zero is undefined. a÷0 undefined
  • Divide whole numbers.
    1. Divide the first digit of the dividend by the divisor.
      If the divisor is larger than the first digit of the dividend, divide the first two digits of the dividend by the divisor, and so on.
    2. Write the quotient above the dividend.
    3. Multiply the quotient by the divisor and write the product under the dividend.
    4. Subtract that product from the dividend.
    5. Bring down the next digit of the dividend.
    6. Repeat from Step 1 until there are no more digits in the dividend to bring down.
    7. Check by multiplying the quotient times the divisor.

Section Exercises

Practice Makes Perfect

Use Division Notation

In the following exercises, translate from math notation to words.

54÷9

Solution

fifty-four divided by nine; the quotient of fifty-four and nine

567

328

Solution

thirty-two divided by eight; the quotient of thirty-two and eight

642

48÷6

Solution

forty-eight divided by six; the quotient of forty-eight and six

639

763

Solution

sixty-three divided by seven; the quotient of sixty-three and seven

72÷8

Model Division of Whole Numbers

In the following exercises, model the division.

15÷5

Solution


Three distinct clusters of grey circles, each enclosed by a blob-like outline, illustrating concepts of grouping or classification.

10÷5

147

Solution


Two irregular shapes enclose groups of grey circles, depicting distinct clusters or collections of items. This image could represent separated data points or cellular groupings.

186

420

Solution


An illustration showing multiple clusters of gray circles, with each group enclosed by an irregular outline, representing data grouping or segmentation.

315

24÷6

Solution


An illustration showing clusters of gray circles, with each group enclosed by an irregular outline, representing data grouping or segmentation.

16÷4

Divide Whole Numbers

In the following exercises, divide. Then check by multiplying.

18÷2

Solution

9

14÷2

273

Solution

9

303

428

Solution

7

436

455

Solution

9

355

72/8

Solution

9

864

357

Solution

5

42÷7

1515

Solution

1

1212

43÷43

Solution

1

37÷37

231

Solution

23

291

19÷1

Solution

19

17÷1

0÷4

Solution

0

0÷8

50

Solution

undefined

90

260

Solution

undefined

320

120

Solution

0

160

72÷3

Solution

24

57÷3

968

Solution

12

786

5465

Solution

93

4528

924÷7

Solution

132

861÷7

5,2266

Solution

871

3,7768

431,324

Solution

7,831

546,855

7,209÷3

Solution

2,403

4,806÷3

5,406÷6

Solution

901

3,208÷4

42,816

Solution

704

63,624

91,8819

Solution

10,209

83,2568

2,470÷7

Solution

352 R6

3,741÷7

855,305

Solution

6,913 R1

951,492

431,1745


Solution

86,234 R4

297,2774

130,016÷3

Solution

43,338 R2

105,609÷2

155,735

Solution

382 R5

4,93321

56,883÷67

Solution

849

43,725/75

30,144314

Solution

96

26,145÷415

273542,195

Solution

1,986 R17

816,243÷462

Mixed Practice

In the following exercises, simplify.

15(204)

Solution

3,060

74·391

256−184

Solution

72

305−262

719+341

Solution

1,060

647+528

25875

Solution

35

1104÷23

Translate Word Phrases to Algebraic Expressions

In the following exercises, translate and simplify.

the quotient of 45 and 15

Solution

45 ÷ 15; 3

the quotient of 64 and 16

the quotient of 288 and 24

Solution

288 ÷ 24; 12

the quotient of 256 and 32

Divide Whole Numbers in Applications

In the following exercises, solve.

Trail mix Ric bought 64 ounces of trail mix. He wants to divide it into small bags, with 2 ounces of trail mix in each bag. How many bags can Ric fill?

Solution

Ric can fill 32 bags.

Crackers Evie bought a 42 ounce box of crackers. She wants to divide it into bags with 3 ounces of crackers in each bag. How many bags can Evie fill?

Astronomy class There are 125 students in an astronomy class. The professor assigns them into groups of 5. How many groups of students are there?

Solution

There are 25 groups.

Flower shop Melissa’s flower shop got a shipment of 152 roses. She wants to make bouquets of 8 roses each. How many bouquets can Melissa make?

Baking One roll of plastic wrap is 48 feet long. Marta uses 3 feet of plastic wrap to wrap each cake she bakes. How many cakes can she wrap from one roll?

Solution

Marta can wrap 16 cakes from 1 roll.

Dental floss One package of dental floss is 54 feet long. Brian uses 2 feet of dental floss every day. How many days will one package of dental floss last Brian?

Mixed Practice

In the following exercises, solve.

Miles per gallon Susana’s hybrid car gets 45 miles per gallon. Her son’s truck gets 17 miles per gallon. What is the difference in miles per gallon between Susana’s car and her son’s truck?

Solution

The difference is 28 miles per gallon.

Distance Mayra lives 53 miles from her mother’s house and 71 miles from her mother-in-law’s house. How much farther is Mayra from her mother-in-law’s house than from her mother’s house?

Field trip The 45 students in a Geology class will go on a field trip, using the college’s vans. Each van can hold 9 students. How many vans will they need for the field trip?

Solution

They will need 5 vans for the field trip

Potting soil Aki bought a 128 ounce bag of potting soil. How many 4 ounce pots can he fill from the bag?

Hiking Bill hiked 8 miles on the first day of his backpacking trip, 14 miles the second day, 11 miles the third day, and 17 miles the fourth day. What is the total number of miles Bill hiked?

Solution

Bill hiked 50 miles

Reading Last night Emily read 6 pages in her Business textbook, 26 pages in her History text, 15 pages in her Psychology text, and 9 pages in her math text. What is the total number of pages Emily read?

Patients LaVonne treats 12 patients each day in her dental office. Last week she worked 4 days. How many patients did she treat last week?

Solution

LaVonne treated 48 patients last week.

Scouts There are 14 boys in Dave’s scout troop. At summer camp, each boy earned 5 merit badges. What was the total number of merit badges earned by Dave’s scout troop at summer camp?

Writing Exercises

Contact lenses Jenna puts in a new pair of contact lenses every 14 days. How many pairs of contact lenses does she need for 365 days?

Solution

Jenna uses 26 pairs of contact lenses, but there is 1 day left over, so she needs 27 pairs for 365 days.

Cat food One bag of cat food feeds Lara’s cat for 25 days. How many bags of cat food does Lara need for 365 days?

Everyday Math

Explain how you use the multiplication facts to help with division.

Solution

Answers may vary. Using multiplication facts can help you check your answers once you’ve finished division.

Oswaldo divided 300 by 8 and said his answer was 37 with a remainder of 4. How can you check to make sure he is correct?

Self Check

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

A self-assessment checklist for division and algebraic expressions, allowing students to rate their understanding as confident, needing some help, or not understanding.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Introduction to Whole Numbers

Identify Counting Numbers and Whole Numbers

In the following exercises, determine which of the following are (a) counting numbers (b) whole numbers.

0,2,99

Solution
  1. ⓐ 2, 99
  2. ⓑ 0, 2, 99

0,3,25

0,4,90

Solution
  1. ⓐ 4, 90
  2. ⓑ 0, 4, 90

0,1,75

Model Whole Numbers

In the following exercises, model each number using base-10 blocks and then show its value using place value notation.

258

Solution


The image shows sets of hundred blocks, sets of blocks of 10, and single blocks. Below, the types of blocks and their total values are organized in a table with a total block values.

104

Identify the Place Value of a Digit

In the following exercises, find the place value of the given digits.

472,981

  1. ⓐ 8
  2. ⓑ 4
  3. ⓒ 1
  4. ⓓ 7
  5. ⓔ 2
Solution
  1. ⓐ tens
  2. ⓑ hundred thousands
  3. ⓒ ones
  4. ⓓ ten thousands
  5. ⓔ thousands

12,403,295

  1. ⓐ 4
  2. ⓑ 0
  3. ⓒ 1
  4. ⓓ 9
  5. ⓔ 3

Use Place Value to Name Whole Numbers

In the following exercises, name each number in words.

5,280

Solution

Five thousand, two hundred eighty

204,614

5,012,582

Solution

Five million, twelve thousand, five hundred eighty-two

31,640,976

Use Place Value to Write Whole Numbers

In the following exercises, write as a whole number using digits.

six hundred two

fifteen thousand, two hundred fifty-three

Solution

15,253

three hundred forty million, nine hundred twelve thousand, sixty-one

Solution

340,912,061

two billion, four hundred ninety-two million, seven hundred eleven thousand, two

Round Whole Numbers

In the following exercises, round to the nearest ten.

412

Solution

410

648

3,556

Solution

3,560

2,734

In the following exercises, round to the nearest hundred.

38,975

Solution

39,000

26,849

81,486

Solution

81,500

75,992

Add Whole Numbers

Use Addition Notation

In the following exercises, translate the following from math notation to words.

4+3

Solution

four plus three; the sum of four and three

25+18

571+629

Solution

five hundred seventy-one plus six hundred twenty-nine; the sum of five hundred seventy-one and six hundred twenty-nine

10,085+3,492

Model Addition of Whole Numbers

In the following exercises, model the addition.

6+7

Solution


A row of light gray vertical panels with subtle patterns, reflecting on a darker surface below, creating an abstract and minimalist composition.

38+14

Add Whole Numbers

In the following exercises, fill in the missing values in each chart.

A table with 11 rows down and 11 rows across. The first row and first column are headers and include the numbers 0 through 9 both across and down, with a plus sign in the first cell. The numbers across in the second row down appear as follows: 0,0,  1, null, 3, 4, null, 6, 7, null, 9. The numbers across in the third row down appear as follows: 1, 1, 2, 3, 4, null, null, 7, 8, 9, null. The numbers in the fourth row down appear across as follows: 2, null, 3,4,5,6,7,8, null, 10, 11. The numbers across in the fifth row down appear as follows: 3, 3, null, 5, null, 7, 8, null, 10, null 12. The numbers across in the sixth row down appear as follows: 4, 4, 5, null, null, 8, 9,null, null, 12, null. The numbers across in the seventh row down appear as follows: 5, 5, null, 7, 8, null null, 11, null, 13, null. The numbers across in the eighth row down appear as follows:6, 6, 7, 8, null, 10, null, null, 13, null, 15. The numbers across in the ninth row down appear as follows: null, null, 9, null, null, 12, 13, null, 15, 16. The numbers across in the tenth row down appear as follows: 8,8,9,null, 11, null, null, 14, null, eleventh row down appear as follows: 9, 9, 10, 11, null, 13, 14, null, null, 17, null.
Solution


A clear and simple addition table showing the sums of numbers from 0 to 9. It's a fundamental tool for understanding basic arithmetic and number relationships in a grid format.

This table is 5 rows and 8 columns. The top row is a header row and includes the numbers 3 through 9, one number to each cell. The rows down include 6, 7, 8, and 9. There is a plus sign in the first cell. All cells are null.

In the following exercises, add.

ⓐ 0+19 ⓑ 19+0

Solution
  1. ⓐ 19
  2. ⓑ 19

ⓐ 0+480 ⓑ 480+0

ⓐ 7+6 ⓑ 6+7

Solution
  1. ⓐ 13
  2. ⓑ 13

ⓐ 23+18 ⓑ 18+23

44+35

Solution

79

63+29

96+58

Solution

154

375+591

7,281+12,546

Solution

19,827

5,280+16,324+9,731

Translate Word Phrases to Math Notation

In the following exercises, translate each phrase into math notation and then simplify.

the sum of 30 and 12

Solution

30 + 12; 42

11 increased by 8

25 more than 39

Solution

39 + 25; 64

total of 15 and 50

Add Whole Numbers in Applications

In the following exercises, solve.

Shopping for an interview Nathan bought a new shirt, tie, and slacks to wear to a job interview. The shirt cost $24, the tie cost $14, and the slacks cost $38. What was Nathan’s total cost?

Solution

$76

Running Jackson ran 4 miles on Monday, 12 miles on Tuesday, 1 mile on Wednesday, 8 miles on Thursday, and 5 miles on Friday. What was the total number of miles Jackson ran?

In the following exercises, find the perimeter of each figure.

An image of a rectangle that is 8 feet tall and 15 feet wide.
Solution

46 feet

An image of a right triangle that has a base of 12 centimeters, height of 5 centimeters, and diagonal hypotenuse of 13 centimeters.

Subtract Whole Numbers

Use Subtraction Notation

In the following exercises, translate the following from math notation to words.

14−5

Solution

fourteen minus five; the difference of fourteen and five

40−15

351−249

Solution

three hundred fifty-one minus two hundred forty-nine; the difference between three hundred fifty-one and two hundred forty-nine

5,724−2,918

Model Subtraction of Whole Numbers

In the following exercises, model the subtraction.

18−4

Solution


The image uses blocks to demonstrate the subtraction problem 18 – 4.

41−29

Subtract Whole Numbers

In the following exercises, subtract and then check by adding.

8−5

Solution

3

12−7

23−9

Solution

14

46−21

82−59

Solution

23

110−87

539−217

Solution

322

415−296

1,020−640

Solution

380

8,355−3,947

10,000−15

Solution

9,985

54,925−35,647

Translate Word Phrases to Math Notation

In the following exercises, translate and simplify.

the difference of nineteen and thirteen

Solution

19 − 13; 6

subtract sixty-five from one hundred

seventy-four decreased by eight

Solution

74 − 8; 66

twenty-three less than forty-one

Subtract Whole Numbers in Applications

In the following exercises, solve.

Temperature The high temperature in Peoria one day was 86 degrees Fahrenheit and the low temperature was 28 degrees Fahrenheit. What was the difference between the high and low temperatures?

Solution

58 degrees Fahrenheit

Savings Lynn wants to go on a cruise that costs $2,485. She has $948 in her vacation savings account. How much more does she need to save in order to pay for the cruise?

Multiply Whole Numbers

Use Multiplication Notation

In the following exercises, translate from math notation to words.

8×5

Solution

eight times five; the product of eight and five

6·14

(10)(95)

Solution

ten times ninety-five; the product of ten and ninety-five

54(72)

Model Multiplication of Whole Numbers

In the following exercises, model the multiplication.

2×4

Solution


The image uses circles to demonstrate the multiplication problem of 2 × 4.

3×8

Multiply Whole Numbers

In the following exercises, fill in the missing values in each chart.

A table with 10 rows down and 10 rows across. The first row and first column are headers and include the numbers 0 through 9 both across and down, with a plus sign in the first cell. The numbers across in the second row down appear as follows: 0, 0, 0, 0, 0, 0, null, 0, null, 0,0. The numbers across in the third row down appear as follows: 1, 0, 1, 2, null, 4, 5, 6, 7, null, 9.  The numbers across in the fourth row down appear as follows: 2, 0, null, 4, null, 8, 10, null, 14, 16, null. The numbers across in the fifth row down appear as follows: 3, null, 3, null, 9, null, null, 18, null, 24, null. The numbers across in the sixth row down appear as follows: 4, 0, 4, 0, 12, null, null, 24, null, null, 36.  The numbers across in the seventh row down appear as follows:5, 0, 5, 10, null, 20, null, 30, 35, 40, 45. The numbers across in the eighth row down appear as follows: 6, null, null, 12, 18, null, null, 36, 42, null, 54.  The numbers across in the ninth row down appear as follows: 7, 0, 7, null, 21, null, 35, null, null, 56, 63. The numbers in the tenth row down appear as follows: 8, 0, 8, 16, null, 32, null, 48, null, 64, null. The numbers in the eleventh row down appear across as follows: 9, null, null, 18, 27, 36, null, null, 63, 72, null.
Solution


A multiplication table from 0x0 to 9x9, displaying the product of each row and column intersection. The table is shaded with alternating light and dark grey rows for readability. Perfect for learning basic math.

An image of a table with 8 columns and 5 rows. The cells in the first row and first column are shaded darker than the other cells. The first row has the values “x; 3; 4; 5; 6; 7; 8; 9”. The first column has the values “x;  6; 7; 8; 9”. All other cells are null.

In the following exercises, multiply.

0·14

Solution

0

(256)0

1·99

Solution

99

(4,789)1

ⓐ 7·4 ⓑ 4·7

Solution
  1. ⓐ 28
  2. ⓑ 28

(25)(6)

9,261×3

Solution

27,783

48·76

64·10

Solution

640

1,000(22)

162×493

Solution

79,866

(601)(943)

3,624×517

Solution

1,873,608

10,538·22

Translate Word Phrases to Math Notation

In the following exercises, translate and simplify.

the product of 15 and 28

Solution

15(28); 420

ninety-four times thirty-three

twice 575

Solution

2(575); 1,150

ten times two hundred sixty-four

Multiply Whole Numbers in Applications

In the following exercises, solve.

Gardening Geniece bought 8 packs of marigolds to plant in her yard. Each pack has 6 flowers. How many marigolds did Geniece buy?

Solution

48 marigolds

Cooking Ratika is making rice for a dinner party. The number of cups of water is twice the number of cups of rice. If Ratika plans to use 4 cups of rice, how many cups of water does she need?

Multiplex There are twelve theaters at the multiplex and each theater has 150 seats. What is the total number of seats at the multiplex?

Solution

1,800 seats

Roofing Lewis needs to put new shingles on his roof. The roof is a rectangle, 30 feet by 24 feet. What is the area of the roof?

Divide Whole Numbers

Use Division Notation

Translate from math notation to words.

54÷9

Solution

fifty-four divided by nine; the quotient of fifty-four and nine

42/7

728

Solution

seventy-two divided by eight; the quotient of seventy-two and eight

648

Model Division of Whole Numbers

In the following exercises, model.

8÷2

Solution


The image shows clusters of grey circles to demonstrate 8 divided by 2.

312

Divide Whole Numbers

In the following exercises, divide. Then check by multiplying.

14÷2

Solution

7

328

52÷4

Solution

13

2626

971

Solution

97

0÷52

100÷0

Solution

undefined

3555

3828÷6

Solution

638

311,519

750525

Solution

300 R5

5,166÷42

Translate Word Phrases to Math Notation

In the following exercises, translate and simplify.

the quotient of 64 and 16

Solution

64 ÷ 16; 4

the quotient of 572 and 52

Divide Whole Numbers in Applications

In the following exercises, solve.

Ribbon One spool of ribbon is 27 feet. Lizbeth uses 3 feet of ribbon for each gift basket that she wraps. How many gift baskets can Lizbeth wrap from one spool of ribbon?

Solution

9 baskets

Juice One carton of fruit juice is 128 ounces. How many 4 ounce cups can Shayla fill from one carton of juice?

Chapter Practice Test

Determine which of the following numbers are

  1. ⓐ counting numbers
  2. ⓑ whole numbers.

0,4,87

Solution
  1. ⓐ 4, 87
  2. ⓑ 0, 4, 87

Find the place value of the given digits in the number 549,362.

  1. ⓐ 9
  2. ⓑ 6
  3. ⓒ 2
  4. ⓓ 5

Write each number as a whole number using digits.

  1. ⓐ six hundred thirteen
  2. ⓑ fifty-five thousand two hundred eight
Solution
  1. ⓐ 613
  2. ⓑ 55,208

Round 25,849 to the nearest hundred.

Simplify.

45+23

Solution

68

65−42

85÷5

Solution

17

1,000×8

90−58

Solution

32

73+89

(0)(12,675)

Solution

0

634+255

09

Solution

0

8128

145−79

Solution

66

299+836

7·475

Solution

3,325

8,528+704

35(14)

Solution

490

260

733−291

Solution

442

4,916−1,538

495÷45

Solution

11

52×983

Translate each phrase to math notation and then simplify.

The sum of 16 and 58

Solution

16 + 58; 74

The product of 9 and 15

The difference of 32 and 18

Solution

32 − 18; 14

The quotient of 63 and 21

Twice 524

Solution

2(524); 1,048

29 more than 32

50 less than 300

Solution

300 − 50; 250

In the following exercises, solve.

LaVelle buys a jumbo bag of 84 candies to make favor bags for her son’s party. If she wants to make 12 bags, how many candies should she put in each bag?

Last month, Stan’s take-home pay was $3,816 and his expenses were $3,472. How much of his take-home pay did Stan have left after he paid his expenses?

Solution

Stan had $344 left.

Each class at Greenville School has 22 children enrolled. The school has 24 classes. How many children are enrolled at Greenville School?

Clayton walked 12 blocks to his mother’s house, 6 blocks to the gym, and 9 blocks to the grocery store before walking the last 3 blocks home. What was the total number of blocks that Clayton walked?

Solution

Clayton walked 30 blocks.

dividend
When dividing two numbers, the dividend is the number being divided.
divisor
When dividing two numbers, the divisor is the number dividing the dividend.
quotient
The quotient is the result of dividing two numbers.

Introduction to the Language of Algebra

Learning Objectives

--
The image shows a collage of mathematical terms such as algebra, decimals, equations, numbers etcetera. The words are written horizontally, vertically, and in different colors.
Algebra has a language of its own. The picture shows just some of the words you may see and use in your study of Prealgebra.

You may not realize it, but you already use algebra every day. Perhaps you figure out how much to tip a server in a restaurant. Maybe you calculate the amount of change you should get when you pay for something. It could even be when you compare batting averages of your favorite players. You can describe the algebra you use in specific words, and follow an orderly process. In this chapter, you will explore the words used to describe algebra and start on your path to solving algebraic problems easily, both in class and in your everyday life.

Use the Language of Algebra

Learning Objectives

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

  • Use variables and algebraic symbols
  • Identify expressions and equations
  • Simplify expressions with exponents
  • Simplify expressions using the order of operations

Before you get started, take this readiness quiz.

Add:43+69.
If you missed this problem, review Example 9 in Add Whole Numbers.

Solution

112

Multiply:(896)201.
If you missed this problem, review Example 10 in Multiply Whole Numbers.

Solution

180,096

Divide:7,263÷9.
If you missed this problem, review Example 9 in Divide Whole Numbers.

Solution

807

Use Variables and Algebraic Symbols

Greg and Alex have the same birthday, but they were born in different years. This year Greg is 20 years old and Alex is 23, so Alex is 3 years older than Greg. When Greg was 12, Alex was 15. When Greg is 35, Alex will be 38. No matter what Greg’s age is, Alex’s age will always be 3 years more, right?

In the language of algebra, we say that Greg’s age and Alex’s age are variable and the three is a constant. The ages change, or vary, so age is a variable. The 3 years between them always stays the same, so the age difference is the constant.

In algebra, letters of the alphabet are used to represent variables. Suppose we call Greg’s age g. Then we could use g+3 to represent Alex’s age. See Table 1.

Greg’s age Alex’s age
12 15
20 23
35 38
g g+3

Letters are used to represent variables. Letters often used for variables are x,y,a,b,andc.

Variables and Constants

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

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

To write algebraically, we need some symbols as well as numbers and variables. There are several types of symbols we will be using. In Whole Numbers, we introduced the symbols for the four basic arithmetic operations: addition, subtraction, multiplication, and division. We will summarize them here, along with words we use for the operations and the result.

Operation Notation Say: The result is…
Addition a+b aplusb the sum of a and b
Subtraction a−b aminusb the difference of a and b
Multiplication a·b,(a)(b),(a)b,a(b) atimesb The product of a and b
Division a÷b,a/b,ab,ba a divided by b The quotient of a and b

In algebra, the cross symbol, ×, is not used to show multiplication because that symbol may cause confusion. Does 3xy mean 3×y (three times y) or 3·x·y (three times xtimesy)? To make it clear, use • or parentheses for multiplication.

We perform these operations on two numbers. When translating from symbolic form to words, or from words to symbolic form, pay attention to the words of or and to help you find the numbers.

  • The sum of 5 and 3 means add 5 plus 3, which we write as 5+3.
  • The difference of 9 and 2 means subtract 9 minus 2, which we write as 9−2.
  • The product of 4 and 8 means multiply 4 times 8, which we can write as 4·8.
  • The quotient of 20 and 5 means divide 20 by 5, which we can write as 20÷5.
Translate from algebra to words:
  1. ⓐ 12+14
  2. ⓑ (30)(5)
  3. ⓒ 64÷8
  4. ⓓ x−y
Solution

Solution

Different ways to express the sum of twelve and fourteen.
ⓐ
12+14
12 plus 14
the sum of twelve and fourteen
This table illustrates various ways to represent the multiplication expression (30)(5).
ⓑ
(30)(5)
30 times 5
the product of thirty and five
Illustrates various ways to express the mathematical operation '64 divided by 8'.
ⓒ
64÷8
64 divided by 8
the quotient of sixty-four and eight
Shows different ways to express the mathematical difference between variables x and y.
ⓓ
x−y
x minus y
the difference of x and y
Translate from algebra to words.
  1. ⓐ 18+11
  2. ⓑ (27)(9)
  3. ⓒ 84÷7
  4. ⓓ p−q
Solution
  1. ⓐ 18 plus 11; the sum of eighteen and eleven
  2. ⓑ 27 times 9; the product of twenty-seven and nine
  3. ⓒ 84 divided by 7; the quotient of eighty-four and seven
  4. ⓓ p minus q; the difference of p and q
Translate from algebra to words.
  1. ⓐ 47−19
  2. ⓑ 72÷9
  3. ⓒ m+n
  4. ⓓ (13)(7)
Solution
  1. ⓐ 47 minus 19; the difference of forty-seven and nineteen
  2. ⓑ 72 divided by 9; the quotient of seventy-two and nine
  3. ⓒ m plus n; the sum of m and n
  4. ⓓ 13 times 7; the product of thirteen and seven

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

Equality Symbol

a=bis readais equal tob

The symbol = is called the equal sign.

An inequality is used in algebra to compare two quantities that may have different values. The number line can help you understand inequalities. Remember that on the number line the numbers get larger as they go from left to right. So if we know that b is greater than a, it means that b is to the right of a on the number line. We use the symbols “<” and “>” for inequalities.

Inequality

a<b is read a is less than b

a is to the left of b on the number line

The figure shows a horizontal number line that begins with the letter a on the left then the letter b to its right.

a>b is read a is greater than b

a is to the right of b on the number line

The figure shows a horizontal number line that begins with the letter b on the left then the letter a to its right.

The expressions a<banda>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.

When we write an inequality symbol with a line under it, such as a≤b, it means a<b or a=b. We read this a is less than or equal to b. Also, if we put a slash through an equal sign, ≠, it means not equal.

We summarize the symbols of equality and inequality in Table 7.

Algebraic Notation Say
a=b a is equal to b
a≠b a is not equal to b
a<b a is less than b
a>b a is greater than b
a≤b a is less than or equal to b
a≥b a is greater than or equal to b

Symbols < and >

The symbols < and > each have a smaller side and a larger side.

smaller side < larger side
larger side > smaller side

The smaller side of the symbol faces the smaller number and the larger faces the larger number.

Translate from algebra to words:
  1. ⓐ 20≤35
  2. ⓑ 11≠15−3
  3. ⓒ 9>10÷2
  4. ⓓ x+2<10
Solution

Solution

This table presents a mathematical inequality (20 ≤ 35) and its equivalent verbal interpretation.
ⓐ
20≤35
20 is less than or equal to 35
This table illustrates a mathematical inequality, 11 \u2260 15-3, presented in both symbolic and verbal forms.
ⓑ
11≠15−3
11 is not equal to 15 minus 3
Presents a mathematical inequality (9 > 10 ÷ 2) and its verbal description.
ⓒ
9>10÷2
9 is greater than 10 divided by 2
Illustrates the inequality x+2<10 using a symbol, its mathematical form, and a verbal description.
ⓓ
x+2<10
x plus 2 is less than 10
Translate from algebra to words.
  1. ⓐ 14≤27
  2. ⓑ 19−2≠8
  3. ⓒ 12>4÷2
  4. ⓓ x−7<1
Solution
  1. ⓐ fourteen is less than or equal to twenty-seven
  2. ⓑ nineteen minus two is not equal to eight
  3. ⓒ twelve is greater than four divided by two
  4. ⓓ x minus seven is less than one
Translate from algebra to words.
  1. ⓐ 19≥15
  2. ⓑ 7=12−5
  3. ⓒ 15÷3<8
  4. ⓓ y-3>6
Solution
  1. ⓐ nineteen is greater than or equal to fifteen
  2. ⓑ seven is equal to twelve minus five
  3. ⓒ fifteen divided by three is less than eight
  4. ⓓ y minus three is greater than six
The information in Figure 1 compares the fuel economy in miles-per-gallon (mpg) of several cars. Write the appropriate symbol =,<,or> in each expression to compare the fuel economy of the cars.
This table has two rows and six columns. The first column is a header column and it labels each row The first row is labeled “Car” and the second “Fuel economy (mpg)”. To the right of the ‘Car’ row are the labels: “Prius”, “Mini Cooper”, “Toyota Corolla”, “Versa”, “Honda Fit”. Each of these columns contains an image of the labeled car model. To the right of the “Fuel economy (mpg)” row are the algebraic equations: the letter p, the equals symbol, the number forty-eight; the letter m, the equals symbol, the number twenty-seven; the letter c, the equals symbol, the number twenty-eight; the letter v, the equals symbol, the number twenty-six; and the letter f, the equals symbol, the number twenty-seven.
(credit: modification of work by Bernard Goldbach, Wikimedia Commons)
  1. ⓐ MPG of Prius_____ MPG of Mini Cooper
  2. ⓑ MPG of Versa_____ MPG of Fit
  3. ⓒ MPG of Mini Cooper_____ MPG of Fit
  4. ⓓ MPG of Corolla_____ MPG of Versa
  5. ⓔ MPG of Corolla_____ MPG of Prius
Solution

Solution

Comparison of Prius and Mini Cooper MPG values, highlighting Prius's superior fuel efficiency.
ⓐ
MPG of Prius____MPG of Mini Cooper
Find the values in the chart. 48____27
Compare. 48 > 27
MPG of Prius > MPG of Mini Cooper
Comparison of fuel efficiency (MPG) between Versa and Fit.
ⓑ
MPG of Versa____MPG of Fit
Find the values in the chart. 26____27
Compare. 26 < 27
MPG of Versa < MPG of Fit
This table illustrates a comparison of fuel efficiency (MPG) for a Mini Cooper and a Fit, indicating both vehicles achieve 27 MPG.
ⓒ
MPG of Mini Cooper____MPG of Fit
Find the values in the chart. 27____27
Compare. 27 = 27
MPG of Mini Cooper = MPG of Fit
This table outlines a step-by-step comparison of Miles Per Gallon (MPG) values for Toyota Corolla and Nissan Versa, including the numerical data and the comparison result.
ⓓ
MPG of Corolla____MPG of Versa
Find the values in the chart. 28____26
Compare. 28 > 26
MPG of Corolla > MPG of Versa
This table compares the MPG of Corolla (28) and Prius (48), showing Prius has higher fuel efficiency.
ⓔ
MPG of Corolla____MPG of Prius
Find the values in the chart. 28____48
Compare. 28 < 48
MPG of Corolla < MPG of Prius
Use Figure 1 to fill in the appropriate symbol,=,<,or>.
  1. ⓐ MPG of Prius_____MPG of Versa
  2. ⓑ MPG of Mini Cooper_____ MPG of Corolla
Solution
  1. ⓐ >
  2. ⓑ <
Use Figure 1 to fill in the appropriate symbol,=,<,or>.
  1. ⓐ MPG of Fit_____ MPG of Prius
  2. ⓑ MPG of Corolla _____ MPG of Fit
Solution
  1. ⓐ <
  2. ⓑ >

Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in written language. They indicate which expressions are to be kept together and separate from other expressions. Table 17 lists three of the most commonly used grouping symbols in algebra.

Common 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]}

Identify Expressions and Equations

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. “Running very fast” is a phrase, but “The football player was running very fast” is a sentence. A sentence has a subject and a verb.

In algebra, we have expressions and equations. An expression is like a phrase. Here are some examples of expressions and how they relate to word phrases:

Expression Words Phrase
3+5 3plus5 the sum of three and five
n−1 n minus one the difference of n and one
6·7 6times7 the product of six and seven
xy x divided by y the quotient of x and y

Notice that the phrases do not form a complete sentence because the phrase does not have a verb. An equation is two expressions linked with 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. Here are some examples of equations:

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

Expressions and Equations

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

An equation is made up of two expressions connected by an equal sign.

Determine if each is an expression or an equation:
  1. ⓐ 16−6=10
  2. ⓑ 4·2+1
  3. ⓒ x÷25
  4. ⓓ y+8=40
Solution

Solution

This table provides examples of mathematical statements, classifying each as either an equation or an expression, based on the presence or absence of an equal sign.
ⓐ 16−6=10 This is an equation—two expressions are connected with an equal sign.
ⓑ 4·2+1 This is an expression—no equal sign.
ⓒ x÷25 This is an expression—no equal sign.
ⓓ y+8=40 This is an equation—two expressions are connected with an equal sign.
Determine if each is an expression or an equation:
  1. ⓐ 23+6=29
  2. ⓑ 7·3−7
Solution
  1. ⓐ equation
  2. ⓑ expression
Determine if each is an expression or an equation:
  1. y÷14
  2. x−6=21
Solution
  1. ⓐ expression
  2. ⓑ equation

Simplify Expressions with Exponents

To simplify a numerical expression means to do all the math possible. For example, to simplify 4·2+1 we’d 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+1
8+1
9

Suppose we have the expression 2·2·2·2·2·2·2·2·2. We could write this more compactly using exponential notation. Exponential notation is used in algebra to represent a quantity multiplied by itself several times. 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 factors of the base we have to multiply.

The image shows the number two with the number three, in superscript, to the right of the two. The number two is labeled as “base” and the number three is labeled as “exponent”.
means multiply three factors of 2

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

Exponential Notation

For any expression an,a is a factor multiplied by itself n times if n is a positive integer.

anmeans multiplynfactors ofa
At the top of the image is the letter a with the letter n, in superscript, to the right of the a. The letter a is labeled as “base” and the letter n is labeled as “exponent”. Below this is the letter a with the letter n, in superscript, to the right of the a set equal to n factors of a.

The expression an is read a to the nth power.

For powers of n=2 and n=3, we have special names.

a2is read as"asquared"a3is read as"acubed"

Table 21 lists some examples of expressions written in exponential notation.

Exponential Notation 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
Write each expression in exponential form:
  1. ⓐ 16·16·16·16·16·16·16
  2. ⓑ 9·9·9·9·9
  3. ⓒ x·x·x·x
  4. ⓓ a·a·a·a·a·a·a·a
Solution

Solution

Illustrates verbal descriptions of repeated factors and their corresponding exponential notation.
ⓐ The base 16 is a factor 7 times. 167
ⓑ The base 9 is a factor 5 times. 95
ⓒ The base x is a factor 4 times. x4
ⓓ The base a is a factor 8 times. a8

Write each expression in exponential form:

41·41·41·41·41

Solution

415

Write each expression in exponential form:

7·7·7·7·7·7·7·7·7

Solution

79

Write each exponential expression in expanded form:
  1. ⓐ 86
  2. ⓑ x5
Solution

Solution

ⓐ The base is 8 and the exponent is 6, so 86 means 8·8·8·8·8·8

ⓑ The base is x and the exponent is 5, so x5 means x·x·x·x·x

Write each exponential expression in expanded form:
  1. ⓐ 48
  2. ⓑ a7
Solution
  1. ⓐ 4 · 4 · 4 · 4 · 4 · 4 · 4 · 4
  2. ⓑ a · a · a · a · a · a · a
Write each exponential expression in expanded form:
  1. ⓐ 88
  2. ⓑ b6
Solution
  1. ⓐ 8 · 8 · 8 · 8 · 8 · 8 · 8 · 8
  2. ⓑ b · b · b · b · b · b

To simplify an exponential expression without using a calculator, we write it in expanded form and then multiply the factors.

Simplify: 34.

Solution

Solution

Steps to evaluate the exponential expression 3^4, showing expansion and multiplication to reach the final result.
34
Expand the expression. 3⋅3⋅3⋅3
Multiply left to right. 9⋅3⋅3
27⋅3
Multiply. 81
Simplify:
  1. ⓐ 53
  2. ⓑ 17
Solution
  1. ⓐ 125
  2. ⓑ 1
Simplify:
  1. ⓐ 72
  2. ⓑ 05
Solution
  1. ⓐ 49
  2. ⓑ 0

Simplify Expressions Using the Order of Operations

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 say it simplifies to 49.Some students say it simplifies to 25.4+3·7Since4+3gives 7.7·7And7·7is 49.494+3·7Since3·7is 21.4+21And21+4makes 25.25

Imagine the confusion that could result if every problem had several different correct answers. The same expression should give the same result. So mathematicians established some guidelines called the order of operations, which outlines the order in which parts of an expression must be simplified.

Order of Operations

When simplifying mathematical expressions perform the operations in the following order:

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.

Order of Operations
Please Parentheses
Excuse Exponents
My Dear Multiplication and Division
Aunt Sally Addition and Subtraction

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.

Doing the Manipulative Mathematics activity Game of 24 will give you practice using the order of operations.
Simplify the expressions:
  1. ⓐ 4+3·7
  2. ⓑ (4+3)·7
Solution

Solution

ⓐ
A mathematical expression showing 4 plus 3 times 7, represented as 4 + 3  . 7.
Are there any parentheses? No.
Are there any exponents? No.
Is there any multiplication or division? Yes.
Multiply first. The image shows the mathematical expression '4 + 3 ⋅ 7'. The numbers '3' and '7' are in red, while '4' and the plus sign are in black.
Add. The mathematical expression '4 + 21' is displayed against a white background.
The number 25 is prominently displayed on a plain white background.
ⓑ
The mathematical expression (4 + 3) multiplied by 7, clearly displaying the order of operations with parentheses around the addition.
Are there any parentheses? Yes. A mathematical expression reads (4 + 3) * 7, with the numbers 4 and 3, and the plus sign in red, enclosed in red parentheses, followed by a black multiplication dot and the number 7.
Simplify inside the parentheses. The image shows a dark gray numerical expression on a white background, featuring the number seven enclosed in parentheses immediately followed by another number seven, appearing as (7)7.
Are there any exponents? No.
Is there any multiplication or division? Yes.
Multiply. The number 49 is prominently displayed in the center of the image against a white background.
Simplify the expressions:
  1. ⓐ 12−5·2
  2. ⓑ (12−5)·2
Solution
  1. ⓐ 2
  2. ⓑ 14
Simplify the expressions:
  1. ⓐ 8+3·9
  2. ⓑ (8+3)·9
Solution
  1. ⓐ 35
  2. ⓑ 99
Simplify:
  1. ⓐ 18÷9·2
  2. ⓑ 18·9÷2
Solution

Solution

ⓐ
A mathematical expression displays the numbers 18, 9, and 2, separated by a division symbol and a multiplication dot, reading as 18 ÷ 9 ⋅ 2.
Are there any parentheses? No.
Are there any exponents? No.
Is there any multiplication or division? Yes.
Multiply and divide from left to right. Divide. The image displays the number '2.2', with the first '2' in a red hue, followed by a black dot and another '2' in black, against a clean white background.
Multiply. The number four.
ⓑ
The mathematical expression '18   9 ÷ 2' is shown, representing the multiplication of eighteen by nine, followed by the division of the product by two. The calculation yields a result of 81.
Are there any parentheses? No.
Are there any exponents? No.
Is there any multiplication or division? Yes.
Multiply and divide from left to right.
Multiply. A mathematical expression displays '162 ÷ 2', with the number 162 rendered in red and the division symbol and the number 2 in black against a white background.
Divide. The number 81 is shown in a bold, black font on a plain white background.

Simplify:

42÷7·3

Solution

18

Simplify:

12·3÷4

Solution

9

Simplify: 18÷6+4(5−2).

Solution

Solution

A mathematical expression featuring division, addition, and multiplication with parentheses: 18 ÷ 6 + 4(5 - 2).
Parentheses? Yes, subtract first. A mathematical expression reads '18 divided by 6 plus 4 times 3', or 18 / 6 + 4(3). The number '3' is highlighted in red, enclosed in parentheses.
Exponents? No.
Multiplication or division? Yes.
Divide first because we multiply and divide left to right. A mathematical expression '3 + 4(3)' is displayed on a white background. The number 3 in '3 +' is colored red, while the rest of the expression is black.
Any other multiplication or division? Yes.
Multiply. A mathematical expression '3 + 12' is displayed on a white background. The number '3' and the plus sign are black, while the number '12' is colored red.
Any other multiplication or division? No.
Any addition or subtraction? Yes. The number 15 is prominently displayed in the center of a plain white background.

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

Solution

A mathematical expression showing the order of operations: 5 + 2^3 + 3[6 - 3(4 - 2)].
Are there any parentheses (or other grouping symbol)? Yes.
Focus on the parentheses that are inside the brackets. A mathematical expression reads 5 + 2^3 + 3[6 - 3(4 - 2)]. The subtraction within the inner parentheses, '4 - 2,' is highlighted in red.
Subtract. A mathematical expression displaying 5 + 2^3 + 3[6 - 3(2)], with the '3(2)' part highlighted in red.
Continue inside the brackets and multiply. A mathematical expression reads '5 + 2^3 + 3[6 - 6]'. The final '6' inside the bracket is highlighted in red.
Continue inside the brackets and subtract. A mathematical expression displaying 5 + 2 cubed + 3 multiplied by 0, with the 0 highlighted in red within brackets.
The expression inside the brackets requires no further simplification.
Are there any exponents? Yes.
Simplify exponents. A mathematical expression: 5 + 2^3 + 3[0]. The term 2^3 is highlighted in red.
Is there any multiplication or division? Yes.
Multiply. A mathematical expression reads '5 + 8 + 3[0]' with the '3[0]' part highlighted in red.
Is there any addition or subtraction? Yes.
Add. A mathematical expression '5 + 8 + 0' is displayed on a white background. The number '5' and the first '+' sign are red. The number '8' and the second '+' sign are also red. The number '0' is black.
Add. The image displays the mathematical expression '13 + 0' rendered in red characters against a plain white background, showing a simple addition problem.
The number 13 is displayed in black text on a plain white background, centered in the frame.

Simplify:

9+53−[ 4(9+3) ]

Solution

86

Simplify:

72−2[4(5+1) ]

Solution

1

Simplify: 23+34÷3−52.

Solution

Solution

A mathematical expression reads '2^3 + 3^4 ÷ 3 - 5^2'.
If an expression has several exponents, they may be simplified in the same step.
Simplify exponents. A mathematical expression displaying powers, addition, division, and subtraction: 2^3 + 3^4 ÷ 3 - 5^2.
Divide. A mathematical expression '8 + 81 : 3 - 25' is displayed. The numbers 81 and 3, along with the division symbol, are highlighted in red.
Add. A mathematical expression '8 + 27 - 25' is displayed, with the numbers 8 and 27, and the plus sign in red, while the minus sign and number 25 are in black.
Subtract. The image displays the numbers 35 and 25 separated by a minus sign, forming the mathematical expression '35 - 25' in a reddish-brown font against a white background.
The number 10 is prominently displayed in the upper right corner of a clean, white background, rendered in a bold, dark font.

Simplify:

32+24÷2+43

Solution

81

Simplify:

62−53÷5+82

Solution

75

ACCESS ADDITIONAL ONLINE RESOURCES

  • Order of Operations
  • Order of Operations – The Basics
  • Ex: Evaluate an Expression Using the Order of Operations
  • Example 3: Evaluate an Expression Using The Order of Operations

Key Concepts

Operation Notation Say: The result is…
Addition a+b aplusb the sum of a and b
Multiplication a·b,(a)(b),(a)b,a(b) atimesb The product of a and b
Subtraction a−b aminusb the difference of a and b
Division a÷b,a/b,ab,ba a divided by b The quotient of a and b
  • Equality Symbol
    • a=b is read as a is equal to b
    • The symbol = is called the equal sign.
  • Inequality
    • a<b is read a is less than b
    • a is to the left of b on the number line
      A number line with two points, 'a' and 'b', marked by vertical ticks. Point 'a' is to the left of point 'b', indicating that 'a' is a smaller value than 'b'.
    • a>b is read a is greater than b
    • a is to the right of b on the number line
      A number line shows points 'b' and 'a' marked by vertical ticks. Point 'b' is to the left of point 'a', indicating that b is less than a.
Algebraic Notation Say
a=b a is equal to b
a≠b a is not equal to b
a<b a is less than b
a>b a is greater than b
a≤b a is less than or equal to b
a≥b a is greater than or equal to b
  • Exponential Notation
    • For any expression an is a factor multiplied by itself n times, if n is a positive integer.
    • an means multiply n factors of a
      This image defines the components of an expression. In the example of a to the power of n, a is the base, while n is the exponent. a multiplied by a several times equals the n factors.
    • The expression of an is read a to the nth power.
Order of Operations When simplifying mathematical expressions perform the operations in the following order:
  • Parentheses and other Grouping Symbols: Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
  • Exponents: Simplify all expressions with exponents.
  • Multiplication and Division: Perform all multiplication and division in order from left to right. These operations have equal priority.
  • Addition and Subtraction: Perform all addition and subtraction in order from left to right. These operations have equal priority.

Practice Makes Perfect

Use Variables and Algebraic Symbols

In the following exercises, translate from algebraic notation to words.

16−9

Solution

16 minus 9, the difference of sixteen and nine

25−7

5·6

Solution

5 times 6, the product of five and six

3·9

28÷4

Solution

28 divided by 4, the quotient of twenty-eight and four

45÷5

x+8

Solution

x plus 8, the sum of x and eight

x+11

(2)(7)

Solution

2 times 7, the product of two and seven

(4)(8)

14<21

Solution

fourteen is less than twenty-one

17<35

36≥19

Solution

thirty-six is greater than or equal to nineteen

42≥27

3n=24

Solution

3 times n equals 24, the product of three and n equals twenty-four

6n=36

y−1>6

Solution

y minus 1 is greater than 6, the difference of y and one is greater than six

y−4>8

2≤18÷6

Solution

2 is less than or equal to 18 divided by 6; 2 is less than or equal to the quotient of eighteen and six

3≤20÷4

a≠7·4

Solution

a is not equal to 7 times 4, a is not equal to the product of seven and four

a≠1·12

Identify Expressions and Equations

In the following exercises, determine if each is an expression or an equation.

9·6=54

Solution

equation

7·9=63

5·4+3

Solution

expression

6·3+5

x+7

Solution

expression

x+9

y−5=25

Solution

equation

y−8=32

Simplify Expressions with Exponents

In the following exercises, write in exponential form.

3·3·3·3·3·3·3

Solution

37

4·4·4·4·4·4

x·x·x·x·x

Solution

x5

y·y·y·y·y·y

In the following exercises, write in expanded form.

53

Solution

5·5·5

83

28

Solution

2·2·2·2·2·2·2·2

105

Simplify Expressions Using the Order of Operations

In the following exercises, simplify.

  1. ⓐ 3+8·5
  2. ⓑ (3+8)·5
Solution
  1. ⓐ 43
  2. ⓑ 55
  1. ⓐ 2+6·3
  2. ⓑ (2+6)·3

23−12÷(9−5)

Solution

5

32−18÷(11−5)

3·8+5·2

Solution

34

4·7+3·5

2+8(6+1)

Solution

58

4+6(3+6)

4·12/8

Solution

6

2·36/6

6+10/2+2

Solution

13

9+12/3+4

(6+10)÷(2+2)

Solution

4

(9+12)÷(3+4)

20÷4+6·5

Solution

35

33÷3+8·2

20÷(4+6)·5

Solution

10

33÷(3+8)·2

42+52

Solution

41

32+72

(4+5)2

Solution

81

(3+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)]

Everyday Math

Basketball In the 2014 NBA playoffs, the San Antonio Spurs beat the Miami Heat. The table below shows the heights of the starters on each team. Use this table to fill in the appropriate symbol (=,<,>).

Spurs Height Heat Height
Tim Duncan 83″ Rashard Lewis 82″
Boris Diaw 80″ LeBron James 80″
Kawhi Leonard 79″ Chris Bosh 83″
Tony Parker 74″ Dwyane Wade 76″
Danny Green 78″ Ray Allen 77″
  1. ⓐ Height of Tim Duncan____Height of Rashard Lewis
  2. ⓑ Height of Boris Diaw____Height of LeBron James
  3. ⓒ Height of Kawhi Leonard____Height of Chris Bosh
  4. ⓓ Height of Tony Parker____Height of Dwyane Wade
  5. ⓔ Height of Danny Green____Height of Ray Allen
Elevation In Colorado there are more than 50 mountains with an elevation of over 14,000feet. The table shows the ten tallest. Use this table to fill in the appropriate inequality symbol.
Mountain Elevation
Mt. Elbert 14,433′
Mt. Massive 14,421′
Mt. Harvard 14,420′
Blanca Peak 14,345′
La Plata Peak 14,336′
Uncompahgre Peak 14,309′
Crestone Peak 14,294′
Mt. Lincoln 14,286′
Grays Peak 14,270′
Mt. Antero 14,269′
  1. ⓐ Elevation of La Plata Peak____Elevation of Mt. Antero
  2. ⓑ Elevation of Blanca Peak____Elevation of Mt. Elbert
  3. ⓒ Elevation of Gray’s Peak____Elevation of Mt. Lincoln
  4. ⓓ Elevation of Mt. Massive____Elevation of Crestone Peak
  5. ⓔ Elevation of Mt. Harvard____Elevation of Uncompahgre Peak

Writing Exercises

Explain the difference between an expression and an equation.

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

Self Check

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

A self-assessment checklist for students to evaluate their understanding of algebraic concepts, including using variables, identifying expressions and equations, and simplifying expressions with exponents and order of operations.

ⓑ 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.

expressions
An expression is a number, a variable, or a combination of numbers and variables and operation symbols.
equation
An equation is made up of two expressions connected by an equal sign.

Evaluate, Simplify, and Translate Expressions

Learning Objectives

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

  • Evaluate algebraic expressions
  • Identify terms, coefficients, and like terms
  • Simplify expressions by combining like terms
  • Translate word phrases to algebraic expressions

Before you get started, take this readiness quiz.

Is n÷5 an expression or an equation?
If you missed this problem, review Example 4 in Use the Language of Algebra.

Solution

expression

Simplify 45.
If you missed this problem, review Example 7 in Use the Language of Algebra.

Solution

1,024

Simplify 1+8⋅9.
If you missed this problem, review Example 8 in Use the Language of Algebra.

Solution

73

Evaluate Algebraic Expressions

In the last section, we simplified expressions using the order of operations. In this section, we’ll evaluate expressions—again following the order of operations.

To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number. To evaluate an expression, we substitute the given number for the variable in the expression and then simplify the expression using the order of operations.

Evaluate x+7 when
  1. ⓐ x=3
  2. ⓑ x=12
Solution

Solution

ⓐ To evaluate, substitute 3 for x in the expression, and then simplify.
The mathematical expression 'x + 7' is displayed in black text on a white background.
Substitute. The image displays a simple mathematical equation, '3 + 7', with the number 3 in red and the plus sign and number 7 in black, all against a white background.
Add. The number 10 is displayed.

When x=3, the expression x+7 has a value of 10.

ⓑ To evaluate, substitute 12 for x in the expression, and then simplify.
A mathematical expression consisting of the variable x, a plus sign, and the number 7, forming 'x + 7' in bold, sans-serif font on a white background.
Substitute. A mathematical expression showing the addition of two numbers, '12' in red and '7' in black, separated by a black plus sign: 12 + 7.
Add. The number 19 is displayed.

When x=12, the expression x+7 has a value of 19.

Notice that we got different results for parts ⓐ and ⓑ even though we started with the same expression. This is because the values used for x were different. When we evaluate an expression, the value varies depending on the value used for the variable.

Evaluate:

y+4when
  1. ⓐ y=6
  2. ⓑ y=15
Solution
  1. ⓐ 10
  2. ⓑ 19

Evaluate:

a−5when
  1. ⓐ a=9
  2. ⓑ a=17
Solution
  1. ⓐ 4
  2. ⓑ 12
Evaluate 9x−2,when
  1. ⓐ x=5
  2. ⓑ x=1
Solution

Solution

Remember ab means a times b, so 9x means 9 times x.

ⓐ To evaluate the expression when x=5, we substitute 5 for x, and then simplify.
A mathematical expression '9x - 2' is displayed in a clear, digital font against a white background.
The text 'Substitute 5 for x.' is shown, with the number 5 highlighted in red and the rest of the text in a dark teal color against a white background. A mathematical expression '9.5-2' is displayed. The number '5' in '9.5' is rendered in red, while the other numbers '9' and '2', as well as the dot and minus sign, are in black.
Multiply. The image displays the mathematical expression '45 - 2' in a simple black font on a white background, indicating a subtraction problem.
Subtract. The number 43 is displayed in black text on a plain white background.
ⓑ To evaluate the expression when x=1, we substitute 1 for x, and then simplify.
The mathematical expression '9x - 2' is displayed in a clear, dark font against a white background.
The text 'Substitute 1 for x.' is shown, with the number 1 highlighted in red, on a plain white background. The image shows a mathematical expression: 9(1) - 2. The number 1 is highlighted in red, indicating it might be a substituted value or a point of focus in the calculation.
Multiply. The mathematical expression '9 - 2' is displayed in black text on a white background, representing a subtraction problem.
Subtract. A number 7 is visible in the bottom right corner of a plain white background.

Notice that in part ⓐ that we wrote 9⋅5 and in part ⓑ we wrote 9(1). Both the dot and the parentheses tell us to multiply.

Evaluate:

8x−3,when
  1. ⓐ x=2
  2. ⓑ x=1
Solution
  1. ⓐ 13
  2. ⓑ 5

Evaluate:

4y−4,when
  1. ⓐ y=3
  2. ⓑ y=5
Solution
  1. ⓐ 8
  2. ⓑ 16

Evaluate x2 when x=10.

Solution

Solution

We substitute 10 for x, and then simplify the expression.
The image displays the mathematical expression 'x^2' with the letter 'x' in lowercase and the number '2' as a superscript, indicating 'x squared'.
The image displays text that instructs the user to 'Substitute 10 for x.' The number 10 is highlighted in red, while the rest of the text is in a dark teal color. The number 10 is shown in red with a black superscript 2, representing 10 squared or 10 to the power of 2.
Use the definition of exponent. A mathematical expression displays '10 . 10', indicating the multiplication of ten by ten, which equals one hundred. The numbers are bold and centered against a white background.
Multiply. The number 100 is displayed in bold black text on a white background.

When x=10, the expression x2 has a value of 100.

Evaluate:

x2whenx=8.

Solution

64

Evaluate:

x3whenx=6.

Solution

216

Evaluate2xwhenx=5.

Solution

Solution

In this expression, the variable is an exponent.
The expression '2x' is shown in black text against a white background.
The text 'Substitute 5 for x.' is shown in a dark teal font on a white background, with the number 5 highlighted in red. The mathematical expression 2^5 is shown with the number 2 in black and the exponent 5 in a distinct red color, indicating 2 raised to the power of 5.
Use the definition of exponent. The mathematical expression showing the number 2 multiplied by itself five times, represented as 2 . 2 . 2 . 2 . 2, which is equal to 32 or 2^5.
Multiply. The number 32 is displayed.

When x=5, the expression 2x has a value of 32.

Evaluate:

2xwhenx=6.

Solution

64

Evaluate:

3xwhenx=4.

Solution

81

Evaluate3x+4y−6whenx=10andy=2.

Solution

Solution

This expression contains two variables, so we must make two substitutions.
The mathematical expression 3x + 4y - 6 is displayed on a white background.
The image shows the text 'Substitute 10 for x and 2 for y.' A mathematical expression reads as three times ten, plus four times two, minus six. The '10' is highlighted in red, and the '2' is highlighted in light blue.
Multiply. A mathematical expression featuring the numbers 30, 8, and 6, connected by a plus sign and a minus sign: 30 + 8 - 6.
Add and subtract left to right. The number 32 in black text on a plain white background.

When x=10 and y=2, the expression 3x+4y−6 has a value of 32.

Evaluate:

2x+5y−4whenx=11andy=3

Solution

33

Evaluate:

5x−2y−9whenx=7andy=8

Solution

10

Evaluate2x2+3x+8whenx=4.

Solution

Solution

We need to be careful when an expression has a variable with an exponent. In this expression, 2x2 means 2⋅x⋅x and is different from the expression (2x)2, which means 2x⋅2x.
The mathematical expression '2x^2 + 3x + 8' is displayed in a clear, dark gray font against a plain white background, appearing as a standard quadratic equation or polynomial.
The image shows text that says 'Substitute 4 for each x.' The number 4 is highlighted in red, while the rest of the text is in a dark blue-grey color. The image shows the numerical evaluation of the expression 2x^2 + 3x + 8 where x is replaced by 4. The number 4 is highlighted in red in both occurrences.
Simplify 42. A mathematical expression showing the sum of products and a single number: 2 multiplied by 16, plus 3 multiplied by 4, plus 8.
Multiply. The image displays the mathematical expression '32 + 12 + 8' in a dark font against a white background. Each number and operator is clearly visible, forming a simple addition problem.
Add. The number 52 is displayed in a black, sans-serif font on a white background.

Evaluate:

3x2+4x+1whenx=3.

Solution

40

Evaluate:

6x2−4x−7whenx=2.

Solution

9

Identify Terms, Coefficients, and 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. Some examples of terms are 7,y,5x2,9a,and13xy.

The constant that multiplies the variable(s) in a term is called the coefficient. We can 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. Table 9 gives the coefficients for each of the terms in the left column.

Term Coefficient
9a 9
y 1
5x2 5

An algebraic expression may consist of one or more terms added or subtracted. In this chapter, we will only work with terms that are added together. Table 10 gives some examples of algebraic expressions with various numbers of terms. Notice that we include the operation before a term with it.

Expression Terms
7 7
y y
x+7 x,7
2x+7y+4 2x,7y,4
3x2+4x2+5y+3 3x2,4x2,5y,3

Identify each term in the expression 9b+15x2+a+6. Then identify the coefficient of each term.

Solution

Solution

The expression has four terms. They are 9b,15x2,a, and 6.

The coefficient of 9b is 9.

The coefficient of 15x2 is 15.

Remember that if no number is written before a variable, the coefficient is 1. So the coefficient of a is 1.

The coefficient of a constant is the constant, so the coefficient of 6 is 6.

Identify all terms in the given expression, and their coefficients:

4x+3b+2

Solution

The terms are 4x, 3b, and 2. The coefficients are 4, 3, and 2.

Identify all terms in the given expression, and their coefficients:

9a+13a2+a3

Solution

The terms are 9a, 13a2, and a3, The coefficients are 9, 13, and 1.

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

5x,7,n2,4,3x,9n2

Which of these terms are like terms?

  • The terms 7 and 4 are both constant terms.
  • The terms 5x and 3x are both terms with x.
  • The terms n2 and 9n2 both have n2.

Terms are called like terms if they have the same variables and exponents. All constant terms are also like terms. So among the terms 5x,7,n2,4,3x,9n2,

7and4are like terms.
5xand3xare like terms.
n2and9n2are like terms.

Like Terms

Terms that are either constants or have the same variables with the same exponents are like terms.

Identify the like terms:

  1. ⓐ y3,7x2,14,23,4y3,9x,5x2
  2. ⓑ 4x2+2x+5x2+6x+40x+8xy
Solution

Solution

ⓐ y3,7x2,14,23,4y3,9x,5x2

Look at the variables and exponents. The expression contains y3,x2,x, and constants.

The terms y3 and 4y3 are like terms because they both have y3.

The terms 7x2 and 5x2 are like terms because they both have x2.

The terms 14 and 23 are like terms because they are both constants.

The term 9x does not have any like terms in this list since no other terms have the variable x raised to the power of 1.

ⓑ 4x2+2x+5x2+6x+40x+8xy

Look at the variables and exponents. The expression contains the terms 4x2,2x,5x2,6x,40x,and8xy

The terms 4x2 and 5x2 are like terms because they both have x2.

The terms 2x,6x,and40x are like terms because they all have x.

The term 8xy has no like terms in the given expression because no other terms contain the two variables xy.

Identify the like terms in the list or the expression:

9,2x3,y2,8x3,15,9y,11y2

Solution

9 and 15; 2x3 and 8x3; y2 and 11y2

Identify the like terms in the list or the expression:

4x3+8x2+19+3x2+24+6x3

Solution

4x3 and 6x3; 8x2 and 3x2; 19 and 24

Simplify Expressions by Combining Like Terms

We can simplify an expression by combining the like terms. What do you think 3x+6x would simplify to? If you thought 9x, you would be right!

We can see why this works by writing both terms as addition problems.

The image shows the expression 3 x plus 6 x. The 3 x represents x plus x plus x. The 6 x represents x plus x plus x plus x plus x plus x. The expression 3 x plus 6 x becomes x plus x plus x plus x plus x plus x plus x plus x plus x. This simplifies to a total of 9 x's or the term 9 x.

Add the coefficients and keep the same variable. It doesn’t matter what x is. If you have 3 of something and add 6 more of the same thing, the result is 9 of them. For example, 3 oranges plus 6 oranges is 9 oranges. We will discuss the mathematical properties behind this later.

The expression 3x+6x has only two terms. When an expression contains more terms, it may be helpful to rearrange the terms so that like terms are together. The Commutative Property of Addition says that we can change the order of addends without changing the sum. So we could rearrange the following expression before combining like terms.

The image shows the expression 3 x plus 4 y plus 2 x plus 6 y. The position of the middle terms, 4 y and 2 x, can be switched so that the expression becomes 3 x plus 2 x plus 4 y plus 6 y. Now the terms containing x are together and the terms containing y are together.

Now it is easier to see the like terms to be combined.

Combine like terms.

  1. Identify like terms.
  2. Rearrange the expression so like terms are together.
  3. Add the coefficients of the like terms.

Simplify the expression: 3x+7+4x+5.

Solution

Solution

A mathematical expression showing the sum of terms: three times x, plus seven, plus four times x, plus five (3x + 7 + 4x + 5). The numbers and the variable 'x' are displayed in black against a white background.
Identify the like terms. An algebraic expression is shown as 3x + 7 + 4x + 5. The terms 3x and 4x are in red, while 7 and 5 are in light blue, separated by black plus signs.
Rearrange the expression, so the like terms are together. A mathematical expression displaying the sum of two variable terms (3x and 4x) and two constant terms (7 and 5), written as 3x + 4x + 7 + 5.
Add the coefficients of the like terms. An algebraic expression being simplified by combining like terms: 3x + 4x becomes 7x, and 7 + 5 becomes 12. The final simplified form is 7x + 12.
The original expression is simplified to... The image displays the mathematical expression '7x + 12' in a clear, dark font on a white background, representing a linear algebraic binomial.

Simplify:

7x+9+9x+8

Solution

16x + 17

Simplify:

5y+2+8y+4y+5

Solution

17y + 7

When any of the terms have negative coefficients, the procedure is the same, except that you have to subtract instead of adding to combine like terms.

Simplify the expression: 7x2+8x–x2–4x.

Solution

Solution

The image shows the mathematical expression 7x^2 + 8x - x^2 - 4x. This expression represents a polynomial with terms involving x squared and x.
Identify the like terms. A mathematical expression: 7x^2 + 8x - x^2 - 4x. Terms with x^2 are red, and terms with x are blue. The expression simplifies to 6x^2 + 4x.
Rearrange the expression so like terms are together. A mathematical expression displaying algebraic terms: 7x^2 - x^2 + 8x - 4x. The terms are color-coded, with x^2 terms in red and x terms in blue, to highlight like terms for simplification.
Add the coefficients of the like terms. The image shows the mathematical expression 6x^2 + 4x, where 6x^2 is colored red and 4x is colored blue, connected by a black plus sign.

These are not like terms and cannot be combined. So 6x2+4x is in simplest form.

Simplify:

3x2+9x+x2+5x

Solution

4x2 + 14x

Simplify:

11y2+8y+y2+7y

Solution

12y2 + 15y

Translate Words to Algebraic Expressions

In the previous section, we listed many operation symbols that are used in algebra, and then we translated expressions and equations into word phrases and sentences. Now we’ll reverse the process and translate word phrases into algebraic expressions. The symbols and variables we’ve talked about will help us do that. They are summarized in Table 13.

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
b subtracted from a
a decreased by b
b less than a
a−b
Multiplication a times b
the product of a and b
a⋅b, ab, a(b), (a)(b)
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 you to operate on two numbers. Look for the words of and and to find the numbers.

Translate each word phrase into an algebraic expression:

  1. ⓐ the difference of 20 and 4
  2. ⓑ the quotient of 10x and 3
Solution

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 differenceof20and420minus420−4

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

the quotient of10xand3divide10xby310x÷3

This can also be written as 10x/3or10x3

Translate the given word phrase into an algebraic expression:

  1. ⓐ the difference of 47 and 41
  2. ⓑ the quotient of 5x and 2
Solution
  1. ⓐ 47 − 41
  2. ⓑ 5x ÷ 2

Translate the given word phrase into an algebraic expression:

  1. ⓐ the sum of 17 and 19
  2. ⓑ the product of 7 and x
Solution
  1. ⓐ 17 + 19
  2. ⓑ 7x

How old will you be in eight years? What age is eight more years than your age now? Did you add 8 to your present age? Eight more than means eight added to your present age.

How old were you seven years ago? This is seven years less than your age now. You subtract 7 from your present age. Seven less than means seven subtracted from your present age.

Translate each word phrase into an algebraic expression:

  1. ⓐ Eight more than y
  2. ⓑ Seven less than 9z
Solution

Solution

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

Eight more thanyEight added toyy+8

ⓑ The key words are less than. They tell us the operation is subtraction. Less than means “subtracted from”.

Seven less than9zSeven subtracted from9z9z−7

Translate each word phrase into an algebraic expression:

  1. ⓐ Eleven more than x
  2. ⓑ Fourteen less than 11a
Solution
  1. ⓐ x + 11
  2. ⓑ 11a − 14

Translate each word phrase into an algebraic expression:

  1. ⓐ 19 more than j
  2. ⓑ 21 less than 2x
Solution
  1. ⓐ j + 19
  2. ⓑ 2x − 21

Translate each word phrase into an algebraic expression:

  1. ⓐ five times the sum of m and n
  2. ⓑ the sum of five times m and n
Solution

Solution

ⓐ There are two operation words: times tells us to multiply and sum tells us to add. Because we are multiplying 5 times the sum, we need parentheses around the sum of m and n.

five times the sum of m and n
5(m+n)

ⓑ 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 five times m and n.

the sum of five times m and n
5m+n

Notice how the use of parentheses changes the result. In part ⓐ , we add first and in part ⓑ , we multiply first.

Translate the word phrase into an algebraic expression:

  1. ⓐ four times the sum of p and q
  2. ⓑ the sum of four times p and q
Solution
  1. ⓐ 4(p + q)
  2. ⓐ 4p + q

Translate the word phrase into an algebraic expression:

  1. ⓐ the difference of two times xand 8
  2. ⓑ two times the difference of xand8
Solution
  1. ⓐ 2x − 8
  2. ⓑ 2(x − 8)

Later in this course, we’ll apply our skills in algebra to solving equations. We’ll usually start by translating a word phrase to an algebraic expression. We’ll need to be clear about what the expression will represent. We’ll see how to do this in the next two examples.

The height of a rectangular window is 6 inches less than the width. Let w represent the width of the window. Write an expression for the height of the window.

Solution

Solution

Step-by-step translation of the verbal phrase '6 less than the width' into its algebraic expression, 'w - 6'.
Write a phrase about the height. 6 less than the width
Substitute w for the width. 6 less than w
Rewrite 'less than' as 'subtracted from'. 6 subtracted from w
Translate the phrase into algebra. w−6

The length of a rectangle is 5 inches less than the width. Let w represent the width of the rectangle. Write an expression for the length of the rectangle.

Solution

w − 5

The width of a rectangle is 2 meters greater than the length. Let l represent the length of the rectangle. Write an expression for the width of the rectangle.

Solution

l + 2

Blanca has dimes and quarters in her purse. The number of dimes is 2 less than 5 times the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes.

Solution

Solution

Step-by-step guide on translating a word phrase comparing the number of dimes and quarters into an algebraic expression.
Write a phrase about the number of dimes. two less than five times the number of quarters
Substitute q for the number of quarters. 2 less than five times q
Translate 5 times q. 2 less than 5q
Translate the phrase into algebra. 5q−2

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

Solution

6q − 7

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

Solution

4n + 8

ACCESS ADDITIONAL ONLINE RESOURCES

  • Algebraic Expression Vocabulary

Key Concepts

  • Combine like terms.
    1. Identify like terms.
    2. Rearrange the expression so like terms are together.
    3. Add the coefficients of the like terms

Practice Makes Perfect

Evaluate Algebraic Expressions

In the following exercises, evaluate the expression for the given value.

7x+8whenx=2

Solution

22

9x+7whenx=3

5x−4whenx=6

Solution

26

8x−6whenx=7

x2whenx=12

Solution

144

x3whenx=5

x5whenx=2

Solution

32

x4whenx=3

3xwhenx=3

Solution

27

4xwhenx=2

x2+3x−7whenx=4

Solution

21

x2+5x−8whenx=6

2x+4y−5whenx=7,y=8

Solution

41

6x+3y−9whenx=6,y=9

(x−y)2whenx=10,y=7

Solution

9

(x+y)2whenx=6,y=9

Solution

225

a2+b2whena=3,b=8

Solution

73

r2−s2whenr=12,s=5

2l+2wwhenl=15,w=12

Solution

54

2l+2wwhenl=18,w=14

Identify Terms, Coefficients, and Like Terms

In the following exercises, list the terms in the given expression.

15x2+6x+2

Solution

15x2, 6x, 2

11x2+8x+5


10y3+y+2

Solution

10y3, y, 2

9y3+y+5

In the following exercises, identify the coefficient of the given term.

8a

Solution

8

13m

5r2

Solution

5

6x3

In the following exercises, identify all sets of like terms.

x3,8x,14,8y,5,8x3

Solution

x3 and 8x3; 14 and 5

6z,3w2,1,6z2,4z,w2

9a,a2,16ab,16b2,4ab,9b2

Solution

16ab and 4ab; 16b2 and 9b2

3,25r2,10s,10r,4r2,3s

Simplify Expressions by Combining Like Terms

In the following exercises, simplify the given expression by combining like terms.

10x+3x

Solution

13x

15x+4x

17a+9a

Solution

26a

18z+9z

4c+2c+c

Solution

7c

6y+4y+y

9x+3x+8

Solution

12x + 8

8a+5a+9

7u+2+3u+1

Solution

10u + 3

8d+6+2d+5

7p+6+5p+4

Solution

12p + 10

8x+7+4x−5

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 English Phrases into Algebraic Expressions

In the following exercises, translate the given word phrase into an algebraic expression.

The sum of 8 and 12

Solution

8 + 12

The sum of 9 and 1

The difference of 14 and 9

Solution

14 − 9

8 less than 19

The product of 9 and 7

Solution

9 ⋅ 7

The product of 8 and 7

The quotient of 36 and 9

Solution

36 ÷ 9

The quotient of 42 and 7

The difference of x and 4

Solution

x − 4

3 less than x

The product of 6 and y

Solution

6y

The product of 9 and y

The sum of 8x and 3x

Solution

8x + 3x

The sum of 13x and 3x

The quotient of y and 3

Solution

y ÷ 3

The quotient of y and 8

Eight times the difference of y and nine

Solution

8 (y − 9)

Seven times the difference of y and one

Five times the sum of x and y

Solution

5 (x + y)

Nine times five less than twice x

In the following exercises, write an algebraic expression.

Adele bought a skirt and a blouse. The skirt cost $15 more than the blouse. Let b represent the cost of the blouse. Write an expression for the cost of the skirt.

Solution

b + 15

Eric has rock and classical CDs in his car. The number of rock CDs is 3 more than the number of classical CDs. Let c represent the number of classical CDs. Write an expression for the number of rock CDs.

The number of girls in a second-grade class is 4 less than the number of boys. Let b represent the number of boys. Write an expression for the number of girls.

Solution

b − 4

Marcella has 6 fewer male cousins than female cousins. Let f represent the number of female cousins. Write an expression for the number of boy cousins.

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

Solution

2n − 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.

Everyday Math

In the following exercises, use algebraic expressions to solve the problem.

Car insurance Justin’s car insurance has a $750 deductible per incident. This means that he pays $750 and his insurance company will pay all costs beyond $750. If Justin files a claim for $2,100, how much will he pay, and how much will his insurance company pay?

Solution

He will pay $750. His insurance company will pay $1350.

Home insurance Pam and Armando’s home insurance has a $2,500 deductible per incident. This means that they pay $2,500 and their insurance company will pay all costs beyond $2,500. If Pam and Armando file a claim for $19,400, how much will they pay, and how much will their insurance company pay?

Writing Exercises

Explain why “the sum of x and y” is the same as “the sum of y and x,” but “the difference of x and y” is not the same as “the difference of y and x.” Try substituting two random numbers for x and y to help you explain.

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

Self Check

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

A self-assessment table for algebraic skills. It includes columns for 'I can...', 'Confidently', 'With some help', and 'No-I don't get it!'. Skills listed are evaluating expressions, identifying terms, simplifying, and translating word phrases.

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

term
A term is a constant or the product of a constant and one or more variables.
coefficient
The constant that multiplies the variable(s) in a term is called the coefficient.
like terms
Terms that are either constants or have the same variables with the same exponents are like terms.
evaluate
To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number.

Solving Equations Using the Subtraction and Addition Properties of Equality

Learning Objectives

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

  • Determine whether a number is a solution of an equation
  • Model the Subtraction Property of Equality
  • Solve equations using the Subtraction Property of Equality
  • Solve equations using the Addition Property of Equality
  • Translate word phrases to algebraic equations
  • Translate to an equation and solve

Before you get started, take this readiness quiz.

Evaluatex+8whenx=11.
If you missed this problem, review Example 1 in Evaluate, Simplify, and Translate Expressions.

Solution

19

Evaluate5x−3whenx=9.
If you missed this problem, review Example 2 in Evaluate, Simplify, and Translate Expressions.

Solution

42

Translate into algebra: the difference of x and 8.
If you missed this problem, review Example 12 in Evaluate, Simplify, and Translate Expressions.

Solution

x−8

When some people hear the word algebra, they think of solving equations. The applications of solving equations are limitless and extend to all careers and fields. In this section, we will begin solving equations. We will start by solving basic equations, and then as we proceed through the course we will build up our skills to cover many different forms of equations.

Determine Whether a Number is a Solution of an Equation

Solving an equation is like discovering the answer to a puzzle. An algebraic equation states that two algebraic expressions are equal. To solve an equation is to determine the values of the variable that make the equation a true statement. Any number that makes the equation true is called a solution of the equation. It is the answer to the puzzle!

Solution of an Equation

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

The process of finding the solution to an equation is called solving the equation.

To find the solution to an equation means to find the value of the variable that makes the equation true. Can you recognize the solution of x+2=7? If you said 5, you’re right! We say 5 is a solution to the equation x+2=7 because when we substitute 5 for x the resulting statement is true.

x+2=75+2=?77=7✓

Since 5+2=7 is a true statement, we know that 5 is indeed a solution to the equation.

The symbol =? asks whether the left side of the equation is equal to the right side. Once we know, we can change to an equal sign (=) or not-equal sign (≠).

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 whetherx=5is a solution of6x−17=16.

Solution

Solution

The image shows a linear algebraic equation: 6x - 17 = 16, typically solved for the variable 'x'. This equation represents a basic problem in algebra where one needs to isolate the variable.
The text 'Substitute 5 for x.' is displayed in a dark teal color, except for the number '5' which is highlighted in red. The background is white. A mathematical equation reads '6 x 5 - 17 ?= 16,' with the number '5' highlighted in red. The expression asks whether (6 times 5) minus 17 is equal to 16. 6x5=30, 30-17=13. 13 does not equal 16.
Multiply. A mathematical equation is displayed on a white background, reading '30 - 17 ?= 16'. The question mark is positioned directly above the equals sign, suggesting an inquiry into the equality of the two sides.
Subtract. The image displays a mathematical inequality, '13 not equal to 16'.

So x=5 is not a solution to the equation 6x−17=16.

Isx=3a solution of4x−7=16?

Solution

no

Isx=2a solution of6x−2=10?

Solution

yes

Determine whethery=2is a solution of6y−4=5y−2.

Solution

Solution

Here, the variable appears on both sides of the equation. We must substitute 2 for each y.
An algebraic equation is displayed, reading '6y - 4 = 5y - 2'.
The image displays the instruction 'Substitute 2 for y.' with the number 2 highlighted in red and the rest of the text in a blue-green color. A math problem asking whether 6(2) - 4 equals 5(2) - 2. Both sides simplify to 8, meaning the equation is true.
Multiply. A mathematical equation reads '12 - 4 ?= 10 - 2,' where the question mark indicates an unknown operator in a comparison between the two expressions. The solution would be 8 = 8, so the '?=' should be an equals sign.
Subtract. The image displays the equation '8 = 8' followed by a checkmark, indicating that the statement is mathematically correct.

Since y=2 results in a true equation, we know that 2 is a solution to the equation 6y−4=5y−2.

Isy=3a solution of9y−2=8y+1?

Solution

yes

Isy=4a solution of5y−3=3y+5?

Solution

yes

Model the Subtraction Property of Equality

We will use a model to help you understand how the process of solving an equation is like solving a puzzle. An envelope represents the variable – since its contents are unknown – and each counter represents one.

Suppose a desk has an imaginary line dividing it in half. We place three counters and an envelope on the left side of desk, and eight counters on the right side of the desk as in Figure 1. Both sides of the desk have the same number of counters, but some counters are hidden in the envelope. Can you tell how many counters are in the envelope?

The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.

What steps are you taking in your mind to figure out how many counters are in the envelope? Perhaps you are thinking “I need to remove the 3 counters from the left side to get the envelope by itself. Those 3 counters on the left match with 3 on the right, so I can take them away from both sides. That leaves five counters on the right, so there must be 5 counters in the envelope.” Figure 2 shows this process.

The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with three counters below it. The 3 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 8 counters. The bottom 3 counters are circled in red with an arrow pointing out of the rectangle. The 3 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 5 counters.

What algebraic equation is modeled by this situation? Each side of the desk represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope x, so the number of counters on the left side of the desk is x+3. On the right side of the desk are 8 counters. We are told that x+3 is equal to 8 so our equation isx+3=8.

The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.
x+3=8

Let’s write algebraically the steps we took to discover how many counters were in the envelope.

A mathematical equation is displayed, showing 'x + 3 = 8' in black text against a white background.
First, we took away three from each side. Isolating 'x' in x + 3 = 8 by subtracting 3 from both sides of the equation, shown as x + 3 - 3 = 8 - 3, with the subtracted 3s highlighted in red.
Then we were left with five. The equation x = 5 is displayed in black text on a white background.

Now let’s check our solution. We substitute 5 for x in the original equation and see if we get a true statement.

The image shows the original equation, x plus 3 equal to 8. Substitute 5 in for x to check. The equation becomes 5 plus 3 equal to 8. Is this true? The left side simplifies by adding 5 and 3 to get 8. Both sides of the equal symbol are 8.

Our solution is correct. Five counters in the envelope plus three more equals eight.

Doing the Manipulative Mathematics activity, “Subtraction Property of Equality” will help you develop a better understanding of how to solve equations by using the Subtraction Property of Equality.

Write an equation modeled by the envelopes and counters, and then solve the equation:

The image is divided in half vertically. On the left side is an envelope with 4 counters below it. On the right side is 5 counters.
Solution

Solution

Illustrates the process of setting up and partially solving a linear equation, x + 4 = 5, using a step-by-step description and its mathematical representation.
On the left, write x for the contents of the envelope, add the 4 counters, so we have x+4. x+4
On the right, there are 5 counters. 5
The two sides are equal. x+4=5
Solve the equation by subtracting 4 counters from each side.
The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with 4 counters below it. The 4 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 5 counters. The bottom 4 counters are circled in red with an arrow pointing out of the rectangle. The 4 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 1 counter.

We can see that there is one counter in the envelope. This can be shown algebraically as:
The image shows the given equation, x plus 4 equal to 5. Take 4 away from both sides of the equation to get x plus 4 minus 4 equal to 5 minus 4. On the left, plus 4 and minus 4 cancel out to leave just x. On the right 5 minus 4 is 1. The equation becomes x equal to 1.

Substitute 1 for x in the equation to check.
The image shows the original equation, x plus 4 equal to 5. Substitute 1 in for x to check. The equation becomes 1 plus 4 equal to 5. Is this true? The left side simplifies by adding 1 and 4 to get 5. Both sides of the equal symbol are 5.

Since x=1 makes the statement true, we know that 1 is indeed a solution.

Write the equation modeled by the envelopes and counters, and then solve the equation:

The image is divided in half vertically. On the left side is an envelope with one counter below it. On the right side is 7 counters.
Solution

x + 1 = 7; x = 6

Write the equation modeled by the envelopes and counters, and then solve the equation:

The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 4 counters.
Solution

x + 3 = 4; x = 1

Solve Equations Using the Subtraction Property of Equality

Our puzzle has given us an idea of what we need to do to solve an equation. The goal is to isolate the variable by itself on one side of the equations. In the previous examples, we used the Subtraction Property of Equality, which states that when we subtract the same quantity from both sides of an equation, we still have equality.

Subtraction Property of Equality

For any numbers a,b, and c, if

a=b

then

a−c=b−c

Think about twin brothers Andy and Bobby. They are 17 years old. How old was Andy 3 years ago? He was 3 years less than 17, so his age was 17−3, or 14. What about Bobby’s age 3 years ago? Of course, he was 14 also. Their ages are equal now, and subtracting the same quantity from both of them resulted in equal ages 3 years ago.

a=ba−3=b−3

Solve an equation using the Subtraction Property of Equality.

  1. Use the Subtraction Property of Equality to isolate the variable.
  2. Simplify the expressions on both sides of the equation.
  3. Check the solution.

Solve: x+8=17.

Solution

Solution

We will use the Subtraction Property of Equality to isolate x.
An algebraic equation is shown, displaying 'x + 8 = 17' in black text on a white background.
Subtract 8 from both sides. The equation x + 8 - 8 = 17 - 8, demonstrating the subtraction property of equality to isolate the variable x.
Simplify. A mathematical equation, x = 9, displayed in black text on a white background.
A basic algebraic equation is displayed, showing 'x + 8 = 17' in black text against a white background. This simple addition equation requires solving for the variable 'x'.
A mathematical equation shows '9 + 8 = 17'. The number '9' is red, while the rest of the equation is black.
The image shows the equation '17 = 17' followed by a checkmark, indicating that the statement is correct and verified.

Since x=9 makes x+8=17 a true statement, we know 9 is the solution to the equation.

Solve:

x+6=19

Solution

x = 13

Solve:

x+9=14

Solution

x = 5

Solve: 100=y+74.

Solution

Solution

To solve an equation, we must always isolate the variable—it doesn’t matter which side it is on. To isolate y, we will subtract 74 from both sides.
A mathematical equation is displayed, showing '100 = y + 74' in black text against a white background.
Subtract 74 from both sides. A mathematical equation displays '100 - 74 = y + 74 - 74', illustrating the subtraction property of equality where 74 is subtracted from both sides to isolate the variable 'y'.
Simplify. The image displays a mathematical equation '26 = y' in a clean, legible font against a plain white background, presenting a simple assignment of the value 26 to the variable y.
Substitute 26 for y to check.
The image shows the original equation,100 equal to y plus 74. Substitute 26 in for y to check. The equation becomes 100 equal to 26 plus 74. Is this true? The right side simplifies by adding 26 and 74 to get 100. Both sides of the equal symbol are 100.

Since y=26 makes 100=y+74 a true statement, we have found the solution to this equation.

Solve:

95=y+67

Solution

y = 28

Solve:

91=y+45

Solution

y = 46

Solve Equations Using the Addition Property of Equality

In all the equations we have solved so far, a number was added to the variable on one side of the equation. We used subtraction to “undo” the addition in order to isolate the variable.

But suppose we have an equation with a number subtracted from the variable, such as x−5=8. We want to isolate the variable, so to “undo” the subtraction we will add the number to both sides.

We use the Addition Property of Equality, which says we can add the same number to both sides of the equation without changing the equality. Notice how it mirrors the Subtraction Property of Equality.

Addition Property of Equality

For any numbers a,b, and c, if

a=b

then

a+c=b+c

Remember the 17-year-old twins, Andy and Bobby? In ten years, Andy’s age will still equal Bobby’s age. They will both be 27.

a=ba+10=b+10

We can add the same number to both sides and still keep the equality.

Solve an equation using the Addition Property of Equality.

  1. Use the Addition Property of Equality to isolate the variable.
  2. Simplify the expressions on both sides of the equation.
  3. Check the solution.

Solve: x−5=8.

Solution

Solution

We will use the Addition Property of Equality to isolate the variable.
A simple algebraic equation is displayed, showing 'x - 5 = 8' in a dark, bold font on a white background.
Add 5 to both sides. A mathematical equation shows 'x minus 5 plus 5 equals 8 plus 5', demonstrating the addition property of equality where 5 is added to both sides of the equation x - 5 = 8.
Simplify. A simple mathematical equation is displayed on a white background, showing 'x = 13'.
The image displays the text 'Now we can check. Let x = 13.', suggesting a step in a mathematical or problem-solving process where a value is substituted for a variable to verify a solution.
A mathematical equation, rendered in black text on a white background, shows 'x - 5 = 8'.
A mathematical equation shows '13 - 5 =? 8', with the question mark above the equals sign. This expression tests if thirteen minus five is indeed equal to eight, which is a true statement as 13 - 5 equals 8.
The mathematical statement '8 = 8' is shown, accompanied by a checkmark, indicating its correctness or verification.

Solve:

x−9=13

Solution

x = 22

Solve:

y−1=3

Solution

y = 4

Solve: 27=a−16.

Solution

Solution

We will add 16 to each side to isolate the variable.
The image displays the mathematical equation 27 = a - 16, centered against a white background. This equation represents a basic algebraic problem where the variable 'a' needs to be solved.
Add 16 to each side. An algebraic equation showing 27 + 16 = a - 16 + 16, with plus signs and the number 16 on the right side highlighted in red against a white background.
Simplify. The equation 43 = a is displayed on a white background, indicating that the value of the variable 'a' is 43.
The text 'Now we can check. Let q = 43.' is displayed on a white background. The number 43 is highlighted in red, while the rest of the text is in a dark blue-grey color. A basic algebraic equation is displayed, showing '27 = a - 16' against a plain white background. The equation requires solving for the variable 'a'.
A math problem displaying '27 ?= 43 - 16' with a question mark above the equals sign, indicating a query about the equality of the two sides of the equation. The number 43 is highlighted in red.
The number 27 is shown to be equal to 27, followed by a checkmark, indicating correctness or agreement.

The solution to 27=a−16 is a=43.

Solve:

19=a−18

Solution

a = 37

Solve:

27=n−14

Solution

n = 41

Translate Word Phrases to Algebraic Equations

Remember, an equation has an equal sign between two algebraic expressions. So if we have a sentence that tells us that two phrases are equal, we can translate it into an equation. We look for clue words that mean equals. Some words that translate to the equal sign are:

  • is equal to
  • is the same as
  • is
  • gives
  • was
  • will be

It may be helpful to put a box around the equals word(s) in the sentence to help you focus separately on each phrase. Then translate each phrase into an expression, and write them on each side of the equal sign.

We will practice translating word sentences into algebraic equations. Some of the sentences will be basic number facts with no variables to solve for. Some sentences will translate into equations with variables. The focus right now is just to translate the words into algebra.

Translate the sentence into an algebraic equation: The sum of 6 and 9 is 15.

Solution

Solution

The word is tells us the equal sign goes between 9 and 15.
Locate the “equals” word(s). Illustrates replacing 'is' with '='. The statement 'The sum of 6 and 9 is 15' becomes the mathematical expression 'The sum of 6 and 9 = 15'.
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression. A simple math problem is displayed on a white background, showing the equation '6 + 9 = ___' with a blank space for the answer, indicating an addition sum to be solved.
Translate the words to the right of the equals word into an algebraic expression. The equation 6 + 9 = 15 is displayed in a dark blue font against a white background.

Translate the sentence into an algebraic equation:

The sum of 7 and 6 gives 13.

Solution

7 + 6 = 13

Translate the sentence into an algebraic equation:

The sum of 8 and 6 is 14.

Solution

8 + 6 = 14

Translate the sentence into an algebraic equation: The product of 8 and 7 is 56.

Solution

Solution

The location of the word is tells us that the equal sign goes between 7 and 56.
Locate the “equals” word(s). This image demonstrates how the word 'is' in a descriptive mathematical statement, like 'The product of 8 and 7 is 56,' is equivalent to the equals sign (=), shown as 'The product of 8 and 7 = 56.'
Write the = sign.
Translate the words to the left of the equals word into an algebraic expression. A mathematical equation is shown with '8 . 7 = _' on a white background, indicating a multiplication problem where the product of 8 and 7 needs to be filled in.
Translate the words to the right of the equals word into an algebraic expression. The image displays the multiplication equation 8 multiplied by 7 equals 56, written as '8 ', '.', ' 7 = 56' in a dark blue or gray font against a white background.

Translate the sentence into an algebraic equation:

The product of 6 and 9 is 54.

Solution

6 ⋅ 9 = 54

Translate the sentence into an algebraic equation:

The product of 21 and 3 gives 63.

Solution

21 ⋅ 3 = 63

Translate the sentence into an algebraic equation: Twice the difference of x and 3 gives 18.

Solution

Solution

Locate the “equals” word(s). The text 'Twice the difference of x and 3 gives 18.' is displayed on a white background, with the word 'gives' enclosed in a thin black rectangle.
Recognize the key words: twice; difference of …. and …. Twice means two times.
Translate. This image translates the word problem 'Twice the difference of x and 3 gives 18' into its algebraic equation form, '2(x - 3) = 18', highlighting corresponding parts.

Translate the given sentence into an algebraic equation:

Twice the difference of x and 5 gives 30.

Solution

2(x − 5) = 30

Translate the given sentence into an algebraic equation:

Twice the difference of y and 4 gives 16.

Solution

2(y − 4) = 16

Translate to an Equation and Solve

Now let’s practice translating sentences into algebraic equations and then solving them. We will solve the equations by using the Subtraction and Addition Properties of Equality.

Translate and solve: Three more than x is equal to 47.

Solution

Solution

Three more than x is equal to 47.
Translate. A simple algebraic equation is displayed, showing 'x + 3 = 47' in a bold, sans-serif font against a white background.
Subtract 3 from both sides of the equation. A mathematical equation shows 'x + 3 - 3 = 47 - 3' with the second '3' on the left side and the '3' on the right side of the equation highlighted in red to emphasize subtraction from both sides.
Simplify. The mathematical equation 'x = 44' is displayed in black text against a white background.
We can check. Let x=44. A mathematical equation is displayed with a variable 'x', showing 'x + 3 = 47' against a plain white background.
An equation displaying 44 + 3 with a question mark over the equals sign before 47, asking for verification.
The number 47 is shown equal to 47, followed by a checkmark, indicating the mathematical statement's correctness or verification.

So x=44 is the solution.

Translate and solve:

Seven more than x is equal to 37.

Solution

x + 7 = 37; x = 30

Translate and solve:

Eleven more than y is equal to 28.

Solution

y + 11 = 28; y = 17

Translate and solve: The difference of y and 14 is 18.

Solution

Solution

The difference of y and 14 is 18.
Translate. A mathematical equation is displayed, showing 'y - 14 = 18' in black text on a white background. This equation represents a basic algebraic problem where 'y' is an unknown variable.
Add 14 to both sides. A mathematical equation showing y - 14 + 14 = 18 + 14, with the number 14 highlighted in red as it is added to both sides of the equation.
Simplify. The image shows the mathematical equation y = 32 in black text on a white background. The variable 'y' is set equal to the numerical value '32', indicating a simple assignment or a constant.
We can check. Let y=32. A mathematical equation is displayed, showing 'y - 14 = 18' in black text on a white background. This is a basic algebraic problem to solve for the variable 'y'.
A mathematical equation shows 32 minus 14 followed by an equals sign with a question mark above it, and then the number 18, asking if 32 - 14 equals 18.
The image shows a mathematical expression '18 = 18' followed by a checkmark, symbolizing correctness or validation of the equality.

So y=32 is the solution.

Translate and solve:

The difference of z and 17 is equal to 37.

Solution

z − 17 = 37; z = 54

Translate and solve:

The difference of x and 19 is equal to 45.

Solution

x − 19 = 45; x = 64

ACCESS ADDITIONAL ONLINE RESOURCES

  • Solving One Step Equations By Addition and Subtraction

Key Concepts

  • 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.
  • Subtraction Property of Equality
    • For any numbers a, b, and c,
      if a=b
      then a−c=b−c
  • Solve an equation using the Subtraction Property of Equality.
    1. Use the Subtraction Property of Equality to isolate the variable.
    2. Simplify the expressions on both sides of the equation.
    3. Check the solution.
  • Addition Property of Equality
    • For any numbers a, b, and c,
      if a=b
      then a+c=b+c
  • Solve an equation using the Addition Property of Equality.
    1. Use the Addition Property of Equality to isolate the variable.
    2. Simplify the expressions on both sides of the equation.
    3. Check the solution.

Practice Makes Perfect

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each given value is a solution to the equation.

x+13=21
  1. ⓐ x=8
  2. ⓑ x=34
Solution
  1. ⓐ yes
  2. ⓑ no
y+18=25
  1. ⓐ y=7
  2. ⓑ y=43
m−4=13
  1. ⓐ m=9
  2. ⓑ m=17
Solution
  1. ⓐ no
  2. ⓑ yes
n−9=6
  1. ⓐ n=3
  2. ⓑ n=15
3p+6=15
  1. ⓐ p=3
  2. ⓑ p=7
Solution
  1. ⓐ yes
  2. ⓑ no
8q+4=20
  1. ⓐ q=2
  2. ⓑ q=3
18d−9=27
  1. ⓐ d=1
  2. ⓑ d=2
Solution
  1. ⓐ no
  2. ⓑ yes
24f−12=60
  1. ⓐ f=2
  2. ⓑ f=3
8u−4=4u+40
  1. ⓐ u=3
  2. ⓑ u=11
Solution
  1. ⓐ no
  2. ⓑ yes
7v−3=4v+36
  1. ⓐ v=3
  2. ⓑ v=11
20h−5=15h+35
  1. ⓐ h=6
  2. ⓑ h=8
Solution
  1. ⓐ no
  2. ⓑ yes
18k−3=12k+33
  1. ⓐ k=1
  2. ⓑ k=6

Model the Subtraction Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve using the subtraction property of equality.

The image is divided in half vertically. On the left side is an envelope with 2 counters below it. On the right side is 5 counters.
Solution

x + 2 = 5; x = 3

The image is divided in half vertically. On the left side is an envelope with 4 counters below it. On the right side is 7 counters.
The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 6 counters.
Solution

x + 3 = 6; x = 3

The image is divided in half vertically. On the left side is an envelope with 5 counters below it. On the right side is 9 counters.

Solve Equations using the Subtraction Property of Equality

In the following exercises, solve each equation using the subtraction property of equality.

a+2=18

Solution

a = 16

b+5=13

p+18=23

Solution

p = 5

q+14=31

r+76=100

Solution

r = 24

s+62=95

16=x+9

Solution

x = 7

17=y+6

93=p+24

Solution

p = 69

116=q+79

465=d+398

Solution

d = 67

932=c+641

Solve Equations using the Addition Property of Equality

In the following exercises, solve each equation using the addition property of equality.

y−3=19

Solution

y = 22

x−4=12

u−6=24

Solution

u = 30

v−7=35

f−55=123

Solution

f = 178

g−39=117

19=n−13

Solution

n = 32

18=m−15

10=p−38

Solution

p = 48

18=q−72

268=y−199

Solution

y = 467

204=z−149

Translate Word Phrase to Algebraic Equations

In the following exercises, translate the given sentence into an algebraic equation.

The sum of 8 and 9 is equal to 17.

Solution

8 + 9 = 17

The sum of 7 and 9 is equal to 16.

The difference of 23 and 19 is equal to 4.

Solution

23 − 19 = 4

The difference of 29 and 12 is equal to 17.

The product of 3 and 9 is equal to 27.

Solution

3 ⋅ 9 = 27

The product of 6 and 8 is equal to 48.

The quotient of 54 and 6 is equal to 9.

Solution

54 ÷ 6 = 9

The quotient of 42 and 7 is equal to 6.

Twice the difference of n and 10 gives 52.

Solution

2(n − 10) = 52

Twice the difference of m and 14 gives 64.

The sum of three times y and 10 is 100.

Solution

3y + 10 = 100

The sum of eight times x and 4 is 68.

Translate to an Equation and Solve

In the following exercises, translate the given sentence into an algebraic equation and then solve it.

Five more than p is equal to 21.

Solution

p + 5 = 21; p = 16

Nine more than q is equal to 40.

The sum of r and 18 is 73.

Solution

r + 18 = 73; r = 55

The sum of s and 13 is 68.

The difference of d and 30 is equal to 52.

Solution

d − 30 = 52; d = 82

The difference of c and 25 is equal to 75.

12 less than u is 89.

Solution

u − 12 = 89; u = 101

19 less than w is 56.

325 less than c gives 799.

Solution

c − 325 = 799; c = 1124

299 less than d gives 850.

Everyday Math

Insurance Vince’s car insurance has a $500 deductible. Find the amount the insurance company will pay, p, for an $1800 claim by solving the equation 500+p=1800.

Solution

$1300

Insurance Marta’s homeowner’s insurance policy has a $750 deductible. The insurance company paid $5800 to repair damages caused by a storm. Find the total cost of the storm damage, d, by solving the equation d−750=5800.

Sale purchase Arthur bought a suit that was on sale for $120 off. He paid $340 for the suit. Find the original price, p, of the suit by solving the equation p−120=340.

Solution

$460

Sale purchase Rita bought a sofa that was on sale for $1299. She paid a total of $1409, including sales tax. Find the amount of the sales tax, t, by solving the equation 1299+t=1409.

Writing Exercises

Is x=1 a solution to the equation 8x−2=16−6x? How do you know?

Write the equation y−5=21 in words. Then make up a word problem for this equation.

Self Check

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

A self-assessment chart for math skills, including determining solutions, modeling and solving equations using addition/subtraction properties, and translating word phrases to algebraic equations.

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

solution of an equation
A solution to an equation is a value of a variable that makes a true statement when substituted into the equation. The process of finding the solution to an equation is called solving the equation.

Find Multiples and Factors

Learning Objectives

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

  • Identify multiples of numbers
  • Use common divisibility tests
  • Find all the factors of a number
  • Identify prime and composite numbers

Before you get started, take this readiness quiz.

Which of the following numbers are counting numbers (natural numbers)?
0,4,215
If you missed this problem, review Example 1 in Introduction to Whole Numbers.

Solution

4 and 215

Find the sum of 3,5, and 7.
If you missed the problem, review Example 1 in Use the Language of Algebra.

Solution

15

Identify Multiples of Numbers

Annie is counting the shoes in her closet. The shoes are matched in pairs, so she doesn’t have to count each one. She counts by twos: 2,4,6,8,10,12. She has 12 shoes in her closet.

The numbers 2,4,6,8,10,12 are called multiples of 2. Multiples of 2 can be written as the product of a counting number and 2. The first six multiples of 2 are given below.

1⋅2=22⋅2=43⋅2=64⋅2=85⋅2=106⋅2=12

A multiple of a number is the product of the number and a counting number. So a multiple of 3 would be the product of a counting number and 3. Below are the first six multiples of 3.

1⋅3=32⋅3=63⋅3=94⋅3=125⋅3=156⋅3=18

We can find the multiples of any number by continuing this process. Table 1 shows the multiples of 2 through 9 for the first twelve counting numbers.

Counting Number 1 2 3 4 5 6 7 8 9 10 11 12
Multiples of2 2 4 6 8 10 12 14 16 18 20 22 24
Multiples of3 3 6 9 12 15 18 21 24 27 30 33 36
Multiples of4 4 8 12 16 20 24 28 32 36 40 44 48
Multiples of5 5 10 15 20 25 30 35 40 45 50 55 60
Multiples of6 6 12 18 24 30 36 42 48 54 60 66 72
Multiples of7 7 14 21 28 35 42 49 56 63 70 77 84
Multiples of8 8 16 24 32 40 48 56 64 72 80 88 96
Multiples of9 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.

Recognizing the patterns for multiples of 2,5,10,and3 will be helpful to you as you continue in this course.

Doing the Manipulative Mathematics activity “Multiples” will help you develop a better understanding of multiples.

Figure 1 shows the counting numbers from 1 to 50. Multiples of 2 are highlighted. Do you notice a pattern?

The image shows a chart with five rows and ten columns. The first row lists the numbers from 1 to 10. The second row lists the numbers from 11 to 20. The third row lists the numbers from 21 to 30. The fourth row lists the numbers from 31 and 40. The fifth row lists the numbers from 41 to 50. All factors of 2 are highlighted in blue.
Multiples of 2 between 1 and 50

The last digit of each highlighted number in Figure 1 is either 0,2,4,6,or8. This is true for the product of 2 and any counting number. So, to tell if any number is a multiple of 2 look at the last digit. If it is 0,2,4,6,or8, then the number is a multiple of 2.

Determine whether each of the following is a multiple of 2:
  1. ⓐ 489
  2. ⓑ 3,714
Solution

Solution

Steps to determine if the number 489 is a multiple of 2, based on checking its last digit.
ⓐ
Is 489 a multiple of 2?
Is the last digit 0, 2, 4, 6, or 8? No.
489 is not a multiple of 2.
This table demonstrates how to determine if 3,714 is a multiple of 2 by checking its last digit against the divisibility rule for 2.
ⓑ
Is 3,714 a multiple of 2?
Is the last digit 0, 2, 4, 6, or 8? Yes.
3,714 is a multiple of 2.

Determine whether each number is a multiple of 2:

  1. ⓐ 678
  2. ⓑ 21,493
Solution
  1. ⓐ yes
  2. ⓑ no

Determine whether each number is a multiple of 2:

  1. ⓐ 979
  2. ⓑ 17,780
Solution
  1. ⓐ no
  2. ⓑ yes

Now let’s look at multiples of 5. Figure 2 highlights all of the multiples of 5 between 1 and 50. What do you notice about the multiples of 5?

The image shows a chart with five rows and ten columns. The first row lists the numbers from 1 to 10. The second row lists the numbers from 11 to 20. The third row lists the numbers from 21 to 30. The fourth row lists the numbers from 31 and 40. The fifth row lists the numbers from 41 to 50. All factors of 5 are highlighted in blue.
Multiples of 5 between 1 and 50

All multiples of 5 end with either 5 or 0. Just like we identify multiples of 2 by looking at the last digit, we can identify multiples of 5 by looking at the last digit.

Determine whether each of the following is a multiple of 5:
  1. ⓐ 579
  2. ⓑ 880
Solution

Solution

This table demonstrates how to determine if 579 is a multiple of 5 by checking its last digit.
ⓐ
Is 579 a multiple of 5?
Is the last digit 5 or 0? No.
579 is not a multiple of 5.
This table illustrates the divisibility rule for 5 by asking if 880 is a multiple of 5 and verifying the condition based on its last digit.
ⓑ
Is 880 a multiple of 5?
Is the last digit 5 or 0? Yes.
880 is a multiple of 5.

Determine whether each number is a multiple of 5.

  1. ⓐ 675
  2. ⓑ 1,578
Solution
  1. ⓐ yes
  2. ⓑ no

Determine whether each number is a multiple of 5.

  1. ⓐ 421
  2. ⓑ 2,690
Solution
  1. ⓐ no
  2. ⓑ yes

Figure 3 highlights the multiples of 10 between 1 and 50. All multiples of 10 all end with a zero.

The image shows a chart with five rows and ten columns. The first row lists the numbers from 1 to 10. The second row lists the numbers from 11 to 20. The third row lists the numbers from 21 to 30. The fourth row lists the numbers from 31 and 40. The fifth row lists the numbers from 41 to 50. All factors of 10 are highlighted in blue.
Multiples of 10 between 1 and 50
Determine whether each of the following is a multiple of 10:
  1. ⓐ 425
  2. ⓑ 350
Solution

Solution

Illustrates the divisibility test for 425 by 10, focusing on checking the last digit.
ⓐ
Is 425 a multiple of 10?
Is the last digit zero? No.
425 is not a multiple of 10.
Example showing how to determine if a number is a multiple of 10 by checking its last digit.
ⓑ
Is 350 a multiple of 10?
Is the last digit zero? Yes.
350 is a multiple of 10.

Determine whether each number is a multiple of 10:

  1. ⓐ 179
  2. ⓑ 3,540
Solution
  1. ⓐ no
  2. ⓑ yes

Determine whether each number is a multiple of 10:

  1. ⓐ 110
  2. ⓑ 7,595
Solution
  1. ⓐ yes
  2. ⓑ no

Figure 4 highlights multiples of 3. The pattern for multiples of 3 is not as obvious as the patterns for multiples of 2,5,and10.

The image shows a chart with five rows and ten columns. The first row lists the numbers from 1 to 10. The second row lists the numbers from 11 to 20. The third row lists the numbers from 21 to 30. The fourth row lists the numbers from 31 and 40. The fifth row lists the numbers from 41 to 50. All factors of 3 are highlighted in blue.
Multiples of 3 between 1 and 50

Unlike the other patterns we’ve examined so far, this pattern does not involve the last digit. The pattern for multiples of 3 is based on the sum of the digits. If the sum of the digits of a number is a multiple of 3, then the number itself is a multiple of 3. See Table 8.

Multiple of 3 3 6 9 12 15 18 21 24
Sum of digits 3 6 9 1+23 1+56 1+89 2+13 2+46

Consider the number 42. The digits are 4 and 2, and their sum is 4+2=6. Since 6 is a multiple of 3, we know that 42 is also a multiple of 3.

Determine whether each of the given numbers is a multiple of 3:
  1. ⓐ 645
  2. ⓑ 10,519
Solution

Solution

ⓐ Is 645 a multiple of 3?

Demonstrating the divisibility rule for 3 with an example, including summing digits and verifying the result.
Find the sum of the digits. 6+4+5=15
Is 15 a multiple of 3? Yes.
If we're not sure, we could add its digits to find out. We can check it by dividing 645 by 3. 645÷3
The quotient is 215. 3⋅215=645

ⓑ Is 10,519 a multiple of 3?

Demonstrates the divisibility rule for 3 by summing digits and verifying with division, using 10,519 as an example.
Find the sum of the digits. 1+0+5+1+9=16
Is 16 a multiple of 3? No.
So 10,519 is not a multiple of 3 either.. 645÷3
We can check this by dividing by 10,519 by 3. 3,506R1310,519

When we divide 10,519 by 3, we do not get a counting number, so 10,519 is not the product of a counting number and 3. It is not a multiple of 3.

Determine whether each number is a multiple of 3:

  1. ⓐ 954
  2. ⓑ 3,742
Solution
  1. ⓐ yes
  2. ⓑ no

Determine whether each number is a multiple of 3:

  1. ⓐ 643
  2. ⓑ 8,379
Solution
  1. ⓐ no
  2. ⓑ yes

Look back at the charts where you highlighted the multiples of 2, of 5, and of 10. Notice that the multiples of 10 are the numbers that are multiples of both 2 and 5. That is because 10=2⋅5. Likewise, since 6=2⋅3, the multiples of 6 are the numbers that are multiples of both 2 and 3.

Use Common Divisibility Tests

Another way to say that 375 is a multiple of 5 is to say that 375 is divisible by 5. In fact, 375÷5 is 75, so 375 is 5⋅75. Notice in Example 4 that 10,519 is not a multiple 3. When we divided 10,519 by 3 we did not get a counting number, so 10,519 is not divisible by 3.

Divisibility

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

Since multiplication and division are inverse operations, the patterns of multiples that we found can be used as divisibility tests. Table 11 summarizes divisibility tests for some of the counting numbers between one and ten.

Divisibility Tests
A number is divisible by
2 if the last digit is 0,2,4,6,or8
3 if the sum of the digits is divisible by 3
5 if the last digit is 5 or 0
6 if divisible by both 2 and 3
10 if the last digit is 0

Determine whether 1,290 is divisible by 2,3,5,and10.

Solution

Solution

Table 12 applies the divisibility tests to 1,290. In the far right column, we check the results of the divisibility tests by seeing if the quotient is a whole number.

Divisible by…? Test Divisible? Check
2 Is last digit 0,2,4,6,or8? Yes. yes 1290÷2=645
3 Is sum of digits divisible by3?
1+2+9+0=12 Yes.
yes 1290÷3=430
5 Is last digit 5 or 0? Yes. yes 1290÷5=258
10 Is last digit 0? Yes. yes 1290÷10=129

Thus, 1,290 is divisible by 2,3,5,and10.

Determine whether the given number is divisible by 2,3,5,and10.

6240

Solution

Divisible by 2, 3, 5, and 10

Determine whether the given number is divisible by 2,3,5,and10.

7248

Solution

Divisible by 2 and 3, not 5 or 10.

Determine whether 5,625 is divisible by 2,3,5,and10.

Solution

Solution

Table 13 applies the divisibility tests to 5,625 and tests the results by finding the quotients.

Divisible by…? Test Divisible? Check
2 Is last digit 0,2,4,6,or8? No. no 5625÷2=2812.5
3 Is sum of digits divisible by3?
5+6+2+5=18 Yes.
yes 5625÷3=1875
5 Is last digit is 5 or 0? Yes. yes 5625÷5=1125
10 Is last digit 0? No. no 5625÷10=562.5

Thus, 5,625 is divisible by 3 and 5, but not 2, or 10.

Determine whether the given number is divisible by2,3,5,and10.

4962

Solution

Divisible by 2 and 3, not 5 or 10.

Determine whether the given number is divisible by2,3,5,and10.

3765

Solution

Divisible by 3 and 5.

Find All the Factors of a Number

There are often several ways to talk about the same idea. So far, we’ve seen that if m is a multiple of n, we can say that m is divisible by n. We know that 72 is the product of 8 and 9, so we can say 72 is a multiple of 8 and 72 is a multiple of 9. We can also say 72 is divisible by 8 and by 9. Another way to talk about this is to say that 8 and 9 are factors of 72. When we write 72=8⋅9 we can say that we have factored 72.

The image shows the equation 8 times 9 equals 72. The 8 and 9 are labeled as factors and the 72 is labeled product.

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 aandb are factors of m, and m is the product of aandb.

In algebra, it can be useful to determine all of the factors of a number. This is called factoring a number, and it can help us solve many kinds of problems.

Doing the Manipulative Mathematics activity “Model Multiplication and Factoring” will help you develop a better understanding of multiplication and factoring.

For example, suppose a choreographer is planning a dance for a ballet recital. There are 24 dancers, and for a certain scene, the choreographer wants to arrange the dancers in groups of equal sizes on stage.

In how many ways can the dancers be put into groups of equal size? Answering this question is the same as identifying the factors of 24. Table 14 summarizes the different ways that the choreographer can arrange the dancers.

Number of Groups Dancers per Group Total Dancers
1 24 1⋅24=24
2 12 2⋅12=24
3 8 3⋅8=24
4 6 4⋅6=24
6 4 6⋅4=24
8 3 8⋅3=24
12 2 12⋅2=24
24 1 24⋅1=24

What patterns do you see in Table 14? Did you notice that the number of groups times the number of dancers per group is always 24? This makes sense, since there are always 24 dancers.

You may notice another pattern if you look carefully at the first two columns. These two columns contain the exact same set of numbers—but in reverse order. They are mirrors of one another, and in fact, both columns list all of the factors of 24, which are:

1,2,3,4,6,8,12,24

We can find all the factors of any counting number by systematically dividing the number by each counting number, starting with 1. If the quotient is also a counting number, then the divisor and the quotient are factors of the number. We can stop when the quotient becomes smaller than the divisor.

Find all the factors of a counting number.

  1. Divide the number by each of the counting numbers, in order, until the quotient is smaller than the divisor.
    • If the quotient is a counting number, the divisor and quotient are a pair of factors.
    • If the quotient is not a counting number, the divisor is not a factor.
  2. List all the factor pairs.
  3. Write all the factors in order from smallest to largest.

Find all the factors of 72.

Solution

Solution

Divide 72 by each of the counting numbers starting with 1. If the quotient is a whole number, the divisor and quotient are a pair of factors.
The figure shows a table with ten rows and four columns. The first row is a header row and labels the rows “Dividend”, “Divisor”, “Quotient”, and “Factors”. Under the “Dividend” column all rows show the number 72. In the second row the “Divisor” column is 1, the “Quotient” column is 72 and the “Factors” column is 1 and 72. In the third row the “Divisor” column is 2, the “Quotient” column is 36 and the “Factors” column is 2 and 36. In the fourth row the “Divisor” column is 3, the “Quotient” column is 24 and the “Factors” column is 3 and 24. In the fifth row the “Divisor” column is 4, the “Quotient” column is 18 and the “Factors” column is 4 and 18. In the sixth row the “Divisor” column is 5, the “Quotient” column is 14.4 and the “Factors” column is blank. In the seventh row the “Divisor” column is 6, the “Quotient” column is 12 and the “Factors” column is 6 and 12. In the eighth row the “Divisor” column is 7, the “Quotient” column is about 10.29 and the “Factors” column is blank. In the ninth row the “Divisor” column is 8, the “Quotient” column is 9 and the “Factors” column is 8 and 9. In the tenth row the “Divisor” column is 9, the “Quotient” column is 8 and the “Factors” column is 9 and 8.

The next line would have a divisor of 9 and a quotient of 8. The quotient would be smaller than the divisor, so we stop. If we continued, we would end up only listing the same factors again in reverse order. Listing all the factors from smallest to greatest, we have

1,2,3,4,6,8,9,12,18,24,36,and72

Find all the factors of the given number:

96

Solution

1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Find all the factors of the given number:

80

Solution

1, 2, 4, 5, 8, 10, 16, 20, 40, 80

Identify Prime and Composite Numbers

Some numbers, like 72, have many factors. Other numbers, such as 7, have only two factors: 1 and the number. A number with only two factors is called a prime number. A number with more than two factors is called a composite number. The number 1 is neither prime nor composite. It has only one factor, itself.

Prime Numbers and Composite Numbers

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

A composite number is a counting number that is not prime.

Figure 5 lists the counting numbers from 2 through 20 along with their factors. The highlighted numbers are prime, since each has only two factors.

This figure shows a table with twenty rows and three columns. The first row is a header row. It labels the columns as “Number”, “Factor” and “Prime or composite?” The second row lists the number 2, in red, under the “Number” column, the numbers 1 and 2 under the “Factors” column and the word prime under the “Prime or Composite?” column. The third row lists the number 3, in red, under the “Number” column, the numbers 1 and 3 under the “Factors” column and the word prime under the “Prime or Composite?” column. The fourth row lists the number 4 under the “Number” column, the numbers 1, 2 and 4 under the “Factors” column and the word composite under the “Prime or Composite?” column. The fifth row lists the number 5, in red, under the “Number” column, the numbers 1 and 5 under the “Factors” column and the word prime under the “Prime or Composite?” column. The sixth row lists the number 6 under the “Number” column, the numbers 1, 2, 3 and 6 under the “Factors” column and the word composite under the “Prime or Composite?” column. The seventh row lists the number 7, in red, under the “Number” column, the numbers 1 and 7 under the “Factors” column and the word prime under the “Prime or Composite?” column. The eighth row lists the number 8 under the “Number” column, the numbers 1, 2, 4 and 8 under the “Factors” column and the word composite under the “Prime or Composite?” column. The ninth row lists the number 9 under the “Number” column, the numbers 1, 3 and 9 under the “Factors” column and the word composite under the “Prime or Composite?” column. The tenth row lists the number 10 under the “Number” column, the numbers 1, 2, 5 and 10 under the “Factors” column and the word composite under the “Prime or Composite?” column. The eleventh row lists the number 11, in red, under the “Number” column, the numbers 1 and 11 under the “Factors” column and the word prime under the “Prime or Composite?” column. The twelfth row lists the number 12 under the “Number” column, the numbers 1, 2, 3, 4, 6 and 12 under the “Factors” column and the word composite under the “Prime or Composite?” column. The thirteenth row lists the number 13, in red, under the “Number” column, the numbers 1 and 13 under the “Factors” column and the word prime under the “Prime or Composite?” column. The fourteenth row lists the number 14 under the “Number” column, the numbers 1, 2, 7 and 14 under the “Factors” column and the word composite under the “Prime or Composite?” column. The fifteenth row lists the number 15 under the “Number” column, the numbers 1, 2, 3, 5 and 15 under the “Factors” column and the word composite under the “Prime or Composite?” column. The sixteenth row lists the number 16 under the “Number” column, the numbers 1, 2, 4, 8 and 16 under the “Factors” column and the word composite under the “Prime or Composite?” column. The seventeenth row lists the number 17, in red, under the “Number” column, the numbers 1 and 17 under the “Factors” column and the word prime under the “Prime or Composite?” column. The eighteenth row lists the number 18 under the “Number” column, the numbers 1, 2, 3, 6, 9 and 18 under the “Factors” column and the word composite under the “Prime or Composite?” column. The nineteenth row lists the number 19, in red, under the “Number” column, the numbers 1 and 19 under the “Factors” column and the word prime under the “Prime or Composite?” column. The twentieth row lists the number 20 under the “Number” column, the numbers 1, 2, 4, 5, 10 and 20 under the “Factors” column and the word composite under the “Prime or Composite?” column.
Factors of the counting numbers from 2 through 20, with prime numbers highlighted

The prime numbers less than 20 are 2,3,5,7,11,13,17,and19. There are many larger prime numbers too. In order to determine whether a number is prime or composite, we need to see if the number has any factors other than 1 and itself. To do this, we can test each of the smaller prime numbers in order to see if it is a factor of the number. If none of the prime numbers are factors, then that number is also prime.

Determine if a number is prime.

  1. Test each of the primes, in order, to see if it is a factor of the number.
  2. Start with 2 and stop when the quotient is smaller than the divisor or when a prime factor is found.
  3. If the number has a prime factor, then it is a composite number. If it has no prime factors, then the number is prime.
Identify each number as prime or composite:
  1. ⓐ 83
  2. ⓑ 77
Solution

Solution

ⓐ Test each prime, in order, to see if it is a factor of 83, starting with 2, as shown. We will stop when the quotient is smaller than the divisor.

Prime Test Factor of 83?
2 Last digit of 83 is not 0,2,4,6,or8. No.
3 8+3=11, and 11 is not divisible by 3. No.
5 The last digit of 83 is not 5 or 0. No.
7 83÷7=11.857…. No.
11 83÷11=7.545… No.

We can stop when we get to 11 because the quotient (7.545…) is less than the divisor.

We did not find any prime numbers that are factors of 83, so we know 83 is prime.

ⓑ Test each prime, in order, to see if it is a factor of 77.

Prime Test Factor of 77?
2 Last digit is not 0,2,4,6,or8. No.
3 7+7=14, and 14 is not divisible by 3. No.
5 the last digit is not 5 or 0. No.
7 77÷7=11 Yes.

Since 77 is divisible by 7, we know it is not a prime number. It is composite.

Identify the number as prime or composite:

91

Solution

composite

Identify the number as prime or composite:

137

Solution

prime

The Links to Literacy activities One Hundred Hungry Ants, Spunky Monkeys on Parade and A Remainder of One will provide you with another view of the topics covered in this section.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Divisibility Rules
  • Factors
  • Ex 1: Determine Factors of a Number
  • Ex 2: Determine Factors of a Number
  • Ex 3: Determine Factors of a Number

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
4 if the last two digits are a number divisible by 4
5 if the last digit is 5 or 0
6 if divisible by both 2 and 3
10 if the last digit is 0
  • Factors If a⋅b=m, then a and b are factors of m, and m is the product of a and b.
  • Find all the factors of a counting number.
    1. Divide the number by each of the counting numbers, in order, until the quotient is smaller than the divisor.
      1. If the quotient is a counting number, the divisor and quotient are a pair of factors.
      2. If the quotient is not a counting number, the divisor is not a factor.
    2. List all the factor pairs.
    3. Write all the factors in order from smallest to largest.
  • Determine if a number is prime.
    1. Test each of the primes, in order, to see if it is a factor of the number.
    2. Start with 2 and stop when the quotient is smaller than the divisor or when a prime factor is found.
    3. If the number has a prime factor, then it is a composite number. If it has no prime factors, then the number is prime.

Practice Makes Perfect

Identify Multiples of Numbers

In the following exercises, list all the multiples less than 50 for the given number.

2

Solution

2, 4, 6, 8, 10 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48

3

4

Solution

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

5

6

Solution

6, 12, 18, 24, 30, 36, 42, 48

7

8

Solution

8, 16, 24, 32, 40, 48

9

10

Solution

10, 20, 30, 40

12

Use Common Divisibility Tests

In the following exercises, use the divisibility tests to determine whether each number is divisible by 2,3,4,5,6,and10.

84

Solution

Divisible by 2, 3, 4, 6

96

75

Solution

Divisible by 3, 5

78

168

Solution

Divisible by 2, 3, 4, 6

264

900

Solution

Divisible by 2, 3, 4, 5, 6, 10

800

896

Solution

Divisible by 2, 4

942

375

Solution

Divisible by 3, 5

750

350

Solution

Divisible by 2, 5, 10

550

1430

Solution

Divisible by 2, 5, 10

1080

22,335

Solution

Divisible by 3, 5

39,075

Find All the Factors of a Number

In the following exercises, find all the factors of the given number.

36

Solution

1, 2, 3, 4, 6, 9, 12, 18, 36

42

60

Solution

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

48

144

Solution

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72,144

200

588

Solution

1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 49, 84, 98, 147, 196, 294, 588

576

Identify Prime and Composite Numbers

In the following exercises, determine if the given number is prime or composite.

43

Solution

prime

67

39

Solution

composite

53

71

Solution

prime

119

481

Solution

composite

221

209

Solution

composite

359

667

Solution

composite

1771

Everyday Math

Banking Frank’s grandmother gave him $100 at his high school graduation. Instead of spending it, Frank opened a bank account. Every week, he added $15 to the account. The table shows how much money Frank had put in the account by the end of each week. Complete the table by filling in the blanks.

Weeks after graduation Total number of dollars Frank put in the account Simplified Total
0 100 100
1 100+15 115
2 100+15⋅2 130
3 100+15⋅3
4 100+15⋅[]
5 100+[]
6
20
x
Solution


This table has nine rows and three columns. The first row is a header row that labels each column. The first column is labeled “Weeks after opening the account”, the second is labeled “Total number of dollars Gina put in the account”, and the last is labeled “Simplified Total”. Under the “Weeks after opening the account” column are the values: 0, 1, 2, 3, 4, 5, 6, 20, and the letter x. Under the “Total number of dollars Gina put in the account” column are the expressions: 75; 75 plus 20; 75 plus 20 times 2; 75 plus 20 times 3; 75 plus 20 times empty set of brackets; 75 plus empty set of brackets; the last three rows are blank. Under the “Simplified Total” column are the values: 75, 95, 115, the last six rows are blank.

Banking In March, Gina opened a Christmas club savings account at her bank. She deposited $75 to open the account. Every week, she added $20 to the account. The table shows how much money Gina had put in the account by the end of each week. Complete the table by filling in the blanks.

Weeks after opening the account Total number of dollars Gina put in the account Simplified Total
0 75 75
1 75+20 95
2 75+20⋅2 115
3 75+20⋅3
4 75+20⋅[]
5 75+[]
6
20
x

Writing Exercises

If a number is divisible by 2 and by 3, why is it also divisible by 6?

What is the difference between prime numbers and composite numbers?

Self Check

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

Self-assessment grid for math skills in number theory: identifying multiples, using divisibility tests, finding factors, and identifying prime and composite numbers. Students rate their 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?

multiple of a number
A number is a multiple of n if it is the product of a counting number and n.
divisibility
If a number m is a multiple of n, then we say that m is divisible by n.
prime number
A prime number is a counting number greater than 1 whose only factors are 1 and itself.
composite number
A composite number is a counting number that is not prime.

Prime Factorization and the Least Common Multiple

Learning Objectives

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

  • Find the prime factorization of a composite number
  • Find the least common multiple (LCM) of two numbers

Before you get started, take this readiness quiz.

Is 810 divisible by 2,3,5,6,or10?
If you missed this problem, review Example 5 in Find Multiples and Factors.

Solution

2,3,5,6,10

Is 127 prime or composite?
If you missed this problem, review Example 8 in Find Multiples and Factors.

Solution

prime

Write 2⋅2⋅2⋅2 in exponential notation.
If you missed this problem, review Example 5 in Use the Language of Algebra.

Solution

24

Find the Prime Factorization of a Composite Number

In the previous section, we found the factors of a number. Prime numbers have only two factors, the number 1 and the prime number itself. Composite numbers have more than two factors, and every composite number can be written as a unique product of primes. This is called the prime factorization of a number. When we write the prime factorization of a number, we are rewriting the number as a product of primes. Finding the prime factorization of a composite number will help you later in this course.

Prime Factorization

The prime factorization of a number is the product of prime numbers that equals the number.

Doing the Manipulative Mathematics activity “Prime Numbers” will help you develop a better sense of prime numbers.

You may want to refer to the following list of prime numbers less than 50 as you work through this section.

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47

Prime Factorization Using the Factor Tree Method

One way to find the prime factorization of a number is to make a factor tree. We start by writing the number, and then writing it as the product of two factors. We write the factors below the number and connect them to the number with a small line segment—a “branch” of the factor tree.

If a factor is prime, we circle it (like a bud on a tree), and do not factor that “branch” any further. If a factor is not prime, we repeat this process, writing it as the product of two factors and adding new branches to the tree.

We continue until all the branches end with a prime. When the factor tree is complete, the circled primes give us the prime factorization.

For example, let’s find the prime factorization of 36. We can start with any factor pair such as 3 and 12. We write 3 and 12 below 36 with branches connecting them.

The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end.

The factor 3 is prime, so we circle it. The factor 12 is composite, so we need to find its factors. Let’s use 3 and 4. We write these factors on the tree under the 12.

The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end. Two more branches are splitting out from under 12. The right branch has the number 4 at the end and the left branch has the number 3 at the end.

The factor 3 is prime, so we circle it. The factor 4 is composite, and it factors into 2·2. We write these factors under the 4. Since 2 is prime, we circle both 2s.

The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end. Two more branches are splitting out from under 12. The right branch has the number 4 at the end and the left branch has the number 3 at the end with a circle around it. Two more branches are splitting out from under 4. Both the left and right branch have the number 2 at the end with a circle around it.

The prime factorization is the product of the circled primes. We generally write the prime factorization in order from least to greatest.

2⋅2⋅3⋅3

In cases like this, where some of the prime factors are repeated, we can write prime factorization in exponential form.

2⋅2⋅3⋅322⋅32

Note that we could have started our factor tree with any factor pair of 36. We chose 12 and 3, but the same result would have been the same if we had started with 2 and 18,4 and 9,or6and6.

Find the prime factorization of a composite number using the tree method.

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

Find the prime factorization of 48 using the factor tree method.

Solution
Solution

We can start our tree using any factor pair of 48. Let's use 2 and 24.

We circle the 2 because it is prime and so that branch is complete.

A factor tree showing the number 48 branching into its factors, with 2 and 24 as its immediate children, and the number 2 circled.
Now we will factor 24. Let's use 4 and 6. A partially completed factor tree illustrating the prime factorization of the number 48, with the prime factor 2 circled, showing branches for 24, 4, and 6.

Neither factor is prime, so we do not circle either.
We factor the 4, using 2 and 2.
We factor 6, using 2 and 3.

We circle the 2s and the 3 since they are prime. Now all of the branches end in a prime.

A factor tree demonstrates the prime factorization of 48, revealing its prime factors as 2, 2, 2, 2, and 3, which multiply to 48. This method breaks a number into its fundamental prime components.
Write the product of the circled numbers. 2⋅2⋅2⋅2⋅3
Write in exponential form. 24⋅3

Check this on your own by multiplying all the factors together. The result should be 48.

Find the prime factorization using the factor tree method: 80

Solution

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5, or 24 ⋅ 5

Find the prime factorization using the factor tree method: 60

Solution

2 ⋅ 2 ⋅ 3 ⋅ 5, or 22 ⋅ 3 ⋅ 5

Find the prime factorization of 84 using the factor tree method.

Solution
Solution

We start with the factor pair 4 and 21.

Neither factor is prime so we factor them further.

A factor tree shows the number 84 branching into its factors 4 and 21.
Now the factors are all prime, so we circle them. A prime factorization tree illustrating how the number 84 breaks down into its prime factors: 2, 2, 3, and 7.
Then we write 84 as the product of all circled primes. 2⋅2⋅3⋅7
22⋅3⋅7

Draw a factor tree of 84.

Find the prime factorization using the factor tree method: 126

Solution

2 ⋅ 3 ⋅ 3 ⋅ 7, or 2 ⋅ 32 ⋅ 7

Find the prime factorization using the factor tree method: 294

Solution

2 ⋅ 3 ⋅ 7 ⋅ 7, or 2 ⋅ 3 ⋅ 72

Prime Factorization Using the Ladder Method

The ladder method is another way to find the prime factors of a composite number. It leads to the same result as the factor tree method. Some people prefer the ladder method to the factor tree method, and vice versa.

To begin building the “ladder,” divide the given number by its smallest prime factor. For example, to start the ladder for 36, we divide 36 by 2, the smallest prime factor of 36.

The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket.

To add a “step” to the ladder, we continue dividing by the same prime until it no longer divides evenly.

The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket. Another division bracket is written around the 18 with a 2 on the outside left of the bracket and a 9 above the 18, outside of the bracket.

Then we divide by the next prime; so we divide 9 by 3.

The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket. Another division bracket is written around the 18 with a 2 on the outside left of the bracket and a 9 above the 18, outside of the bracket. Another division bracket is written around the 9 with a 3 on the outside left of the bracket and a 3 above the 9, outside of the bracket.

We continue dividing up the ladder in this way until the quotient is prime. Since the quotient, 3, is prime, we stop here.

Do you see why the ladder method is sometimes called stacked division?

The prime factorization is the product of all the primes on the sides and top of the ladder.

2⋅2⋅3⋅322⋅32

Notice that the result is the same as we obtained with the factor tree method.

Find the prime factorization of a composite number using the ladder method.

  1. Divide the number by the smallest prime.
  2. Continue dividing by that prime until it no longer divides evenly.
  3. Divide by the next prime until it no longer divides evenly.
  4. Continue until the quotient is a prime.
  5. Write the composite number as the product of all the primes on the sides and top of the ladder.

Find the prime factorization of 120 using the ladder method.

Solution
Solution
Divide the number by the smallest prime, which is 2. A long division problem showing 120 divided by 2, with the quotient 60 written above the division bar. The divisor is 2, the dividend is 120, and the result of the division is 60.
Continue dividing by 2 until it no longer divides evenly. A series of long division calculations demonstrating repeated division by 2: 120 divided by 2 equals 60, then 60 divided by 2 equals 30, and 30 divided by 2 equals 15.
Divide by the next prime, 3. Decomposing 120 into its prime factors using repeated division, resulting in 2, 2, 2, 3, and 5.
The quotient, 5, is prime, so the ladder is complete. Write the prime factorization of 120. 2⋅2⋅2⋅3⋅5
23⋅3⋅5

Check this yourself by multiplying the factors. The result should be 120.

Find the prime factorization using the ladder method: 80

Solution

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5, or 24 ⋅ 5

Find the prime factorization using the ladder method: 60

Solution

2 ⋅ 2 ⋅ 3 ⋅ 5, or 22 ⋅ 3 ⋅ 5

Find the prime factorization of 48 using the ladder method.

Solution
Solution
Divide the number by the smallest prime, 2. A long division problem showing 48 divided by 2, with the quotient 24 written above the divisor, indicating that 24 is the result of 48 divided by 2.
Continue dividing by 2 until it no longer divides evenly. A step-by-step illustration of repeated division by 2, showing 48 divided by 2 repeatedly until it reaches 3. The divisions shown are 48/2=24, 24/2=12, 12/2=6, and 6/2=3.
The quotient, 3, is prime, so the ladder is complete. Write the prime factorization of 48. 2⋅2⋅2⋅2⋅3
24⋅3

Find the prime factorization using the ladder method. 126

Solution

2 ⋅ 3 ⋅ 3 ⋅ 7, or 2 ⋅ 32 ⋅ 7

Find the prime factorization using the ladder method. 294

Solution

2 ⋅ 3 ⋅ 7 ⋅ 7, or 2 ⋅ 3 ⋅ 72

Find the Least Common Multiple (LCM) of Two Numbers

One of the reasons we look at multiples and 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.

Listing Multiples Method

A common multiple of two numbers is a number that is a multiple of both numbers. Suppose we want to find common multiples of 10 and 25. We can list the first several multiples of each number. Then we look for multiples that are common to both lists—these are the common multiples.

10:10,20,30,40,50,60,70,80,90,100,110,… 25:25,50,75,100,125,…

We see that 50 and 100 appear in both lists. They are common multiples of 10 and 25. We would find more common multiples if we continued the list of multiples for each.

The smallest number that is a multiple of two numbers is called the least common multiple (LCM). So the least LCM of 10 and 25 is 50.

Find the least common multiple (LCM) of two numbers by listing multiples.

  1. List the first several multiples of each number.
  2. Look for multiples common to both lists. If there are no common multiples in the lists, write out additional multiples for each number.
  3. Look for the smallest number that is common to both lists.
  4. This number is the LCM.

Find the LCM of 15 and 20 by listing multiples.

Solution
Solution

List the first several multiples of 15 and of 20. Identify the first common multiple.

15:15,30,45,60,75,90,105,12020:20,40,60,80,100,120,140,160

The smallest number to appear on both lists is 60, so 60 is the least common multiple of 15 and 20.

Notice that 120 is on both lists, too. It is a common multiple, but it is not the least common multiple.

Find the least common multiple (LCM) of the given numbers: 9and12

Solution

36

Find the least common multiple (LCM) of the given numbers: 18and24

Solution

72

Prime Factors Method

Another way to find the least common multiple of two numbers is to use their prime factors. We’ll use this method to find the LCM of 12 and 18.

We start by finding the prime factorization of each number.

12=2⋅2⋅318=2⋅3⋅3

Then we write each number as a product of primes, matching primes vertically when possible.

12=2⋅2⋅3 18=2⋅3⋅3

Now we bring down the primes in each column. The LCM is the product of these factors.

The image shows the prime factorization of 12 written as the equation 12 equals 2 times 2 times 3. Below this equation is another showing the prime factorization of 18 written as the equation 18 equals 2 times 3 times 3. The two equations line up vertically at the equal symbol. The first 2 in the prime factorization of 12 aligns with the 2 in the prime factorization of 18. Under the second 2 in the prime factorization of 12 is a gap in the prime factorization of 18. Under the 3 in the prime factorization of 12 is the first 3 in the prime factorization of 18. The second 3 in the prime factorization has no factors above it from the prime factorization of 12. A horizontal line is drawn under the prime factorization of 18. Below this line is the equation LCM equal to 2 times 2 times 3 times 3. Arrows are drawn down vertically from the prime factorization of 12 through the prime factorization of 18 ending at the LCM equation. The first arrow starts at the first 2 in the prime factorization of 12 and continues down through the 2 in the prime factorization of 18. Ending with the first 2 in the LCM. The second arrow starts at the next 2 in the prime factorization of 12 and continues down through the gap in the prime factorization of 18. Ending with the second 2 in the LCM. The third arrow starts at the 3 in the prime factorization of 12 and continues down through the first 3 in the prime factorization of 18. Ending with the first 3 in the LCM. The last arrow starts at the second 3 in the prime factorization of 18 and points down to the second 3 in the LCM.

Notice that the prime factors of 12 and the prime factors of 18 are included in the LCM. By matching up the common primes, each common prime factor is used only once. This ensures that 36 is the least common multiple.

Find the LCM using the prime factors method.

  1. Find the prime factorization of each number.
  2. Write each number as a product of primes, matching primes vertically when possible.
  3. Bring down the primes in each column.
  4. Multiply the factors to get the LCM.

Find the LCM of 15 and 18 using the prime factors method.

Solution
Solution
Write each number as a product of primes. The image displays the prime factorization of two numbers: 15 is shown as 3 times 5, and 18 is shown as 2 times 3 times 3.
Write each number as a product of primes, matching primes vertically when possible. The prime factorization of 15 (3 * 5) and 18 (2 * 3 * 3) is shown in a clear, concise format.
Bring down the primes in each column. Finding the Least Common Multiple (LCM) of 15 and 18 is shown using prime factorization. 15 = 3 * 5 and 18 = 2 * 3 * 3. The LCM combines these factors to 2 * 3 * 3 * 5.
Multiply the factors to get the LCM. LCM=2⋅3⋅3⋅5
The LCM of 15 and 18 is 90.

Find the LCM using the prime factors method. 15and20

Solution

60

Find the LCM using the prime factors method. 15and35

Solution

105

Find the LCM of 50 and 100 using the prime factors method.

Solution
Solution
Write the prime factorization of each number. The image shows the prime factorization of 50 as 2 x 5 x 5, and the prime factorization of 100 as 2 x 2 x 5 x 5.
Write each number as a product of primes, matching primes vertically when possible. The prime factorization of 50 (2*5*5) and 100 (2*2*5*5) is displayed on a white background.
Bring down the primes in each column. The image demonstrates the calculation of the Least Common Multiple (LCM) for 50 and 100 using prime factorization, showing 50 as 2 x 5 x 5 and 100 as 2 x 2 x 5 x 5, resulting in an LCM of 2 x 2 x 5 x 5.
Multiply the factors to get the LCM. LCM=2⋅2⋅5⋅5
The LCM of 50 and 100 is 100.

Find the LCM using the prime factors method: 55,88

Solution

440

Find the LCM using the prime factors method: 60,72

Solution

360

ACCESS ADDITIONAL ONLINE RESOURCES

  • Ex 1: Prime Factorization
  • Ex 2: Prime Factorization
  • Ex 3: Prime Factorization
  • Ex 1: Prime Factorization Using Stacked Division
  • Ex 2: Prime Factorization Using Stacked Division
  • The Least Common Multiple
  • Example: Determining the Least Common Multiple Using a List of Multiples
  • Example: Determining the Least Common Multiple Using Prime Factorization

Key Concepts

  • Find the prime factorization of a composite number using the tree method.
    1. Find any factor pair of the given number, and use these numbers to create two branches.
    2. If a factor is prime, that branch is complete. Circle the prime.
    3. If a factor is not prime, write it as the product of a factor pair and continue the process.
    4. Write the composite number as the product of all the circled primes.
  • Find the prime factorization of a composite number using the ladder method.
    1. Divide the number by the smallest prime.
    2. Continue dividing by that prime until it no longer divides evenly.
    3. Divide by the next prime until it no longer divides evenly.
    4. Continue until the quotient is a prime.
    5. Write the composite number as the product of all the primes on the sides and top of the ladder.
  • Find the LCM by listing multiples.
    1. List the first several multiples of each number.
    2. Look for multiples common to both lists. If there are no common multiples in the lists, write out additional multiples for each number.
    3. Look for the smallest number that is common to both lists.
    4. This number is the LCM.
  • Find the LCM using the prime factors method.
    1. Find the prime factorization of each number.
    2. Write each number as a product of primes, matching primes vertically when possible.
    3. Bring down the primes in each column.
    4. Multiply the factors to get the LCM.

Section Exercises

Practice Makes Perfect

Find the Prime Factorization of a Composite Number

In the following exercises, find the prime factorization of each number using the factor tree method.

86

Solution

2 ⋅ 43

78

132

Solution

2 ⋅ 2 ⋅ 3 ⋅ 11

455

693

Solution

3 ⋅ 3 ⋅ 7 ⋅ 11

420

115

Solution

5 ⋅ 23

225

2475

Solution

3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 11

1560

In the following exercises, find the prime factorization of each number using the ladder method.

56

Solution

2 ⋅ 2 ⋅ 2 ⋅ 7

72

168

Solution

2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7

252

391

Solution

17 ⋅ 23

400

432

Solution

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3

627

2160

Solution

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 5

2520

In the following exercises, find the prime factorization of each number using any method.

150

Solution

2 ⋅ 3 ⋅ 5 ⋅ 5

180

525

Solution

3 ⋅ 5 ⋅ 5 ⋅ 7

444

36

Solution

2 ⋅ 2 ⋅ 3 ⋅ 3

50

350

Solution

2 ⋅ 5 ⋅ 5 ⋅ 7

144

Find the Least Common Multiple (LCM) of Two Numbers

In the following exercises, find the least common multiple (LCM) by listing multiples.

8,12

Solution

24

4,3

6,15

Solution

30

12,16

30,40

Solution

120

20,30

60,75

Solution

300

44,55

In the following exercises, find the least common multiple (LCM) by using the prime factors method.

8,12

Solution

24

12,16

24,30

Solution

120

28,40

70,84

Solution

420

84,90

In the following exercises, find the least common multiple (LCM) using any method.

6,21

Solution

42

9,15

24,30

Solution

120

32,40

Everyday Math

Grocery shopping Hot dogs are sold in packages of ten, but hot dog buns come in packs of eight. What is the smallest number of hot dogs and buns that can be purchased if you want to have the same number of hot dogs and buns? (Hint: it is the LCM!)

Solution

40

Grocery shopping Paper plates are sold in packages of 12 and party cups come in packs of 8. What is the smallest number of plates and cups you can purchase if you want to have the same number of each? (Hint: it is the LCM!)

Writing Exercises

Do you prefer to find the prime factorization of a composite number by using the factor tree method or the ladder method? Why?

Do you prefer to find the LCM by listing multiples or by using the prime factors method? Why?

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: prime factorization and least common multiple (LCM). Users can rate their understanding as Confidently, With some help, or No-I don't get it!

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Use the Language of Algebra

Use Variables and Algebraic Symbols

In the following exercises, translate from algebra to English.

3⋅8

Solution

3 times 8, the product of three and eight.

12−x

24÷6

Solution

24 divided by 6, the quotient of twenty-four and six.

9+2a

50≥47

Solution

50 is greater than or equal to 47

3y<15

n+4=13

Solution

The sum of n and 4 is equal to 13

32−k=7

Identify Expressions and Equations

In the following exercises, determine if each is an expression or equation.

5+u=84

Solution

equation

36−6s

4y−11

Solution

expression

10x=120

Simplify Expressions with Exponents

In the following exercises, write in exponential form.

2⋅2⋅2

Solution

23

a⋅a⋅a⋅a⋅a

x⋅x⋅x⋅x⋅x⋅x

Solution

x6

10⋅10⋅10

In the following exercises, write in expanded form.

84

Solution

8 ⋅ 8 ⋅ 8 ⋅ 8

36

y5

Solution

y ⋅ y ⋅ y ⋅ y ⋅ y

n4

In the following exercises, simplify each expression.

34

Solution

81

106

27

Solution

128

43

Simplify Expressions Using the Order of Operations

In the following exercises, simplify.

10+2⋅5

Solution

20

(10+2)⋅5

(30+6)÷2

Solution

18

30+6÷2

72+52

Solution

74

(7+5)2

4+3(10−1)

Solution

31

(4+3)(10−1)

Evaluate, Simplify, and Translate Expressions

Evaluate an Expression

In the following exercises, evaluate the following expressions.

9x−5whenx=7

Solution

58

y3wheny=5

3a−4bwhena=10,b=1

Solution

26

bhwhenb=7,h=8

Identify Terms, Coefficients and Like Terms

In the following exercises, identify the terms in each expression.

12n2+3n+1

Solution

12n2,3n, 1

4x3+11x+3

In the following exercises, identify the coefficient of each term.

6y

Solution

6

13x2

In the following exercises, identify the like terms.

5x2,3,5y2,3x,x,4

Solution

3 and 4; 3x and x

8,8r2,8r,3r,r2,3s

Simplify Expressions by Combining Like Terms

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

15a+9a

Solution

24a

12y+3y+y

4x+7x+3x

Solution

14x

6+5c+3

8n+2+4n+9

Solution

12n + 11

19p+5+4p−1+3p

7y2+2y+11+3y2−8

Solution

10y2 + 2y + 3

13x2−x+6+5x2+9x

Translate English Phrases to Algebraic Expressions

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

the difference of x and 6

Solution

x − 6

the sum of 10 and twice a

the product of 3n and 9

Solution

3n ⋅ 9

the quotient of s and 4

5 times the sum of y and 1

Solution

5(y + 1)

10 less than the product of 5 and z

Jack bought a sandwich and a coffee. The cost of the sandwich was $3 more than the cost of the coffee. Call the cost of the coffee c. Write an expression for the cost of the sandwich.

Solution

c + 3

The number of poetry books on Brianna’s bookshelf is 5 less than twice the number of novels. Call the number of novels n. Write an expression for the number of poetry books.

Solve Equations Using the Subtraction and Addition Properties of Equality

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each number is a solution to the equation.

y+16=40
  1. ⓐ 24
  2. ⓑ 56
Solution
  1. ⓐ yes
  2. ⓑ no
d−6=21
  1. ⓐ 15
  2. ⓑ 27
4n+12=36
  1. ⓐ 6
  2. ⓑ 12
Solution
  1. ⓐ yes
  2. ⓑ no
20q−10=70
  1. ⓐ 3
  2. ⓑ 4
15x−5=10x+45
  1. ⓐ 2
  2. ⓑ 10
Solution
  1. ⓐ no
  2. ⓑ yes
22p−6=18p+86
  1. ⓐ 4
  2. ⓑ 23

Model the Subtraction Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve the equation using the subtraction property of equality.

This image is divided into two parts: the first part shows an envelope and 3 blue counters and the next to it, the second part shows five counters.
Solution

x + 3 = 5; x = 2

This image is divided into two parts: the first part shows an envelope and 4 blue counters and next to it, the second part shows 9 counters.

Solve Equations using the Subtraction Property of Equality

In the following exercises, solve each equation using the subtraction property of equality.

c+8=14

Solution

c = 6

v+8=150

23=x+12

Solution

x = 11

376=n+265

Solve Equations using the Addition Property of Equality

In the following exercises, solve each equation using the addition property of equality.

y−7=16

Solution

y = 23

k−42=113

19=p−15

Solution

p = 34

501=u−399

Translate English Sentences to Algebraic Equations

In the following exercises, translate each English sentence into an algebraic equation.

The sum of 7 and 33 is equal to 40.

Solution

7 + 33 = 40

The difference of 15 and 3 is equal to 12.

The product of 4 and 8 is equal to 32.

Solution

4 ⋅ 8 = 32

The quotient of 63 and 9 is equal to 7.

Twice the difference of n and 3 gives 76.

Solution

2(n − 3) = 76

The sum of five times y and 4 is 89.

Translate to an Equation and Solve

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

Eight more than x is equal to 35.

Solution

x + 8 = 35; x = 27

21 less than a is 11.

The difference of q and 18 is 57.

Solution

q − 18 = 57; q = 75

The sum of m and 125 is 240.

Mixed Practice

In the following exercises, solve each equation.

h−15=27

Solution

h = 42

k−11=34

z+52=85

Solution

z = 33

x+93=114

27=q+19

Solution

q = 8

38=p+19

31=v−25

Solution

v = 56

38=u−16

Find Multiples and Factors

Identify Multiples of Numbers

In the following exercises, list all the multiples less than 50 for each of the following.

3

Solution

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48

2

8

Solution

8, 16, 24, 32, 40, 48

10

Use Common Divisibility Tests

In the following exercises, using the divisibility tests, determine whether each number is divisible by 2,by3,by5,by6,and by10.

96

Solution

2, 3, 6

250

420

Solution

2, 3, 5, 6, 10

625

Find All the Factors of a Number

In the following exercises, find all the factors of each number.

30

Solution

1, 2, 3, 5, 6, 10, 15, 30

70

180

Solution

1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180

378

Identify Prime and Composite Numbers

In the following exercises, identify each number as prime or composite.

19

Solution

prime

51

121

Solution

composite

219

Prime Factorization and the Least Common Multiple

Find the Prime Factorization of a Composite Number

In the following exercises, find the prime factorization of each number.

84

Solution

2 ⋅ 2 ⋅ 3 ⋅ 7

165

350

Solution

2 ⋅ 5 ⋅ 5 ⋅ 7

572

Find the Least Common Multiple of Two Numbers

In the following exercises, find the least common multiple of each pair of numbers.

9,15

Solution

45

12,20

25,35

Solution

175

18,40

Everyday Math

Describe how you have used two topics from The Language of Algebra chapter in your life outside of your math class during the past month.

Solution

Answers will vary

Chapter Practice Test

In the following exercises, translate from an algebraic equation to English phrases.

6⋅4

15−x

Solution

15 minus x, the difference of fifteen and x.

In the following exercises, identify each as an expression or equation.

5⋅8+10

x+6=9

Solution

equation

3⋅11=33

  1. ⓐ Write n⋅n⋅n⋅n⋅n⋅n in exponential form.
  2. ⓑ Write 35 in expanded form and then simplify.
Solution
  1. ⓐ n6
  2. ⓑ 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3 = 243

In the following exercises, simplify, using the order of operations.

4+3⋅5

(8+1)⋅4

Solution

36

1+6(3−1)

(8+4)÷3+1

Solution

5

(1+4)2

5[2+7(9−8)]

Solution

45

In the following exercises, evaluate each expression.

8x−3whenx=4

y3wheny=5

Solution

125

6a−2bwhena=5,b=7

hwwhenh=12,w=3

Solution

36

Simplify by combining like terms.
  1. ⓐ 6x+8x
  2. ⓑ 9m+10+m+3

In the following exercises, translate each phrase into an algebraic expression.

5 more than x

Solution

x + 5

the quotient of 12 and y

three times the difference of aandb

Solution

3(a − b)

Caroline has 3 fewer earrings on her left ear than on her right ear. Call the number of earrings on her right ear, r. Write an expression for the number of earrings on her left ear.

In the following exercises, solve each equation.

n−6=25

Solution

n = 31

x+58=71

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

15 less than y is 32.

Solution

y − 15 = 32; y = 47

the sum of a and 129 is 164.

List all the multiples of 4 that are less than 50.

Solution

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

Find all the factors of 90.

Find the prime factorization of 1080.

Solution

23 ⋅ 33 ⋅ 5

Find the LCM (Least Common Multiple) of 24 and 40.

least common multiple
The smallest number that is a multiple of two numbers is called the least common multiple (LCM).
prime factorization
The prime factorization of a number is the product of prime numbers that equals the number.

Introduction to Integers

Photo shows a snow-covered mountain range.
The peak of Mount Everest. (credit: Gunther Hagleitner, Flickr)

At over 29,000 feet, Mount Everest stands as the tallest peak on land. Located along the border of Nepal and China, Mount Everest is also known for its extreme climate. Near the summit, temperatures never rise above freezing. Every year, climbers from around the world brave the extreme conditions in an effort to scale the tremendous height. Only some are successful. Describing the drastic change in elevation the climbers experience and the change in temperatures requires using numbers that extend both above and below zero. In this chapter, we will describe these kinds of numbers and operations using them.

Introduction to Integers

Learning Objectives

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

  • Locate positive and negative numbers on the number line
  • Order positive and negative numbers
  • Find opposites
  • Simplify expressions with absolute value
  • Translate word phrases to expressions with integers

Before you get started, take this readiness quiz.

Plot 0,1,and3 on a number line.
If you missed this problem, review Example 1 in Introduction to Whole Numbers.

Solution

A horizontal number line shows integers from -5 to 5. Orange dots are placed at 0, 1, and 3.

Fill in the appropriate symbol: (=, <, or >):2___4
If you missed this problem, review Example 3 in Use the Language of Algebra.

Solution

<

Locate Positive and Negative Numbers on the Number Line

Do you live in a place that has very cold winters? Have you ever experienced a temperature below zero? If so, you are already familiar with negative numbers. A negative number is a number that is less than 0. Very cold temperatures are measured in degrees below zero and can be described by negative numbers. For example, −1°F (read as “negative one degree Fahrenheit”) is 1degree below 0. A minus sign is shown before a number to indicate that it is negative. Figure 1 shows −20°F, which is 20degrees below 0.

This figure is a thermometer scaled in degrees Fahrenheit. The thermometer has a reading of 20 degrees.
Temperatures below zero are described by negative numbers.

Temperatures are not the only negative numbers. A bank overdraft is another example of a negative number. If a person writes a check for more than he has in his account, his balance will be negative.

Elevations can also be represented by negative numbers. The elevation at sea level is 0 feet. Elevations above sea level are positive and elevations below sea level are negative. The elevation of the Dead Sea, which borders Israel and Jordan, is about 1,302feet below sea level, so the elevation of the Dead Sea can be represented as −1,302feet. See Figure 2.

This figure is a drawing of a side view of the coast of Israel, showing different elevations. The Mediterranean Sea is labeled 0 feet elevation and the Dead Sea is labeled negative 1302 feet elevation. The country of Jordan is also labeled in the figure.
The surface of the Mediterranean Sea has an elevation of 0ft. The diagram shows that nearby mountains have higher (positive) elevations whereas the Dead Sea has a lower (negative) elevation.

Depths below the ocean surface are also described by negative numbers. A submarine, for example, might descend to a depth of 500feet. Its position would then be −500feet as labeled in Figure 3.

This figure is a drawing of a submarine underwater. In the water is also a vertical number line, scaled in feet. The number line has 0 feet at the surface and negative 500 feet below the water where the submarine is located.
Depths below sea level are described by negative numbers. A submarine 500ft below sea level is at −500ft.

Both positive and negative numbers can be represented on a number line. Recall that the number line created in Add Whole Numbers started at 0 and showed the counting numbers increasing to the right as shown in Figure 4. The counting numbers (1, 2, 3, …) on the number line are all positive. We could write a plus sign, +, before a positive number such as +2 or +3, but it is customary to omit the plus sign and write only the number. If there is no sign, the number is assumed to be positive.

This figure is a number line scaled from 0 to 6.

Now we need to extend the number line to include negative numbers. We mark several units to the left of zero, keeping the intervals the same width as those on the positive side. We label the marks with negative numbers, starting with −1 at the first mark to the left of 0,−2 at the next mark, and so on. See Figure 5.

This figure is a number line with 0 in the middle. Then, the scaling has positive numbers 1 to 4 to the right of 0 and negative numbers, negative 1 to negative 4 to the left of 0.
On a number line, positive numbers are to the right of zero. Negative numbers are to the left of zero. What about zero? Zero is neither positive nor negative.

The arrows at either end of the line indicate that the number line extends forever in each direction. There is no greatest positive number and there is no smallest negative number.

Doing the Manipulative Mathematics activity "Number Line-part 2" will help you develop a better understanding of integers.

Plot the numbers on a number line:
  1. ⓐ 3
  2. ⓑ −3
  3. ⓒ −2
Solution

Solution

Draw a number line. Mark 0 in the center and label several units to the left and right.

  1. ⓐ To plot 3, start at 0 and count three units to the right. Place a point as shown in Figure 6.
    This figure is a number line scaled from negative 4 to 4, with the point 3 labeled with a dot.
  2. ⓑ To plot −3, start at 0 and count three units to the left. Place a point as shown in Figure 7.
    This figure is a number line scaled from negative 4 to 4, with the point negative 3 labeled with a dot.
  3. ⓒ To plot −2, start at 0 and count two units to the left. Place a point as shown in Figure 8.
    This figure is a number line scaled from negative 4 to 4, with the point negative 2 labeled with a dot.
Plot the numbers on a number line.
  1. ⓐ 1
  2. ⓑ −1
  3. ⓒ −4
Solution


This figure is a number line. The point negative 4 is labeled with the letter c, the point negative 1 is labeled with the letter b, and the point 1 is labeled with the letter a.

Plot the numbers on a number line.
  1. ⓐ −4
  2. ⓑ 4
  3. ⓐ −1
Solution


This figure is a number line. The point negative 4 is labeled with the letter a, the point negative 1 is labeled with the letter c, and the point 4 is labeled with the letter b.

Order Positive and Negative Numbers

We can use the number line to compare and order positive and negative numbers. Going from left to right, numbers increase in value. Going from right to left, numbers decrease in value. See Figure 9.

This figure is a number line. Above the number line there is an arrow pointing to the right labeled increasing. Below the number line there is an arrow pointing to the left labeled decreasing.

Just as we did with positive numbers, we can use inequality symbols to show the ordering of positive and negative numbers. Remember that we use the notation a<b (read a is less than b) when a is to the left of b on the number line. We write a>b (read a is greater than b) when a is to the right of b on the number line. This is shown for the numbers 3 and 5 in Figure 10.

This figure is a number line with points 3 and 5 labeled with dots. Below the number line is the statements 3 is less than 5 and 5 is greater than 3.
The number 3 is to the left of 5 on the number line. So 3 is less than 5, and 5 is greater than 3.

The numbers lines to follow show a few more examples.

ⓐ
This figure is a number line with points 1 and 4 labeled with dots.

4 is to the right of 1 on the number line, so 4>1.

1 is to the left of 4 on the number line, so 1<4.

ⓑ
This figure is a number line with points negative 2 and 1 labeled with dots.

−2 is to the left of 1 on the number line, so −2<1.

1 is to the right of −2 on the number line, so 1>−2.

ⓒ
This figure is a number line with points negative 3 and negative 1 labeled with dots.

−1 is to the right of −3 on the number line, so −1>−3.

−3 is to the left of −1 on the number line, so −3<−1.

Order each of the following pairs of numbers using < or >:
  1. ⓐ 14___6
  2. ⓑ −1___9
  3. ⓒ −1___−4
  4. ⓓ 2___−20
Solution

Solution

Begin by plotting the numbers on a number line as shown in Figure 11.
This figure is a number line with points negative 20, negative 4, negative 1, 2, 6, 9, and 14 labeled with dots.
Compares 14 and 6, illustrating the context and resulting mathematical inequality using a number line explanation.
ⓐ Compare 14 and 6. 14___6
14 is to the right of 6 on the number line. 14>6
Comparison of -1 and 9 using number line reasoning and mathematical inequality.
ⓑ Compare −1 and 9. −1___9
−1 is to the left of 9 on the number line. −1<9
Comparison of negative integers, demonstrating how their positions on a number line determine their inequality.
ⓒ Compare −1 and −4. −1___−4
−1 is to the right of −4 on the number line. −1>−4
Compares 2 and -20, detailing the descriptive logic and the corresponding mathematical inequality, demonstrating how to compare positive and negative numbers.
ⓓ Compare 2 and −20. 2___−20
2 is to the right of −20 on the number line. 2>−20
Order each of the following pairs of numbers using < or >.
  1. ⓐ 15___7
  2. ⓑ −2___5
  3. ⓒ −3___−7
  4. ⓓ 5___−17
Solution
  1. ⓐ >
  2. ⓑ <
  3. ⓒ >
  4. ⓓ >

Order each of the following pairs of numbers using < or >.

  1. ⓐ 8___13
  2. ⓑ 3___−4
  3. ⓒ −5___−2
  4. ⓓ 9___−21
Solution
  1. ⓐ <
  2. ⓑ >
  3. ⓒ <
  4. ⓓ >

Find Opposites

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, they are called opposites. The opposite of 2 is −2, and the opposite of −2 is 2 as shown in Figure 12(a). Similarly, 3 and −3 are opposites as shown in Figure 12(b).

This figure shows two number lines. The first has points negative 2 and positive 2 labeled. Below the first line the statement is the numbers negative 2 and 2 are opposites. The second number line has the points negative 3 and 3 labeled. Below the number line is the statement negative 3 and 3 are opposites.

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.

Find the opposite of each number:
  1. ⓐ 7
  2. ⓑ −10
Solution

Solution

  1. ⓐ The number −7 is the same distance from 0 as 7, but on the opposite side of 0. So −7 is the opposite of 7 as shown in Figure 13.
    This figure is a number line. The points negative 7 and 7 are labeled. Above the line it is shown the distance from 0 to negative 7 and the distance from 0 to 7 are both 7.
  2. ⓑ The number 10 is the same distance from 0 as −10, but on the opposite side of 0. So 10 is the opposite of −10 as shown in Figure 14.
    This figure is a number line. The points negative 10 and 10 are labeled. Above the line it is shown the distance from 0 to negative 10 and the distance from 0 to 10 are both 10.
Find the opposite of each number:
  1. ⓐ 4
  2. ⓑ −3
Solution
  1. ⓐ −4
  2. ⓑ 3
Find the opposite of each number:
  1. ⓐ 8
  2. ⓑ −5
Solution
  1. ⓐ −8
  2. ⓑ 5

Opposite Notation

Just as the same word in English can have different meanings, the same symbol in algebra can have different meanings. The specific meaning becomes clear by looking at how it is used. You have seen the symbol “−”, in three different ways.

This table explains the various uses of the minus/negative symbol ('-') in mathematical contexts, detailing its meaning in subtraction, negative numbers, and expressing opposites.
10−4 Between two numbers, the symbol indicates the operation of subtraction.
We read 10−4 as 10 minus 4.
−8 In front of a number, the symbol indicates a negative number.
We read −8 as negative eight.
−x In front of a variable or a number, it indicates the opposite.
We read−x as the opposite of x.
−(−2) Here we have two signs. The sign in the parentheses indicates that the number is negative 2.
The sign outside the parentheses indicates the opposite. We read −(−2) as the opposite of −2.

Opposite Notation

−a means the opposite of the number a

The notation −a is read the opposite of a.

Simplify: −(−6).

Solution
Solution
Illustrates the concept of finding the opposite of a negative number, showing that -(-6) equals 6.
−(−6)
The opposite of −6 is 6. 6

Simplify:

−(−1)

Solution

1

Simplify:

−(−5)

Solution

5

Integers

The set of counting numbers, their opposites, and 0 is the set of integers.

Integers

Integers are counting numbers, their opposites, and zero.

…−3,−2,−1,0,1,2,3…

We must be very careful with the signs when evaluating the opposite of a variable.

Evaluate −x:
  1. ⓐ when x=8
  2. ⓑ when x=−8.
Solution
ⓐ To evaluate −x when x=8, substitute 8 for x.
−x
Substitute 8 for x. A mathematical expression showing a negative sign followed by the number 8 enclosed in parentheses, with the '8' colored in red.
Simplify. −8
ⓑ To evaluate −x when x=−8, substitute −8 for x.
−x
The image shows text that says 'Substitute -8 for x.' The text '-8' is highlighted in red, while the rest of the text is a dark teal color. An image of the math problem -(-8) where the '8' and the inner negative sign are colored red, while the outermost negative sign and parentheses are black.
Simplify. 8
Evaluate −n:
  1. ⓐ whenn=4
  2. ⓑ whenn=−4
Solution
  1. ⓐ −4
  2. ⓑ 4
Evaluate: −m:
  1. ⓐ whenm=11
  2. ⓑ whenm=−11
Solution
  1. ⓐ −11
  2. ⓑ 11

Simplify Expressions with Absolute Value

We saw that numbers such as 5 and −5 are opposites because they are the same distance from 0 on the number line. They are both five units from 0. The distance between 0 and any number on the number line is called the absolute value of that number. Because distance is never negative, the absolute value of any number is never negative.

The symbol for absolute value is two vertical lines on either side of a number. So the absolute value of 5 is written as |5|, and the absolute value of −5 is written as |−5| as shown in Figure 15.

This figure is a number line. The points negative 5 and 5 are labeled. Above the number line the distance from negative 5 to 0 is labeled as 5 units. Also above the number line the distance from 0 to 5 is labeled as 5 units.

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|.

|n|≥0for all numbers
Simplify:
  1. ⓐ |3|
  2. ⓑ |−44|
  3. ⓒ |0|
Solution

Solution

This table illustrates the absolute value concept, showing that 3 is 3 units from zero, represented mathematically as |3| = 3.
ⓐ
|3|
3 is 3 units from zero. 3
This table demonstrates the concept of absolute value using the example of -44, showing its mathematical expression and verbal explanation.
ⓑ
|−44|
−44 is 44 units from zero. 44
A minimal table showcasing a copyright symbol, the absolute value of zero, and a related textual statement.
ⓒ
|0|
0 is already at zero. 0
Simplify:
  1. ⓐ |12|
  2. ⓑ −|−28|
Solution
  1. ⓐ 12
  2. ⓑ −28
Simplify:
  1. ⓐ |9|
  2. ⓑ −|37|
Solution
  1. ⓐ 9
  2. ⓑ −37

We treat absolute value bars just like we treat parentheses in the order of operations. We simplify the expression inside first.

Evaluate:
  1. ⓐ |x|whenx=−35
  2. ⓑ |−y|wheny=−20
  3. ⓒ −|u|whenu=12
  4. ⓓ −|p|whenp=−14
Solution

Solution

ⓐ To find |x| when x=−35:
|x|
The image shows the text 'Substitute -35 for x.' in a light blue-grey font, with '-35' highlighted in red, against a white background. An image displaying the mathematical notation for the absolute value of -35.
Take the absolute value. 35
ⓑ To find |−y| when y=−20:
|−y|
The text 'Substitute -20 for y.' is displayed on a white background, with '-20' highlighted in red and the rest of the text in a dark blue-grey color. The absolute value expression |-(-(-20))| is displayed, with the number -20 highlighted in red.
Simplify. |20|
Take the absolute value. 20
ⓒ To find −|u| when u=12:
−|u|
The text reads 'Substitute 12 for u.', where '12' is highlighted in red. An image featuring the mathematical expression -|12|, which evaluates to negative twelve.
Take the absolute value. −12
ⓓ To find −|p| when p=−14:
−|p|
The text reads 'Substitute -14 for p.' where '-14' is highlighted in red, indicating it is the value to be substituted for the variable 'p'. A mathematical expression showing the negative of the absolute value of negative fourteen, written as -|-14|.
Take the absolute value. −14

Notice that the result is negative only when there is a negative sign outside the absolute value symbol.

Evaluate:
  1. ⓐ |x|whenx=−17
  2. ⓑ |−y|wheny=−39
  3. ⓒ −|m|whenm=22
  4. ⓓ −|p|whenp=−11
Solution
  1. ⓐ 17
  2. ⓑ 39
  3. ⓒ −22
  4. ⓓ −11
  1. ⓐ |y|wheny=−23
  2. ⓑ |−y|wheny=−21
  3. ⓒ −|n|whenn=37
  4. ⓓ −|q|whenq=−49
Solution
  1. ⓐ 23
  2. ⓑ 21
  3. ⓒ −37
  4. ⓓ −49

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

  1. ⓐ |−5|___−|−5|
  2. ⓑ 8___−|−8|
  3. ⓒ −9___−|−9|
  4. ⓓ −|−7|___−7
Solution

Solution

To compare two expressions, simplify each one first. Then compare.

This table illustrates the step-by-step simplification and ordering of an absolute value expression, culminating in the comparison of 5 and -5.
ⓐ
|−5|___−|−5|
Simplify. 5___−5
Order. 5>−5
Steps to simplify and order a mathematical expression involving absolute values and negative numbers.
ⓑ
8___−|−8|
Simplify. 8___−8
Order. 8>−8
This table demonstrates the process of comparing -9 with the negative absolute value of -9, illustrating simplification steps.
ⓒ
−9___−|−9|
Simplify. −9___−9
Order. −9=−9
Illustrates steps to simplify and order an expression involving a negative absolute value.
ⓓ
−|−7|___−7
Simplify. −7___−7
Order. −7=−7
Fill in <,>,or=for each of the following:
  1. ⓐ |−9|___−|−9|
  2. ⓑ 2___−|−2|
  3. ⓒ −8___|−8|
  4. ⓓ −|−5|___−5
Solution
  1. ⓐ >
  2. ⓑ >
  3. ⓒ <
  4. ⓓ =
Fill in <,>,or= for each of the following:
  1. ⓐ 7___−|−7|
  2. ⓑ −|−11|___−11
  3. ⓒ |−4|___−|−4|
  4. ⓓ −1___|−1|
Solution
  1. ⓐ >
  2. ⓑ =
  3. ⓒ >
  4. ⓓ <

Absolute value bars act like grouping symbols. First simplify inside the absolute value bars as much as possible. Then take the absolute value of the resulting number, and continue with any operations outside the absolute value symbols.

Simplify:
  1. ⓐ |9−3|
  2. ⓑ 4|−2|
Solution

Solution

For each expression, follow the order of operations. Begin inside the absolute value symbols just as with parentheses.

Illustrates the step-by-step process of simplifying an absolute value expression, showing intermediate and final results.
ⓐ
|9−3|
Simplify inside the absolute value sign. |6|
Take the absolute value. 6
This table illustrates the step-by-step evaluation of the absolute value expression 4|-2|.
ⓑ
4|−2|
Take the absolute value. 4⋅2
Multiply. 8
Simplify:
  1. ⓐ |12−9|
  2. ⓑ 3|−6|
Solution
  1. ⓐ 3
  2. ⓑ 18
Simplify:
  1. ⓐ |27−16|
  2. ⓑ 9|−7|
Solution
  1. ⓐ 11
  2. ⓑ 63

Simplify: |8+7|−|5+6|.

Solution

Solution

For each expression, follow the order of operations. Begin inside the absolute value symbols just as with parentheses.

Demonstrates the step-by-step simplification of an absolute value expression, from the initial problem to its final numerical result.
|8+7|−|5+6|
Simplify inside each absolute value sign. |15|−|11|
Subtract. 4

Simplify: |1+8|−|2+5|

Solution

2

Simplify: |9−5|−|7−6|

Solution

3

Simplify: 24−|19−3(6−2)|.

Solution

Solution

We use the order of operations. Remember to simplify grouping symbols first, so parentheses inside absolute value symbols would be first.

This table illustrates the step-by-step simplification of a mathematical expression using the order of operations and absolute values.
24−|19−3(6−2)|
Simplify in the parentheses first. 24−|19−3(4)|
Multiply 3(4). 24−|19−12|
Subtract inside the absolute value sign. 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

Translate Word Phrases into Expressions with Integers

Now we can translate word phrases into expressions with integers. Look for words that indicate a negative sign. For example, the word negative in “negative twenty” indicates −20. So does the word opposite in “the opposite of 20.”

Translate each phrase into an expression with integers:
  1. ⓐ the opposite of positive fourteen
  2. ⓑ the opposite of −11
  3. ⓒ negative sixteen
  4. ⓓ two minus negative seven
Solution

Solution

  1. ⓐ the opposite of fourteen
    −14
  2. ⓑ the opposite of −11
    −(−11)=11
  3. ⓒ negative sixteen
    −16
  4. ⓓ two minus negative seven
    2−(−7)
Translate each phrase into an expression with integers:
  1. ⓐ the opposite of positive nine
  2. ⓑ the opposite of −15
  3. ⓒ negative twenty
  4. ⓓ eleven minus negative four
Solution
  1. ⓐ −9
  2. ⓑ 15
  3. ⓒ −20
  4. ⓓ 11−(−4)
Translate each phrase into an expression with integers:
  1. ⓐ the opposite of negative nineteen
  2. ⓑ the opposite of twenty-two
  3. ⓒ negative nine
  4. ⓓ negative eight minus negative five
Solution
  1. ⓐ 19
  2. ⓑ −22
  3. ⓒ −9
  4. ⓓ −8−(−5)

As we saw at the start of this section, negative numbers are needed to describe many real-world situations. We’ll look at some more applications of negative numbers in the next example.

Translate into an expression with integers:
  1. ⓐ The temperature is 12degrees Fahrenheit below zero.
  2. ⓑ The football team had a gain of 3yards.
  3. ⓒ The elevation of the Dead Sea is 1,302feet below sea level.
  4. ⓓ A checking account is overdrawn by $40.
Solution

Solution

Look for key phrases in each sentence. Then look for words that indicate negative signs. Don’t forget to include units of measurement described in the sentence.

Illustrates the interpretation and numerical representation of phrases describing negative temperatures.
ⓐ The temperature is 12 degrees Fahrenheit below zero.
Below zero tells us that 12 is a negative number. −12ºF
This table illustrates a football gain scenario, linking its description to a positive numerical value.
ⓑ The football team had a gain of 3 yards.
A gain tells us that 3 is a positive number. 3 yards
Table illustrating the concept of 'below sea level' as a negative value, exemplified by the Dead Sea's -1,302 feet elevation.
ⓒ The elevation of the Dead Sea is 1,302 feet below sea level.
Below sea level tells us that 1,302 is a negative number. −1,302 feet
Example of an overdrawn checking account represented as a negative number.
ⓓ A checking account is overdrawn by $40.
Overdrawn tells us that 40 is a negative number. −$40

Translate into an expression with integers:

The football team had a gain of 5yards.

Solution

5 yards

Translate into an expression with integers:

The scuba diver was 30feet below the surface of the water.

Solution

−30 feet

ACCESS ADDITIONAL ONLINE RESOURCES

  • Introduction to Integers
  • Simplifying the Opposites of Negative Integers
  • Comparing Absolute Value of Integers
  • Comparing Integers Using Inequalities

Key Concepts

  • Opposite Notation
    • −a means the opposite of the number a
    • The notation −a is read the opposite of a.
  • Absolute Value Notation
    • The absolute value of a number n is written as |n|.
    • |n|≥0 for all numbers.

Practice Makes Perfect

Locate Positive and Negative Numbers on the Number Line

For the following exercises, draw a number line and locate and label the given points on that number line.

  1. ⓐ 2
  2. ⓑ −2
  3. ⓒ −5
Solution


This figure is a number line. Negative 5 is labeled with c, two units to the left of 0 is labeled b, and two units to the right of 0 is labeled a.

  1. ⓐ 5
  2. ⓑ −5
  3. ⓒ −2
  1. ⓐ −8
  2. ⓑ 8
  3. ⓒ −6
Solution


This figure is a number line. Negative 8 is labeled a, negative 6 is labeled c, and 5 is labeled b.

  1. ⓐ −7
  2. ⓑ 7
  3. ⓒ −1

Order Positive and Negative Numbers on the Number Line

In the following exercises, order each of the following pairs of numbers, using < or >.

  1. ⓐ 9__4
  2. ⓑ −3__6
  3. ⓒ −8__−2
  4. ⓓ 1__−10
Solution
  1. ⓐ >
  2. ⓑ <
  3. ⓒ <
  4. ⓓ >
  1. ⓐ 6__2;
  2. ⓑ −7__4;
  3. ⓒ −9__−1;
  4. ⓓ 9__−3
  1. ⓐ −5__1;
  2. ⓑ −4__−9;
  3. ⓒ 6__10;
  4. ⓓ 3__−8
Solution
  1. ⓐ <
  2. ⓑ >
  3. ⓒ <
  4. ⓓ >
  1. ⓐ −7__3;
  2. ⓑ −10__−5;
  3. ⓒ 2__−6;
  4. ⓓ 8__9

Find Opposites

In the following exercises, find the opposite of each number.

  1. ⓐ 2
  2. ⓑ −6
Solution
  1. ⓐ −2
  2. ⓑ 6
  1. ⓐ 9
  2. ⓑ −4
  1. ⓐ −8
  2. ⓑ 1
Solution
  1. ⓐ 8
  2. ⓑ −1
  1. ⓐ −2
  2. ⓑ 6

In the following exercises, simplify.

−(−4)

Solution

4

−(−8)

−(−15)

Solution

15

−(−11)

In the following exercises, evaluate.

−mwhen
  1. ⓐ m=3
  2. ⓑ m=−3
Solution
  1. ⓐ −3
  2. ⓑ 3
−pwhen
  1. ⓐ p=6
  2. ⓑ p=−6
−cwhen
  1. ⓐ c=12
  2. ⓑ c=−12
Solution
  1. ⓐ −12;
  2. ⓑ 12
−dwhen
  1. ⓐ d=21
  2. ⓑ d=−21

Simplify Expressions with Absolute Value

In the following exercises, simplify each absolute value expression.

  1. ⓐ |7|
  2. ⓑ |−25|
  3. ⓒ|0|
Solution
  1. ⓐ 7
  2. ⓑ 25
  3. ⓒ 0
  1. ⓐ |5|
  2. ⓑ |20|
  3. ⓒ |−19|
  1. ⓐ |−32|
  2. ⓑ |−18|
  3. ⓒ |16|
Solution
  1. ⓐ 32
  2. ⓑ 18
  3. ⓒ 16
  1. ⓐ |−41|
  2. ⓑ |−40|
  3. ⓒ |22|

In the following exercises, evaluate each absolute value expression.

  1. ⓐ |x|whenx=−28
  2. ⓑ |−u|whenu=−15
Solution
  1. ⓐ 28
  2. ⓑ 15
  1. ⓐ |y|wheny=−37
  2. ⓑ |−z|whenz=−24
  1. ⓐ −|p|whenp=19
  2. ⓑ −|q|whenq=−33
Solution
  1. ⓐ −19
  2. ⓑ −33
  1. ⓐ −|a|whena=60
  2. ⓑ −|b|whenb=−12

In the following exercises, fill in <,>,or= to compare each expression.

  1. ⓐ −6__|−6|
  2. ⓑ −|−3|__−3
Solution
  1. ⓐ <
  2. ⓑ =
  1. ⓐ −8__|−8|
  2. ⓑ −|−2|__−2
  1. ⓐ |−3|__−|−3|
  2. ⓑ 4__−|−4|
Solution
  1. ⓐ >
  2. ⓑ >
  1. ⓐ |−5|__−|−5|
  2. ⓑ9__−|−9|

In the following exercises, simplify each expression.

|8−4|

Solution

4

|9−6|

8|−7|

Solution

56

5|−5|

|15−7|−|14−6|

Solution

0

|17−8|−|13−4|

18−|2(8−3)|

Solution

8

15−|3(8−5)|

8(14−2|−2|)

Solution

80

6(13−4|−2|)

Translate Word Phrases into Expressions with Integers

Translate each phrase into an expression with integers. Do not simplify.

  1. ⓐ the opposite of 8
  2. ⓑ the opposite of −6
  3. ⓒ negative three
  4. ⓓ 4 minus negative 3
Solution
  1. ⓐ −8
  2. ⓑ −(−6), or 6
  3. ⓒ −3
  4. ⓓ 4−(−3)
  1. ⓐ the opposite of 11
  2. ⓑ the opposite of −4
  3. ⓒ negative nine
  4. ⓓ 8 minus negative 2
  1. ⓐ the opposite of 20
  2. ⓑ the opposite of −5
  3. ⓒ negative twelve
  4. ⓓ 18 minus negative 7
Solution
  1. ⓐ −20
  2. ⓑ −(−5), or 5
  3. ⓒ −12
  4. ⓓ 18−(−7)
  1. ⓐ the opposite of 15
  2. ⓑ the opposite of −9
  3. ⓒ negative sixty
  4. ⓓ 12 minus 5

a temperature of 6degrees below zero

Solution

−6 degrees

a temperature of 14degrees below zero

an elevation of 40feet below sea level

Solution

−40 feet

an elevation of 65feet below sea level

a football play loss of 12yards

Solution

−12 yards

a football play gain of 4yards

a stock gain of $3

Solution

$3

a stock loss of $5

a golf score one above par

Solution

+1

a golf score of 3 below par

Everyday Math

Elevation The highest elevation in the United States is Mount McKinley, Alaska, at 20,320feet above sea level. The lowest elevation is Death Valley, California, at 282feet below sea level. Use integers to write the elevation of:
  1. ⓐ Mount McKinley
  2. ⓑ Death Valley
Solution
  1. ⓐ 20,320 feet
  2. ⓑ −282 feet
Extreme temperatures The highest recorded temperature on Earth is 57° Celsius. The lowest recorded temperature is 90° below 0° Celsius. Use integers to write the:
  1. ⓐ highest recorded temperature
  2. ⓑ lowest recorded temperature
State budgets In June, 2011, the state of Pennsylvania estimated it would have a budget surplus of $540 million. That same month, Texas estimated it would have a budget deficit of $27 billion. Use integers to write the budget:
  1. ⓐ surplus
  2. ⓑ deficit
Solution
  1. ⓐ $540 million
  2. ⓑ −$27 billion
College enrollments Across the United States, community college enrollment grew by 1,400,000 students from 2007 to 2010. In California, community college enrollment declined by 110,171 students from 2009 to 2010. Use integers to write the change in enrollment:
  1. ⓐ growth
  2. ⓑ decline

Writing Exercises

Give an example of a negative number from your life experience.

Solution

Sample answer: I have experienced negative temperatures.

What are the three uses of the “−” sign in algebra? Explain how they differ.

Self Check

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

A self-assessment checklist for math skills, covering locating and ordering positive/negative numbers, finding opposites, simplifying absolute value expressions, and translating word phrases to integer expressions.

ⓑ 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.

absolute value
The absolute value of a number is its distance from 0 on the number line.
integers
Integers are counting numbers, their opposites, and zero ... –3, –2, –1, 0, 1, 2, 3 ...
negative number
A negative number is less than zero.
opposites
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.

Add Integers

Learning Objectives

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

  • Model addition of integers
  • Simplify expressions with integers
  • Evaluate variable expressions with integers
  • Translate word phrases to algebraic expressions
  • Add integers in applications

Before you get started, take this readiness quiz.

Evaluate x+8 when x=6.
If you missed this problem, review Example 1 in Evaluate, Simplify, and Translate Expressions.

Solution

14

Simplify: 8+2(5+1).
If you missed this problem, review Example 8 in Use the Language of Algebra.

Solution

20

Translate the sum of 3 and negative 7 into an algebraic expression.
If you missed this problem, review Table 13 in Evaluate, Simplify, and Translate Expressions

Solution

3+(−7)

Model Addition of Integers

Now that we have located positive and negative numbers on the number line, it is time to discuss arithmetic operations with integers.

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 difficult. This difficulty relates to the way the brain learns.

The brain learns best by working with objects in the real world and then generalizing to abstract concepts. Toddlers learn quickly that if they have two cookies and their older brother steals one, they have only one left. This is a concrete example of 2−1. Children learn their basic addition and subtraction facts from experiences in their everyday lives. Eventually, they know the number facts without relying on cookies.

Addition and subtraction of negative numbers have fewer real world examples that are meaningful to us. Math teachers have several different approaches, such as number lines, banking, temperatures, and so on, to make these concepts real.

We will model addition and subtraction of negatives with two color counters. We let a blue counter represent a positive and a red counter will represent a negative.

This figure has a blue circle labeled positive and a red circle labeled negative.

If we have one positive and one negative counter, the value of the pair is zero. They form a neutral pair. The value of this neutral pair is zero as summarized in Figure 1.

This figure has a blue circle over a red circle. Beside them is the statement 1 plus negative 1 equals 0.
A blue counter represents +1. A red counter represents −1. Together they add to zero.
Doing the Manipulative Mathematics activity "Addition of signed Numbers" will help you develop a better understanding of adding integers.

We will model four addition facts using the numbers 5,−5and3,−3.

5+3−5+(−3)−5+35+(−3)

Model: 5+3.

Solution

Solution

Interpret the expression. 5+3 means the sum of 5 and 3.
Model the first number. Start with 5 positives. Five light blue circles are horizontally aligned, with the number '5' positioned directly beneath the central circle, representing the total count of the circles shown.
Model the second number. Add 3 positives. Two groups of light blue circles are shown against a white background. The first group contains five circles, with the number '5' written below them. The second group consists of three circles, with the number '3' below.
Count the total number of counters. Eight light blue circles, arranged in a row, with the text '8 positives' written directly below them, illustrating a concept likely related to mathematics or data representation.
The sum of 5 and 3 is 8. 5+3=8

Model the expression.

2+4

Solution


This figure has six pink circles in a row, representing positive counters. The first two circles are separated from the following four circles.
6

Model the expression.

2+5

Solution


This figure has seven pink circles in a row, representing positive counters. Two circles are separated from the following 5 circles.
7

Model: −5+(−3).

Solution

Solution

Interpret the expression. −5+(−3) means the sum of −5 and −3.
Model the first number. Start with 5 negatives. Five light red circles are arranged horizontally on a white background, with the number '-5' centered beneath them.
Model the second number. Add 3 negatives. Five red circles are grouped together representing -5, while three red circles are grouped separately representing -3 on a white background.
Count the total number of counters. Eight light red circles with red outlines are arranged in a row. Below them, the text '8 negatives' is written in dark blue, indicating they represent negative values.
The sum of −5 and −3 is −8. −5+−3=−8

Model the expression.

−2+(−4)

Solution


This figure shows a row of 6 dark pink circles, representing negative counters. They are grouped by 2 circles followed by 4 circles.
−6

Model the expression.

−2+(−5)

Solution


This figure shows a row of 7 dark pink circles, representing negative counters. They are grouped by 2 circles followed by 5 circles.
−7

Example 1 and Example 2 are very similar. The first example adds 5 positives and 3 positives—both positives. The second example adds 5 negatives and 3 negatives—both negatives. In each case, we got a result of 8—either8 positives or 8 negatives. When the signs are the same, the counters are all the same color.

Now let’s see what happens when the signs are different.

Model: −5+3.

Solution

Solution

Interpret the expression. −5+3 means the sum of −5 and 3.
Model the first number. Start with 5 negatives. Five light red circles, uniformly spaced, are arranged in a horizontal line against a plain white background.
Model the second number. Add 3 positives. Five red circles are arranged in a horizontal row above three light blue circles, also in a horizontal row, set against a plain white background.
Remove any neutral pairs. Three magenta ovals, each containing a red and a blue circle, are shown alongside two separate red circles.
Count the result. A diagram displays two red-outlined pink circles, side-by-side, with the label '2 negatives' written below them on a white background, representing negative values or concepts.
The sum of −5 and 3 is −2. −5+3=−2

Notice that there were more negatives than positives, so the result is negative.

Model the expression, and then simplify:

2+(−4)

Solution


This figure shows two rows of counter circles. The first row has 2 dark pink circles, representing negative counters. The second row has 4 light pink circles, representing positive counters.
−2

Model the expression, and then simplify:

2+(−5)

Solution


This figure shows two rows of counter circles. The first row has 2 light pink circles, representing positive counters. The second row has 5 dark pink circles, representing negative counters.
−3

Model: 5+(−3).

Solution

Solution

Interpret the expression. 5+(−3) means the sum of 5 and −3.
Model the first number. Start with 5 positives. Five light blue circles with a darker blue outline are arranged in a horizontal line on a white background, resembling a star rating system or a progress indicator.
Model the second number. Add 3 negatives. An illustration showing five light blue circles in the top row and three red circles in the bottom row.
Remove any neutral pairs. Three pairs of light blue and red circles are grouped within magenta ovals, followed by two standalone light blue circles, illustrating a mix of grouped and individual elements.
Count the result. Two light blue circles are displayed above the text '2 positives' on a white background, indicating a count of two positive results or items.
The sum of 5 and −3 is 2. 5+(−3)=2

Model the expression, and then simplify:

(−2)+4

Solution


This figure shows two rows of counter circles. The first row has 2 light pink circles, representing positive counters. The second row has 4 dark pink circles, representing negative counters.
2

Model the expression:

(−2)+5

Solution


This figure shows two rows of counters. The first row shows 2 dark pink circles, representing negative counters. The second row has 5 light pink circles, representing negative counters. The neutral pairs of one positive and one negative counter are circled leaving three positive counters.
3

Modeling Addition of Positive and Negative Integers

Model each addition.

  1. ⓐ 4 + 2
  2. ⓑ −3 + 6
  3. ⓒ 4 + (−5)
  4. ⓓ -2 + (−3)
Solution
This table illustrates the addition problem 4 + 2, detailing each step with a description and its corresponding mathematical or visual representation.
ⓐ
4+2
Start with 4 positives. Four light blue circles with blue outlines are arranged horizontally on a white background, suggesting a selection or progress indicator.
Add two positives. Six light blue circles with blue outlines are arranged in a horizontal row, separated into a group of four and a group of two, on a white background.
How many do you have? 4+2=6
Visual step-by-step demonstration of integer addition for -3 + 6, showing the process and result.
ⓑ
−3+6
Start with 3 negatives. Three light red circles with dark red outlines are arranged horizontally on a white background.
Add 6 positives. An image displaying three red circles in the top row and six blue circles in the bottom row, representing a 3:6 ratio or comparison.
Remove neutral pairs. An array of circles: 3 red and 3 blue circles encircled by an arrowed loop, suggesting repetition. This is followed by 3 more blue circles and an ellipsis indicating continuation.
How many are left? Three light blue circles with a blue outline, arranged horizontally on a white background.
3. −3+6=3
Visual demonstration of adding 4 and -5, resulting in -1, using a step-by-step approach with illustrations.
ⓒ
4+(−5)
Start with 4 positives. Four light blue circles, outlined in a slightly darker blue, are arranged horizontally on a white background.
Add 5 negatives. Two rows of circles are displayed on a white background: four light blue circles are in the top row, and five red circles are in the bottom row.
Remove neutral pairs. Eight circles, four blue and four red, are grouped by an oval, while one red circle is separate.
How many are left? A single, plain light pink or pale red circle with a darker red outline is centered against a pristine white background. The simple, minimalist design highlights the single geometric shape.
−1. 4+(−5)=−1
Step-by-step demonstration of adding negative integers (-2 + -3 = -5) using descriptions, math, and visuals.
ⓓ
−2+(−3)
Start with 2 negatives. Two light red circles with thin dark red outlines and subtle shadows are positioned horizontally close to each other on a plain white background, appearing as simple graphic elements.
Add 3 negatives. A simple graphic featuring two light red circles on the left, separated by space from three light red circles on the right, all against a white background.
How many do you have? −5. −2+(−3)=−5
Model each addition.
  1. ⓐ 3 + 4
  2. ⓑ −1 + 4
  3. ⓒ 4 + (−6)
  4. ⓓ −2 + (−2)
Solution
  1. ⓐ
    Two distinct groups of dark-outlined gray circles are displayed on a white background: a set of three on the left and a set of five on the right.
  2. ⓑ
    A dark blue circle stands alone, followed by four lighter grey circles in a row, all aligned horizontally on a white background.
  3. ⓒ
    An abstract image featuring two rows of dark grey oval shapes, with three on the left and five on the right, against a stark white background.
  4. ⓓ
    Four dark gray circles, arranged in two pairs, against a white background, suggesting a simple pattern or abstract design.
  1. ⓐ 5 + 1
  2. ⓑ −3 + 7
  3. ⓒ 2 + (−8)
  4. ⓓ −3 + (−4)
Solution
  1. ⓐ
    Six gray circles, five connected in a row and one separate, suggesting a multi-step process or selectable items.
  2. ⓑ
    A row of dark grey and light grey oval shapes on a white background, with a gap separating the first two darker shapes from the subsequent lighter shapes.
  3. ⓒ
    An abstract pattern featuring a row of oval shapes. Two lighter grey ovals are on the left, separated by a white space from nine darker, blue-grey ovals that continue to the right.
  4. ⓓ
    An abstract image showing eight dark grey-blue circles in two sets of four, with subtle connections, on a white background.

Simplify Expressions with Integers

Now that you have modeled adding small positive and negative integers, you can visualize the model in your mind to simplify expressions with any integers.

For example, if you want to add 37+(−53), you don’t have to count out 37 blue counters and 53 red counters.

Picture 37 blue counters with 53 red counters lined up underneath. Since there would be more negative counters than positive counters, the sum would be negative. Because 53−37=16, there are 16 more negative counters.

37+(−53)=−16

Let’s try another one. We’ll add −74+(−27). Imagine 74 red counters and 27 more red counters, so we have 101 red counters all together. This means the sum is −101.

−74+(−27)=−101

Look again at the results of Example 1 - Example 4.

Addition of Positive and Negative Integers
5+3 −5+(−3)
both positive, sum positive both negative, sum negative
When the signs are the same, the counters would be all the same color, so add them.
−5+3 5+(−3)
different signs, more negatives different signs, more positives
Sum negative sum positive
When the signs are different, some counters would make neutral pairs; subtract to see how many are left.
Simplify:
  1. ⓐ 19+(−47)
  2. ⓑ −32+40
Solution

Solution

ⓐ Since the signs are different, we subtract 19 from 47. The answer will be negative because there are more negatives than positives.

19+(−47)−28

ⓑ The signs are different so we subtract 32 from 40. The answer will be positive because there are more positives than negatives

−32+408

Simplify each expression:

  1. ⓐ 15+(−32)
  2. ⓑ −19+76
Solution
  1. ⓐ −17
  2. ⓑ 57

Simplify each expression:

  1. ⓐ −55+9
  2. ⓑ 43+(−17)
Solution
  1. ⓐ −46
  2. ⓑ 26

Simplify: −14+(−36).

Solution

Solution

Since the signs are the same, we add. The answer will be negative because there are only negatives.

−14+(−36)−50

Simplify the expression:

−31+(−19)

Solution

−50

Simplify the expression:

−42+(−28)

Solution

−70

The techniques we have used up to now extend to more complicated expressions. Remember to follow the order of operations.

Simplify: −5+3(−2+7).

Solution

Solution

This table illustrates the step-by-step simplification of the mathematical expression -5 + 3(-2 + 7) following the order of operations.
−5+3(−2+7)
Simplify inside the parentheses. −5+3(5)
Multiply. −5+15
Add left to right. 10

Simplify the expression:

−2+5(−4+7)

Solution

13

Simplify the expression:

−4+2(−3+5)

Solution

0

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 when evaluating expressions.

Evaluate x+7when
  1. ⓐ x=−2
  2. ⓑ x=−11.
Solution

Solution

ⓐ Evaluate x+7 when x=−2
A mathematical expression showing '-2 + 7'. The number -2 is in red, while the addition sign and the number 7 are in black.
The image shows the text 'Substitute -2 for x.' with '-2' highlighted in red, indicating a mathematical instruction to replace the variable x with the value -2. A mathematical expression showing '-2 + 7'. The number -2 is in red, while the addition sign and the number 7 are in black.
Simplify. The number 5, written in black, appears in the top right corner of a plain white background.
ⓑ Evaluate x+7 when x=−11
The image displays the mathematical expression 'x + 7' in a clear, sans-serif font against a white background.
The image shows the mathematical instruction, 'Substitute -11 for x.' The number -11 is highlighted in red, indicating its importance in the substitution. The image displays a mathematical expression showing the sum of negative eleven and positive seven, written as -11 + 7, with the negative number in red and the positive number and plus sign in black.
Simplify. A close-up view shows the number '-4' in black text against a plain white background.

Evaluate each expression for the given values:

x+5when
  1. ⓐ x=−3and
  2. ⓑ x=−17
Solution
  1. ⓐ 2
  2. ⓑ −12
Evaluate each expression for the given values: y+7 when
  1. ⓐ y=−5
  2. ⓑ y=−8
Solution
  1. ⓐ 2
  2. ⓑ −1
When n=−5, evaluate
  1. ⓐ n+1
  2. ⓑ −n+1.
Solution

Solution

ⓐ Evaluate n+1 when n=−5
The mathematical expression 'n + 1' is shown in a black, bold font on a white background.
The text 'Substitute -5 for n.' is displayed, instructing the viewer to replace the variable 'n' with the value -5. The image displays the mathematical expression '-5 + 1' with the number -5 in red and +1 in black, representing an arithmetic problem involving integers.
Simplify. The number negative four, or -4, displayed prominently on a white background.
ⓑ Evaluate −n+1 when n=−5
The mathematical expression -n + 1 is displayed in a clear, bold font against a white background.
The text reads 'Substitute -5 for n.' with '-5' highlighted in red, indicating a specific instruction for a mathematical or variable substitution. A mathematical expression shows the operation minus, followed by an opening parenthesis, then minus five, a closing parenthesis, then plus one. The number '5' and the minus sign before it are in red.
Simplify. The image displays a simple mathematical expression '5+1' in black text against a white background, representing an addition problem.
Add. A stark white image with a barely discernible '6' in the top right corner.
When n=−8, evaluate
  1. ⓐ n+2
  2. ⓑ −n+2
Solution
  1. ⓐ −6
  2. ⓑ 10
Wheny=−9,evaluate
  1. ⓐ y+8
  2. ⓑ −y+8.
Solution
  1. ⓐ −1
  2. ⓑ 17

Next we'll evaluate an expression with two variables.

Evaluate 3a+b when a=12 and b=−30.

Solution

Solution

The mathematical expression '3a + b' is shown against a white background.
The text 'Substitute 12 for a and -30 for b.' is displayed on a white background, with '12' in red and '-30' in light blue, and the rest of the text in dark gray. The mathematical expression 3(12) + (-30) is displayed, with the number 12 highlighted in red and -30 highlighted in blue.
Multiply. The image displays the mathematical expression '36 + (-30)' in black text against a white background.
Add. A plain white background featuring the number '6' subtly positioned in the top right corner.

Evaluate the expression:

a+2bwhena=−19andb=14.

Solution

9

Evaluate the expression:

5p+qwhenp=4andq=−7.

Solution

13

Evaluate (x+y)2 when x=−18 and y=24.

Solution

Solution

This expression has two variables. Substitute −18 for x and 24 for y.
(x+y)2
The text reads: 'Substitute -18 for x and 24 for y.' The number -18 is highlighted in red, and 24 is highlighted in light blue. (−18+24)2
Add inside the parentheses. (6)2
Simplify 36

Evaluate:

(x+y)2 when x=−15 and y=29.

Solution

196

Evaluate:

(x+y)3 when x=−8 and y=10.

Solution

8

Translate Word Phrases to Algebraic Expressions

All our earlier work translating word phrases to algebra also applies to expressions that include both positive and negative numbers. Remember that the phrase the sum indicates addition.

Translate and simplify: the sum of −9 and 5.

Solution

Solution

Steps to translate a verbal mathematical phrase into an expression and then simplify it.
The sum of −9 and 5 indicates addition. the sum of −9 and 5
Translate. −9+5
Simplify. −4

Translate and simplify the expression:

the sum of −7 and 4

Solution

−7 + 4 = −3

Translate and simplify the expression:

the sum of −8 and −6

Solution

−8 + (−6) = −14

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

Solution

Solution

The phrase increased by indicates addition.

Steps to translate and simplify a mathematical phrase into its numerical result.
The sum of 8 and −12, increased by 3
Translate. [8+(−12)]+3
Simplify. −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

Add Integers in Applications

Recall that we were introduced to some situations in everyday life that use positive and negative numbers, such as temperatures, banking, and sports. For example, a debt of $5 could be represented as −$5. Let’s practice translating and solving a few applications.

Solving applications is easy if we have a plan. First, we determine what we are looking for. Then we write a phrase that gives the information to find it. We translate the phrase into math notation and then simplify to get the answer. Finally, we write a sentence to answer the question.

The temperature in Buffalo, NY, one morning started at 7degrees below zero Fahrenheit. By noon, it had warmed up 12degrees. What was the temperature at noon?

Solution

Solution

We are asked to find the temperature at noon.

This table illustrates the step-by-step process of solving a temperature word problem, including phrasing, translation to math notation, simplification, and final answer sentence.
Write a phrase for the temperature. The temperature warmed up 12 degrees from 7 degrees below zero.
Translate to math notation. −7 + 12
Simplify. 5
Write a sentence to answer the question. The temperature at noon was 5 degrees Fahrenheit.

The temperature in Chicago at 5 A.M. was 10degrees below zero Celsius. Six hours later, it had warmed up 14 degrees Celsius. What is the temperature at 11 A.M.?

Solution

4 degrees Celsius

A scuba diver was swimming 16 feet below the surface and then dove down another 17 feet. What is her new depth?

Solution

−33 feet

A football team took possession of the football on their 42-yard line. In the next three plays, they lost 6 yards, gained 4 yards, and then lost 8 yards. On what yard line was the ball at the end of those three plays?

Solution

Solution

We are asked to find the yard line the ball was on at the end of three plays.

Illustrates problem-solving steps: converting a word phrase to math notation, simplifying, and stating the final answer for a ball's position.
Write a word phrase for the position of the ball. Start at 42, then lose 6, gain 4, lose 8.
Translate to math notation. 42 − 6 + 4 − 8
Simplify. 32
Write a sentence to answer the question. At the end of the three plays, the ball is on the 32-yard line.

The Bears took possession of the football on their 20-yard line. In the next three plays, they lost 9 yards, gained 7 yards, then lost 4 yards. On what yard line was the ball at the end of those three plays?

Solution

14-yard line

The Chargers began with the football on their 25-yard line. They gained 5 yards, lost 8 yards and then gained 15 yards on the next three plays. Where was the ball at the end of these plays?

Solution

37-yard line

ACCESS ADDITIONAL ONLINE RESOURCES

  • Adding Integers with Same Sign Using Color Counters
  • Adding Integers with Different Signs Using Counters
  • Ex1: Adding Integers
  • Ex2: Adding Integers

Key Concepts

  • Addition of Positive and Negative Integers
    This table demonstrates the rules for adding integers, illustrating examples and results for sums involving numbers with both same and different signs.
    5+3 −5+(−3)
    both positive, sum positive both negative, sum negative
    When the signs are the same, the counters would be all the same color, so add them.
    −5+3 5+(−3)
    different signs, more negatives different signs, more positives
    Sum negative sum positive
    When the signs are different, some counters would make neutral pairs; subtract to see how many are left.

Practice Makes Perfect

Model Addition of Integers

In the following exercises, model the expression to simplify.

7+4

Solution


This figure shows a row of 11 light pink circles, representing positive counters. They are separated into a group of  seven and a group of four.
11

8+5

−6+(−3)

Solution


This figure shows a row of 9 dark pink circles, representing negative counters. They are separated into a group of six and a group of three.
−9

−5+(−5)

−7+5

Solution


This figure shows two rows of  circles. The top row shows 7 dark pink circles, representing negative counters. The bottom row shows 5 light pink circles, representing positive counters.
−2

−9+6

8+(−7)

Solution


This figure shows two rows of circles. The top row shows 8 light pink circles, representing positive counters. The bottom row shows 7 light pink circles, representing negative counters.
1

9+(−4)

Simplify Expressions with Integers

In the following exercises, simplify each expression.

−21+(−59)

Solution

−80

−35+(−47)

48+(−16)

Solution

32

34+(−19)

−200+65

Solution

−135

−150+45

2+(−8)+6

Solution

0

4+(−9)+7

−14+(−12)+4

Solution

−22

−17+(−18)+6

135+(−110)+83

Solution

108

140+(−75)+67

−32+24+(−6)+10

Solution

−4

−38+27+(−8)+12

19+2(−3+8)

Solution

29

24+3(−5+9)

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

x+8 when
  1. ⓐ x=−26
  2. ⓑ x=−95
Solution
  1. ⓐ −18
  2. ⓑ −87
y+9 when
  1. ⓐ y=−29
  2. ⓑ y=−84
y+(−14) when
  1. ⓐ y=−33
  2. ⓑ y=30
Solution
  1. ⓐ −47
  2. ⓑ 16
x+(−21) when
  1. ⓐ x=−27
  2. ⓑ x=44
When a=−7, evaluate:
  1. ⓐ a+3
  2. ⓑ −a+3
Solution
  1. ⓐ −4
  2. ⓑ 10
When b=−11, evaluate:
  1. ⓐ b+6
  2. ⓑ −b+6
When c=−9, evaluate:
  1. ⓐ c+(−4)
  2. ⓑ −c+(−4)
Solution
  1. ⓐ −13
  2. ⓑ 5
When d=−8, evaluate:
  1. ⓐ d+(−9)
  2. ⓑ −d+(−9)

m+n when, m=−15, n=7

Solution

−8

p+q when, p=−9, q=17

r−3s when, r=16, s=2

Solution

10

2t+u when, t=−6, u=−5

(a+b)2 when, a=−7, b=15

Solution

64

(c+d)2 when, c=−5, d=14

(x+y)2 when, x=−3, y=14

Solution

121

(y+z)2 when, y=−3, z=15

Translate Word Phrases to Algebraic Expressions

In the following exercises, translate each phrase into an algebraic expression and then simplify.

The sum of −14 and 5

Solution

−14 + 5 = −9

The sum of −22 and 9

8 more than −2

Solution

−2 + 8 = 6

5 more than −1

−10 added to −15

Solution

−15 + (−10) = −25

−6 added to −20

6 more than the sum of −1 and −12

Solution

[−1 + (−12)] + 6 = −7

3 more than the sum of −2 and −8

the sum of 10 and −19, increased by 4

Solution

[10 + (−19)] + 4 = −5

the sum of 12 and −15, increased by 1

Add Integers in Applications

In the following exercises, solve.

Temperature The temperature in St. Paul, Minnesota was −19°F at sunrise. By noon the temperature had risen 26°F. What was the temperature at noon?

Solution

7°F

Temperature The temperature in Chicago was −15°F at 6 am. By afternoon the temperature had risen 28°F. What was the afternoon temperature?

Credit Cards Lupe owes $73 on her credit card. Then she charges $45 more. What is the new balance?

Solution

−$118

Credit Cards Frank owes $212 on his credit card. Then he charges $105 more. What is the new balance?

Football A team lost 3 yards the first play. Then they lost 2 yards, gained 1 yard, and then lost 4 yards. What was the change in overall yardage over the four plays?

Solution

−8 yards

Card Games April lost 5 cards the first turn. Over the next three turns, she lost 3 cards, gained 2 cards, and then lost 1 card. What was the change in cards over the four turns?

Football The Rams took possession of the football on their own 35-yard line. In the next three plays, they lost 12 yards, gained 8 yards, then lost 6 yards. On what yard line was the ball at the end of those three plays?

Solution

25-yard line

Football The Cowboys began with the ball on their own 20-yard line. They gained 15 yards, lost 3 yards and then gained 6 yards on the next three plays. Where was the ball at the end of these plays?

Scuba Diving A scuba diver swimming 8 feet below the surface dove 17 feet deeper; the pressure got to them and they rose five feet. What is their new depth?

Solution

20 feet

Gas Consumption: Ozzie rode their motorcycle for 30 minutes, using 168 fluid ounces of gas. Then they stopped and got 140-fluid ounces of gas. Represent the change in gas amount as an integer.

Everyday Math

Stock Market The week of September 15, 2008, was one of the most volatile weeks ever for the U.S. stock market. The change in the Dow Jones Industrial Average each day was:

Monday−504Tuesday+142Wednesday−449Thursday+410Friday+369

What was the overall change for the week?

Solution

−32

Stock Market During the week of June 22, 2009, the change in the Dow Jones Industrial Average each day was:

Monday−201Tuesday−16Wednesday−23Thursday+172Friday−34

What was the overall change for the week?

Writing Exercises

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

Solution

Sample answer: In the first case, there are more negatives so the sum is negative. In the second case, there are more positives so the sum is positive.

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

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 students, allowing them to rate their confidence in skills such as modeling integer addition, simplifying expressions, evaluating variables, and translating word phrases.

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

Subtract Integers

Learning Objectives

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

  • Model subtraction of integers
  • Simplify expressions with integers
  • Evaluate variable expressions with integers
  • Translate words phrases to algebraic expressions
  • Subtract integers in applications

Before you get started, take this readiness quiz.

Simplify: 12−(8−1).
If you missed this problem, review Example 8 in Use the Language of Algebra.

Solution

5

Translate the difference of 20 and −15 into an algebraic expression.
If you missed this problem, review Example 11 in Subtract Whole Numbers.

Solution

20−(−15)

Add: −18+7.
If you missed this problem, review Example 7 in Add Integers.

Solution

-11

Model Subtraction of Integers

Remember the story in the last section about the toddler and the cookies? Children learn how to subtract numbers through their everyday experiences. Real-life experiences serve as models for subtracting positive numbers, and in some cases, such as temperature, for adding negative as well as positive numbers. But it is difficult to relate subtracting negative numbers to common life experiences. Most people do not have an intuitive understanding of subtraction when negative numbers are involved. Math teachers use several different models to explain subtracting negative numbers.

We will continue to use counters to model subtraction. Remember, the blue counters represent positive numbers and the red counters represent negative numbers.

Perhaps when you were younger, you read 5−3 as five take away three. When we use counters, we can think of subtraction the same way.

Doing the Manipulative Mathematics activity "Subtraction of Signed Numbers" will help you develop a better understanding of subtracting integers.

We will model four subtraction facts using the numbers 5 and 3.

5−3−5−(−3)−5−35−(−3)

Model: 5−3.

Solution

Solution

Interpret the expression. 5−3 means 5 take away 3.
Model the first number. Start with 5 positives. Five light blue circles are arranged in a horizontal line, with the number '5' positioned directly below the middle circle.
Take away the second number. So take away 3 positives. Four light blue circles are enclosed by a pink oval and a left-pointing arrow. The text '2 positives' is written below the grouping, likely denoting a count or selection within the four.
Find the counters that are left. Two light blue circles side-by-side on a white background, resembling abstract bubbles or subtle decorative elements.
5−3=2.
The difference between 5 and 3 is 2.

Model the expression:

6−4

Solution


This figure shows a row of 6 light pink circles, representing positive counters. The first four are circled.
2

Model the expression:

7−4

Solution


This figure shows a row of 7 light pink circles representing positive counters. The first four counters are circled.
3

Model: −5−(−3).

Solution

Solution

Interpret the expression. −5−(−3) means −5 take away −3.
Model the first number. Start with 5 negatives. Five red circles are displayed in a row, with the number -5 written directly below the middle circles, illustrating a visual representation of the integer negative five.
Take away the second number. So take away 3 negatives. A diagram shows four red circles in a row. The first three are encircled by a purple oval with an arrow. The text '2 negatives' is positioned below the fourth circle.
Find the number of counters that are left. Two light red circles with darker red outlines are shown against a white background.
−5−(−3)=−2.
The difference between −5 and −3 is −2.

Model the expression:

−6−(−4)

Solution


This figure shows a row of 6 dark pink circles, representing negative counters. The last four counters are circled.
−2

Model the expression:

−7−(−4)

Solution


This figure is a row of 7 dark pink circles representing negative counters. The first four counters are circled.
−3

Notice that Example 1 and Example 2 are very much alike.

  • First, we subtracted 3 positives from 5 positives to get 2 positives.
  • Then we subtracted 3 negatives from 5 negatives to get 2 negatives.

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

This figure has a row of 5 blue circles. The first three are circled. Above the row is 5 minus 3 equals 2. Next to this is a row of 5 red circles. The first three are circled. Above the row is negative 5 minus negative 3 equals negative 2.

Now let’s see what happens when we subtract one positive and one negative number. We will need to use both positive and negative counters and sometimes some neutral pairs, too. Adding a neutral pair does not change the value.

Model: −5−3.

Solution

Solution

Interpret the expression. −5−3 means −5 take away 3.
Model the first number. Start with 5 negatives. Five light red circles with dark red outlines are arranged horizontally, with the number -5 positioned directly below the third circle from the left.
Take away the second number.
So we need to take away 3 positives.
But there are no positives to take away.
Add neutral pairs until you have 3 positives.
Eight red circles and three light blue circles are shown, with five red circles on the left and a group of three red circles above three blue circles on the right.
Now take away 3 positives. An illustration of subtraction, showing three blue circles highlighted with an oval and an arrow, representing a quantity removed from a larger group of red circles.
Count the number of counters that are left. A diagram shows a row of eight light red circles with a darker red outline. Below the circles, the text '8 negatives' is written in a dark teal color.
−5−3=−8.
The difference of −5 and 3 is −8.

Model the expression:

−6−4

Solution


This figure shows a row of 10 dark pink circles, representing negative counters. The first six counters are separated from the last four. Below the dark pink circles are four light pink circles, representing positive counters.  These four positive counters are circled.
−10

Model the expression:

−7−4

Solution


This figure shows a row of 11 dark pink circles, representing negative counters. The first seven counters are separated from the last four. Below the dark pink circles are four light pink circles, representing positive counters.  These four positive counters are circled.
−11

Model: 5−(−3).

Solution

Solution

Interpret the expression. 5−(−3) means 5 take away −3.
Model the first number. Start with 5 positives. Five light blue circles with a darker blue outline are arranged horizontally on a white background, resembling a progress indicator or a rating system.
Take away the second number, so take away 3 negatives.
But there are no negatives to take away.
Add neutral pairs until you have 3 negatives.
Eight light blue circles and three red circles are displayed against a white background, arranged in two rows with a gap between groups.
Then take away 3 negatives. Three red circles are grouped by a magenta oval and arrow, signifying their removal or separation from below three of seven light blue circles, likely demonstrating a subtraction or counting concept.
Count the number of counters that are left. Eight light blue circles with a slightly darker blue outline are arranged horizontally on a white background, with the text '8 positives' centered beneath them.
The difference of 5 and −3 is 8.
5−(−3)=8

Model the expression:

6−(−4)

Solution


This figure shows a row of 10 light pink circles, representing positive counters. The first six counters are separated from the last four. Below the light pink circles are four dark pink circles, representing negative counters.  These four negative counters are circled.
10

Model the expression:

7−(−4)

Solution


This figure shows a row of 11 light pink circles, representing positive counters. The first seven counters are separated from the last four. Below the light pink circles are four dark pink circles, representing negative counters.  These four negative counters are circled.
11

Model each subtraction.
  1. ⓐ 8 − 2
  2. ⓑ −5 − 4
  3. ⓒ 6 − (−6)
  4. ⓓ −8 − (−3)
Solution
This table illustrates the step-by-step process of solving the subtraction problem 8 - 2.
ⓐ
8−2
This means 8 take away 2.
Start with 8 positives. Eight light blue circles with a subtle blue outline and shadow are arranged horizontally on a white background.
Take away 2 positives. The first two light blue circles in a sequence are encircled by a magenta oval, with a curving arrow pointing back towards them, signifying a pair or a loop.
How many are left? 6
8−2=6
Visual demonstration of integer subtraction, showing the steps to calculate -5 - 4 using a chip model.
ⓑ
−5−4
This means −5 take away 4.
Start with 5 negatives. Five light red circles, each outlined in a darker red, are arranged horizontally in a row against a plain white background.
You need to take away 4 positives.
Add 4 neutral pairs to get 4 positives.
Eight light red circles with darker red borders, arranged in two groups of five and three, separated by a gap.
Four light blue circles with blue outlines, arranged horizontally on a white background, potentially indicating a loading state or steps in a process.
Take away 4 positives. Four light blue circles are grouped by a pink oval and arrow, indicating their position beneath four of the seven red circles shown above.
How many are left? A series of nine light pink circles with red outlines are arranged horizontally, separated into a group of five on the left and a group of four on the right by a small gap.
−9
−5−4=−9
Step-by-step visual illustration of how to calculate 6 - (-6) using positive and negative integer representations.
ⓒ
6−(−6)
This means 6 take away −6.
Start with 6 positives. Five light blue circles, outlined in a slightly darker blue, are arranged horizontally across a white background.
Add 6 neutrals to get 6 negatives to take away. Two rows of circles, with 10 light blue circles on top and 6 red circles below, right-aligned, against a white background.
Remove 6 negatives. An illustration showing eleven light blue circles in a row, with five red circles directly below them. A purple oval encircles the red circles, with a curved arrow inside pointing left, suggesting a cyclical process or a moving window.
How many are left? A horizontal row of eleven light blue circles, each with a thin cyan outline, is arranged against a plain white background.
12
6−(−6)=12
Illustrates the subtraction of -3 from -8 using a step-by-step visual method with images and mathematical representation.
ⓓ
−8−(−3)
This means −8 take away −3.
Start with 8 negatives. Eight pale red outlined circles in a row.
Take away 3 negatives. A magenta oval groups the first three circles in a row of eight red circles, with a magenta arrow curving from the oval to the left, indicating a grouping or counting action.
How many are left? Five identical light red circles with a darker red outline, arranged horizontally on a plain white background.
−5
−8−(−3)=−5
Model each subtraction.
  1. ⓐ 7 - (-8)
  2. ⓑ -7 - (-2)
  3. ⓒ 4 - 1
  4. ⓓ -6 - 8
Solution

ⓐ
An illustration featuring two rows of circles. The bottom row shows seven dark gray circles enclosed by an oval with a left-pointing arrow, indicating a selected group or a process.
ⓑ
A row of seven grey circles, with the last two circled together and an arrow indicating a loop back to them, symbolizing repetition or a recurring cycle in a sequence.
ⓒ
A circular arrow highlights one grey circle in a row of four, indicating focus or a looping action on that specific element.
ⓓ
A diagram showing a row of 7 light grey circles encircled by an oval with a looping arrow, beneath two rows of 7 dark grey circles each.

Model each subtraction.
  1. ⓐ 4 - (-6)
  2. ⓑ -8 - (-1)
  3. ⓒ 7 - 3
  4. ⓓ -4 - 2
Solution

ⓐ
The image shows ten light grey circles in the top row, separated into groups of four and six. Below them are six darker grey circles, which are enclosed by an oval and have a curved arrow beneath.
ⓑ
An illustration of a finite automaton with eight states, where the final state has a self-loop, often representing an accepting or terminal state.
ⓒ
A sequence of gray circles, with the last three enclosed in an oval loop, suggesting a repeating pattern or iterative process. Four circles appear sequentially, followed by three that are part of a cycle.
ⓓ
An illustration featuring a row of dark grey circles, with two lighter grey circles below, encircled by an oval with a curved arrow pointing left.

Model each subtraction expression:

  1. ⓐ 2−8
  2. ⓑ −3−(−8)
Solution

Solution

ⓐ
We start with 2 positives.
Two light blue circles are displayed horizontally, with the numeral '2' printed directly beneath them, indicating a count of two.
We need to take away 8 positives, but we have only 2.
Add neutral pairs until there are 8 positives to take away. An array of 8 blue circles and 6 red circles, illustrating a simple numerical comparison with two blue circles separated from the main group.
Then take away eight positives. Two rows of circles, 8 blue on top with a pink oval and left arrow, and 6 orange on the bottom, likely showing counting or a sequence.
Find the number of counters that are left.
There are 6 negatives.
Six red circles are aligned horizontally above the text '6 negatives' on a white background, likely illustrating a concept related to negative numbers or values.
2−8=−6
ⓑ
We start with 3 negatives.
Three red circles are arranged horizontally above the number 3, all set against a plain white background.
We need to take away 8 negatives, but we have only 3.
Add neutral pairs until there are 8 negatives to take away. A graphic displaying red and blue circles on a white background. Eight red circles are arranged in two groups on the top row, while five blue circles form a single row below them.
Then take away the 8 negatives. Eight red circles are enclosed by an oval with a leftward arrow, positioned above six blue circles, illustrating two distinct sets of objects.
Find the number of counters that are left.
There are 5 positives.
Five blue-gray circles are arranged in a horizontal line above the text '5 positives' on a white background, suggesting a count or representation of five positive items.
−3−(−8)=5

Model each subtraction expression.

  1. ⓐ 7−9
  2. ⓑ −5−(−9)
Solution

ⓐ
This figure shows a row of 9 light pink circles, representing positive counters. The first seven are separated from the last two. The entire row is circled.  Below the last two light pink circles is a row of two dark pink  circles, representing negative counters.
−2
ⓑ
This figure shows a row of 9 dark pink circles, representing negative counters. The first five are separated from the last four. The entire row is circled.  Below the last four dark pink circles is a row of four light pink  circles, representing positive counters.
4

Model each subtraction expression.

  1. ⓐ 4−7
  2. ⓑ −7−(−10)
Solution

ⓐ
This figure shows a row of seven light pink circles, representing positive counters. The first four are separated from the last three. The entire row is circled.  Below the last three light pink circles is a row of three dark pink  circles, representing negative counters.
−3
ⓑ
This figure shows a row of 10 dark pink circles, representing negative counters. The first seven are separated from the last three. The entire row is circled.  Below the last three dark pink circles is a row of three light pink circles, representing positive counters.
3

Simplify Expressions with Integers

Do you see a pattern? Are you ready to subtract integers without counters? Let’s do two more subtractions. We’ll think about how we would model these with counters, but we won’t actually use the counters.

  • Subtract −23−7.
    Think: We start with 23 negative counters.
    We have to subtract 7 positives, but there are no positives to take away.
    So we add 7 neutral pairs to get the 7 positives. Now we take away the 7 positives.
    So what’s left? We have the original 23 negatives plus 7 more negatives from the neutral pair. The result is 30 negatives.
    −23−7=−30

    Notice, that to subtract 7, we added 7 negatives.
  • Subtract 30−(−12).
    Think: We start with 30 positives.
    We have to subtract 12 negatives, but there are no negatives to take away.
    So we add 12 neutral pairs to the 30 positives. Now we take away the 12 negatives.
    What’s left? We have the original 30 positives plus 12 more positives from the neutral pairs. The result is 42 positives.
    30−(−12)=42

    Notice that to subtract −12, we added 12.

While we may not always use the counters, especially when we work with large numbers, practicing with them first gave us a concrete way to apply the concept, so that we can visualize and remember how to do the subtraction without the counters.

Have you noticed that subtraction of signed numbers can be done by adding the opposite? You will often see the idea, the Subtraction Property, written as follows:

Subtraction Property

a−b=a+(−b)

Look at these two examples.

This figure has two columns. The first column has 6 minus 4. Underneath, there is a row of 6 blue circles, with the first 4 separated from the last 2. The first 4 are circled. Under this row there is 2. The second column has 6 plus negative 4. Underneath there is a row of 6 blue circles with the first 4 separated from the last 2. The first 4 are circled. Under the first four is a row of 4 red circles. Under this there is 2.

We see that 6−4 gives the same answer as 6+(−4).

Of course, when we have a subtraction problem that has only positive numbers, like the first example, we just do the subtraction. We already knew how to subtract 6−4 long ago. But knowing that 6−4 gives the same answer as 6+(−4) helps when we are subtracting negative numbers.

Simplify:
  1. ⓐ 13−8and13+(−8)
  2. ⓑ −17−9and−17+(−9)
Solution

Solution

This table demonstrates that subtracting a number is equivalent to adding its opposite, using 13 - 8 and 13 + (-8) as an example.
ⓐ
13−8 and 13+(−8)
Subtract to simplify. 13−8=5
Add to simplify. 13+(−8)=5
Subtracting 8 from 13 is the same as adding −8 to 13.
This table illustrates that subtracting a number is equivalent to adding its negative counterpart, demonstrated with an example.
ⓑ
−17−9 and −17+(−9)
Subtract to simplify. −17−9=−26
Add to simplify. −17+(−9)=−26
Subtracting 9 from −17 is the same as adding −9 to −17.

Simplify each expression:

  1. ⓐ 21−13and21+(−13)
  2. ⓑ −11−7and−11+(−7)
Solution
  1. ⓐ 8, 8
  2. ⓑ −18, −18

Simplify each expression:

  1. ⓐ 15−7and15+(−7)
  2. ⓑ −14−8and−14+(−8)
Solution
  1. ⓐ 8, 8
  2. ⓑ −22, −22

Now look what happens when we subtract a negative.

This figure has two columns. The first column has 8 minus negative 5. Underneath, there is a row of 13 blue  circles. The first 8 are separated from the next 5. Under the last 5 blue circles there is a row of 5 red circles. They are circled. Under this there is 13. The second column has 8 plus 5. Underneath there is a row of 13 blue circles. The first 8 are separated from the last 5. Under this there is 13.

We see that 8−(−5) gives the same result as 8+5. Subtracting a negative number is like adding a positive.

Simplify:
  1. ⓐ 9−(−15)and9+15
  2. ⓑ −7−(−4)and−7+4
Solution

Solution

Demonstrates that subtracting a negative number is equivalent to adding its positive counterpart, using 9 - (-15) = 9 + 15 = 24 as an example.
ⓐ
9−(−15) and 9+15
Subtract to simplify. 9−(−15)=24
Add to simplify. 9+15=24
Subtracting −15 from 9 is the same as adding 15 to 9.
This table illustrates that subtracting a negative number is equivalent to adding a positive number, using the example of -7 - (-4) = -7 + 4.
ⓑ
−7−(−4) and −7+4
Subtract to simplify. −7−(−4)=−3
Add to simplify. −7+4=−3
Subtracting −4 from −7 is the same as adding 4 to −7

Simplify each expression:

  1. ⓐ 6−(−13)and6+13
  2. ⓑ −5−(−1)and−5+1
Solution
  1. ⓐ 19, 19
  2. ⓑ −4, −4

Simplify each expression:

  1. ⓐ 4−(−19)and4+19
  2. ⓐ −4−(−7)and−4+7
Solution
  1. ⓐ 23, 23
  2. ⓑ 3, 3

Look again at the results of Example 1 - Example 4.

Subtraction of Integers
5–3 –5–(–3)
2 –2
2 positives 2 negatives
When there would be enough counters of the color to take away, subtract.
–5–3 5–(–3)
–8 8
5 negatives, want to subtract 3 positives 5 positives, want to subtract 3 negatives
need neutral pairs need neutral pairs
When there would not be enough of the counters to take away, add neutral pairs.

Simplify: −74−(−58).

Solution

Solution

This table illustrates the process of subtracting negative numbers, presenting both a verbal description and its corresponding mathematical expression and result.
We are taking 58 negatives away from 74 negatives. −74−(−58)
Subtract. −16

Simplify the expression:

−67−(−38)

Solution

−29

Simplify the expression:

−83−(−57)

Solution

−26

Simplify: 7−(−4−3)−9.

Solution

Solution

We use the order of operations to simplify this expression, performing operations inside the parentheses first. Then we subtract from left to right.

Illustrates the step-by-step simplification of a mathematical expression using textual descriptions and visual representations.
Simplify inside the parentheses first. A mathematical expression reads '7 minus open parenthesis minus 4 minus 3 close parenthesis minus 9.'
Subtract from left to right. A mathematical expression showing 7 minus negative 7 minus 9.
Subtract. The mathematical expression '14 - 9' is displayed in a clear, dark font against a plain white background, ready for calculation.
The black numeral '5' is prominently displayed against a plain white background.

Simplify the expression:

8−(−3−1)−9

Solution

3

Simplify the expression:

12−(−9−6)−14

Solution

13

Simplify: 3·7−4·7−5·8.

Solution

Solution

We use the order of operations to simplify this expression. First we multiply, and then subtract from left to right.

This table illustrates the sequential steps and their corresponding visual changes when evaluating a mathematical expression, demonstrating the order of operations.
Multiply first. The image shows the mathematical expression '3.7 - 4.7 - 5.8' in a horizontal line, rendered in a plain, clear font against a white background.
Subtract from left to right. The numbers 21, 28, and 40 are displayed in a sequence, separated by hyphens, against a white background.
Subtract. The image shows the mathematical expression -7 - 40, centered against a white background.
The number -47 is displayed in black text on a white background.

Simplify the expression:

6·2−9·1−8·9.

Solution

−69

Simplify the expression:

2·5−3·7−4·9

Solution

−47

Evaluate Variable Expressions with Integers

Now we’ll practice evaluating expressions that involve subtracting negative numbers as well as positive numbers.

Evaluate x−4when
  1. ⓐ x=3
  2. ⓑ x=−6.
Solution

Solution

ⓐ To evaluate x−4 when x=3, substitute 3 for x in the expression.
The image displays the mathematical expression 'x - 4' in black text against a white background. The variable 'x' is followed by a minus sign, and then the numeral '4' completes the simple algebraic expression.
The image displays the text 'Substitute 3 for x.' in a bold, sans-serif font against a white background. The number '3' is highlighted in red, while the rest of the text is a dark, desaturated teal color. The mathematical expression '3 - 4' is displayed, showing a subtraction operation between the numbers three and four.
Subtract. The image displays the number '-1' in black font against a white background.
ⓑ To evaluate x−4 when x=−6, substitute −6 for x in the expression.
The mathematical expression 'x - 4' is displayed in black font against a white background.
The text 'Substitute -6 for x.' is displayed in a blue-green gradient font on a white background, with the '-6' highlighted in red. A mathematical expression displays '-6 - 4' in red and black text on a white background, representing the subtraction of 4 from -6.
Subtract. The number negative 10, or minus 10, is displayed in black text on a white background.

Evaluate each expression:

y−7when
  1. ⓐ y=5
  2. ⓑ y=−8
Solution
  1. ⓐ −2
  2. ⓑ −15

Evaluate each expression:

m−3when
  1. ⓐ m=1
  2. ⓑ m=−4
Solution
  1. ⓐ −2
  2. ⓑ −7
Evaluate 20−zwhen
  1. ⓐ z=12
  2. ⓑ z=−12
Solution

Solution

ⓐ To evaluate 20−zwhenz=12, substitute 12 for z in the expression.
A mathematical expression displays '20 - z' in a simple, clear font against a white background.
The image shows the text 'Substitute 12 for z.' The word 'Substitute' is dark blue-green, the number '12' is red, and 'for z.' is dark blue-green. The text suggests a mathematical or algebraic instruction. A mathematical expression showing '20 - 12', with the number '12' highlighted in red.
Subtract. A single black numeral '8' is displayed against a stark white background.
ⓑ To evaluate 20−zwhenz=−12,substitute−12forzin the expression.
The image displays the algebraic expression '20 - z' in a simple, clear font against a white background.
The text reads 'Substitute -12 for z.', with '-12' highlighted in red. A mathematical expression displays '20 - (-12)' in black and red text against a white background.
Subtract. The number 32 is displayed prominently in a gray font on a plain white background, centered within the image frame. The digits appear to be a standard, sans-serif typeface, clean and legible.

Evaluate each expression:

17−kwhen
  1. ⓐ k=19
  2. ⓑ k=−19
Solution
  1. ⓐ −2
  2. ⓑ 36

Evaluate each expression:

−5−bwhen
  1. ⓐ b=14
  2. ⓑ b=−14
Solution
  1. ⓐ −19
  2. ⓑ 9

Translate Word Phrases to Algebraic Expressions

When we first introduced the operation symbols, we saw that the expression a−b may be read in several ways as shown below.

This table has six rows. The first row has a - b. The second row states a minus b. The third row states the difference of a and b. The fourth row states subtract b from a. The fifth row states b subtracted from a. The sixth row states b less than a.

Be careful to get a and b in the right order!

Translate and then simplify:

  1. ⓐ the difference of 13 and −21
  2. ⓑ subtract 24 from −19
Solution

Solution

ⓐ A difference means subtraction. Subtract the numbers in the order they are given.
The text 'the difference of 13 and -21' is displayed, with the word 'difference' highlighted in red.
Translate. A mathematical expression '13 - (-21)' is displayed on a white background, demonstrating the subtraction of a negative number.
Simplify. The number 34 is prominently displayed in the upper right portion of a white background. The digits are in a dark color, providing a clear contrast against the bright, plain surface.
ⓑ Subtract means to take 24 away from −19.
The text 'subtract 24 from -19' is displayed, instructing a mathematical operation.
Translate. The image displays a mathematical expression '-19 - 24' in a dark gray font on a plain white background, presenting a simple arithmetic problem.
Simplify. The number -43 is displayed in black text on a plain white background.

Translate and simplify:

  1. ⓐ the difference of 14 and −23
  2. ⓑ subtract 21 from −17
Solution
  1. ⓐ 14 − (−23) = 37
  2. ⓑ −17 − 21 = −38

Translate and simplify:

  1. ⓐ the difference of 11 and −19
  2. ⓑ subtract 18 from −11
Solution
  1. ⓐ 11 − (−19) = 30
  2. ⓑ −11 − 18 = −29

Subtract Integers in 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 problem, we first need to determine what we are asked to find. Then we can 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.

Solve Application Problems.

  1. Identify what you are asked to find.
  2. Write a phrase that gives the information to find it.
  3. Translate the phrase to an expression.
  4. Simplify the expression.
  5. Answer the question with a complete sentence.

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

Solution

Solution

This table illustrates the step-by-step process of solving a temperature difference problem, from identifying the unknown to stating the final answer.
Step 1. Identify what we are asked to find. the difference between the morning and afternoon temperatures
Step 2. Write a phrase that gives the information to find it. the difference of 11 and −9
Step 3. Translate the phrase to an expression.
The word difference indicates subtraction.
11−(−9)
Step 4. Simplify the expression. 20
Step 5. Write a complete sentence that answers the question. The difference in temperature was 20 degrees Fahrenheit.

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

Solution

45 degrees Fahrenheit

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

Solution

9 degrees Fahrenheit

Geography provides another application of negative numbers with the elevations of places below sea level.

Dinesh hiked from Mt. Whitney, the highest point in California, to Death Valley, the lowest point. The elevation of Mt. Whitney is 14,497 feet above sea level and the elevation of Death Valley is 282 feet below sea level. What is the difference in elevation between Mt. Whitney and Death Valley?

Solution

Solution

Step-by-step solution for calculating the elevation difference between Mt. Whitney and Death Valley.
Step 1. What are we asked to find? The difference in elevation between Mt. Whitney and Death Valley
Step 2. Write a phrase. elevation of Mt. Whitney−elevation of Death Valley
Step 3. Translate. 14,497−(−282)
Step 4. Simplify. 14,779
Step 5. Write a complete sentence that answers the question. The difference in elevation is 14,779 feet.

One day, John hiked to the 10,023 foot summit of Haleakala volcano in Hawaii. The next day, while scuba diving, he dove to a cave 80 feet below sea level. What is the difference between the elevation of the summit of Haleakala and the depth of the cave?

Solution

10,103 feet

The submarine Nautilus is at 340 feet below the surface of the water and the submarine Explorer is 573 feet below the surface of the water. What is the difference in the position of the Nautilus and the Explorer?

Solution

233 feet

Managing your money can involve both positive and negative numbers. You might have overdraft protection on your checking account. This means the bank lets you write checks for more money than you have in your account (as long as they know they can get it back from you!)

Leslie has $25 in her checking account and she writes a check for $8.
  1. ⓐ What is the balance after she writes the check?
  2. ⓑ She writes a second check for $20. What is the new balance after this check?
  3. ⓒ Leslie’s friend told her that she had lost a check for $10 that Leslie had given her with her birthday card. What is the balance in Leslie’s checking account now?
Solution

Solution

This table outlines the steps, from question to final answer, for solving a basic mathematical word problem involving an account balance.
ⓐ
What are we asked to find? The balance of the account
Write a phrase. $25 minus $8
Translate The image displays a mathematical subtraction problem, written as '$25 - $8', indicating an operation to find the difference between twenty-five dollars and eight dollars.
Simplify. The number '$17' is displayed in black text on a plain white background, indicating a monetary value of seventeen dollars.
Write a sentence answer. The balance is $17.
Step-by-step solution for calculating a bank balance after a withdrawal, illustrating an overdraft scenario.
ⓑ
What are we asked to find? The new balance
Write a phrase. $17 minus $20
Translate A numerical text image displays a price range of '$17 - $20' in black font on a white background.
Simplify. The image displays '-$3' in black text against a plain white background, indicating a negative financial value or a cost of three dollars.
Write a sentence answer. She is overdrawn by $3.
Step-by-step solution demonstrating how to calculate a new financial balance, from identifying the problem to stating the final answer.
ⓒ
What are we asked to find? The new balance
Write a phrase. $10 more than −$3
Translate The image displays a mathematical expression showing a sum: -$3 + $10. This represents adding a positive $10 to a negative $3, resulting in a net positive value of $7.
Simplify. The image shows the text $7, indicating a price or monetary value of seven dollars.
Write a sentence answer. The balance is now $7.
Araceli has $75 in her checking account and writes a check for $27.
  1. ⓐ What is the balance after she writes the check?
  2. ⓑ She writes a second check for $50. What is the new balance?
  3. ⓒ The check for $20 that she sent a charity was never cashed. What is the balance in Araceli’s checking account now?
Solution
  1. ⓐ $48
  2. ⓑ −$2
  3. ⓒ $18
Genevieve’s bank account was overdrawn and the balance is −$78.
  1. ⓐ She deposits a check for $24 that she earned babysitting. What is the new balance?
  2. ⓑ She deposits another check for $49. Is she out of debt yet? What is her new balance?
Solution
  1. ⓐ −$54
  2. ⓑ No, −$5
The Links to Literacy activity "Elevator Magic" will provide you with another view of the topics covered in this section.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Adding and Subtracting Integers
  • Subtracting Integers with Color Counters
  • Subtracting Integers Basics
  • Subtracting Integers
  • Integer Application

Key Concepts

  • Subtraction of Integers
    This table illustrates integer subtraction methods, covering direct subtraction and cases requiring neutral pairs, with examples and conceptual steps.
    5–3 –5–(–3)
    2 –2
    2 positives 2 negatives
    When there would be enough counters of the color to take away, subtract.
    –5–3 5–(–3)
    –8 8
    5 negatives, want to subtract 3 positives 5 positives, want to subtract 3 negatives
    need neutral pairs need neutral pairs
    When there would not be enough of the counters to take away, add neutral pairs.
  • Subtraction Property
    • a−b=a+(−b)
    • a−(−b)=a+b
  • Solve Application Problems
    • Step 1. Identify what you are asked to find.
    • Step 2. Write a phrase that gives the information to find it.
    • Step 3. Translate the phrase to an expression.
    • Step 4. Simplify the expression.
    • Step 5. Answer the question with a complete sentence.

Practice Makes Perfect

Model Subtraction of Integers

In the following exercises, model each expression and simplify.

8−2

Solution


This figure shows a row of 8 light pink circles, representing positive counters. The first 2 are circles and are separated from the last 6.
6

9−3

−5−(−1)

Solution


This figure ishows a row of 5 dark pink  circles. The first one is circled.
−4

−6−(−4)

−5−4

Solution


This figure has a row of 9 dark pink circles representing negative counters. The first 5 are separated from the last 4. Below the last 4 is a row of 4 light pink circles, representing positive counters. These four positive counters are circled.
−9

−7−2

8−(−4)

Solution


This figure has a row of 12 light pink circles, representing positive counters. The first 8 are separated from the last 4. Below the last 4 is a row of 4 dark pink circles, representing negative counters. These four negative counters are circled.
12

7−(−3)

Simplify Expressions with Integers

In the following exercises, simplify each expression.

  1. ⓐ 15−6
  2. ⓑ 15+(−6)
Solution
  1. ⓐ 9
  2. ⓑ 9
  1. ⓐ 12−9
  2. ⓑ 12+(−9)
  1. ⓐ 44−28
  2. ⓑ 44+(−28)
Solution
  1. ⓐ 16
  2. ⓑ 16
  1. ⓐ 35−16
  2. ⓑ 35+(−16)
  1. ⓐ 8−(−9)
  2. ⓑ 8+9
Solution
  1. ⓐ 17
  2. ⓑ 17
  1. ⓐ 4−(−4)
  2. ⓑ 4+4
  1. ⓐ 27−(−18)
  2. ⓑ 27+18
Solution
  1. ⓐ 45
  2. ⓑ 45
  1. ⓐ 46−(−37)
  2. ⓑ 46+37

In the following exercises, simplify each expression.

15−(−12)

Solution

27

14−(−11)

10−(−19)

Solution

29

11−(−18)

48−87

Solution

−39

45−69

31−79

Solution

−48

39−81

−31−11

Solution

−42

−32−18

−17−42

Solution

−59

−19−46

−103−(−52)

Solution

−51

−105−(−68)

−45−(−54)

Solution

9

−58−(−67)

8−3−7

Solution

−2

9−6−5

−5−4+7

Solution

−2

−3−8+4

−14−(−27)+9

Solution

22

−15−(−28)+5

71+(−10)−8

Solution

53

64+(−17)−9

−16−(−4+1)−7

Solution

−20

−15−(−6+4)−3

(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)]

6⋅3−4⋅3−7⋅2

Solution

–8

5⋅7−8⋅2−4⋅9

52−62

Solution

−11

62−72

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression for the given values.

x−6when
  1. ⓐ x=3
  2. ⓑ x=−3
Solution
  1. ⓐ −3
  2. ⓑ −9
x−4when
  1. ⓐ x=5
  2. ⓑ x=−5
5−ywhen
  1. ⓐ y=2
  2. ⓑ y=−2
Solution
  1. ⓐ 3
  2. ⓑ 7
8−ywhen
  1. ⓐ y=3
  2. ⓑ y=−3

4x2−15x+1whenx=3

Solution

−8

5x2−14x+7whenx=2

−12−5x2whenx=6

Solution

−192

−19−4x2whenx=5

Translate Word Phrases to Algebraic Expressions

In the following exercises, translate each phrase into an algebraic expression and then simplify.

  1. ⓐ The difference of 3 and −10
  2. ⓑ Subtract −20 from 45
Solution
  1. ⓐ 3 − (−10) = 13
  2. ⓑ 45 − (−20) = 65
  1. ⓐ The difference of 8 and −12
  2. ⓑ Subtract −13 from 50
  1. ⓐ The difference of −6 and 9
  2. ⓑ Subtract −12 from −16
Solution
  1. ⓐ −6 − 9 = −15
  2. ⓑ −16 − (−12) = −4
  1. ⓐ The difference of −8 and 9
  2. ⓑ Subtract −15 from −19
  1. ⓐ 8 less than −17
  2. ⓑ −24 minus 37
Solution
  1. ⓐ −17 − 8 = −25
  2. ⓑ −24 − 37 = −61
  1. ⓐ 5 less than −14
  2. ⓑ −13 minus 42
  1. ⓐ 21 less than6
  2. ⓑ 31 subtracted from −19
Solution
  1. ⓐ 6 − 21 = −15
  2. ⓑ −19 − 31 = −50
  1. ⓐ 34 less than7
  2. ⓑ 29 subtracted from −50

Subtract Integers in Applications

In the following exercises, solve the following applications.

Temperature One morning, the temperature in Urbana, Illinois, was 28° Fahrenheit. By evening, the temperature had dropped 38° Fahrenheit. What was the temperature that evening?

Solution

−10°

Temperature On Thursday, the temperature in Spincich Lake, Michigan, was 22° Fahrenheit. By Friday, the temperature had dropped 35° Fahrenheit. What was the temperature on Friday?

Temperature On January 15, the high temperature in Anaheim, California, was 84° Fahrenheit. That same day, the high temperature in Embarrass, Minnesota was −12° Fahrenheit. 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 At the first down, the Warriors football team had the ball on their 30-yard line. On the next three downs, they gained 2 yards, lost 7 yards, and lost 4 yards. What was the yard line at the end of the third down?

Solution

21-yard line

Football At the first down, the Barons football team had the ball on their 20-yard line. On the next three downs, they lost 8 yards, gained 5 yards, and lost 6 yards. What was the yard line at the end of the third down?

Checking Account John has $148 in his checking account. He writes a check for $83. What is the new balance in his checking account?

Solution

$65

Checking Account Ellie has $426 in her checking account. She writes a check for $152. What is the new balance in her checking account?

Checking Account Gina has $210 in her checking account. She writes a check for $250. What is the new balance in her checking account?

Solution

−$40

Checking Account Frank has $94 in his checking account. He writes a check for $110. What is the new balance in his checking account?

Checking Account Bill has a balance of −$14 in his checking account. He deposits $40 to the account. What is the new balance?

Solution

$26

Checking Account Patty has a balance of −$23 in her checking account. She deposits $80 to the account. What is the new balance?

Everyday Math

Camping Rene is on an Alpine hike. The temperature is−7°. Rene’s sleeping bag is rated “comfortable to −20°”. How much can the temperature change before it is too cold for Rene’s sleeping bag?

Solution

13°

Scuba Diving Shelly’s scuba watch is guaranteed to be watertight to −100feet. She is diving at −45feet on the face of an underwater canyon. By how many feet can she change her depth before her watch is no longer guaranteed?

Writing Exercises

Explain why the difference of 9 and −6 is 15.

Solution

Sample answer: On a number line, 9 is 15 units away from −6.

Why is the result of subtracting 3−(−4) the same as the result of adding 3+4?

Self Check

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

A self-assessment checklist for mathematics skills related to integers. The table lists five 'I can...' statements (e.g., model subtraction of integers) and three columns for assessment: Confidently, With some help, and 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?

Multiply and Divide Integers

Learning Objectives

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

  • Multiply integers
  • Divide integers
  • Simplify expressions with integers
  • Evaluate variable expressions with integers
  • Translate word phrases to algebraic expressions

Before you get started, take this readiness quiz.

Translate the quotient of 20 and 13 into an algebraic expression.
If you missed this problem, review Example 12 in Divide Whole Numbers.

Solution

20÷13

Add: −5+(−5)+(−5).
If you missed this problem, review Example 8 in Add Integers.

Solution

−15

Evaluaten+4whenn=−7.
If you missed this problem, review Example 10 in Add Integers.

Solution

−3

Multiply Integers

Since multiplication is mathematical shorthand for repeated addition, our counter 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.

We remember that a·b means add a,b times. Here, we are using the model shown in Figure 1 just to help us discover the pattern.

This image has two columns. The first column has 5 times 3. Underneath, it states add 5, 3 times. Under this there are 3 rows of 5 blue circles labeled 15 positives and 5 times 3 equals 15. The second column has negative 5 times 3. Underneath it states add negative 5, 3 times. Under this there are 3 rows of 5 red circles labeled 15 negatives and negative 5 times 3 equals 15.

Now consider what it means 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 as shown in Figure 2.

This figure has 2 columns. The first column has 5 times negative 3. Underneath it states take away 5, 3 times. Under this there are 3 rows of 5 red circles. A downward arrow points to six rows of alternating colored circles in rows of fives. The first row includes 5 red circles, followed by five blue circles, then 5 red, five blue, five red, and five blue. All of the rows of blue circles are circled. The non-circled rows are labeled 15 negatives.  Under the label is 5 times negative 3 equals negative 15. The second column has negative 5 times negative 3. Underneath it states take away negative 5, 3 times. Then there are 6 rows of 5 circles alternating in color. The first row is 5 blue circles followed by 5 red circles. All of the red rows are circled. The non-circles rows are labeled 15 positives. Under the label is negative 5 times negative 3 equals 15.

In both cases, we started with 15 neutral pairs. In the case on the left, we took away 5,3 times and the result was −15. To multiply (−5)(−3), we took away −5,3 times and the result was 15. So we found that

5·3=15−5(3)=−155(−3)=−15(−5)(−3)=15

Notice that for multiplication of two signed numbers, when the signs are the same, the product is positive, and when the signs are different, the product is negative.

Multiplication of Signed Numbers

The sign of the product of two numbers depends on their signs.

Same signs Product
•Two positives
•Two negatives
Positive
Positive
Different signs Product
•Positive • negative
•Negative • positive
Negative
Negative
Multiply each of the following:
  1. ⓐ −9·3
  2. ⓑ −2(−5)
  3. ⓒ 4(−8)
  4. ⓓ 7·6
Solution

Solution

Worked example: Multiplication of a negative and positive integer, showing the problem, explanation, and final result.
ⓐ
–9⋅3
Multiply, noting that the signs are different and so the product is negative. –27
Example demonstrating the multiplication of two negative numbers, showing that a positive product results when signs are the same.
ⓑ
–2(–5)
Multiply, noting that the signs are the same and so the product is positive. 10
Example showing multiplication of integers with different signs, yielding a negative product.
ⓒ
4(–8)
Multiply, noting that the signs are different and so the product is negative. –32
Illustrates the multiplication of 7 and 6, showing the positive product and the reason for its sign.
ⓓ
7⋅6
The signs are the same, so the product is positive. 42

Multiply:

  1. ⓐ −6·8
  2. ⓑ −4(−7)
  3. ⓒ 9(−7)
  4. ⓓ 5·12
Solution
  1. ⓐ −48
  2. ⓑ 28
  3. ⓒ −63
  4. ⓓ 60

Multiply:

  1. ⓐ −8·7
  2. ⓑ −6(−9)
  3. ⓒ 7(−4)
  4. ⓓ 3·13
Solution
  1. ⓐ −56
  2. ⓑ 54
  3. ⓒ −28
  4. ⓓ 39

When we multiply a number by 1, the result is the same number. What happens when we multiply a number by−1? Let’s multiply a positive number and then a negative number by −1 to see what we get.

−1·4−1(−3)−43−4is the opposite of43is the opposite of−3

Each time we multiply a number by −1, we get its opposite.

Multiplication by −1

Multiplying a number by −1 gives its opposite.

−1a=−a
Multiply each of the following:
  1. ⓐ −1·7
  2. ⓑ −1(−11)
Solution

Solution

Explains properties of negative numbers, covering multiplication with different signs and opposites, with mathematical examples.
ⓐ
The signs are different, so the product will be negative. −1⋅7
Notice that −7 is the opposite of 7. −7
Demonstrates rules for multiplying negative numbers, showing negative times negative equals positive, and the concept of opposites.
ⓑ
The signs are the same, so the product will be positive. −1(−11)
Notice that 11 is the opposite of −11. 11
Multiply.
  1. ⓐ −1·9
  2. ⓑ −1·(−17)
Solution
  1. ⓐ −9
  2. ⓑ 17
Multiply.
  1. ⓐ −1·8
  2. ⓑ −1·(−16)
Solution
  1. ⓐ −8
  2. ⓑ 16

Divide Integers

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 we look at some examples of multiplying integers, we might figure out the rules for dividing integers.

5·3=15so15÷3=5−5(3)=−15so−15÷3=−5(−5)(−3)=15so15÷(−3)=−55(−3)=−15so−15÷−3=5

Division of signed numbers follows the same rules as multiplication. When the signs are the same, the quotient is positive, and when the signs are different, the quotient is negative.

Division of Signed Numbers

The sign of the quotient of two numbers depends on their signs.

Same signs Quotient
•Two positives
•Two negatives
Positive
Positive
Different signs Quotient
•Positive & negative
•Negative & positive
Negative
Negative

Remember, you can always check the answer to a division problem by multiplying.

Divide each of the following:
  1. ⓐ −27÷3
  2. ⓑ −100÷(−4)
Solution

Solution

Example of integer division with a detailed explanation and the resulting quotient.
ⓐ
–27÷3
Divide, noting that the signs are different and so the quotient is negative. –9
Demonstration of dividing negative integers, illustrating the problem, an explanatory step, and the final positive result.
ⓑ
–100÷(–4)
Divide, noting that the signs are the same and so the quotient is positive. 25

Divide:

  1. ⓐ −42÷6
  2. ⓑ −117÷(−3)
Solution
  1. ⓐ −7
  2. ⓑ 39

Divide:

  1. ⓐ −63÷7
  2. ⓑ −115÷(−5)
Solution
  1. ⓐ −9
  2. ⓑ 23

Just as we saw with multiplication, when we divide a number by 1, the result is the same number. What happens when we divide a number by −1? Let’s divide a positive number and then a negative number by −1 to see what we get.

8÷(−1)−9÷(−1)−89−8 is the opposite of 89 is the opposite of −9

When we divide a number by, −1 we get its opposite.

Division by −1

Dividing a number by −1 gives its opposite.

a÷(−1)=−a
Divide each of the following:
  1. ⓐ 16÷(−1)
  2. ⓑ −20÷(−1)
Solution

Solution

This table illustrates the division of 16 by -1, explaining the rule that dividing a number by -1 results in its opposite.
ⓐ
16÷(–1)
The dividend, 16, is being divided by –1. –16
Dividing a number by –1 gives its opposite.
Notice that the signs were different, so the result was negative.
This table illustrates the rule and provides an example of dividing a number by -1 to obtain its opposite.
ⓑ
–20÷(–1)
The dividend, –20, is being divided by –1. 20
Dividing a number by –1 gives its opposite.

Notice that the signs were the same, so the quotient was positive.

Divide:

  1. ⓐ 6÷(−1)
  2. ⓑ −36÷(−1)
Solution
  1. ⓐ −6
  2. ⓑ 36

Divide:

  1. ⓐ 28÷(−1)
  2. ⓑ −52÷(−1)
Solution
  1. ⓐ −28
  2. ⓑ 52

Simplify Expressions with Integers

Now we’ll simplify expressions that use all four operations–addition, subtraction, multiplication, and division–with integers. Remember to follow the order of operations.

Simplify:7(−2)+4(−7)−6.

Solution

Solution

We use the order of operations. Multiply first and then add and subtract from left to right.

Step-by-step simplification of a mathematical expression involving multiplication, addition, and subtraction of integers.
7(−2)+4(−7)−6
Multiply first. −14+(−28)−6
Add. −42−6
Subtract. −48

Simplify:

8(−3)+5(−7)−4

Solution

−63

Simplify:

9(−3)+7(−8)−1

Solution

−84

Simplify:
  1. ⓐ (−2)4
  2. ⓑ −24
Solution

Solution

The exponent tells how many times to multiply the base.

ⓐ The exponent is 4 and the base is −2. We raise −2 to the fourth power.

Illustrates the step-by-step process of calculating (-2)^4, showing its expanded form and sequential multiplication.
(−2)4
Write in expanded form. (−2)(−2)(−2)(−2)
Multiply. 4(−2)(−2)
Multiply. −8(−2)
Multiply. 16

ⓑ The exponent is 4 and the base is 2. We raise 2 to the fourth power and then take the opposite.

Step-by-step calculation of -2^4, showing the expansion and sequential multiplication to determine the final value.
−24
Write in expanded form. −(2⋅2⋅2⋅2)
Multiply. −(4⋅2⋅2)
Multiply. −(8⋅2)
Multiply. −16

Simplify:

  1. ⓐ (−3)4
  2. ⓑ −34
Solution
  1. ⓐ 81
  2. ⓑ −81

Simplify:

  1. ⓐ (−7)2
  2. ⓑ −72
Solution
  1. ⓐ 49
  2. ⓑ −49

Simplify:12−3(9−12).

Solution

Solution

According to the order of operations, we simplify inside parentheses first. Then we will multiply and finally we will subtract.

Step-by-step solution demonstrating the order of operations for the expression 12 - 3(9 - 12).
12−3(9−12)
Subtract the parentheses first. 12−3(−3)
Multiply. 12−(−9)
Subtract. 21

Simplify:

17−4(8−11)

Solution

29

Simplify:

16−6(7−13)

Solution

52

Simplify: 8(−9)÷(−2)3.

Solution

Solution

We simplify the exponent first, then multiply and divide.

This table demonstrates the step-by-step simplification of the mathematical expression 8(-9) ÷ (-2)^3, culminating in the result of 9.
8(−9)÷(−2)3
Simplify the exponent. 8(−9)÷(−8)
Multiply. −72÷(−8)
Divide. 9

Simplify:

12(−9)÷(−3)3

Solution

4

Simplify:

18(−4)÷(−2)3

Solution

9

Simplify:−30÷2+(−3)(−7).

Solution

Solution

First we will multiply and divide from left to right. Then we will add.

This table demonstrates the step-by-step evaluation of a mathematical expression following the order of operations.
−30÷2+(−3)(−7)
Divide. −15+(−3)(−7)
Multiply. −15+21
Add. 6

Simplify:

−27÷3+(−5)(−6)

Solution

21

Simplify:

−32÷4+(−2)(−7)

Solution

6

Evaluate Variable Expressions with Integers

Now we can evaluate expressions that include multiplication and division with integers. Remember that to evaluate an expression, substitute the numbers in place of the variables, and then simplify.

Evaluate2x2−3x+8whenx=−4.

Solution

Solution

A mathematical expression, 2x^2 - 3x + 8, is shown against a white background.
The text 'Substitute -4 for x.' is displayed, with '-4' highlighted in red, indicating a numerical substitution instruction. A mathematical expression: 2(-4)^2 - 3(-4) + 8, with the number -4 highlighted in red within parentheses.
Simplify exponents. A mathematical expression showing the calculation 2(16) - 3(-4) + 8.
Multiply. A mathematical expression showing '32 minus negative 12 plus 8' on a white background. This problem involves integer operations, specifically subtraction of a negative number which becomes addition, followed by addition.
Subtract. The mathematical expression '44 + 8' is displayed in black text against a plain white background.
Add. The number '52' is displayed in black text on a plain white background, positioned towards the right side of the image.

Keep in mind that when we substitute −4 for x, we use parentheses to show the multiplication. Without parentheses, it would look like 2·−42−3·−4+8.

Evaluate:

3x2−2x+6whenx=−3

Solution

39

Evaluate:

4x2−x−5whenx=−2

Solution

13

Evaluate3x+4y−6whenx=−1andy=2.

Solution

Solution

The mathematical expression '3x + 4y - 6' is displayed in black text on a white background.
Substitute x=−1 and y=2. A mathematical expression reads 3 multiplied by negative 1, plus 4 multiplied by 2, minus 6. The number -1 is highlighted in red, and 2 is highlighted in blue.
Multiply. A mathematical expression showing the addition and subtraction of integers: -3 + 8 - 6.
Simplify. The mathematical expression '-1' is displayed in a dark gray font against a plain white background.

Evaluate:

7x+6y−12whenx=−2andy=3

Solution

−8

Evaluate:

8x−6y+13whenx=−3andy=−5

Solution

19

Translate Word Phrases to Algebraic Expressions

Once again, all our prior work translating words to algebra transfers to phrases that include both multiplying and dividing integers. Remember that the key word for multiplication is product and for division is quotient.

Translate to an algebraic expression and simplify if possible: the product of −2 and 14.

Solution

Solution

The word product tells us to multiply.

Steps to translate a verbal expression into a mathematical product and simplify it.
the product of −2 and 14
Translate. (−2)(14)
Simplify. −28

Translate to an algebraic expression and simplify if possible:

the product of −5 and 12

Solution

−5 (12) = −60

Translate to an algebraic expression and simplify if possible:

the product of 8 and −13

Solution

8 (−13) = −104

Translate to an algebraic expression and simplify if possible: the quotient of −56 and −7.

Solution

Solution

The word quotient tells us to divide.

Step-by-step solution demonstrating the translation and simplification of a verbal division problem.
the quotient of −56 and −7
Translate. −56÷(−7)
Simplify. 8

Translate to an algebraic expression and simplify if possible:

the quotient of −63 and −9

Solution

−63 ÷ −9 = 7

Translate to an algebraic expression and simplify if possible:

the quotient of −72 and −9

Solution

−72 ÷ −9 = 8

ACCESS ADDITIONAL ONLINE RESOURCES

  • Multiplying Integers Using Color Counters
  • Multiplying Integers Using Color Counters With Neutral Pairs
  • Multiplying Integers Basics
  • Dividing Integers Basics
  • Ex. Dividing Integers
  • Multiplying and Dividing Signed Numbers

Key Concepts

  • Multiplication of Signed Numbers
    • To determine the sign of the product of two signed numbers:
      Same Signs Product
      Two positives
      Two negatives
      Positive
      Positive

      Different Signs Product
      Positive • negative
      Negative • positive
      Negative
      Negative
  • Division of Signed Numbers
    • To determine the sign of the quotient of two signed numbers:
      Same Signs Quotient
      Two positives
      Two negatives
      Positive
      Positive

      Different Signs Quotient
      Positive • negative
      Negative • Positive
      Negative
      Negative
  • Multiplication by −1
    • Multiplying a number by −1 gives its opposite: −1a=−a
  • Division by −1
    • Dividing a number by −1 gives its opposite: a÷(−1)=−a

Practice Makes Perfect

Multiply Integers

In the following exercises, multiply each pair of integers.

−4·8

Solution

−32

−3·9

−5(7)

Solution

−35

−8(6)

−18(−2)

Solution

36

−10(−6)

9(−7)

Solution

−63

13(−5)

−1·6

Solution

−6

−1·3

−1(−14)

Solution

14

−1(−19)

Divide Integers

In the following exercises, divide.

−24÷6

Solution

−4

−28÷7

56÷(−7)

Solution

−8

35÷(−7)

−52÷(−4)

Solution

13

−84÷(−6)

−180÷15

Solution

−12

−192÷12

49÷(−1)

Solution

−49

62÷(−1)

Simplify Expressions with Integers

In the following exercises, simplify each expression.

5(−6)+7(−2)−3

Solution

−47

8(−4)+5(−4)−6

−8(−2)−3(−9)

Solution

43

−7(−4)−5(−3)

(−5)3

Solution

−125

(−4)3

(−2)6

Solution

64

(−3)5

−42

Solution

−16

−62

−3(−5)(6)

Solution

90

−4(−6)(3)

−4·2·11

Solution

−88

−5·3·10

(8−11)(9−12)

Solution

9

(6−11)(8−13)

26−3(2−7)

Solution

41

23−2(4−6)

−10(−4)÷(−8)

Solution

−5

−8(−6)÷(−4)

65÷(−5)+(−28)÷(−7)

Solution

−9

52÷(−4)+(−32)÷(−8)

9−2[3−8(−2)]

Solution

−29

11−3[7−4(−2)]

(−3)2−24÷(8−2)

Solution

5

(−4)2−32÷(12−4)

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

−2x+17when
  1. ⓐ x=8
  2. ⓑ x=−8
Solution
  1. ⓐ 1
  2. ⓑ 33
−5y+14when
  1. ⓐ y=9
  2. ⓑ y=−9
10−3mwhen
  1. ⓐ m=5
  2. ⓑ m=−5
Solution
  1. ⓐ −5
  2. ⓑ 25
18−4nwhen
  1. ⓐ n=3
  2. ⓑ n=−3

p2−5p+5whenp=−1

Solution

11

q2−2q+9 when q=−2

2w2−3w+7 when w=−2

Solution

21

3u2−4u+5 when u=−3

6x−5y+15 when x=3 and y=−1

Solution

38

3p−2q+9 when p=8 and q=−2

9a−2b−8 when a=−6 and b=−3

Solution

−56

7m−4n−2 when m=−4 and n=−9

Translate Word Phrases to Algebraic Expressions

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

The product of −3 and 15

Solution

−3·15 = −45

The product of −4 and 16

The quotient of −60 and −20

Solution

−60 ÷ (−20) = 3

The quotient of −40 and −20

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

Solution

−6a+b

The quotient of −7 and the sum of m and n

The product of −10 and the difference of pandq

Solution

−10 (p − q)

The product of −13 and the difference of candd



Everyday Math

Stock market Javier owns 300 shares of stock in one company. On Tuesday, the stock price dropped $12 per share. What was the total effect on Javier’s portfolio?

Solution

−$3,600

Weight loss In the first week of a diet program, eight women lost an average of 3 pounds each. What was the total weight change for the eight women?

Writing Exercises

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

Solution

Sample answer: Multiplying two integers with the same sign results in a positive product. Multiplying two integers with different signs results in a negative product.

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

Why is −24≠(−2)4?

Solution

Sample answer: In the first expression the base is positive and after you raise it to the power you should take the opposite. Then in the second expression the base is negative so you simply raise it to the power.

Why is −42≠(−4)2?

Self Check

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

A self-assessment chart for math skills including multiplying and dividing integers, simplifying expressions, evaluating variable expressions, and translating word phrases to algebraic expressions.

ⓑ 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 Equations Using Integers; The Division Property of Equality

Learning Objectives

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

  • Determine whether an integer is a solution of an equation
  • Solve equations with integers using the Addition and Subtraction Properties of Equality
  • Model the Division Property of Equality
  • Solve equations using the Division Property of Equality
  • Translate to an equation and solve

Before you get started, take this readiness quiz.

Evaluatex+4whenx=−4.
If you missed this problem, review Example 9 in Add Integers.

Solution

0

Solve:y−6=10.
If you missed this problem, review Example 6 in Solving Equations Using the Subtraction and Addition Properties of Equality.

Solution

16

Translate into an algebraic expression 5 less than x.
If you missed this problem, review Table 12 in Subtract Whole Numbers.

Solution

x−5

Determine Whether a Number is a Solution of an Equation

In Solve Equations with the Subtraction and Addition Properties of Equality, we saw that a solution of an equation is a value of a variable that makes a true statement when substituted into that equation. In that section, we found solutions that were whole numbers. Now that we’ve worked with integers, we’ll find integer solutions to equations.

The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number or an integer.

How to 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 each of the following is a solution of 2x−5=−13:
  1. ⓐ x=4
  2. ⓑ x=−4
  3. ⓒ x=−9.
Solution

Solution

ⓐ Substitute 4 for x in the equation to determine if it is true.
A mathematical equation is displayed: 2x - 5 = -13.
The text 'Substitute 4 for x.' is displayed in a dark teal-like font, with the number 4 highlighted in red. A mathematical equation displays '2(4) - 5 = -13,' with the number 4 highlighted in red, indicating a specific value in the problem.
Multiply. The image presents the mathematical expression '8 - 5 ? = -13', with a question mark placed above the equals sign, indicating an inquiry into the truth of the statement.
Subtract. A mathematical expression displays '3  eq -13', indicating that 3 is not equal to -13.

Since x=4 does not result in a true equation, 4 is not a solution to the equation.

ⓑ Substitute −4 for x in the equation to determine if it is true. A mathematical equation, 2x - 5 = -13, is displayed in a dark gray font against a plain white background.
The text 'Substitute -4 for x.' is displayed in the image. The word 'Substitute' is in dark blue-gray, '-4' is in red, and 'for x.' is in dark blue-gray. The background is white. The image displays the mathematical expression 2(-4) - 5 ?= -13, where the question mark asks to verify the equality. Calculating the left side, 2 multiplied by -4 is -8, and then subtracting 5 yields -13. Thus, the equality holds true.
Multiply. The image displays a math problem questioning if -8 minus 5 is equal to -13. The calculation confirms that -8 - 5 does indeed equal -13.
Subtract. The mathematical equation -13 = -13 with a checkmark, indicating it is correct.

Since x=−4 results in a true equation, −4 is a solution to the equation.

ⓒ Substitute −9 for x in the equation to determine if it is true.
The image shows a mathematical equation, 2x - 5 = -13, in black text on a white background.
Substitute −9 for x. A math equation `2(-9) - 5 ?= -13` is shown. The question mark in the equals sign asks to verify the equality. The number -9 is highlighted in red, drawing attention to its negative value.
Multiply. A mathematical expression is displayed, '-18 - 5 ?= -13,' where the question mark above the equality sign indicates a verification of whether the left side is equal to the right side. -18 minus 5 equals -23.
Subtract. A mathematical inequality states that -23 is not equal to -13, displayed in black text on a white background.

Since x=−9 does not result in a true equation, −9 is not a solution to the equation.

Determine whether each of the following is a solution of 2x−8=−14:
  1. ⓐ x=−11
  2. ⓑ x=11
  3. ⓒ x=−3
Solution
  1. ⓐ no
  2. ⓑ no
  3. ⓒ yes
Determine whether each of the following is a solution of 2y+3=−11:
  1. ⓐ y=4
  2. ⓑ y=−4
  3. ⓒ y=−7
Solution
  1. ⓐ no
  2. ⓑ no
  3. ⓒ yes

Solve Equations with Integers Using the Addition and Subtraction Properties of Equality

In Solve Equations with the Subtraction and Addition Properties of Equality, we solved equations similar to the two shown here using the Subtraction and Addition Properties of Equality. Now we can use them again with integers.

This figure has two columns. The first column has the equation x plus 4 equals 12. Underneath there is x plus 4 minus 4 equals 12 minus 4. Under this there is x equals 8. The second column has the equation y minus 5 equals 9. Underneath there is the equation y minus 5 plus 5 equals 9 plus 5. Under this there is y equals 14.

When you add or subtract the same quantity from both sides of an equation, you still have equality.

Properties of Equalities

Subtraction Property of Equality Addition Property of Equality
For any numbersa,b,c,
ifa=bthena−c=b−c.
For any numbersa,b,c,
ifa=bthena+c=b+c.

Solve: y+9=5.

Solution

Solution

A mathematical equation is displayed, showing 'y + 9 = 5' in a clear, black font against a white background.
Subtract 9 from each side to undo the addition. The mathematical equation y + 9 - 9 = 5 - 9 is displayed, demonstrating a step-by-step approach to isolate the variable y.
Simplify. The equation y = -4 is displayed in black text on a white background.

Check the result by substituting −4 into the original equation.

Demonstration of substituting a value into an algebraic equation to verify its solution.
y+9=5
Substitute −4 for y −4+9=?5
5=5✓

Since y=−4 makes y+9=5 a true statement, we found the solution to this equation.

Solve:

y+11=7

Solution

−4

Solve:

y+15=−4

Solution

−19

Solve: a−6=−8

Solution

Solution

A mathematical equation displays 'a - 6 = -8' in black font on a white background. This is a linear equation with one variable 'a', where 6 is subtracted from 'a', resulting in -8.
Add 6 to each side to undo the subtraction. A mathematical equation showing 'a - 6 + 6 = -8 + 6', where the '+ 6' on both sides is highlighted in red, illustrating the addition property of equality to solve for 'a'.
Simplify. The mathematical equation 'a = -2' is displayed in a bold, black font on a plain white background, presenting a simple algebraic assignment of the value negative two to the variable 'a'.
Check the result by substituting −2 into the original equation: The image shows the equation 'a - 6 = -8' in black text on a white background.
Substitute −2 for a The mathematical expression '-2 - 6 = -8' is presented, with a question mark over the equal sign, indicating a query about its validity or prompting verification of the sum.
A mathematical equation shows '-8 = -8' with a black checkmark symbol, indicating the equality is correct. The numbers and symbols are rendered in a slightly bold, shadowed style against a white background.

The solution to a−6=−8 is −2.

Since a=−2 makes a−6=−8 a true statement, we found the solution to this equation.

Solve:

a−2=−8

Solution

−6

Solve:

n−4=−8

Solution

−4

Model the Division Property of Equality

All of the equations we have solved so far have been of the form x+a=b or x−a=b. We were able to isolate the variable by adding or subtracting the constant term. Now we’ll see how to solve equations that involve division.

We will model an equation with envelopes and counters in Figure 1.

This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed.

Here, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?

To determine the number, separate the counters on the right side into 2 groups of the same size. So 6 counters divided into 2 groups means there must be 3 counters in each group (since 6÷2=3).

What equation models the situation shown in Figure 2? There are two envelopes, and each contains x counters. Together, the two envelopes must contain a total of 6 counters. So the equation that models the situation is 2x=6.

This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed. Under the figure is two times x equals 6.

We can divide both sides of the equation by 2 as we did with the envelopes and counters.

This figure has two rows. The first row has the equation 2x divided by 2 equals 6 divided by 2. The second row has the equation x equals 3.

We found that each envelope contains 3 counters. Does this check? We know 2·3=6, so it works. Three counters in each of two envelopes does equal six.

Figure 3 shows another example.

This image has two columns. In the first column are three envelopes. In the second column there are four rows of  three blue circles. Underneath the image is the equation 3x equals 12.

Now we have 3 identical envelopes and 12 counters. How many counters are in each envelope? We have to separate the 12 counters into 3 groups. Since 12÷3=4, there must be 4 counters in each envelope. See Figure 4.

This image has two columns. In the first column are four envelopes. In the second column there are twelve blue circles.

The equation that models the situation is 3x=12. We can divide both sides of the equation by 3.

This image shows the equation 3x divided by 3 equals 12 divided by 3. Below this equation is the equation x equals 4.

Does this check? It does because 3·4=12.

Doing the Manipulative Mathematics activity “Division Property of Equality” will help you develop a better understanding of how to solve equations using the Division Property of Equality.

Write an equation modeled by the envelopes and counters, and then solve it.

This image has two columns. In the first column are four envelopes. In the second column there are 8 blue circles.
Solution

Solution

There are 4 envelopes, or 4 unknown values, on the left that match the 8 counters on the right. Let’s call the unknown quantity in the envelopes x.
Write the equation. A simple algebraic equation '4x = 8' is displayed in black font on a white background.
Divide both sides by 4. An algebraic step demonstrating the division of both sides of the equation 4x = 8 by 4, to isolate x. This simplifies to x = 2.
Simplify. A mathematical expression on a white background reads 'x = 2' in a dark, bold font.

There are 2 counters in each envelope.

Write the equation modeled by the envelopes and counters. Then solve it.
This image has two columns. In the first column are four envelopes. In the second column there are 12 blue circles.

Solution

4x = 12; x = 3

Write the equation modeled by the envelopes and counters. Then solve it.
This image has two columns. In the first column are three envelopes. In the second column there are six blue circles.

Solution

3x = 6; x = 2

Solve Equations Using the Division Property of Equality

The previous examples lead to the Division Property of Equality. When you divide both sides of an equation by any nonzero number, you still have equality.

Division Property of Equality

For any numbers a,b,c,and c≠0, If a=bthen ac=bc.

Solve:7x=−49.

Solution

Solution

To isolate x, we need to undo multiplication.
A mathematical equation is displayed, showing '7x = -49' in bold black characters on a white background. This is a linear equation where the variable x is multiplied by 7 and set equal to -49.
Divide each side by 7. A mathematical equation showing the division of both sides by 7 to solve for x, represented as 7x/7 = -49/7. The number 7 in the denominator on both sides is highlighted in red.
Simplify. A mathematical equation shows 'x = -7' rendered in a bold, sans-serif style font against a clean white background. The variable 'x' is set equal to the negative integer '7'.

Check the solution.

Verification of a linear equation solution through substitution.
7x=−49
Substitute −7 for x. 7(−7)=?−49
−49=−49✓

Therefore, −7 is the solution to the equation.

Solve:

8a=56

Solution

7

Solve:

11n=121

Solution

11

Solve: −3y=63.

Solution

Solution

To isolate y, we need to undo the multiplication.
A mathematical equation reads '-3y = 63' on a white background.
Divide each side by −3. A mathematical equation showing the division of -3y by -3 on the left side, which equals the division of 63 by -3 on the right side. The -3 in the denominator is highlighted in red on both sides.
Simplify The image displays a mathematical equation in black text on a white background, which states 'y = -21'.

Check the solution.

Demonstrates checking the solution y = -21 for the equation -3y = 63, verifying the equality of both sides.
−3y=63
Substitute −21 for y. −3(−21)=?63
63=63✓

Since this is a true statement, y=−21 is the solution to the equation.

Solve:

−8p=96

Solution

−12

Solve:

−12m=108

Solution

−9

Translate to an Equation and Solve

In the past several examples, we were given an equation containing a variable. In the next few examples, we’ll have to first translate word sentences into equations with variables and then we will solve the equations.

Translate and solve: five more than x is equal to −3.

Solution

Solution

This table illustrates the step-by-step process of translating a verbal statement into a linear equation and solving it.
five more than x is equal to −3
Translate x+5=−3
Subtract 5 from both sides. x+5−5=−3−5
Simplify. x=−8

Check the answer by substituting it into the original equation.

x+5=−3−8+5=?−3−3=−3✓

Translate and solve:

Seven more than x is equal to −2.

Solution

x + 7 = −2; x = −9

Translate and solve:

Eleven more thanyis equal to 2.

Solution

y + 11 = 2; y = −9

Translate and solve: the difference of n and 6 is −10.

Solution

Solution

This table illustrates the step-by-step process of translating a verbal problem into a linear equation and solving for the variable 'n'.
the difference of n and 6 is −10
Translate. n−6=−10
Add 6 to each side. n−6+6=−10+6
Simplify. n=−4

Check the answer by substituting it into the original equation.

n−6=−10−4−6=?−10−10=−10✓

Translate and solve:

The difference of p and 2 is −4.

Solution

p − 2 = −4; p = −2

Translate and solve:

The difference of q and 7 is −3.

Solution

q − 7 = −3; q = 4

Translate and solve: the number 108 is the product of −9 and y.

Solution

Solution

Detailed steps to solve the linear equation 108 = -9y, starting from its word problem translation.
the number of 108 is the product of −9 and y
Translate. 108=−9y
Divide by −9. 108−9=−9y−9
Simplify. −12=y

Check the answer by substituting it into the original equation.

108=−9y108=?−9(−12)108=108✓

Translate and solve:

The number 132 is the product of −12 and y.

Solution

132 = −12y; y = −11

Translate and solve:

The number 117 is the product of −13 and z.

Solution

117 = −13z; z = −9

ACCESS ADDITIONAL ONLINE RESOURCES

  • One-Step Equations With Adding Or Subtracting
  • One-Step Equations With Multiplying Or Dividing

Key Concepts

  • How to determine whether a number is a solution to an equation.
    • Step 1. Substitute the number for the variable in the equation.
    • Step 2. Simplify the expressions on both sides of the equation.
    • Step 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.
  • Properties of Equalities
    Subtraction Property of Equality Addition Property of Equality
    For any numbersa,b,c,
    ifa=bthena−c=b−c.
    For any numbersa,b,c,
    ifa=bthena+c=b+c.
  • Division Property of Equality
    • For any numbers a,b,c, and c≠0
      If a=b, then ac=bc.

Section Exercises

Practice Makes Perfect

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each number is a solution of the given equation.

4x−2=6
  1. ⓐ x=−2
  2. ⓑ x=−1
  3. ⓒ x=2
Solution
  1. ⓐ no
  2. ⓑ no
  3. ⓒ yes
4y−10=−14
  1. ⓐ y=−6
  2. ⓑ y=−1
  3. ⓒ y=1
9a+27=−63
  1. ⓐ a=6
  2. ⓑ a=−6
  3. ⓒ a=−10
Solution
  1. ⓐ no
  2. ⓑ no
  3. ⓒ yes
7c+42=−56
  1. ⓐ c=2
  2. ⓑ c=−2
  3. ⓒ c=−14

Solve Equations Using the Addition and Subtraction Properties of Equality

In the following exercises, solve for the unknown.

n+12=5

Solution

n = −7

m+16=2

p+9=−8

Solution

p = −17

q+5=−6

u−3=−7

Solution

u = −4

v−7=−8

h−10=−4

Solution

h = 6

k−9=−5

x+(−2)=−18

Solution

x = −16

y+(−3)=−10

r−(−5)=−9

Solution

r = −14

s−(−2)=−11

Model the Division Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve it.

Three light blue envelopes are stacked vertically on the left side of a white background, while six light yellow circular tokens are arranged in two columns of three on the right.
Solution

3x = 6; x = 2

Two light blue envelopes are paired with ten light yellow circles arranged in two columns of five on a white background.
Two light blue envelopes are paired with eight light yellow circles arranged in two columns of four on a white background.
Solution

2x = 8; x = 4

Three light blue envelopes are displayed on the left, while nine yellow-rimmed circles are arranged in a 3x3 grid on the right, all on a white background.

Solve Equations Using the Division Property of Equality

In the following exercises, solve each equation using the division property of equality and check the solution.

5x=45

Solution

x = 9

4p=64

−7c=56

Solution

c = −8

−9x=54

−14p=−42

Solution

p = 3

−8m=−40

−120=10q

Solution

q = −12

−75=15y

24x=480

Solution

x = 20

18n=540

−3z=0

Solution

z = 0

4u=0

Translate to an Equation and Solve

In the following exercises, translate and solve.

Four more than n is equal to 1.

Solution

n + 4 = 1; n = −3

Nine more than m is equal to 5.

The sum of eight and p is −3.

Solution

8 + p = −3; p = −11

The sum of two and q is −7.

The difference of a and three is −14.

Solution

a − 3 = −14; a = −11

The difference of b and 5 is −2.

The number −42 is the product of −7 and x.

Solution

−42 = −7x; x = 6

The number −54 is the product of −9 and y.

The product of -15 and f is 75.

Solution

−15f = 75; f = −5

The product of −18 and g is 36.

−6 plus c is equal to 4.

Solution

−6 + c = 4; c = 10

−2 plus d is equal to 1.

Nine less than m is −4.

Solution

m − 9 = −4; m = 5

Thirteen less than n is −10.

Mixed Practice

In the following exercises, solve.

  1. ⓐ x+2=10
  2. ⓑ 2x=10
Solution
  1. ⓐ x = 8
  2. ⓑ x = 5
  1. ⓐ y+6=12
  2. ⓑ 6y=12
  1. ⓐ −3p=27
  2. ⓑ p−3=27
Solution
  1. ⓐ p = −9
  2. ⓑ p = 30
  1. ⓐ −2q=34
  2. ⓑ q−2=34

a−4=16

Solution

a = 20

b−1=11

−8m=−56

Solution

m = 7

−6n=−48

−39=u+13

Solution

u = −52

−100=v+25

11r=−99

Solution

r = −9

15s=−300

100=20d

Solution

d = 5

250=25n

−49=x−7

Solution

x = −42

64=y−4

Everyday Math

Cookie packaging A package of 51 cookies has 3 equal rows of cookies. Find the number of cookies in each row, c, by solving the equation 3c=51.

Solution

17 cookies

Kindergarten class Connie’s kindergarten class has 24 children. She wants them to get into 4 equal groups. Find the number of children in each group, g, by solving the equation 4g=24.

Writing Exercises

Is modeling the Division Property of Equality with envelopes and counters helpful to understanding how to solve the equation 3x=15? Explain why or why not.

Solution

Sample answer: It is helpful because it shows how the counters can be divided among the envelopes.

Suppose you are using envelopes and counters to model solving the equations x+4=12 and 4x=12. Explain how you would solve each equation.

Frida started to solve the equation −3x=36 by adding 3 to both sides. Explain why Frida’s method will not solve the equation.

Solution

Sample answer: The operation used in the equation is multiplication. The inverse of multiplication is division, not addition.

Raoul started to solve the equation 4y=40 by subtracting 4 from both sides. Explain why Raoul’s method will not solve the equation.

Self Check

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

A self-assessment checklist for math skills, asking users to rate their ability to solve equations with integers and understand properties of equality using options like 'Confidently', 'With some help', or 'No-I don't get it!'.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Introduction to Integers

Locate Positive and Negative Numbers on the Number Line

In the following exercises, locate and label the integer on the number line.

5

Solution


This figure is a number line. It is scaled from negative 10 to 10 in increments of 2. There is a point at 5.

−5

−3

Solution


This figure is a number line. It is scaled from negative 10 to 10 in increments of 2. There is a point at negative 3.

3

−8

Solution


This figure is a number line. It is scaled from negative 10 to 10 in increments of 2. There is a point at negative 8.

−7

Order Positive and Negative Numbers

In the following exercises, order each of the following pairs of numbers, using < or >.

4__8

Solution

<

−6__3

−5__−10

Solution

>

−9__−4

2__−7

Solution

>

−3__1

Find Opposites

In the following exercises, find the opposite of each number.

6

Solution

−6

−2

−4

Solution

4

3

In the following exercises, simplify.

  1. ⓐ −(8)
  2. ⓑ −(−8)
Solution
  1. ⓐ −8
  2. ⓑ 8
  1. ⓐ −(9)
  2. ⓑ −(−9)

In the following exercises, evaluate.

−x,when
  1. ⓐ x=32
  2. ⓑ x=−32
Solution
  1. ⓐ −32
  2. ⓑ 32
−n,when
  1. ⓐ n=20
  2. ⓐ n=−20

Simplify Absolute Values

In the following exercises, simplify.

|−21|

Solution

21

|−42|

|36|

Solution

36

−|15|

|0|

Solution

0

−|−75|

In the following exercises, evaluate.

|x|whenx=−14

Solution

14

−|r|whenr=27

−|−y|wheny=33

Solution

−33

|−n|whenn=−4

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

−|−4|__4

Solution

<

−2__|−2|

−|−6|__−6

Solution

=

−|−9|__|−9|

In the following exercises, simplify.

−(−55)and−|−55|

Solution

55; −55

−(−48)and−|−48|

|12−5|

Solution

7

|9+7|

6|−9|

Solution

54

|14−8|−|−2|

|9−3|−|5−12|

Solution

−1

5+4|15−3|

Translate Phrases to Expressions with Integers

In the following exercises, translate each of the following phrases into expressions with positive or negative numbers.

the opposite of 16

Solution

−16

the opposite of −8

negative 3

Solution

−3

19 minus negative 12

a temperature of 10 below zero

Solution

−10°

an elevation of 85 feet below sea level

Add Integers

Model Addition of Integers

In the following exercises, model the following to find the sum.

3+7

Solution

10

−2+6

5+(−4)

Solution

1

−3+(−6)

Simplify Expressions with Integers

In the following exercises, simplify each expression.

14+82

Solution

96

−33+(−67)

−75+25

Solution

−50

54+(−28)

11+(−15)+3

Solution

−1

−19+(−42)+12

−3+6(−1+5)

Solution

21

10+4(−3+7)

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

n+4when
  1. ⓐ n=−1
  2. ⓑ n=−20
Solution
  1. ⓐ 3
  2. ⓑ −16
x+(−9)when
  1. ⓐ x=3
  2. ⓑ x=−3

(x+y)3whenx=−4,y=1

Solution

−27

(u+v)2whenu=−4,v=11

Translate Word Phrases to Algebraic Expressions

In the following exercises, translate each phrase into an algebraic expression and then simplify.

the sum of −8 and 2

Solution

−8 + 2 = −6

4 more than −12

10 more than the sum of −5 and −6

Solution

10 + [−5 + (−6)] = −1

the sum of3and−5,increased by 18

Add Integers in Applications

In the following exercises, solve.

Temperature On Monday, the high temperature in Denver was −4 degrees. Tuesday’s high temperature was 20 degrees more. What was the high temperature on Tuesday?

Solution

16 degrees

Credit Frida owed $75 on her credit card. Then she charged $21 more. What was her new balance?

Subtract Integers

Model Subtraction of Integers

In the following exercises, model the following.

6−1

Solution


This figure is a row of 6 light pink circles, representing positive counters. The first one is circled.
5

−4−(−3)

2−(−5)

Solution


This figure shows 2 rows. The first row shows 7 light pink circles, representing positive counters. The second row shows 5 dark pink circles, representing negative counters. The entire second row is circled.
7

−1−4

Simplify Expressions with Integers

In the following exercises, simplify each expression.

24−16

Solution

8

19−(−9)

−31−7

Solution

−38

−40−(−11)

−52−(−17)−23

Solution

−58

25−(−3−9)

(1−7)−(3−8)

Solution

−1

32−72

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

x−7when
  1. ⓐ x=5
  2. ⓑ x=−4
Solution
  1. ⓐ −2
  2. ⓑ −11
10−ywhen
  1. ⓐ y=15
  2. ⓑ y=−16

2n2−n+5whenn=−4

Solution

41

−15−3u2whenu=−5

Translate Phrases to Algebraic Expressions

In the following exercises, translate each phrase into an algebraic expression and then simplify.

the difference of −12and5

Solution

−12 − 5 = −17

subtract 23 from −50

Subtract Integers in Applications

In the following exercises, solve the given applications.

Temperature One morning the temperature in Bangor, Maine was 18 degrees. By afternoon, it had dropped 20 degrees. What was the afternoon temperature?

Solution

−2 degrees

Temperature On January 4, the high temperature in Laredo, Texas was 78 degrees, and the high in Houlton, Maine was −28degrees. What was the difference in temperature of Laredo and Houlton?

Multiply and Divide Integers

Multiply Integers

In the following exercises, multiply.

−9⋅4

Solution

−36

5(−7)

(−11)(−11)

Solution

121

−1⋅6

Divide Integers

In the following exercises, divide.

56÷(−8)

Solution

−7

−120÷(−6)

−96÷12

Solution

−8

96÷(−16)

45÷(−1)

Solution

−45

−162÷(−1)

Simplify Expressions with Integers

In the following exercises, simplify each expression.

5(−9)−3(−12)

Solution

−9

(−2)5

−34

Solution

−81

(−3)(4)(−5)(−6)

42−4(6−9)

Solution

54

(8−15)(9−3)

−2(−18)÷9

Solution

4

45÷(−3)−12

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

7x−3whenx=−9

Solution

−66

16−2nwhenn=−8

5a+8bwhena=−2,b=−6

Solution

−58

x2+5x+4whenx=−3

Translate Word Phrases to Algebraic Expressions

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

the product of −12 and 6

Solution

−12(6) = −72

the quotient of 3 and the sum of −7 and s

Solve Equations using Integers; The Division Property of Equality

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each number is a solution of the given equation.

5x−10=−35
  1. ⓐ x=−9
  2. ⓑ x=−5
  3. ⓒ x=5
Solution
  1. ⓐ no
  2. ⓑ yes
  3. ⓒ no
8u+24=−32
  1. ⓐ u=−7
  2. ⓑ u=−1
  3. ⓒ u=7

Using the Addition and Subtraction Properties of Equality

In the following exercises, solve.

a+14=2

Solution

−12

b−9=−15

c+(−10)=−17

Solution

−7

d−(−6)=−26

Model the Division Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters. Then solve it.

This image has two columns. In the first column there are three envelopes. In the second column there are two vertical rows. The first row includes five blue circles, the second row includes four blue circles.
Solution

3x = 9; x = 3

This figure has two columns. In the first column there are  two envelopes. In the second column there are two vertical rows, each includes four blue circles.

Solve Equations Using the Division Property of Equality

In the following exercises, solve each equation using the division property of equality and check the solution.

8p=72

Solution

9

−12q=48

−16r=−64

Solution

4

−5s=−100

Translate to an Equation and Solve.

In the following exercises, translate and solve.

The product of −6 andyis−42

Solution

−6y = −42; y = 7

The difference ofzand −13 is −18.

Four more than m is −48.

Solution

m + 4 = −48; m = −52

The product of −21 andnis 63.

Everyday Math

Describe how you have used two topics from this chapter in your life outside of your math class during the past month.

Solution

Answers will vary.

Chapter Practice Test

Locate and label 0,2,−4, and −1 on a number line.

In the following exercises, compare the numbers, using <or>or=.

  1. ⓐ −6__3
  2. ⓑ −1__−4
Solution
  1. ⓐ <
  2. ⓑ >
  1. ⓐ −5__|−5|
  2. ⓑ −|−2|__−2

In the following exercises, find the opposite of each number.

  1. ⓐ −7
  2. ⓑ 8
Solution
  1. ⓐ 7
  2. ⓑ −8

In the following exercises, simplify.

−(−22)

|4−9|

Solution

5

−8+6

−15+(−12)

Solution

−27

−7−(−3)

10−(5−6)

Solution

11

−3⋅8

−6(−9)

Solution

54

70÷(−7)

(−2)3

Solution

−8

−42

16−3(5−7)

Solution

22

|21−6|−|−8|

In the following exercises, evaluate.

35−awhena=−4

Solution

39

(−2r)2whenr=3

3m−2nwhenm=6,n=−8

Solution

34

−|−y|wheny=17

In the following exercises, translate each phrase into an algebraic expression and then simplify, if possible.

the difference of −7 and −4

Solution

−7 − (−4) = −3

the quotient of 25 and the sum of m and n.

In the following exercises, solve.

Early one morning, the temperature in Syracuse was −8°F. By noon, it had risen 12°. What was the temperature at noon?

Solution

4°F

Collette owed $128 on her credit card. Then she charged $65. What was her new balance?

In the following exercises, solve.

n+6=5

Solution

n = −1

p−11=−4

−9r=−54

Solution

r = 6

In the following exercises, translate and solve.

The product of 15 andxis 75.

Eight less thanyis −32.

Solution

y − 8 = −32; y = −24

Introduction to Fractions

A photo of several bakers at work on a table in a classroom setting.
Bakers combine ingredients to make delicious breads and pastries. (credit: Agustín Ruiz, Flickr)

Often in life, whole amounts are not exactly what we need. A baker must use a little more than a cup of milk or part of a teaspoon of sugar. Similarly a carpenter might need less than a foot of wood and a painter might use part of a gallon of paint. In this chapter, we will learn about numbers that describe parts of a whole. These numbers, called fractions, are very useful both in algebra and in everyday life. You will discover that you are already familiar with many examples of fractions!

Visualize Fractions

Learning Objectives

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

  • Understand the meaning of fractions
  • Model improper fractions and mixed numbers
  • Convert between improper fractions and mixed numbers
  • Model equivalent fractions
  • Find equivalent fractions
  • Locate fractions and mixed numbers on the number line
  • Order fractions and mixed numbers

Before you get started, take this readiness quiz.

Simplify: 5·2+1.
If you missed this problem, review Example 8 in Use the Language of Algebra.

Solution

11

Fill in the blank with < or >:−2__−5
If you missed this problem, review Example 2 in Introduction to Integers.

Solution

−2>−5

Understand the Meaning of Fractions

Andy and Bobby love pizza. On Monday night, they share a pizza equally. How much of the pizza does each one get? Are you thinking that each boy gets half of the pizza? That’s right. There is one whole pizza, evenly divided into two parts, so each boy gets one of the two equal parts.

In math, we write 12 to mean one out of two parts.

An image of a round pizza sliced vertically down the center, creating two equal pieces. Each piece is labeled as one half.

On Tuesday, Andy and Bobby share a pizza with their parents, Fred and Christy, with each person getting an equal amount of the whole pizza. How much of the pizza does each person get? There is one whole pizza, divided evenly into four equal parts. Each person has one of the four equal parts, so each has 14 of the pizza.

An image of a round pizza sliced vertically and horizontally, creating four equal pieces. Each piece is labeled as one fourth.

On Wednesday, the family invites some friends over for a pizza dinner. There are a total of 12 people. If they share the pizza equally, each person would get 112 of the pizza.

An image of a round pizza sliced into twelve equal wedges. Each piece is labeled as one twelfth.

Fractions

A fraction is written ab, where a and b are integers and b≠0. In a fraction, a is called the numerator and b is called the denominator.

A fraction is a way to represent parts of a whole. The denominator b represents the number of equal parts the whole has been divided into, and the numerator a represents how many parts are included. The denominator, b, cannot equal zero because division by zero is undefined.

In Figure 1, the circle has been divided into three parts of equal size. Each part represents 13 of the circle. This type of model is called a fraction circle. Other shapes, such as rectangles, can also be used to model fractions.

A circle is divided into three equal wedges. Each piece is labeled as one third.
Doing the Manipulative Mathematics activity Model Fractions will help you develop a better understanding of fractions, their numerators and denominators.

What does the fraction 23 represent? The fraction 23 means two of three equal parts.

A circle is divided into three equal wedges. Two of the wedges are shaded.

Name the fraction of the shape that is shaded in each of the figures.

In part “a”, a circle is divided into eight equal wedges. Five of the wedges are shaded. In part “b”, a square is divided into nine equal pieces. Two of the pieces are shaded.
Solution

Solution

We need to ask two questions. First, how many equal parts are there? This will be the denominator. Second, of these equal parts, how many are shaded? This will be the numerator.

ⓐ
How many equal parts are there?There are eight equal parts.How many are shaded?Five parts are shaded.

Five out of eight parts are shaded. Therefore, the fraction of the circle that is shaded is 58.

ⓑ
How many equal parts are there?There are nine equal parts.How many are shaded?Two parts are shaded.

Two out of nine parts are shaded. Therefore, the fraction of the square that is shaded is 29.

Name the fraction of the shape that is shaded in each figure:

In part “a”, a circle is divided into eight equal wedges. Three of the wedges are shaded. In part “b”, a square is divided into nine equal pieces. Four of the pieces are shaded.
Solution
  1. ⓐ 38
  2. ⓑ 49

Name the fraction of the shape that is shaded in each figure:

In part “a”, a circle is divided into five equal wedges. Three of the wedges are shaded. In part “b”, a square is divided into four equal pieces. Three of the pieces are shaded.
Solution
  1. ⓐ 35
  2. ⓑ 34

Shade 34 of the circle.

An image of a circle.
Solution

Solution

The denominator is 4, so we divide the circle into four equal parts ⓐ.

The numerator is 3, so we shade three of the four parts ⓑ.

In “a”, a circle is shown divided into four equal pieces. An arrow points from “a” to “b”. In “b”, the same image is shown with three of the pieces shaded.

34 of the circle is shaded.

Shade 68 of the circle.

A circle is divided into eight equal pieces.
Solution


A circle is shown divided into 8 pieces, of which 6 are shaded.

Shade 25 of the rectangle.

A rectangle is divided vertically into five equal pieces.
Solution


A rectangle is divided into 5 sections, of which 2 are shaded.

In Example 1 and Example 2, we used circles and rectangles to model fractions. Fractions can also be modeled as manipulatives called fraction tiles, as shown in Figure 2. Here, the whole is modeled as one long, undivided rectangular tile. Beneath it are tiles of equal length divided into different numbers of equally sized parts.

One long, undivided rectangular tile is shown, labeled “1”. Below it is a rectangular tile of the same size and shape that has been divided vertically into two equal pieces, each labeled as one half. Below that is another rectangular tile that has been divided into three equal pieces, each labeled as one third. Below that is another rectangular tile that has been divided into four equal pieces, each labeled as one fourth. Below that is another rectangular tile that has been divided into six pieces, each labeled as one sixth.

We’ll be using fraction tiles to discover some basic facts about fractions. Refer to Figure 2 to answer the following questions:

This table demonstrates how various fractional parts (e.g., halves, thirds) combine to form a whole, illustrating basic fraction concepts through questions and answers.
How many 12 tiles does it take to make one whole tile? It takes two halves to make a whole, so two out of two is 22=1.
How many 13 tiles does it take to make one whole tile? It takes three thirds, so three out of three is 33=1.
How many 14 tiles does it take to make one whole tile? It takes four fourths, so four out of four is 44=1.
How many 16 tiles does it take to make one whole tile? It takes six sixths, so six out of six is 66=1.
What if the whole were divided into 24 equal parts? (We have not shown fraction tiles to represent this, but try to visualize it in your mind.) How many 124 tiles does it take to make one whole tile? It takes 24 twenty-fourths, so 2424=1.

It takes 24 twenty-fourths, so 2424=1.

This leads us to the Property of One.

Property of One

Any number, except zero, divided by itself is one.

aa=1(a≠0)
Doing the Manipulative Mathematics activity "Fractions Equivalent to One" will help you develop a better understanding of fractions that are equivalent to one

Use fraction circles to make wholes using the following pieces:

  1. ⓐ 4 fourths
  2. ⓑ 5 fifths
  3. ⓒ 6 sixths
Solution

Solution

Three circles are shown. The circle on the left is divided into four equal pieces. The circle in the middle is divided into five equal pieces. The circle on the right is divided into six equal pieces. Each circle says “Form 1 whole” beneath it.

Use fraction circles to make wholes with the following pieces: 3 thirds.

Solution


A circle is shown. It is divided into 3 equal pieces. All 3 pieces are shaded.

Use fraction circles to make wholes with the following pieces: 8 eighths.

Solution


A circle is divided into 8 sections, of which all are shaded.

What if we have more fraction pieces than we need for 1 whole? We’ll look at this in the next example.

Use fraction circles to make wholes using the following pieces:

  1. ⓐ 3 halves
  2. ⓑ 8 fifths
  3. ⓒ 7 thirds
Solution

Solution

ⓐ 3 halves make 1 whole with 1 half left over.

Two circles are shown, both divided into two equal pieces. The circle on the left has both pieces shaded and is labeled as “1”. The circle on the right has one piece shaded and is labeled as one half.

ⓑ 8 fifths make 1 whole with 3 fifths left over.

Two circles are shown, both divided into five equal pieces. The circle on the left has all five pieces shaded and is labeled as “1”. The circle on the right has three pieces shaded and is labeled as three fifths.

ⓒ 7 thirds make 2 wholes with 1 third left over.

Three circles are shown, all divided into three equal pieces. The two circles on the left have all three pieces shaded and are labeled with ones. The circle on the right has one piece shaded and is labeled as one third.

Use fraction circles to make wholes with the following pieces: 5 thirds.

Solution


Two circles are shown. Each is divided into three sections. All of the first circle is shaded. 2 out of 3 sections of the second circle are shaded.

Use fraction circles to make wholes with the following pieces: 5 halves.

Solution


Three circles are shown. Each is divided into two sections. The first two circles are completely shaded. Half of the third circle is shaded.

Model Improper Fractions and Mixed Numbers

In Example 4 (b), you had eight equal fifth pieces. You used five of them to make one whole, and you had three fifths left over. Let us use fraction notation to show what happened. You had eight pieces, each of them one fifth, 15, so altogether you had eight fifths, which we can write as 85. The fraction 85 is one whole, 1, plus three fifths, 35, or 135, which is read as one and three-fifths.

The number 135 is called a mixed number. A mixed number consists of a whole number and a fraction.

Mixed Numbers

A mixed number consists of a whole number a and a fraction bc where c≠0. It is written as follows.

abcc≠0

Fractions such as 54,32,55, and 73 are called improper fractions. In an improper fraction, the numerator is greater than or equal to the denominator, so its value is greater than or equal to one. When a fraction has a numerator that is smaller than the denominator, it is called a proper fraction, and its value is less than one. Fractions such as 12,37, and 1118 are proper fractions.

Proper and Improper Fractions

The fraction ab is a proper fraction if a<b and an improper fraction if a≥b.

Doing the Manipulative Mathematics activity "Model Improper Fractions" and "Mixed Numbers" will help you develop a better understanding of how to convert between improper fractions and mixed numbers.

Name the improper fraction modeled. Then write the improper fraction as a mixed number.

Two circles are shown, both divided into three equal pieces. The circle on the left has all three pieces shaded. The circle on the right has one piece shaded.
Solution

Solution

Each circle is divided into three pieces, so each piece is 13 of the circle. There are four pieces shaded, so there are four thirds or 43. The figure shows that we also have one whole circle and one third, which is 113. So, 43=113.

Name the improper fraction. Then write it as a mixed number.

Two circles are shown, both divided into three equal pieces. The circle on the left has all three pieces shaded. The circle on the right has two pieces shaded.
Solution

53=123

Name the improper fraction. Then write it as a mixed number.

Two circles are shown, both divided into eight equal pieces. The circle on the left has all eight pieces shaded. The circle on the right has five pieces shaded.
Solution

138=158

Draw a figure to model 118.

Solution

Solution

The denominator of the improper fraction is 8. Draw a circle divided into eight pieces and shade all of them. This takes care of eight eighths, but we have 11 eighths. We must shade three of the eight parts of another circle.

Two circles are shown, both divided into eight equal pieces. The circle on the left has all eight pieces shaded and is labeled as eight eighths. The circle on the right has three pieces shaded and is labeled as three eighths. The diagram indicates that eight eighths plus three eighths is one plus three eighths.

So, 118=138.

Draw a figure to model 76.

Solution


Two circles are shown. Each is divided into six sections. All of the first circle is shaded and one section of the second circle is shaded.

Draw a figure to model 65.

Solution


Two circles are shown. Each is divided into five sections. All of the first circle is shaded and one section of the second circle is shaded.

Use a model to rewrite the improper fraction 116 as a mixed number.

Solution

Solution

We start with 11 sixths (116). We know that six sixths makes one whole.

66=1

That leaves us with five more sixths, which is 56(11sixths minus6sixths is5sixths).

So, 116=156.

Two circles are shown, both divided into six equal pieces. The circle on the left has all six pieces shaded and is labeled as six sixths. The circle on the right has five pieces shaded and is labeled as five sixths. Below the circles, it says one plus five sixths, then six sixths plus five sixths equals eleven sixths, and one plus five sixths equals one and five sixths. It then says that eleven sixths equals one and five sixths.

Use a model to rewrite the improper fraction as a mixed number: 97.

Solution

127

Two circles, each divided into eight equal sectors. The left circle is entirely shaded grey, representing 8/8. The right circle has 3 of its 8 sectors shaded grey, representing 3/8.

Use a model to rewrite the improper fraction as a mixed number: 74.

Solution

134

Two circles divided into four equal quadrants. The circle on the left is entirely shaded gray, representing 4/4. The circle on the right has three out of its four quadrants shaded gray, representing 3/4, with one quadrant remaining white.

Use a model to rewrite the mixed number 145 as an improper fraction.

Solution

Solution

The mixed number 145 means one whole plus four fifths. The denominator is 5, so the whole is 55. Together five fifths and four fifths equals nine fifths.

So, 145=95.

Two circles are shown, both divided into five equal pieces. The circle on the left has all five pieces shaded and is labeled as 5 fifths. The circle on the right has four pieces shaded and is labeled as 4 fifths. It then says that 5 fifths plus 4 fifths equals 9 fifths and that 9 fifths is equal to one plus 4 fifths.

Use a model to rewrite the mixed number as an improper fraction: 138.

Solution

118

Two circles divided into eight segments. The left circle is fully shaded gray. The right circle has 3 of its 8 segments shaded gray, with the remaining 5 white.

Use a model to rewrite the mixed number as an improper fraction: 156.

Solution

116

Two circles are shown, each divided into six equal segments. The circle on the left has all six segments shaded gray. The circle on the right has five segments shaded gray and one segment white.

Convert between Improper Fractions and Mixed Numbers

In Example 7, we converted the improper fraction 116 to the mixed number 156 using fraction circles. We did this by grouping six sixths together to make a whole; then we looked to see how many of the 11 pieces were left. We saw that 116 made one whole group of six sixths plus five more sixths, showing that 116=156.

The division expression 116 (which can also be written as 611) tells us to find how many groups of 6 are in 11. To convert an improper fraction to a mixed number without fraction circles, we divide.

Convert 116 to a mixed number.

Solution

Solution

116
Divide the denominator into the numerator. Remember 116 means 11÷6.
A long division example of 11 divided by 6, illustrating the divisor (6), quotient (1), and remainder (5).
Identify the quotient, remainder and divisor.
Write the mixed number as quotientremainderdivisor. 156
So, 116=156

Convert the improper fraction to a mixed number: 137.

Solution

167.

Convert the improper fraction to a mixed number: 149.

Solution

159

Convert an improper fraction to a mixed number.

  1. Divide the denominator into the numerator.
  2. Identify the quotient, remainder, and divisor.
  3. Write the mixed number as quotient remainderdivisor.

Convert the improper fraction 338 to a mixed number.

Solution

Solution

338
Divide the denominator into the numerator. Remember, 338 means 833.
Identify the quotient, remainder, and divisor. An image illustrating long division, showing 33 divided by 8, resulting in a quotient of 4 and a remainder of 1. The terms divisor, quotient, and remainder are labeled with arrows.
Write the mixed number as quotient remainderdivisor. 418
So, 338=418

Convert the improper fraction to a mixed number: 237.

Solution

327

Convert the improper fraction to a mixed number: 4811.

Solution

4411

In Example 8, we changed 145 to an improper fraction by first seeing that the whole is a set of five fifths. So we had five fifths and four more fifths.

55+45=95

Where did the nine come from? There are nine fifths—one whole (five fifths) plus four fifths. Let us use this idea to see how to convert a mixed number to an improper fraction.

Convert the mixed number 423 to an improper fraction.

Solution

Solution

423
Multiply the whole number by the denominator.
The whole number is 4 and the denominator is 3. An incomplete fraction with a numerator of '4 times 3 plus a missing value' and a denominator of 'a missing value'.
Simplify. A mathematical expression showing a fraction with 12 plus an empty box in the numerator, all divided by another empty box in the denominator, indicating missing values or a puzzle.
Add the numerator to the product.
The numerator of the mixed number is 2. A fraction with '12 + 2' in the numerator and an empty square representing an unknown variable in the denominator.
Simplify. A fraction with 14 as the numerator and an empty box as the denominator, representing an incomplete mathematical expression or a problem to be solved.
Write the final sum over the original denominator.
The denominator is 3. 143

Convert the mixed number to an improper fraction: 357.

Solution

267

Convert the mixed number to an improper fraction: 278.

Solution

238

Convert a mixed number to an improper fraction.

  1. Multiply the whole number by the denominator.
  2. Add the numerator to the product found in Step 1.
  3. Write the final sum over the original denominator.

Convert the mixed number 1027 to an improper fraction.

Solution

Solution

1027
Multiply the whole number by the denominator.
The whole number is 10 and the denominator is 7. A fraction with a numerator of '10 times 7 plus an empty square' and a denominator of 'an empty square'.
Simplify. A mathematical expression featuring a fraction with '70 + blank square' in the numerator and a 'blank square' in the denominator, set up like a fill-in-the-blanks problem.
Add the numerator to the product.
The numerator of the mixed number is 2. A mathematical expression showing 70 + 2 as the numerator of a fraction, with an empty square box as the denominator, indicating a missing value.
Simplify. A fraction with 72 in the numerator and a placeholder square in the denominator.
Write the final sum over the original denominator.
The denominator is 7. 727

Convert the mixed number to an improper fraction: 4611.

Solution

5011

Convert the mixed number to an improper fraction: 1113.

Solution

343

Model Equivalent Fractions

Let’s think about Andy and Bobby and their favorite food again. If Andy eats 12 of a pizza and Bobby eats 24 of the pizza, have they eaten the same amount of pizza? In other words, does 12=24? We can use fraction tiles to find out whether Andy and Bobby have eaten equivalent parts of the pizza.

Equivalent Fractions

Equivalent fractions are fractions that have the same value.

Fraction tiles serve as a useful model of equivalent fractions. You may want to use fraction tiles to do the following activity. Or you might make a copy of Figure 2 and extend it to include eighths, tenths, and twelfths.

Start with a 12 tile. How many fourths equal one-half? How many of the 14 tiles exactly cover the 12 tile?

One long, undivided rectangle is shown. Below it is a rectangle divided vertically into two pieces, each labeled as one half. Below that is a rectangle divided vertically into four pieces, each labeled as one fourth.

Since two 14 tiles cover the 12 tile, we see that 24 is the same as 12, or 24=12.

How many of the 16 tiles cover the 12 tile?

One long, undivided rectangle is shown. Below it is a rectangle divided vertically into two pieces, each labeled as one half. Below that is a rectangle divided vertically into six pieces, each labeled as one sixth.

Since three 16 tiles cover the 12 tile, we see that 36 is the same as 12.

So, 36=12. The fractions are equivalent fractions.

Doing the activity "Equivalent Fractions" will help you develop a better understanding of what it means when two fractions are equivalent.

Use fraction tiles to find equivalent fractions. Show your result with a figure.
  1. ⓐ How many eighths equal one-half?
  2. ⓑ How many tenths equal one-half?
  3. ⓒ How many twelfths equal one-half?
Solution

Solution

ⓐ It takes four 18 tiles to exactly cover the 12 tile, so 48=12.

One long, undivided rectangle is shown, labeled 1. Below it is an identical rectangle divided vertically into two pieces, each labeled 1 half. Below that is an identical rectangle divided vertically into eight pieces, each labeled 1 eighth.

ⓑ It takes five 110 tiles to exactly cover the 12 tile, so 510=12.

One long, undivided rectangle is shown. Below it is a rectangle divided vertically into two pieces, each labeled as one half. Below that is a rectangle divided vertically into ten pieces, each labeled as one tenth.

ⓒ It takes six 112 tiles to exactly cover the 12 tile, so 612=12.

One long, undivided rectangle is shown. Below it is a rectangle divided vertically into two pieces, each labeled as one half. Below that is a rectangle divided vertically into twelve pieces, each labeled as one twelfth.

Suppose you had tiles marked 120. How many of them would it take to equal 12? Are you thinking ten tiles? If you are, you’re right, because 1020=12.

We have shown that 12,24,36,48,510,612, and 1020 are all equivalent fractions.

Use fraction tiles to find equivalent fractions: How many eighths equal one-fourth?

Solution

2

Use fraction tiles to find equivalent fractions: How many twelfths equal one-fourth?

Solution

3

Find Equivalent Fractions

We used fraction tiles to show that there are many fractions equivalent to 12. For example, 24,36, and 48 are all equivalent to 12. When we lined up the fraction tiles, it took four of the 18 tiles to make the same length as a 12 tile. This showed that 48=12. See Example 13.

We can show this with pizzas, too. Figure 3(a) shows a single pizza, cut into two equal pieces with 12 shaded. Figure 3(b) shows a second pizza of the same size, cut into eight pieces with 48 shaded.

Two pizzas are shown. The pizza on the left is divided into 2 equal pieces. 1 piece is shaded. The pizza on the right is divided into 8 equal pieces. 4 pieces are shaded.

This is another way to show that 12 is equivalent to 48.

How can we use mathematics to change 12 into 48? How could you take a pizza that is cut into two pieces and cut it into eight pieces? You could cut each of the two larger pieces into four smaller pieces! The whole pizza would then be cut into eight pieces instead of just two. Mathematically, what we’ve described could be written as:

1 times 4 over 2 times 4 is written with the 4s in red. This is set equal to 4 over 8.

These models lead to the Equivalent Fractions Property, which states that if we multiply the numerator and denominator of a fraction by the same number, the value of the fraction does not change.

Equivalent Fractions Property

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

ab=a·cb·c

When working with fractions, it is often necessary to express the same fraction in different forms. To find equivalent forms of a fraction, we can use the Equivalent Fractions Property. For example, consider the fraction one-half.

The top line says that 1 times 3 over 2 times 3 equals 3 over 6, so one half equals 3 sixths. The next line says that 1 times 2 over 2 times 2 equals 2 over 4, so one half equals 2 fourths. The last line says that 1 times 10 over 2 times 10 equals 10 over 20, so one half equals 10 twentieths.

So, we say that 12,24,36, and 1020 are equivalent fractions.

Find three fractions equivalent to 25.

Solution

Solution

To find a fraction equivalent to 25, we multiply the numerator and denominator by the same number (but not zero). Let us multiply them by 2,3, and 5.

On the left, we see that 2 times 2 over 5 times 2 equals 4 over 10. In the middle, we see that 2 times 3 over 5 times 3 equals 6 over 15. On the right, we see that 2 times 5 over 5 times 5 equals 10 over 25.

So, 410,615, and 1025 are equivalent to 25.

Find three fractions equivalent to 35.

Solution

Correct answers include 610,915, and 1220.

Find three fractions equivalent to 45.

Solution

Correct answers include 810,1215, and 1620.

Find a fraction with a denominator of 21 that is equivalent to 27.

Solution

Solution

To find equivalent fractions, we multiply the numerator and denominator by the same number. In this case, we need to multiply the denominator by a number that will result in 21.

Since we can multiply 7 by 3 to get 21, we can find the equivalent fraction by multiplying both the numerator and denominator by 3.

2 over 7 equals 2 time 3 over 7 times 3. The 3s are shown in red. This is set equal to 6 over 21.

Find a fraction with a denominator of 21 that is equivalent to 67.

Solution

1821

Find a fraction with a denominator of 100 that is equivalent to 310.

Solution

30100

Locate Fractions and Mixed Numbers on the Number Line

Now we are ready to plot fractions on a number line. This will help us visualize fractions and understand their values.

Doing the Manipulative Mathematics activity "Number Line Part 3" will help you develop a better understanding of the location of fractions on the number line.

Let us locate 15,45,3,313,74,92,5, and 83 on the number line.

We will start with the whole numbers 3 and 5 because they are the easiest to plot.

A number line is shown with the numbers 3, 4, and 5. There are red dots at 3 and at 5.

The proper fractions listed are 15 and 45. We know proper fractions have values less than one, so 15 and 45 are located between the whole numbers 0 and 1. The denominators are both 5, so we need to divide the segment of the number line between 0 and 1 into five equal parts. We can do this by drawing four equally spaced marks on the number line, which we can then label as 15,25,35, and 45.

Now plot points at 15 and 45.

A number line is shown. It shows 0, 1 fifth, 2 fifths, 3 fifths, 4 fifths, and 1. There are red dots at 1 fifth and at 4 fifths.

The only mixed number to plot is 313. Between what two whole numbers is 313? Remember that a mixed number is a whole number plus a proper fraction, so 313>3. Since it is greater than 3, but not a whole unit greater, 313 is between 3 and 4. We need to divide the portion of the number line between 3 and 4 into three equal pieces (thirds) and plot 313 at the first mark.

A number line is shown with whole number 0 through 5. Between 3 and 4, 3 and 1 third and 3 and 2 thirds are labeled. There is a red dot at 3 and 1 third.

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

74=134,92=412,83=223

Here is the number line with all the points plotted.

A number line is shown with whole numbers 0 through 6. Between 0 and 1, 1 fifth and 4 fifths are labeled and shown with red dots. Between 1 and 2, 7 fourths is labeled and shown with a red dot. Between 2 and 3, 8 thirds is labeled and shown with a red dot. Between 3 and 4, 3 and 1 third is labeled and shown with a red dot. Between 4 and 5, 9 halves is labeled and shown with a red dot.

Locate and label the following on a number line: 34,43,53,415, and 72.

Solution

Solution

Start by locating the proper fraction 34. It is between 0 and 1. To do this, divide the distance between 0 and 1 into four equal parts. Then plot 34.

A number line is shown. It shows 0, 1 fourth, 2 fourths, 3 fourths, and 1. There is a red dot at 3 fourths.

Next, locate the mixed number 415. It is between 4 and 5 on the number line. Divide the number line between 4 and 5 into five equal parts, and then plot 415 one-fifth of the way between 4 and 5.

A number line is shown. It shows 4, 4 and 1 fifth, 4 and 2 fifths, 4 and 3 fifths, 4 and 4 fifths, and 5. There is a red dot at 4 and 1 fifth.

Now locate the improper fractions 43 and 53.

It is easier to plot them if we convert them to mixed numbers first.

43=113,53=123

Divide the distance between 1 and 2 into thirds.

A number line is shown. It shows 1, 1 and 1 third, 1 and 2 thirds, and 2. Below 1 it says 3 thirds. Below 1 and 1 third it says 4 thirds. Below 1 and 2 thirds it says 5 thirds. Below 2 it says 6 thirds. There are red dots at 1 and 1 third and 1 and 2 thirds.

Next let us plot 72. We write it as a mixed number, 72=312. Plot it between 3 and 4.

A number line is shown. It shows 3, 3 and 1 half, and 4. Below 3 it says 6 halves. Below 3 and 1 half it says 7 halves. Below 4 it says 8 halves. There is a red dot at 3 and 1 half.

The number line shows all the numbers located on the number line.

A number line is shown. It shows the whole numbers 0 through 5. Between any 2 numbers are 10 tick marks. Between 0 and 1, between the 7th and 8th tick mark, 3 fourths is labeled and shown with a red dot. Between 1 and 2, 4 thirds and 5 thirds are labeled and shown with red dots. Between 3 and 4, 7 halves is labeled and shown with a red dot. Between 4 and 5, 4 and 1 fifth is labeled and shown with a red dot.

Locate and label the following on a number line: 13,54,74,235,92.

Solution


A number line is shown. It shows the whole numbers 0 through 6. Between 0 and 1, 1 third is labeled and shown with a red dot. Between 1 and 2, 5 fourths and 7 fourths are labeled and shown with red dots. Between 2 and 3, 2 and 3 fifths is labeled and shown with a red dot. Between 4 and 5, 9 halves is labeled and shown with a red dot.

Locate and label the following on a number line: 23,52,94,114,325.

Solution


A number line is shown. It shows the whole numbers 0 through 6. Between 0 and 1, 2 thirds is labeled and shown with a red dot. Between 2 and 3, 9 fourths, 5 halves, and 11 fourths are labeled and shown with red dots. Between 3 and 4, 3 and 2 fifths is labeled and shown with a red dot.

In Introduction to Integers, we defined the opposite of a number. It is the number that is the same distance from zero on the number line but on the opposite side of zero. We saw, for example, that the opposite of 7 is −7 and the opposite of −7 is 7.

A number line is shown. It shows the numbers negative 7, 0 and 7. There are red dots at negative 7 and 7. The space between negative 7 and 0 is labeled as 7 units. The space between 0 and 7 is labeled as 7 units.

Fractions have opposites, too. The opposite of 34 is −34. It is the same distance from 0 on the number line, but on the opposite side of 0.

A number line is shown. It shows the numbers negative 1, negative 3 fourths, 0, 3 fourths, and 1. There are red dots at negative 3 fourths and 3 fourths. The space between negative 3 fourths and 0 is labeled as 3 fourths of a unit. The space between 0 and 3 fourths is labeled as 3 fourths of a unit.

Thinking of negative fractions as the opposite of positive fractions will help us locate them on the number line. To locate −158 on the number line, first think of where 158 is located. It is an improper fraction, so we first convert it to the mixed number 178 and see that it will be between 1 and 2 on the number line. So its opposite, −158, will be between −1 and −2 on the number line.

A number line is shown. It shows the numbers negative 2, negative 1, 0, 1, and 2. Between negative 2 and negative 1, negative 1 and 7 eighths is labeled and marked with a red dot. The distance between negative 1 and 7 eighths and 0 is marked as 15 eighths units. Between 1 and 2, 1 and 7 eighths is labeled and marked with a red dot. The distance between 0 and 1 and 7 eighths is marked as 15 eighths units.

Locate and label the following on the number line: 14,−14,113,−113,52, and −52.

Solution

Solution

Draw a number line. Mark 0 in the middle and then mark several units to the left and right.

To locate 14, divide the interval between 0 and 1 into four equal parts. Each part represents one-quarter of the distance. So plot 14 at the first mark.

A number line is shown. It shows the numbers negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3, and 4. There are 4 tick marks between negative 1 and 0. There are 4 tick marks between 0 and 1. The first tick mark between 0 and 1 is labeled as 1 fourth and marked with a red dot.

To locate −14, divide the interval between 0 and −1 into four equal parts. Plot −14 at the first mark to the left of 0.

A number line is shown. It shows the numbers negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3, and 4. There are 4 tick marks between negative 1 and 0. There are 4 tick marks between 0 and 1. The first tick mark between 0 and 1 is labeled as 1 fourth and marked with a red dot. The first tick mark between 0 and negative 1 is labeled as negative 1 fourth and marked with a red dot.

Since 113 is between 1 and 2, divide the interval between 1 and 2 into three equal parts. Plot 113 at the first mark to the right of 1. Then since −113 is the opposite of 113 it is between −1 and −2. Divide the interval between −1 and −2 into three equal parts. Plot −113 at the first mark to the left of −1.

A number line is shown. The integers from negative 2 to 2 are labeled. Between negative 2 and negative 1, negative 1 and 1 third is labeled and marked with a red dot. Between 1 and 2, 1 and 1 third is labeled and marked with a red dot.

To locate 52 and −52, it may be helpful to rewrite them as the mixed numbers 212 and −212.

Since 212 is between 2 and 3, divide the interval between 2 and 3 into two equal parts. Plot 52 at the mark. Then since −212 is between −2 and −3, divide the interval between −2 and −3 into two equal parts. Plot −52 at the mark.

A number line is shown. The integers from negative 4 to 4 are labeled. Between negative 3 and negative 2, negative 5 halves is labeled and marked with a red dot. Between 2 and 3, 5 halves is labeled and marked with a red dot.

Locate and label each of the given fractions on a number line:

23,−23,214,−214,32,−32

Solution


A number line is shown. The integers from negative 5 to 5 are labeled. Between negative 3 and negative 2, negative 2 and 1 fourth is labeled and marked with a red dot. Between negative 2 and negative 1, negative 3 halves is labeled and marked with a red dot. Between negative 1 and 0, negative 2 thirds is labeled and marked with a red dot. Between 0 and 1, 2 thirds is labeled and marked with a red dot. Between 1 and 2, 3 halves is labeled and marked with a red dot. Between 2 and 3, 2 and 1 fourth is labeled and marked with a red dot.

Locate and label each of the given fractions on a number line:

34,−34,112,−112,73,−73

Solution


A number line is shown. The integers from negative 5 to 5 are labeled. Between negative 3 and negative 2, negative 7 thirds is labeled and marked with a red dot. Between negative 2 and negative 1, negative 1 and 1 half is labeled and marked with a red dot. Between negative 1 and 0, negative 3 fourths is labeled and marked with a red dot. Between 0 and 1, 3 fourths is labeled and marked with a red dot. Between 1 and 2, 1 and 1 half is labeled and marked with a red dot. Between 2 and 3, 7 thirds is labeled and marked with a red dot.

Order Fractions and Mixed Numbers

We can use the inequality symbols to order fractions. Remember that a>b means that a is to the right of b on the number line. As we move from left to right on a number line, the values increase.

Order each of the following pairs of numbers, using < or >:
  1. ⓐ −23____−1
  2. ⓑ −312____−3
  3. ⓒ −37____−38
  4. ⓓ −2____−169
Solution

Solution

ⓐ −23>−1

A number line is shown. The integers from negative 3 to 3 are labeled. Negative 1 is marked with a red dot. Between negative 1 and 0, negative 2 thirds is labeled and marked with a red dot.

ⓑ −312<−3

A number line is shown. The integers from negative 4 to 4 are labeled. There is a red dot at negative 3. Between negative 4 and negative 3, negative 3 and one half is labeled and marked with a red dot.

ⓒ −37<−38

A number line is shown. The numbers negative 3, negative 2, negative 1, 0, 1, 2, and 3 are labeled. Between negative 1 and 0, negative 3 sevenths and negative 3 eighths are labeled and marked with red dots.

ⓓ −2<−169

A number line is shown. The numbers negative 3, negative 2, negative 1, 0, 1, 2, and 3 are labeled. There is a red dot at negative 2. Between negative 2 and negative 1, negative 16 over 9 is labeled and marked with a red dot.

Order each of the following pairs of numbers, using < or >:

  1. ⓐ −13__−1
  2. ⓑ −112__−2
  3. ⓒ −23__−13
  4. ⓓ −3__−73
Solution
  1. ⓐ >
  2. ⓑ >
  3. ⓒ <
  4. ⓓ <

Order each of the following pairs of numbers, using < or >:

  1. ⓐ −3__−175
  2. ⓑ −214__−2
  3. ⓒ −35__−45
  4. ⓓ −4__−103
Solution
  1. ⓐ >
  2. ⓑ <
  3. ⓒ >
  4. ⓓ <

ACCESS ADDITIONAL ONLINE RESOURCES

  • Introduction to Fractions
  • Identify Fractions Using Pattern Blocks

Key Concepts

  • Property of One
    • Any number, except zero, divided by itself is one.
      aa=1, where a≠0.
  • Mixed Numbers
    • A mixed number consists of a whole number a and a fraction bc where c≠0.
    • It is written as follows: abcc≠0
  • Proper and Improper Fractions
    • The fraction ab is a proper fraction if a<b and an improper fraction if a≥b.
  • Convert an improper fraction to a mixed number.
    1. Divide the denominator into the numerator.
    2. Identify the quotient, remainder, and divisor.
    3. Write the mixed number as quotientremainderdivisor.
  • Convert a mixed number to an improper fraction.
    1. Multiply the whole number by the denominator.
    2. Add the numerator to the product found in Step 1.
    3. Write the final sum over the original denominator.
  • Equivalent Fractions Property
    • If a, b, and c are numbers where b≠0, c≠0, then ab=a⋅cb⋅c.

Practice Makes Perfect

In the following exercises, name the fraction of each figure that is shaded.

In part “a”, a circle is divided into 4 equal pieces. 1 piece is shaded. In part “b”, a circle is divided into 4 equal pieces. 3 pieces are shaded. In part “c”, a circle is divided into 8 equal pieces. 3 pieces are shaded. In part “d”, a circle is divided into 8 equal pieces. 5 pieces are shaded.
Solution
  1. ⓐ 14
  2. ⓑ 34
  3. ⓒ 38
  4. ⓓ 58
In part “a”, a circle is divided into 12 equal pieces. 7 pieces are shaded. In part “b”, a circle is divided into 12 equal pieces. 5 pieces are shaded. In part “c”, a square is divided into 9 equal pieces. 4 of the pieces are shaded. In part “d”, a square is divided into 9 equal pieces. 5 pieces are shaded.

In the following exercises, shade parts of circles or squares to model the following fractions.

12

Solution


A circle is shown. It is divided into 2 equal pieces. 1 piece is shaded.

13

34

Solution


A circle is shown. It is divided into 4 equal pieces. 3 pieces are shaded.

25

56

Solution


A circle is shown. It is divided into 6 equal pieces. 5 pieces are shaded.

78

58

Solution


A circle is shown. It is divided into 8 equal pieces. 5 pieces are shaded.

710

In the following exercises, use fraction circles to make wholes using the following pieces.

3 thirds

Solution


A circle is shown. It is divided into 3 equal pieces. All 3 pieces are shaded.

8 eighths

7 sixths

Solution


Two circles are shown. Each is divided into 6 equal pieces. All 6 pieces are shaded in the circle on the left. 1 piece is shaded in the circle on the right.

4 thirds

7 fifths

Solution


Two circles are shown. Each is divided into 5 equal pieces. All 5 pieces are shaded in the circle on the left. 2 pieces are shaded in the circle on the right.

7 fourths

In the following exercises, name the improper fractions. Then write each improper fraction as a mixed number.

In part “a”, two circles are shown. Each is divided into 4 equal pieces. The circle on the left has all 4 pieces shaded. The circle on the right has 1 piece shaded. In part “b”, two circles are shown. Each is divided into 4 equal pieces. The circle on the left has all 4 pieces shaded. The circle on the right has 3 pieces shaded. In part “c”, two circles are shown. Each is divided into 8 equal pieces. The circle on the left has all 8 pieces shaded. The circle on the right has 3 pieces shaded.
Solution
  1. ⓐ 54=114
  2. ⓑ 74=134
  3. ⓒ 118=138
In part “a”, 2 circles are shown. Each is divided into 8 equal pieces. The circle on the left has all 8 pieces shaded. The circle on the right has 1 piece shaded. In part “b”, two squares are shown. Each is divided into 4 equal pieces. The square on the left has all 4 pieces shaded. The circle on the right has 1 piece shaded. In part “c”, two squares are shown. Each is divided into 9 equal pieces. The square on the left has all 9 pieces shaded. The square on the right has 2 pieces shaded.
In part “a”, 3 circles are shown. Each is divided into 4 equal pieces. The first two circles have all 4 pieces shaded. The third circle has 3 pieces shaded. In part “b”, 3 circles are shown. Each is divided into 8 equal pieces. The first two circles have all 8 pieces shaded. The third circle has 3 pieces shaded.
Solution
  1. ⓐ 114=234
  2. ⓑ 198=238

In the following exercises, draw fraction circles to model the given fraction.

33

44

Solution


A circle is shown. It is divided into 4 equal pieces. All 4 pieces are shaded.

74

53

Solution


Two circles are shown. Each is divided into 3 equal pieces. All 3 pieces are shaded in the circle on the left. 2 pieces are shaded in the circle on the right.

116

138

Solution


Two circles are shown. Each is divided into 8 equal pieces. All 8 pieces are shaded in the circle on the left. 5 pieces are shaded in the circle on the right.

103

94

Solution


Three circles are shown. Each is divided into 4 equal pieces. All 4 pieces are shaded in the two circles on the left. 1 piece is shaded in the circle on the right.

In the following exercises, rewrite the improper fraction as a mixed number.

32

53

Solution

123

114

135

Solution

235

256

289

Solution

319

4213

4715

Solution

3215

In the following exercises, rewrite the mixed number as an improper fraction.

123

125

Solution

75

214

256

Solution

176

279

257

Solution

197

347

359

Solution

329

In the following exercises, use fraction tiles or draw a figure to find equivalent fractions.

How many sixths equal one-third?

How many twelfths equal one-third?

Solution

4

How many eighths equal three-fourths?

How many twelfths equal three-fourths?

Solution

9

How many fourths equal three-halves?

How many sixths equal three-halves?

Solution

9

In the following exercises, find three fractions equivalent to the given fraction. Show your work, using figures or algebra.

14

13

Solution

Answers may vary. Correct answers include 26,39,412.

38

56

Solution

Answers may vary. Correct answers include 1012,1518,2024.

27

59

Solution

Answers may vary. Correct answers include 1018,1527,2036.

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

23,54,125

13,74,135

Solution


A number line is shown. The numbers 0, 1, 2, 3, 4, 5, and 6 are labeled. Between 0 and 1, 1 third is labeled and shown with a red dot. Between 1 and 2, 7 fourths is labeled and shown with a red dot. Between 2 and 3, 13 fifths is labeled and shown with a red dot.

14,95,113

710,52,138,3

Solution


A number line is shown. The numbers 0, 1, 2, 3, 4, 5, and 6 are labeled. Between 0 and 1, 7 tenths is labeled and shown with a red dot. Between 1 and 2, 13 eighths is labeled and shown with a red dot. Between 2 and 3, 5 halves is labeled and shown with a red dot. 3 is labeled and shown with a red dot.

213,−213

134,−135

Solution


A number line is shown. The numbers negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3, and 4 are labeled. Between negative 3 and negative 2, negative 2 and 1 third is labeled and shown with a red dot. Between 2 and 3, 2 and 1 third is labeled and shown with a red dot.

34,−34,123,−123,52,−52

25,−25,134,−134,83,−83

Solution


A number line is shown. The numbers negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3, and 4 are labeled. Between negative 3 and negative 2, negative 8 thirds is labeled and shown with a red dot. Between negative 2 and negative 1, negative 1 and 3 fourths is labeled and shown with a red dot. Between negative 1 and 0, negative 2 fifths is labeled and shown with a red dot. Between 0 and 1, 2 fifths is labeled and shown with a red dot. Between 1 and 2, 1 and 3 fourths is labeled and shown with a red dot. Between 2 and 3, 8 thirds is labeled and shown with a red dot.

In the following exercises, order each of the following pairs of numbers, using < or >.

−1__−14

−1__−13

Solution

<

−212__−3

−134__−2

Solution

>

−512__−712

−910__−310

Solution

<

−3__−135

−4__−236

Solution

<

Everyday Math

Music Measures A choreographed dance is broken into counts. A 11 count has one step in a count, a 12 count has two steps in a count and a 13 count has three steps in a count. How many steps would be in a 15 count? What type of count has four steps in it?

Music Measures Fractions are used often in music. In 44 time, there are four quarter notes in one measure.
  1. ⓐ How many measures would eight quarter notes make?
  2. ⓑ The song “Happy Birthday to You” has 25 quarter notes. How many measures are there in “Happy Birthday to You?”
Solution
  1. ⓐ 2
  2. ⓑ 614
Baking Nina is making five pans of fudge to serve after a music recital. For each pan, she needs 12 cup of walnuts.
  1. ⓐ How many cups of walnuts does she need for five pans of fudge?
  2. ⓑ Do you think it is easier to measure this amount when you use an improper fraction or a mixed number? Why?

Writing Exercises

Give an example from your life experience (outside of school) where it was important to understand fractions.

Solution

Answers will vary.

Explain how you locate the improper fraction 214 on a number line on which only the whole numbers from 0 through 10 are marked.

Self Check

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

This table helps students assess their fraction knowledge across topics like understanding, modeling, converting, locating, and ordering fractions, with options for confident, some help, or not getting 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.

equivalent fractions
Equivalent fractions are two or more fractions that have the same value.
fraction
A fraction is written ab. In a fraction, 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.
mixed number
A mixed number consists of a whole number a and a fraction bc where c≠0. It is written as abc, where c≠0.
proper and improper fractions
The fraction ab is proper if a<b and improper if a≥b.

Multiply and Divide Fractions

Learning Objectives

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

  • Simplify fractions
  • Multiply fractions
  • Find reciprocals
  • Divide fractions

Before you get started, take this readiness quiz.

Find the prime factorization of 48.
If you missed this problem, review Example 1 in Prime Factorization and the Least Common Multiple.

Solution

2⋅2⋅2⋅2⋅3

Draw a model of the fraction 34.
If you missed this problem, review Example 2 in Visualize Fractions.

Solution

A circle is shown divided into 4 pieces, of which 3 are shaded.

Find two fractions equivalent to 56.
If you missed this problem, review Example 14 in Visualize Fractions.

Solution

Answers may vary. Acceptable answers include 1012,1518,5060, etc.

Simplify Fractions

In working with equivalent fractions, you saw that there are many ways to write fractions that have the same value, or represent the same part of the whole. How do you know which one to use? Often, we’ll use the fraction that is in simplified form.

A fraction is considered simplified if there are no common factors, other than 1, in the numerator and denominator. If a fraction does have common factors in the numerator and denominator, we can reduce the fraction to its simplified form by removing the common factors.

Simplified Fraction

A fraction is considered simplified if there are no common factors in the 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.

The process of simplifying a fraction is often called reducing the fraction. In the previous section, we used the Equivalent Fractions Property to find equivalent fractions. We can also use the Equivalent Fractions Property in reverse to simplify fractions. We rewrite the property to show both forms together.

Equivalent Fractions Property

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

ab=a·cb·canda·cb·c=ab.

Notice that c is a common factor in the numerator and denominator. Anytime we have a common factor in the numerator and denominator, it can be removed.

Simplify a fraction.

  1. Rewrite the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers.
  2. Simplify, using the equivalent fractions property, by removing common factors.
  3. Multiply any remaining factors.

Simplify: 1015.

Solution

Solution

To simplify the fraction, we look for any common factors in the numerator and the denominator.

Notice that 5 is a factor of both 10 and 15. 1015
Factor the numerator and denominator. A fraction showing (2 multiplied by 5) divided by (3 multiplied by 5), where the common factor '5' is highlighted in red, demonstrating potential simplification.
Remove the common factors. This image illustrates the simplification of a fraction by canceling out a common factor 'x' from both the numerator (2x) and the denominator (3x), resulting in 2/3.
Simplify. 23

Simplify: 812.

Solution

23

Simplify: 1216.

Solution

34

To simplify a negative fraction, we use the same process as in Example 1. Remember to keep the negative sign.

Simplify: −1824.

Solution

Solution

We notice that 18 and 24 both have factors of 6. −1824
Rewrite the numerator and denominator showing the common factor. A mathematical fraction: -(3*6)/(4*6). The '6' in the numerator and denominator is highlighted in red, demonstrating a common factor for simplification.
Remove common factors. Simplifying a fraction by canceling out the common factor 'x' (represented by the crossed-out 6s) from both the numerator and denominator, resulting in -3/4. This illustrates algebraic cancellation.
Simplify. −34

Simplify: −2128.

Solution

−34

Simplify: −1624.

Solution

−23

After simplifying a fraction, it is always important to check the result to make sure that the numerator and denominator do not have any more factors in common. Remember, the definition of a simplified fraction: a fraction is considered simplified if there are no common factors in the numerator and denominator.

When we simplify an improper fraction, there is no need to change it to a mixed number.

Simplify: −5632.

Solution

Solution

−5632
Rewrite the numerator and denominator, showing the common factors, 8. A mathematical fraction with 7 multiplied by 8 in the numerator and 4 multiplied by 8 in the denominator, with the number 8 highlighted in red.
Remove common factors. A visual example of fraction simplification, where the common factor '8' is canceled from both the numerator and denominator, leaving 7/4.
Simplify. −74

Simplify: −5442.

Solution

−97

Simplify: −8145.

Solution

−95

Simplify a fraction.

  1. Rewrite the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers.
  2. Simplify, using the equivalent fractions property, by removing common factors.
  3. Multiply any remaining factors

Sometimes it may not be easy to find common factors of the numerator and denominator. A good idea, then, is to factor the numerator and the denominator into prime numbers. (You may want to use the factor tree method to identify the prime factors.) Then divide out the common factors using the Equivalent Fractions Property.

Simplify: 210385.

Solution

Solution

Use factor trees to factor the numerator and denominator. 210385
Two factor trees demonstrating the prime factorization of 210 (2x3x5x7) and 385 (5x7x11), visually breaking down composite numbers into their prime components.
Rewrite the numerator and denominator as the product of the primes. 210385=2⋅3⋅5⋅75⋅7⋅11
Remove the common factors. Simplification of a mathematical fraction where common factors 5 and 7 are crossed out in both the numerator (2*3*5*7) and denominator (5*7*11).
Simplify. 2⋅311
Multiply any remaining factors. 611

Simplify: 69120.

Solution

2340

Simplify: 120192.

Solution

58

We can also simplify fractions containing variables. If a variable is a common factor in the numerator and denominator, we remove it just as we do with an integer factor.

Simplify: 5xy15x.

Solution

Solution

This table demonstrates the step-by-step simplification of an algebraic rational expression by factoring and canceling common terms.
5xy15x
Rewrite numerator and denominator showing common factors. 5·x·y3·5·x
Remove common factors. 5·x·y3·5·x
Simplify. y3

Simplify: 7x7y.

Solution

xy

Simplify: 9a9b.

Solution

ab

Multiply Fractions

A model may help you understand multiplication of fractions. We will use fraction tiles to model 12·34. To multiply 12 and 34, think 12 of 34.

Start with fraction tiles for three-fourths. To find one-half of three-fourths, we need to divide them into two equal groups. Since we cannot divide the three 14 tiles evenly into two parts, we exchange them for smaller tiles.

A rectangle is divided vertically into three equal pieces. Each piece is labeled as one fourth. There is a an arrow pointing to an identical rectangle divided vertically into six equal pieces. Each piece is labeled as one eighth. There are braces showing that three of these rectangles represent three eighths.

We see 68 is equivalent to 34. Taking half of the six 18 tiles gives us three 18 tiles, which is 38.

Therefore,

12·34=38
Doing the Manipulative Mathematics activity "Model Fraction Multiplication" will help you develop a better understanding of how to multiply fractions.

Use a diagram to model 12·34.

Solution

Solution

First shade in 34 of the rectangle.

A rectangle is shown, divided vertically into four equal pieces. Three of the pieces are shaded.

We will take 12 of this 34, so we heavily shade 12 of the shaded region.

A rectangle is shown, divided vertically into four equal pieces. Three of the pieces are shaded. The rectangle is divided by a horizontal line, creating eight equal pieces. Three of the eight pieces are darkly shaded.

Notice that 3 out of the 8 pieces are heavily shaded. This means that 38 of the rectangle is heavily shaded.

Therefore, 12 of 34 is 38, or 12·34=38.

Use a diagram to model: 12·35.

Solution

310

A rectangle is shown, divided vertically into five equal pieces. Three of the pieces are shaded. The first three pieces of the rectangle are divided by a horizontal line, creating six equal pieces. Three of the six pieces are darkly shaded.

Use a diagram to model: 12·56.

Solution

512

A rectangle is shown, divided vertically into six equal pieces. Five of the pieces are shaded.

Look at the result we got from the model in Example 6. We found that 12·34=38. Do you notice that we could have gotten the same answer by multiplying the numerators and multiplying the denominators?

Step-by-step guide on how to multiply two fractions.
12·34
Multiply the numerators, and multiply the denominators. 12·34
Simplify. 38

This leads to the definition of fraction multiplication. To multiply fractions, we multiply the numerators and multiply the denominators. Then we write the fraction in simplified form.

Fraction Multiplication

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

ab·cd=acbd

Multiply, and write the answer in simplified form: 34·15.

Solution

Solution

Steps and mathematical expressions demonstrating the multiplication of fractions 3/4 and 1/5.
34·15
Multiply the numerators; multiply the denominators. 3·14·5
Simplify. 320

There are no common factors, so the fraction is simplified.

Multiply, and write the answer in simplified form: 13·25.

Solution

215

Multiply, and write the answer in simplified form: 35·78.

Solution

2140

When multiplying fractions, the properties of positive and negative numbers still apply. It is a good idea to determine the sign of the product as the first step. In Example 4.26 we will multiply two negatives, so the product will be positive.

Multiply, and write the answer in simplified form: −58(−23).

Solution

Solution

−58(−23)
The signs are the same, so the product is positive. Multiply the numerators, multiply the denominators. 5⋅28⋅3
Simplify. 1024
Look for common factors in the numerator and denominator. Rewrite showing common factors. The fraction 5x/12x, illustrating the simplification of an algebraic expression by canceling out the common factor 'x' highlighted in red in both the numerator and denominator.
Remove common factors. 512

Another way to find this product involves removing common factors earlier.

−58(−23)
Determine the sign of the product. Multiply. 5⋅28⋅3
Show common factors and then remove them. A mathematical fraction with a numerator of 5 multiplied by 2, and a denominator of 4 multiplied by 2 multiplied by 3. The '2' in both the numerator and denominator is crossed out in red, illustrating cancellation.
Multiply remaining factors. 512

We get the same result.

Multiply, and write the answer in simplified form: −47(−58).

Solution

514

Multiply, and write the answer in simplified form: −712(−89).

Solution

1427

Multiply, and write the answer in simplified form: −1415·2021.

Solution

Solution

−1415·2021
Determine the sign of the product; multiply. −1415·2021
Are there any common factors in the numerator and the denominator?
We know that 7 is a factor of 14 and 21, and 5 is a factor of 20 and 15.
Rewrite showing common factors. A fraction with a negative sign, where common factors 7 and 5 are shown crossed out from the numerator (2*7*4*5) and the denominator (3*5*3*7), illustrating simplification.
Remove the common factors. −2·43·3
Multiply the remaining factors. −89

Multiply, and write the answer in simplified form: −1028·815.

Solution

−421

Multiply, and write the answer in simplified form: −920·512.

Solution

−316

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, 3=31, for example.

Multiply, and write the answer in simplified form:

ⓐ 17·56

ⓑ 125(−20x)

Solution

Solution

Step-by-step calculation demonstrating how to multiply the fraction 1/7 by the whole number 56 to get the result 8.
ⓐ
17·56
Write 56 as a fraction. 17·561
Determine the sign of the product; multiply. 567
Simplify. 8
ⓑ
125(−20x)
Write −20x as a fraction. 125(−20x1)
Determine the sign of the product; multiply. −12·20·x5·1
Show common factors and then remove them. A mathematical expression showing a fraction with 12, 4, and 5x multiplied in the numerator, and 5 and 1 multiplied in the denominator, with the '5's in both numerator and denominator canceled out, preceded by a negative sign.
Multiply remaining factors; simplify. −48x

Multiply, and write the answer in simplified form:

  1. ⓐ 18·72
  2. ⓑ 113(−9a)
Solution
  1. ⓐ 9
  2. ⓑ −33a

Multiply, and write the answer in simplified form:

  1. ⓐ 38·64
  2. ⓑ 16x·1112
Solution
  1. ⓐ 24
  2. ⓑ 44x3

Find Reciprocals

The fractions 23 and 32 are related to each other in a special way. So are −107 and −710. Do you see how? Besides looking like upside-down versions of one another, if we were to multiply these pairs of fractions, the product would be 1.

23·32=1and−107(−710)=1

Such pairs of numbers are called reciprocals.

Reciprocal

The reciprocal of the fraction ab is ba, where a≠0 and b≠0,

A number and its reciprocal have a product of 1.

ab·ba=1

To find the reciprocal of a fraction, we invert the fraction. This means that we place the numerator in the denominator and the denominator in the numerator.

To get a positive result when multiplying two numbers, the numbers must have the same sign. So reciprocals must have the same sign.

“a” over “b” multiplied by “b” over “a” equals positive one.

To find the reciprocal, keep the same sign and invert the fraction. The number zero does not have a reciprocal. Why? A number and its reciprocal multiply to 1. Is there any number r so that 0·r=1? No. So, the number 0 does not have a reciprocal.

Find the reciprocal of each number. Then check that the product of each number and its reciprocal is 1.

  1. ⓐ 49
  2. ⓑ −16
  3. ⓒ −145
  4. ⓓ 7
Solution

Solution

To find the reciprocals, we keep the sign and invert the fractions.

This table provides a step-by-step example of finding the reciprocal of a fraction and verifying the result.
ⓐ
Find the reciprocal of 49. The reciprocal of 49 is 94.
Check:
Multiply the number and its reciprocal. 49⋅94
Multiply numerators and denominators. 3636
Simplify. 1✓
This table illustrates finding the reciprocal of a negative fraction, simplifying the result, and performing a check.
ⓑ
Find the reciprocal of -16. -61
Simplify. -6
Check: -16⋅(-6)
1✓
Demonstrates finding the reciprocal of a fraction and verifying the result.
ⓒ
Find the reciprocal of -145. -514
Check: -145⋅(-514)
7070
1✓
This table demonstrates the step-by-step process of finding the reciprocal of the number 7, including the verification.
ⓓ
Find the reciprocal of 7.
Write 7 as a fraction. 71
Write the reciprocal of 71. 17
Check: 7⋅(17)
1✓

Find the reciprocal:

  1. ⓐ 57
  2. ⓑ −18
  3. ⓒ −114
  4. ⓓ 14
Solution
  1. ⓐ 75
  2. ⓑ −8
  3. ⓒ −411
  4. ⓓ 114

Find the reciprocal:

  1. ⓐ 37
  2. ⓑ −112
  3. ⓒ −149
  4. ⓓ 21
Solution
  1. ⓐ 73
  2. ⓑ −12
  3. ⓒ −914
  4. ⓓ 121

In a previous chapter, we worked with opposites and absolute values. Table 17 compares opposites, absolute values, and reciprocals.

Opposite Absolute Value Reciprocal
has opposite sign is never negative has same sign, fraction inverts

Fill in the chart for each fraction in the left column:

Number Opposite Absolute Value Reciprocal
−38
12
95
−5
Solution

Solution

To find the opposite, change the sign. To find the absolute value, leave the positive numbers the same, but take the opposite of the negative numbers. To find the reciprocal, keep the sign the same and invert the fraction.

Number Opposite Absolute Value Reciprocal
−38 38 38 −83
12 −12 12 2
95 −95 95 59
−5 5 5 −15

Fill in the chart for each number given:

Number Opposite Absolute Value Reciprocal
−58
14
83
−8
Solution


The image is of a table with four columns and 5 rows. The top header row reads: number, opposite, absolute value, and reciprocal. The rest of the cells contain answers to the problem.

Fill in the chart for each number given:

Number Opposite Absolute Value Reciprocal
−47
18
94
−1
Solution


This table demonstrates how to find the opposite, absolute value, and reciprocal for various numbers, including fractions and integers, as fundamental mathematical concepts.

Divide Fractions

Why is 12÷3=4? We previously modeled this with counters. How many groups of 3 counters can be made from a group of 12 counters?

Four red ovals are shown. Inside each oval are three grey circles.

There are 4 groups of 3 counters. In other words, there are four 3s in 12. So, 12÷3=4.

What about dividing fractions? Suppose we want to find the quotient: 12÷16. We need to figure out how many 16s there are in 12. We can use fraction tiles to model this division. We start by lining up the half and sixth fraction tiles as shown in Figure 1. Notice, there are three 16 tiles in 12, so 12÷16=3.

A rectangle is shown, labeled as one half. Below it is an identical rectangle split into three equal pieces, each labeled as one sixth.
Doing the Manipulative Mathematics activity "Model Fraction Division" will help you develop a better understanding of dividing fractions.

Model: 14÷18.

Solution

Solution

We want to determine how many 18s are in 14. Start with one 14 tile. Line up 18 tiles underneath the 14 tile.

A rectangle is shown, labeled one fourth. Below it is an identical rectangle split into two equal pieces, each labeled as one eighth.

There are two 18s in 14.

So, 14÷18=2.

Model: 13÷16.

Solution

2


A rectangle is shown, labeled as one third. Below it is an identical rectangle split into two equal pieces, each labeled as one sixth.

Model: 12÷14.

Solution

2


A rectangle is shown, labeled as one half. Below it is an identical rectangle split into two equal pieces, each labeled as one fourth.

Model: 2÷14.

Solution

Solution

We are trying to determine how many 14s there are in 2. We can model this as shown.

Two rectangles are shown, each labeled as 1. Below it are two identical rectangle, each split into four pieces. Each of the eight pieces is labeled as one fourth.

Because there are eight 14s in 2,2÷14=8.

Model: 2÷13

Solution

6


Two rectangles are shown, each labeled as 1. Below it are two identical rectangle, each split into three pieces. Each of the six pieces is labeled as one third.

Model: 3÷12

Solution

6


Three rectangles are shown, each labeled as 1. Below are three identical rectangles, each split into 2 equal pieces. Each of these six pieces is labeled as one half.

Let’s use money to model 2÷14 in another way. We often read 14 as a ‘quarter’, and we know that a quarter is one-fourth of a dollar as shown in Figure 2. So we can think of 2÷14 as, “How many quarters are there in two dollars?” One dollar is 4 quarters, so 2 dollars would be 8 quarters. So again, 2÷14=8.

A picture of a United States quarter is shown.
The U.S. coin called a quarter is worth one-fourth of a dollar.

Using fraction tiles, we showed that 12÷16=3. Notice that 12·61=3 also. How are 16 and 61 related? They are reciprocals. This leads us to the procedure for fraction division.

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, 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.

Divide, and write the answer in simplified form: 25÷(−37).

Solution

Solution

Illustrates the step-by-step process of dividing fractions, from the initial problem to the final simplified result.
25÷(−37)
Multiply the first fraction by the reciprocal of the second. 25(−73)
Multiply. The product is negative. −1415

Divide, and write the answer in simplified form: 37÷(−23).

Solution

−914

Divide, and write the answer in simplified form: 23÷(−75).

Solution

−1021

Divide, and write the answer in simplified form: 23÷n5.

Solution

Solution

This table illustrates the step-by-step process of dividing two fractions, showing how to convert division into multiplication by a reciprocal.
23÷n5
Multiply the first fraction by the reciprocal of the second. 23·5n
Multiply. 103n

Divide, and write the answer in simplified form: 35÷p7.

Solution

215p

Divide, and write the answer in simplified form: 58÷q3.

Solution

158q

Divide, and write the answer in simplified form: −34÷(−78).

Solution

Solution

Steps to divide two negative fractions, illustrating the procedure from initial expression to simplified result.
−34÷(−78)
Multiply the first fraction by the reciprocal of the second. −34·(−87)
Multiply. Remember to determine the sign first. 3·84·7
Rewrite to show common factors. 3·4·24·7
Remove common factors and simplify. 67

Divide, and write the answer in simplified form: −23÷(−56).

Solution

45

Divide, and write the answer in simplified form: −56÷(−23).

Solution

54

Divide, and write the answer in simplified form: 718÷1427.

Solution

Solution

Step-by-step solution for dividing two fractions, illustrating each stage from the initial problem to the simplified result.
718÷1427
Multiply the first fraction by the reciprocal of the second. 718·2714
Multiply. 7·2718·14
Rewrite showing common factors. A mathematical fraction displays the cancellation of common factors. The terms 7 and 9 are crossed out from both the numerator and denominator, simplifying the expression.
Remove common factors. 32·2
Simplify. 34

Divide, and write the answer in simplified form: 727÷3536.

Solution

415

Divide, and write the answer in simplified form: 514÷1528.

Solution

23

ACCESS ADDITIONAL ONLINE RESOURCES

  • Simplifying Fractions
  • Multiplying Fractions (Positive Only)
  • Multiplying Signed Fractions
  • Dividing Fractions (Positive Only)
  • Dividing Signed Fractions

Key Concepts

  • Equivalent Fractions Property
    • If a, b, c are numbers where b≠0, c≠0, then ab=a⋅cb⋅c and a⋅cb⋅c=ab.
  • Simplify a fraction.
    1. Rewrite the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers.
    2. Simplify, using the equivalent fractions property, by removing common factors.
    3. Multiply any remaining factors.
  • Fraction Multiplication
    • If a, b, c, and d are numbers where b≠0and d≠0, then ab⋅cd=acbd.
  • Reciprocal
    • A number and its reciprocal have a product of 1. ab⋅ba=1
    • Opposite Absolute Value Reciprocal
      has opposite sign is never negative has same sign, fraction inverts
  • 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, multiply the first fraction by the reciprocal of the second.

Practice Makes Perfect

Simplify Fractions

In the following exercises, simplify each fraction. Do not convert any improper fractions to mixed numbers.

721

Solution

13

824

1520

Solution

34

1218

−4088

Solution

−511

−6399

−10863

Solution

−127

−10448

120252

Solution

1021

182294

−168192

Solution

−78

−140224

11x11y

Solution

xy

15a15b

−3x12y

Solution

−x4y

−4x32y

14x221y

Solution

2x23y

24a32b2

Multiply Fractions

In the following exercises, use a diagram to model.

12·23

Solution

13
The image shows green blocks to demonstrate the fraction problem given.

12·58

13·56

Solution

518
The image shows green blocks to demonstrate the fraction problem given.

13·25

In the following exercises, multiply, and write the answer in simplified form.

25·13

Solution

215

12·38

34·910

Solution

2740

45·27

−23(−38)

Solution

14

−34(−49)

−59·310

Solution

−16

−38·415

712(−821)

Solution

−29

512(−815)

(−1415)(920)

Solution

−2150

(−910)(2533)

(−6384)(−4490)

Solution

1130

(−3360)(−4088)

4·511

Solution

2011

5·83

37·21n

Solution

9n

56·30m

−28p(−14)

Solution

7p

−51q(−13)

−8(174)

Solution

−34

145(−15)

−1(−38)

Solution

38

(−1)(−67)

(23)3

Solution

827

(45)2

(65)4

Solution

1296625

(47)4

Find Reciprocals

In the following exercises, find the reciprocal.

34

Solution

43

23

−517

Solution

−175

−619

118

Solution

811

−13

−19

Solution

−119

−1

1

Solution

1

Fill in the chart.

Opposite Absolute Value Reciprocal
−711
45
107
−8
Solution


This table demonstrates how to find the opposite, absolute value, and reciprocal for various numbers, including fractions and integers, as fundamental mathematical concepts.

Fill in the chart.

Opposite Absolute Value Reciprocal
−313
914
157
−9
Solution


A table is shown with four columns and five rows. The first row reads Number, Opposite, Absolute Value, and Reciprocal. The second row reads negative three thirteenths, three thirteenths, three thirteenths, negative thirteen thirds. The third row reads nine fourteenths, negative nine fourteenths, nine fourteenths, and fourteen ninths. The fourth row reads fifteen sevenths, negative fifteen sevenths, fifteen sevenths, and seven fifteenths. The last row reads negative nine, nine, nine, negative one ninth.

Divide Fractions

In the following exercises, model each fraction division.

12÷14

12÷18

Solution

4
The image shows blocks to demonstrate the fraction division problem given.

2÷15

3÷14

Solution

12
The image shows blocks to demonstrate the fraction division problem given.

In the following exercises, divide, and write the answer in simplified form.

12÷14

12÷18

Solution

4

34÷23

45÷34

Solution

1615

−45÷47

−34÷35

Solution

−54

−79÷(−79)

−56÷(−56)

Solution

1

34÷x11

25÷y9

Solution

185y

58÷a10

56÷c15

Solution

252c

518÷(−1524)

718÷(−1427)

Solution

−34

7p12÷21p8

5q12÷15q8

Solution

29

8u15÷12v25

12r25÷18s35

Solution

14r15s

−5÷12

−3÷14

Solution

−12

34÷(−12)

25÷(−10)

Solution

−125

−18÷(−92)

−15÷(−53)

Solution

9

12÷(−34)÷78

112÷78·211

Solution

87

Everyday Math

Baking A recipe for chocolate chip cookies calls for 34 cup brown sugar. Imelda wants to double the recipe.

ⓐ How much brown sugar will Imelda need? Show your calculation. Write your result as an improper fraction and as a mixed number.

ⓑ Measuring cups usually come in sets of 18,14,13,12, and 1 cup. Draw a diagram to show two different ways that Imelda could measure the brown sugar needed to double the recipe.

Baking Nina is making 4 pans of fudge to serve after a music recital. For each pan, she needs 23 cup of condensed milk.

  1. ⓐ How much condensed milk will Nina need? Show your calculation. Write your result as an improper fraction and as a mixed number.
  2. ⓑ Measuring cups usually come in sets of 18,14,13,12, and 1 cup. Draw a diagram to show two different ways that Nina could measure the condensed milk she needs.
Solution
  • ⓐ 4·23 cups=83 cups=223 cups
  • ⓑ Answers will vary.

Portions Don purchased a bulk package of candy that weighs 5 pounds. He wants to sell the candy in little bags that hold 14 pound. How many little bags of candy can he fill from the bulk package?

Portions Kristen has 34 yards of ribbon. She wants to cut it into equal parts to make hair ribbons for her daughter’s 6 dolls. How long will each doll’s hair ribbon be?

Solution

18 yard

Writing Exercises

Explain how you find the reciprocal of a fraction.

Explain how you find the reciprocal of a negative fraction.

Solution

Answers will vary.

Rafael wanted to order half a medium pizza at a restaurant. The waiter told him that a medium pizza could be cut into 6 or 8 slices. Would he prefer 3 out of 6 slices or 4 out of 8 slices? Rafael replied that since he wasn’t very hungry, he would prefer 3 out of 6 slices. Explain what is wrong with Rafael’s reasoning.

Give an example from everyday life that demonstrates how 12·23 is 13.

Solution

Answers will vary.

Self Check

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

A student self-evaluation chart for fractions skills: simplifying, multiplying, reciprocals, and dividing, rated as '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?

reciprocal
The reciprocal of the fraction ab is ba where a≠0 and b≠0.
simplified fraction
A fraction is considered simplified if there are no common factors in the numerator and denominator.

Multiply and Divide Mixed Numbers and Complex Fractions

Learning Objectives

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

  • Multiply and divide mixed numbers
  • Translate phrases to expressions with fractions
  • Simplify complex fractions
  • Simplify expressions written with a fraction bar

Before you get started, take this readiness quiz.

Divide and reduce, if possible: (4+5)÷(10−7).
If you missed this problem, review Example 8 in Add Integers.

Solution

3

Multiply and write the answer in simplified form: 18·23.
If you missed this problem, review Example 7 in Multiply and Divide Fractions.

Solution

112

Convert 235 into an improper fraction.
If you missed this problem, review Example 11 in Visualize Fractions.

Solution

135

Multiply and Divide Mixed Numbers

In the previous section, you learned how to multiply and divide fractions. All of the examples there used either proper or improper fractions. What happens when you are asked to multiply or divide mixed numbers? Remember that we can convert a mixed number to an improper fraction. And you learned how to do that in Visualize Fractions.

Multiply: 313·58

Solution

Solution

Step-by-step guide demonstrating the multiplication of a mixed number by a fraction, from conversion to simplification.
313·58
Convert 313 to an improper fraction. 103·58
Multiply. 10·53·8
Look for common factors. 2̸·5·53·2̸·4
Remove common factors. 5·53·4
Simplify. 2512

Notice that we left the answer as an improper fraction, 2512, and did not convert it to a mixed number. In algebra, it is preferable to write answers as improper fractions instead of mixed numbers. This avoids any possible confusion between 2112 and 2·112.

Multiply, and write your answer in simplified form: 523·617.

Solution

2

Multiply, and write your answer in simplified form: 37·514.

Solution

94

Multiply or divide mixed numbers.

  1. Convert the mixed numbers to improper fractions.
  2. Follow the rules for fraction multiplication or division.
  3. Simplify if possible.

Multiply, and write your answer in simplified form: 245(−178).

Solution

Solution

This table demonstrates the step-by-step process for multiplying mixed numbers, showing each operation and its corresponding mathematical expression.
245(−178)
Convert mixed numbers to improper fractions. 145(−158)
Multiply. − 14·155·8
Look for common factors. − 2̸·7·5̸·35̸·2̸·4
Remove common factors. − 7·34
Simplify. − 214

Multiply, and write your answer in simplified form. 557(−258).

Solution

−15

Multiply, and write your answer in simplified form. −325·416.

Solution

−856

Divide, and write your answer in simplified form: 347÷5.

Solution

Solution

Step-by-step solution for dividing a mixed number by a whole number, illustrating the mathematical transformations.
347÷5
Convert mixed numbers to improper fractions. 257÷51
Multiply the first fraction by the reciprocal of the second. 257·15
Multiply. 25·17·5
Look for common factors. 5̸·5·17·5̸
Remove common factors. 5·17
Simplify. 57

Divide, and write your answer in simplified form: 438÷7.

Solution

58

Divide, and write your answer in simplified form: 258÷3.

Solution

78

Divide: 212÷114.

Solution

Solution

This table illustrates the step-by-step process for dividing mixed numbers, from conversion to simplification.
212÷114
Convert mixed numbers to improper fractions. 52÷54
Multiply the first fraction by the reciprocal of the second. 52·45
Multiply. 5·42·5
Look for common factors. 5̸·2̸·22̸·1·5̸
Remove common factors. 21
Simplify. 2

Divide, and write your answer in simplified form: 223÷113.

Solution

2

Divide, and write your answer in simplified form: 334÷112.

Solution

52

Translate Phrases to Expressions with Fractions

The words quotient and ratio are often used to describe fractions. In Subtract Whole Numbers, we defined quotient as the result of division. The quotient of a and b is the result you get from dividing a by b, or ab. Let’s practice translating some phrases into algebraic expressions using these terms.

Translate the phrase into an algebraic expression: “the quotient of 3x and 8.”

Solution

Solution

The keyword is quotient; it tells us that the operation is division. Look for the words of and and to find the numbers to divide.

The quotientof3xand8.

This tells us that we need to divide 3x by 8. 3x8

Translate the phrase into an algebraic expression: the quotient of 9s and 14.

Solution

9s14

Translate the phrase into an algebraic expression: the quotient of 5y and 6.

Solution

5y6

Translate the phrase into an algebraic expression: the quotient of the difference of m and n, and p.

Solution

Solution

We are looking for the quotient of the difference of m and n, and p. This means we want to divide the difference of m and n by p.

m−np

Translate the phrase into an algebraic expression: the quotient of the difference of a and b, and cd.

Solution

a−bcd

Translate the phrase into an algebraic expression: the quotient of the sum of p and q, and r.

Solution

p+qr

Simplify Complex Fractions

Our work with fractions so far has included proper fractions, improper fractions, and mixed numbers. Another kind of fraction is called complex fraction, which 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. So the complex fraction 3458 can be written as 34÷58.

Simplify: 3458.

Solution

Solution

Illustration of the step-by-step process to divide fractions, exemplified by (3/4) / (5/8).
3458
Rewrite as division. 34÷58
Multiply the first fraction by the reciprocal of the second. 34·85
Multiply. 3·84·5
Look for common factors. 3·4̸·24̸·5
Remove common factors and simplify. 65

Simplify: 2356.

Solution

45

Simplify: 37611.

Solution

1114

Simplify a complex fraction.

  1. Rewrite the complex fraction as a division problem.
  2. Follow the rules for dividing fractions.
  3. Simplify if possible.

Simplify: −673.

Solution

Solution

This table demonstrates the step-by-step process of dividing the fraction -6/7 by 3, simplifying to -2/7.
−673
Rewrite as division. −67÷3
Multiply the first fraction by the reciprocal of the second. −67·13
Multiply; the product will be negative. −6·17·3
Look for common factors. −3̸·2·17·3̸
Remove common factors and simplify. −27

Simplify: −874.

Solution

−27

Simplify: −3910.

Solution

−103

Simplify: x2xy6.

Solution

Solution

Step-by-step simplification of a complex rational expression involving variables.
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. x̸·3·2̸2̸·x̸·y
Remove common factors and simplify. 3y

Simplify: a8ab6.

Solution

34b

Simplify: p2pq8.

Solution

4q

Simplify: 23418.

Solution

Solution

Step-by-step example demonstrating the division of a mixed number by a fraction, including conversion and simplification.
23418
Rewrite as division. 234÷18
Change the mixed number to an improper fraction. 114÷18
Multiply the first fraction by the reciprocal of the second. 114·81
Multiply. 11·84·1
Look for common factors. 11·4̸·24̸·1
Remove common factors and simplify. 22

Simplify: 57125.

Solution

2549.

Simplify: 85315.

Solution

12

Simplify Expressions with a Fraction Bar

Where does the negative sign go in a fraction? Usually, the negative sign is placed in front of the fraction, but you will sometimes see a fraction with a negative numerator or denominator. Remember that fractions represent division. The fraction −13 could be the result of dividing −13, a negative by a positive, or of dividing 1−3, a positive by a negative. When the numerator and denominator have different signs, the quotient is negative.

Negative 1 over positive 3 is equal to negative one third. Negative over positive equals negative. Positive 1 over negative 3 is equal to negative one third. Positive over negative equals negative.

If both the numerator and denominator are negative, then the fraction itself is positive because we are dividing a negative by a negative.

−1−3=13negativenegative=positive

Placement of Negative Sign in a Fraction

For any positive numbers a and b,

−ab=a−b=−ab

Which of the following fractions are equivalent to 7−8?

−7−8,−78,78,−78
Solution

Solution

The quotient of a positive and a negative is a negative, so 7−8 is negative. Of the fractions listed, −78 and −78 are also negative.

Which of the following fractions are equivalent to −35?

−3−5,35,−35,3−5

Solution

−35,3−5

Which of the following fractions are equivalent to −27?

−2−7,−27,27,2−7

Solution

−27,2−7

Fraction bars act as grouping symbols. The expressions above and below the fraction bar should be treated as if they were in parentheses. For example, 4+85−3 means (4+8)÷(5−3). The order of operations tells us to simplify the numerator and the denominator first—as if there were parentheses—before we divide.

We’ll add fraction bars to our set of grouping symbols from Use the Language of Algebra to have a more complete set here.

Grouping Symbols


Parentheses, brackets, braces, an absolute value sign, and a fraction bar are shown.

Simplify an expression with a fraction bar.

  1. Simplify the numerator.
  2. Simplify the denominator.
  3. Simplify the fraction.

Simplify: 4+85−3.

Solution

Solution

Illustrates the step-by-step simplification of a mathematical fraction.
4+85−3
Simplify the expression in the numerator. 125−3
Simplify the expression in the denominator. 122
Simplify the fraction. 6

Simplify: 4+611−2.

Solution

109

Simplify: 3+518−2.

Solution

12

Simplify: 4−2(3)22+2.

Solution

Solution

Step-by-step simplification of a mathematical expression using the order of operations.
4−2(3)22+2
Use the order of operations. Multiply in the numerator and use the exponent in the denominator. 4−64+2
Simplify the numerator and the denominator. −26
Simplify the fraction. -13

Simplify: 6−3(5)32+3.

Solution

−34

Simplify: 4−4(6)33+3.

Solution

−23

Simplify: (8−4)282−42.

Solution

Solution

Step-by-step simplification of a mathematical fraction using the order of operations, demonstrating the transformation from the initial expression to its final reduced form.
(8−4)282−42
Use the order of operations (parentheses first, then exponents). (4)264−16
Simplify the numerator and denominator. 1648
Simplify the fraction. 13

Simplify: (11−7)2112−72.

Solution

29

Simplify: (6+2)262+22.

Solution

85

Simplify: 4(−3)+6(−2)−3(2)−2.

Solution

Solution

This table shows the step-by-step simplification of a complex rational expression, detailing each operation from multiplication to final division.
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

ACCESS ADDITIONAL ONLINE RESOURCES

  • Division Involving Mixed Numbers
  • Evaluate a Complex Fraction

Key Concepts

  • Multiply or divide mixed numbers.
    1. Convert the mixed numbers to improper fractions.
    2. Follow the rules for fraction multiplication or division.
    3. Simplify if possible.
  • Simplify a complex fraction.
    1. Rewrite the complex fraction as a division problem.
    2. Follow the rules for dividing fractions.
    3. Simplify if possible.
  • Placement of negative sign in a fraction.
    • For any positive numbers a and b, -ab=a-b=-ab.
  • Simplify an expression with a fraction bar.
    1. Simplify the numerator.
    2. Simplify the denominator.
    3. Simplify the fraction.

Practice Makes Perfect

Multiply and Divide Mixed Numbers

In the following exercises, multiply and write the answer in simplified form.

438·710

249·67

Solution

4421

1522·335

2536·6310

Solution

358

423(−118)

225(−229)

Solution

−163

−449·51316

−1720·21112

Solution

−6316

In the following exercises, divide, and write your answer in simplified form.

513÷4

1312÷9

Solution

32

−12÷3311

−7÷514

Solution

−43

638÷218

215÷1110

Solution

2

−935÷(−135)

−1834÷(−334)

Solution

5

Translate Phrases to Expressions with Fractions

In the following exercises, translate each English phrase into an algebraic expression.

the quotient of 5u and 11

the quotient of 7v and 13

Solution

7v13

the quotient of p and q

the quotient of a and b

Solution

ab

the quotient of r and the sum of s and 10

the quotient of A and the difference of 3 and B

Solution

A3−B

Simplify Complex Fractions

In the following exercises, simplify the complex fraction.

2389

45815

Solution

32

−8211235

−9163340

Solution

−1522

−452

−9103

Solution

−310

258

5310

Solution

16

m3n2

r5s3

Solution

3r5s

−x6−89

−38−y12

Solution

92y

245110

42316

Solution

28

79−245

38−634

Solution

−118

Simplify Expressions with a Fraction Bar

In the following exercises, identify the equivalent fractions.

Which of the following fractions are equivalent to 5−11?
−5−11,−511,511,−511

Which of the following fractions are equivalent to −49?
−4−9,−49,49,−49

Solution

−49,−49

Which of the following fractions are equivalent to −113?
−113,113,−11−3,11−3

Which of the following fractions are equivalent to −136?
136,13−6,−13−6,−136

Solution

13−6,−136

In the following exercises, simplify.

4+118

9+37

Solution

127

22+310

19−46

Solution

52

4824−15

464+4

Solution

234

−6+68+4

−6+317−8

Solution

−13

22−1419−13

15+918+12

Solution

45

5⋅8−10

3⋅4−24

Solution

−12

4⋅36⋅6

6⋅69⋅2

Solution

2

42−125

72+160

Solution

56

8⋅3+2⋅914+3

9⋅6−4⋅722+3

Solution

2625

15⋅5−522⋅10

12⋅9−323⋅18

Solution

116

5⋅6−3⋅44⋅5−2⋅3

8⋅9−7⋅65⋅6−9⋅2

Solution

52

52−323−5

62−424−6

Solution

−10

2+4(3)−3−22

7+3(5)−2−32

Solution

−2

7⋅4−2(8−5)9⋅3−3⋅5

9⋅7−3(12−8)8⋅7−6⋅6

Solution

5120

9(8−2)−3(15−7)6(7−1)−3(17−9)

8(9−2)−4(14−9)7(8−3)−3(16−9)

Solution

187

Everyday Math

Baking A recipe for chocolate chip cookies calls for 214 cups of flour. Graciela wants to double the recipe.

  1. ⓐ How much flour will Graciela need? Show your calculation. Write your result as an improper fraction and as a mixed number.
  2. ⓑ Measuring cups usually come in sets with cups for 18,14,13,12, and 1 cup. Draw a diagram to show two different ways that Graciela could measure out the flour needed to double the recipe.

Baking A booth at the county fair sells fudge by the pound. Their award winning “Chocolate Overdose” fudge contains 223 cups of chocolate chips per pound.

  1. ⓐ How many cups of chocolate chips are in a half-pound of the fudge?
  2. ⓑ The owners of the booth make the fudge in 10-pound batches. How many chocolate chips do they need to make a 10-pound batch? Write your results as improper fractions and as a mixed numbers.
Solution
  1. ⓐ 43=113 cups
  2. ⓑ 803=2623 cups

Writing Exercises

Explain how to find the reciprocal of a mixed number.

Explain how to multiply mixed numbers.

Solution

Answers will vary.

Randy thinks that 312·514 is 1518. Explain what is wrong with Randy’s thinking.

Explain why −12,−12, and 1−2 are equivalent.

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 understanding of fraction skills, including multiplying mixed numbers, translating phrases to expressions, and simplifying complex fractions.

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

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

Add and Subtract Fractions with Common Denominators

Learning Objectives

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

  • Model fraction addition
  • Add fractions with a common denominator
  • Model fraction subtraction
  • Subtract fractions with a common denominator

Before you get started, take this readiness quiz.

Simplify: 2x+9+3x−4.
If you missed this problem, review Example 10 in Evaluate, Simplify, and Translate Expressions.

Solution

5x+5

Draw a model of the fraction 34.
If you missed this problem, review Example 2 in Visualize Fractions.

Solution

A circle is shown divided into 4 pieces, of which 3 are shaded.

Simplify: 3+26.
If you missed this problem, review Example 12 in Multiply and Divide Mixed Numbers and Complex Fractions.

Solution

56

Model Fraction Addition

How many quarters are pictured? One quarter plus 2 quarters equals 3 quarters.

Three U.S. quarters are shown. One is shown on the left, and two are shown on the right.

Remember, quarters are really fractions of a dollar. Quarters are another way to say fourths. So the picture of the coins shows that

142434one quarter+two quarters=three quarters

Let’s use fraction circles to model the same example, 14+24.

Start with one 14 piece. A light salmon-colored quarter circle with a red outline is positioned on the left side of a white background, suggesting a geometric shape or part of a larger circle. The mathematical fraction 1/4, representing one-fourth, is clearly displayed on a white background.
Add two more 14pieces. A light orange semi-circle, divided in half vertically, with a small black plus sign to its left, rests above a thin horizontal grey line against a white background. A mathematical expression featuring a plus sign followed by the fraction two over four, with a blank line or underscore beneath it, indicating an operation or a problem to be solved.
The result is 34. A circle is divided into four equal sections. Three sections are shaded in a light orange-red color, illustrating three-quarters of the circle being filled, with one section remaining white. A white background with the fraction 3/4 written in black text on the right side.

So again, we see that

14+24=34
Doing the Manipulative Mathematics activity "Model Fraction Addition" will help you develop a better understanding of adding fractions

Use a model to find the sum 38+28.

Solution

Solution

Start with three 18 pieces. A quarter-circle shape, filled with a pale orange color and outlined in red, is sectioned into three distinct segments by two lines extending from the central point of the arc. The image displays the fraction 3/8, written vertically with the numeral 3 above a horizontal line and the numeral 8 below it, centered on a white background.
Add two 18pieces. An abstract image featuring a light orange, fan-like shape divided vertically, with a plus sign to its left and a horizontal line underneath, suggesting a mathematical or symbolic representation. A mathematical expression shows a plus sign followed by the fraction two-eighths (+ 2/8) with a horizontal line underneath, likely indicating part of a larger sum or equation.
How many 18pieces are there? A circle is divided into eight equal segments, with five of these segments shaded in a light orange color and three segments unshaded. The fraction 5/8 is displayed against a plain white background.

There are five 18 pieces, or five-eighths. The model shows that 38+28=58.

Use a model to find each sum. Show a diagram to illustrate your model.

18+48

Solution

58
A circle divided into 8 sections, 5 of which are shaded.

Use a model to find each sum. Show a diagram to illustrate your model.

16+46

Solution

56
A circle divided into 6 sections, 5 of which are shaded.

Add Fractions with a Common Denominator

Example 1 shows that to add the same-size pieces—meaning that the fractions have the same denominator—we just add the number of pieces.

Fraction Addition

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

ac+bc=a+bc

To add fractions with a common denominator, add the numerators and place the sum over the common denominator.

Find the sum: 35+15.

Solution

Solution

Step-by-step guide for adding fractions with a common denominator.
35+15
Add the numerators and place the sum over the common denominator. 3+15
Simplify. 45

Find each sum: 36+26.

Solution

56

Find each sum: 310+710.

Solution

1

Find the sum: x3+23.

Solution

Solution

Illustrates adding fractions with a common denominator, showing the instructional step and the resulting expression.
x3+23
Add the numerators and place the sum over the common denominator. x+23

Note that we cannot simplify this fraction any more. Since x and 2 are not like terms, we cannot combine them.

Find the sum: x4+34.

Solution

x+34

Find the sum: y8+58.

Solution

y+58

Find the sum: −9d+3d.

Solution

Solution

We will begin by rewriting the first fraction with the negative sign in the numerator.

−ab=−ab

Illustrates the step-by-step simplification of -9/d + 3/d, showing the process of adding fractions with a common denominator.
−9d+3d
Rewrite the first fraction with the negative in the numerator. −9d+3d
Add the numerators and place the sum over the common denominator. −9+3d
Simplify the numerator. −6d
Rewrite with negative sign in front of the fraction. −6d

Find the sum: −7d+8d.

Solution

1d

Find the sum: −6m+9m.

Solution

3m

Find the sum: 2n11+5n11.

Solution

Solution

This table illustrates the step-by-step simplification of an algebraic expression involving fractions with a common denominator.
2n11+5n11
Add the numerators and place the sum over the common denominator. 2n+5n11
Combine like terms. 7n11

Find the sum: 3p8+6p8.

Solution

9p8

Find the sum: 2q5+7q5.

Solution

9q5

Find the sum: −312+(−512).

Solution

Solution

Steps to add and simplify two negative fractions with a common denominator.
−312+(−512)
Add the numerators and place the sum over the common denominator. −3+(−5)12
Add. −812
Simplify the fraction. −23

Find each sum: −415+(−615).

Solution

−23

Find each sum: −521+(−921).

Solution

−23

Model Fraction Subtraction

Subtracting two fractions with common denominators is much like adding fractions. Think of a pizza that was cut into 12 slices. Suppose five pieces are eaten for dinner. This means that, after dinner, there are seven pieces (or 712 of the pizza) left in the box. If Leonardo eats 2 of these remaining pieces (or 212 of the pizza), how much is left? There would be 5 pieces left (or 512 of the pizza).

712−212=512

Let’s use fraction circles to model the same example, 712−212.

Start with seven 112 pieces. Take away two 112 pieces. How many twelfths are left?

The bottom reads 7 twelfths minus 2 twelfths equals 5 twelfths. Above 7 twelfths, there is a circle divided into 12 equal pieces, with 7 pieces shaded in orange. Above 2 twelfths, the same circle is shown, but 2 of the 7 pieces are shaded in grey. Above 5 twelfths, the 2 grey pieces are no longer shaded, so there is a circle divided into 12 pieces with 5 of the pieces shaded in orange.

Again, we have five twelfths, 512.

Doing the Manipulative Mathematics activity "Model Fraction Subtraction" will help you develop a better understanding of subtracting fractions.

Use fraction circles to find the difference: 45−15.

Solution

Solution

Start with four 15 pieces. Take away one 15 piece. Count how many fifths are left. There are three 15 pieces left.

The bottom reads 4 fifths minus 1 fifth equals 3 fifths. Above 4 fifths, there is a circle divided into 5 equal pieces, with 4 pieces shaded in orange. Above 1 fifth, the same circle is shown, but 1 of the 4 shaded pieces is shaded in grey. Above 3 fifths, the 1 grey piece is no longer shaded, so there is a circle divided into 5 pieces with 3 of the pieces shaded in orange.

Use a model to find each difference. Show a diagram to illustrate your model.

78−48

Solution

38, models may differ.

Use a model to find each difference. Show a diagram to illustrate your model.

56−46

Solution

16, models may differ

Subtract Fractions with a Common Denominator

We subtract fractions with a common denominator in much the same way as we add fractions with a common denominator.

Fraction Subtraction

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

ac−bc=a−bc

To subtract fractions with a common denominator, we subtract the numerators and place the difference over the common denominator.

Find the difference: 2324−1424.

Solution

Solution

Steps for subtracting fractions with a common denominator and simplifying the result.
2324−1424
Subtract the numerators and place the difference over the common denominator. 23−1424
Simplify the numerator. 924
Simplify the fraction by removing common factors. 38

Find the difference: 1928−728.

Solution

37

Find the difference: 2732−1132.

Solution

12

Find the difference: y6−16.

Solution

Solution

Simplification of a fractional expression by subtracting numerators over a common denominator.
y6−16
Subtract the numerators and place the difference over the common denominator. y−16

The fraction is simplified because we cannot combine the terms in the numerator.

Find the difference: x7−27.

Solution

x−27

Find the difference: y14−1314.

Solution

y−1314

Find the difference: −10x−4x.

Solution

Solution

Remember, the fraction −10x can be written as −10x.

Steps demonstrating the subtraction and simplification of an algebraic expression involving fractions with a common denominator.
−10x−4x
Subtract the numerators. −10−4x
Simplify. −14x
Rewrite with the negative sign in front of the fraction. −14x

Find the difference: −9x−7x.

Solution

−16x

Find the difference: −17a−5a.

Solution

−22a

Now lets do an example that involves both addition and subtraction.

Simplify: 38+(−58)−18.

Solution

Solution

Step-by-step solution for adding and subtracting fractions with common denominators.
38+(−58)−18
Combine the numerators over the common denominator. 3+(−5)−18
Simplify the numerator, working left to right. −2−18
Subtract the terms in the numerator. −38
Rewrite with the negative sign in front of the fraction. −38

Simplify: 25+(−45)−35.

Solution

−1

Simplify: 59+(−49)−79.

Solution

−23

ACCESS ADDITIONAL ONLINE RESOURCES

  • Adding Fractions With Pattern Blocks
  • Adding Fractions With Like Denominators
  • Subtracting Fractions With Like Denominators

Key Concepts

  • Fraction Addition
    • If a, b, and c are numbers where c≠0, then ac+bc=a+bc.
    • To add fractions, add the numerators and place the sum over the common denominator.
  • Fraction Subtraction
    • If a, b, and c are numbers where c≠0, then ac-bc=a-bc.
    • To subtract fractions, subtract the numerators and place the difference over the common denominator.

Practice Makes Perfect

Model Fraction Addition

In the following exercises, use a model to add the fractions. Show a diagram to illustrate your model.

25+15

310+410

Solution


A circle is divided into 10 equal pieces. 7 of the pieces are shaded.
710

16+36

38+38

Solution


A circle is divided into 8 equal pieces. 6 of the pieces are shaded.
34

Add Fractions with a Common Denominator

In the following exercises, find each sum.

49+19

29+59

Solution

79

613+713

915+715

Solution

1615

x4+34

y3+23

Solution

y+23

7p+9p

8q+6q

Solution

14q

8b9+3b9

5a7+4a7

Solution

9a7

−12y8+3y8

−11x5+7x5

Solution

−4x5

−18+(−38)

−18+(−58)

Solution

−34

−316+(−716)

−516+(−916)

Solution

−78

−817+1517

−919+1719

Solution

819

613+(−1013)+(−1213)

512+(−712)+(−1112)

Solution

−1312

Model Fraction Subtraction

In the following exercises, use a model to subtract the fractions. Show a diagram to illustrate your model.

58−28

56−26

Solution


A circle is divided into eight sections, three of which are shaded.
12

Subtract Fractions with a Common Denominator

In the following exercises, find the difference.

45−15

45−35

Solution

15

1115−715

913−413

Solution

513

1112−512

712−512

Solution

16

421−1921

−89−169

Solution

−83

y17−917

x19−819

Solution

x−819

5y8−78

11z13−813

Solution

11z−813

−8d−3d

−7c−7c

Solution

−14c

−23u−15u

−29v−26v

Solution

−55v

6c7−5c7

12d11−9d11

Solution

3d11

−4r13−5r13

−7s3−7s3

Solution

−14s3

−35−(−45)

−37−(−57)

Solution

27

−79−(−59)

−811−(−511)

Solution

−311

Mixed Practice

In the following exercises, perform the indicated operation and write your answers in simplified form.

−518·910

−314·712

Solution

−18

n5−45

611−s11

Solution

6−s11

−724+224

−518+118

Solution

−29

815÷125

712÷928

Solution

4927

Everyday Math

Trail Mix Jacob is mixing together nuts and raisins to make trail mix. He has 610 of a pound of nuts and 310 of a pound of raisins. How much trail mix can he make?

Baking Janet needs 58 of a cup of flour for a recipe she is making. She only has 38 of a cup of flour and will ask to borrow the rest from her next-door neighbor. How much flour does she have to borrow?

Solution

14 cup

Writing Exercises

Greg dropped his case of drill bits and three of the bits fell out. The case has slots for the drill bits, and the slots are arranged in order from smallest to largest. Greg needs to put the bits that fell out back in the case in the empty slots. Where do the three bits go? Explain how you know.
Bits in case: 116, 18, ___, ___, 516, 38, ___, 12, 916, 58.
Bits that fell out: 716, 316, 14.

After a party, Lupe has 512 of a cheese pizza, 412 of a pepperoni pizza, and 412 of a veggie pizza left. Will all the slices fit into 1 pizza box? Explain your reasoning.

Solution

No, adding up the number of pieces gives 1312, which is greater than 1. (Answers may vary.)

Self Check

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

Fraction skills self-assessment table. Users rate their ability to model and add/subtract fractions with common denominators: '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?

Add and Subtract Fractions with Different Denominators

Learning Objectives

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

  • Find the least common denominator (LCD)
  • Convert fractions to equivalent fractions with the LCD
  • Add and subtract fractions with different denominators
  • Identify and use fraction operations
  • Use the order of operations to simplify complex fractions
  • Evaluate variable expressions with fractions

Before you get started, take this readiness quiz.

Find two fractions equivalent to 56.
If you missed this problem, review Example 14 in Visualize Fractions.

Solution

1012,1518

Simplify: 1+5·322+4.
If you missed this problem, review Example 12 in Multiply and Divide Mixed Numbers and Complex Fractions.

Solution

2

Find the Least Common Denominator

In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators?

Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals 25 cents and one dime equals 10 cents, so the sum is 35 cents. See Figure 1.

A quarter and a dime are shown. Below them, it reads 25 cents plus 10 cents. Below that, it reads 35 cents.
Together, a quarter and a dime are worth 35 cents, or 35100 of a dollar.

Similarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is 100. Since there are 100 cents in one dollar, 25 cents is 25100 and 10 cents is 10100. So we add 25100+10100 to get 35100, which is 35 cents.

You have practiced adding and subtracting fractions with common denominators. Now let’s see what you need to do with fractions that have different denominators.

First, we will use fraction tiles to model finding the common denominator of 12 and 13.

We’ll start with one 12 tile and 13 tile. We want to find a common fraction tile that we can use to match both 12 and 13 exactly.

If we try the 14 pieces, 2 of them exactly match the 12 piece, but they do not exactly match the 13 piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into two pieces, each labeled 1 fourth. Underneath the second rectangle are two pieces, each labeled 1 fourth. These rectangles together are longer than the rectangle labeled as 1 third.

If we try the 15 pieces, they do not exactly cover the 12 piece or the 13 piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into three pieces, each labeled 1 sixth. Underneath the second rectangle is an equally sized rectangle split vertically into 2 pieces, each labeled 1 sixth.

If we try the 16 pieces, we see that exactly 3 of them cover the 12 piece, and exactly 2 of them cover the 13 piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle are three smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 half rectangle. Below the 1 third rectangle are two smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 third rectangle.

If we were to try the 112 pieces, they would also work.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into 6 pieces, each labeled 1 twelfth. Underneath the second rectangle is an equally sized rectangle split vertically into 4 pieces, each labeled 1 twelfth.

Even smaller tiles, such as 124 and 148, would also exactly cover the 12 piece and the 13 piece.

The denominator of the largest piece that covers both fractions is the least common denominator (LCD) of the two fractions. So, the least common denominator of 12 and 13 is 6.

Notice that all of the tiles that cover 12 and 13 have something in common: Their denominators are common multiples of 2 and 3, the denominators of 12 and 13. The least common multiple (LCM) of the denominators is 6, and so we say that 6 is the least common denominator (LCD) of the fractions 12 and 13.

Doing the Manipulative Mathematics activity "Finding the Least Common Denominator" will help you develop a better understanding of the LCD.

Least Common Denominator

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

To find the LCD of two fractions, we will find the LCM of their denominators. We follow the procedure we used earlier to find the LCM of two numbers. We only use the denominators of the fractions, not the numerators, when finding the LCD.

Find the LCD for the fractions 712 and 518.

Solution

Solution

Factor each denominator into its primes. Two factor trees illustrate prime factorization. One tree shows 12 factoring into 3, 2, 2. The other shows 18 factoring into 3, 2, 3. The circled numbers are prime factors.
List the primes of 12 and the primes of 18 lining them up in columns when possible. Prime factorization of 12 and 18, showing 12 = 2 x 2 x 3 and 18 = 2 x 3 x 3, a setup for finding the greatest common factor or least common multiple.
Bring down the columns. This image illustrates how to calculate the Least Common Multiple (LCM) of 12 and 18 using their prime factorization. It shows 12 = 2  as 2   * 2   * 3 and 18 = 2   * 3   * 3. The LCM is derived by taking all prime factors with their highest powers from either number, resulting in LCM = 2   * 2   * 3   * 3.
Multiply the factors. The product is the LCM. LCM=36
The LCM of 12 and 18 is 36, so the LCD of 712 and 518 is 36. LCD of 712 and 518 is 36.

Find the least common denominator for the fractions: 712 and 1115.

Solution

60

Find the least common denominator for the fractions: 1315 and 175.

Solution

15

To find the LCD of two fractions, find the LCM of their denominators. Notice how the steps shown below are similar to the steps we took to find the LCM.

Find the least common denominator (LCD) of two fractions.

  1. Factor each denominator into its primes.
  2. List the primes, matching primes in columns when possible.
  3. Bring down the columns.
  4. Multiply the factors. The product is the LCM of the denominators.
  5. The LCM of the denominators is the LCD of the fractions.

Find the least common denominator for the fractions 815 and 1124.

Solution

Solution

To find the LCD, we find the LCM of the denominators.

Find the LCM of 15 and 24.

The top line shows 15 equals 3 times 5. The next line shows 24 equals 2 times 2 times 2 times 3. The 3s are lined up vertically. The next line shows LCM equals 2 times 2 times 2 times 3 times 5. The last line shows LCM equals 120.

The LCM of 15 and 24 is 120. So, the LCD of 815 and 1124 is 120.

Find the least common denominator for the fractions: 1324 and 1732.

Solution

96

Find the least common denominator for the fractions: 928 and 2132.

Solution

224

Convert Fractions to Equivalent Fractions with the LCD

Earlier, we used fraction tiles to see that the LCD of 14 when 16 is 12. We saw that three 112 pieces exactly covered 14 and two 112 pieces exactly covered 16, so

14=312 and 16=212.
On the left is a rectangle labeled 1 fourth. Below it is an identical rectangle split vertically into 3 equal pieces, each labeled 1 twelfth. On the right is a rectangle labeled 1 sixth. Below it is an identical rectangle split vertically into 2 equal pieces, each labeled 1 twelfth.

We say that 14 and 312 are equivalent fractions and also that 16 and 212 are equivalent fractions.

We can use the Equivalent Fractions Property to algebraically change a fraction to an equivalent one. Remember, two fractions are equivalent if they have the same value. The Equivalent Fractions Property is repeated below for reference.

Equivalent Fractions Property

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

ab=a·cb·canda·cb·c=ab

To add or subtract fractions with different denominators, we will first have to convert each fraction to an equivalent fraction with the LCD. Let’s see how to change 14 and 16 to equivalent fractions with denominator 12 without using models.

Convert 14 and 16 to equivalent fractions with denominator 12, their LCD.

Solution

Solution

Find the LCD. The LCD of 14 and 16 is 12.
Find the number to multiply 4 to get 12. A multiplication equation displays 4 multiplied by 3 (highlighted in red), equaling 12. This is a basic arithmetic problem, demonstrating a fundamental concept in mathematics.
Find the number to multiply 6 to get 12. A mathematical equation shows '6  2 = 12' on a white background, with the number 2 highlighted in red. The dot symbol signifies multiplication, indicating that six multiplied by two equals twelve.
Use the Equivalent Fractions Property to convert each fraction to an equivalent fraction with the LCD, multiplying both the numerator and denominator of each fraction by the same number. Fractions 1/4 and 1/6 are shown being converted to equivalent fractions by multiplying their numerators and denominators by 3 and 2, respectively, to find a common denominator.
Simplify the numerators and denominators. Two fractions, 3/12 and 2/12, are displayed on a white background.

We do not reduce the resulting fractions. If we did, we would get back to our original fractions and lose the common denominator.

Change to equivalent fractions with the LCD:

34 and 56, LCD =12

Solution

912,1012

Change to equivalent fractions with the LCD:

−712 and 1115, LCD =60

Solution

−3560,4460

Convert two fractions to equivalent fractions with their LCD as the common denominator.

  1. Find the LCD.
  2. For each fraction, determine the number needed to multiply the denominator to get the LCD.
  3. Use the Equivalent Fractions Property to multiply both the numerator and denominator by the number you found in Step 2.
  4. Simplify the numerator and denominator.

Convert 815 and 1124 to equivalent fractions with denominator 120, their LCD.

Solution

Solution

The LCD is 120. We will start at Step 2.
Find the number that must multiply 15 to get 120. A mathematical equation showing 15 multiplied by 8 equals 120. The number '8' is highlighted in red, indicating it might be a specific element or variable in a larger problem.
Find the number that must multiply 24 to get 120. The mathematical equation '24 * 5 = 120' is displayed on a white background, with the number '5' highlighted in red.
Use the Equivalent Fractions Property. Fractions illustrating that multiplying both the numerator and denominator by the same non-zero number (highlighted in red) results in an equivalent fraction.
Simplify the numerators and denominators. Two fractions, 64 over 120 and 55 over 120, are displayed on a white background.

Change to equivalent fractions with the LCD:

1324 and 1732, LCD 96

Solution

5296,5196

Change to equivalent fractions with the LCD:

928 and 2732, LCD 224

Solution

72224,189224

Add and Subtract Fractions with Different Denominators

Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators.

Add or subtract fractions with different denominators.

  1. Find the LCD.
  2. Convert each fraction to an equivalent form with the LCD as the denominator.
  3. Add or subtract the fractions.
  4. Write the result in simplified form.

Add: 12+13.

Solution

Solution

12+13
Find the LCD of 2, 3.
Calculating the Least Common Denominator (LCD) for 2 and 3, showing the product 2x3 equals 6.
Change into equivalent fractions with the LCD 6. A math problem showing the addition of 1/2 and 1/3 by finding a common denominator, represented as (1*3)/(2*3) + (1*2)/(3*2).
Simplify the numerators and denominators. 36+26
Add. 56

Remember, always check to see if the answer can be simplified. Since 5 and 6 have no common factors, the fraction 56 cannot be reduced.

Add: 14+13.

Solution

712

Add: 12+15.

Solution

710

Subtract: 12−(−14).

Solution

Solution

12−(−14)
Find the LCD of 2 and 4.
The image demonstrates finding the Least Common Denominator (LCD) for 2 and 4, detailing their prime factorizations (2 and 2*2) and concluding that the LCD is 4.
Rewrite as equivalent fractions using the LCD 4. A mathematical expression featuring a fraction (1 times 2, over 2 times 2, with the '2's highlighted in red) subtracted by a negative fraction (-1/4).
Simplify the first fraction. 24−(−14)
Subtract. 2−(−1)4
Simplify. 34

One of the fractions already had the least common denominator, so we only had to convert the other fraction.

Simplify: 12−(−18).

Solution

58

Simplify: 13−(−16).

Solution

12

Add: 712+518.

Solution

Solution

712+518
Find the LCD of 12 and 18.
Calculating the Least Common Denominator (LCD) of 12 and 18 using prime factorization, resulting in 36.
Rewrite as equivalent fractions with the LCD. An arithmetic expression showing the addition of two fractions, 7/12 and 5/18, where each fraction is multiplied by a factor (3/3 and 2/2 respectively) to achieve a common denominator for summation.
Simplify the numerators and denominators. 2136+1036
Add. 3136

Because 31 is a prime number, it has no factors in common with 36. The answer is simplified.

Add: 712+1115.

Solution

7960

Add: 1315+1720.

Solution

10360

When we use the Equivalent Fractions Property, there is a quick way to find the number you need to multiply by to get the LCD. Write the factors of the denominators and the LCD just as you did to find the LCD. The “missing” factors of each denominator are the numbers you need.

The first line says 12 equals 2 times 2 times 3. There is a blank space next to the 3. The next line says 18 equals 2 times 3 times 3. There is a blank space between the 2 and the first 3. There are red lines drawn from the blank spaces. This is labeled as missing factors. There is a horizontal line. Below the line, it says LCD equals 2 times 2 times 3 times 3. Below this, it says LCD equals 36.

The LCD, 36, has 2 factors of 2 and 2 factors of 3.

Twelve has two factors of 2, but only one of 3—so it is ‘missing‘ one 3. We multiplied the numerator and denominator of 712 by 3 to get an equivalent fraction with denominator 36.

Eighteen is missing one factor of 2—so you multiply the numerator and denominator 518 by 2 to get an equivalent fraction with denominator 36. We will apply this method as we subtract the fractions in the next example.

Subtract: 715−1924.

Solution

Solution

715−1924
Find the LCD.
Calculation of the Least Common Denominator (LCD) for 15 and 24 using prime factorization, demonstrating how 15 = 3x5 and 24 = 2x2x2x3 lead to an LCD of 2x2x2x3x5 = 120.
15 is 'missing' three factors of 2
24 is 'missing' a factor of 5
Rewrite as equivalent fractions with the LCD. A math problem displaying the subtraction of two fractions, (7 * 8) / (15 * 8) and (19 * 5) / (24 * 5). Common factors 8 and 5 in the numerators and denominators are highlighted in red, suggesting simplification.
Simplify each numerator and denominator. 56120−95120
Subtract. −39120
Rewrite showing the common factor of 3. −13·340·3
Remove the common factor to simplify. −1340

Subtract: 1324−1732.

Solution

196

Subtract: 2132−928.

Solution

75224

Add: −1130+2342.

Solution

Solution

−1130+2342
Find the LCD.
The image illustrates the calculation of the Least Common Denominator (LCD) for 30 and 42 using prime factorization, showing 30 = 2*3*5, 42 = 2*3*7, and the resulting LCD = 2*3*5*7 = 210.
Rewrite as equivalent fractions with the LCD. Math expression for adding fractions: -11/30 and 23/42. Each fraction's numerator and denominator are multiplied by a distinct factor (7 or 5, in red) to find a common denominator.
Simplify each numerator and denominator. −77210+115210
Add. 38210
Rewrite showing the common factor of 2. 19·2105·2
Remove the common factor to simplify. 19105

Add: −1342+1735.

Solution

37210

Add: −1924+1732.

Solution

−2596

In the next example, one of the fractions has a variable in its numerator. We follow the same steps as when both numerators are numbers.

Add: 35+x8.

Solution

Solution

The fractions have different denominators.

35+x8
Find the LCD.
A step-by-step calculation showing how to find the Least Common Denominator (LCD) of 5 and 8. The prime factorization of each number is displayed, leading to an LCD of 2 x 2 x 2 x 5, which equals 40.
Rewrite as equivalent fractions with the LCD. A mathematical expression featuring two fractions being added. The first fraction is (3 times 8) divided by (5 times 8). The second fraction is (x times 5) divided by (8 times 5).
Simplify the numerators and denominators. 2440+5x40
Add. 24+5x40

We cannot add 24 and 5x since they are not like terms, so we cannot simplify the expression any further.

Add: y6+79.

Solution

3y+1418

Add: x6+715.

Solution

5x+1430

Identify and Use Fraction Operations

By now in this chapter, you have practiced multiplying, dividing, adding, and subtracting fractions. The following table summarizes these four fraction operations. Remember: You need a common denominator to add or subtract fractions, but not to multiply or divide fractions

Summary of Fraction Operations

Fraction multiplication: Multiply the numerators and multiply the denominators.

ab·cd=acbd

Fraction division: Multiply the first fraction by the reciprocal of the second.

ab÷cd=ab·dc

Fraction addition: Add the numerators and place the sum over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.

ac+bc=a+bc

Fraction subtraction: Subtract the numerators and place the difference over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.

ac−bc=a−bc
Simplify:
  1. ⓐ −14+16
  2. ⓑ −14÷16
Solution

Solution

First we ask ourselves, “What is the operation?”

ⓐ The operation is addition.

Do the fractions have a common denominator? No.

−14+16
Find the LCD.
Calculation of the Least Common Denominator (LCD) for numbers 4 and 6, showing prime factorization to arrive at an LCD of 12.
Rewrite each fraction as an equivalent fraction with the LCD. A mathematical expression showing the addition of two fractions: - (1 * 3) / (4 * 3) + (1 * 2) / (6 * 2). The numbers 3 and 2, used to adjust fractions to a common denominator, are highlighted in red.
Simplify the numerators and denominators. −312+212
Add the numerators and place the sum over the common denominator. −112
Check to see if the answer can be simplified. It cannot.

ⓑ The operation is division. We do not need a common denominator.

Step-by-step process for dividing fractions, showing the conversion to multiplication by the reciprocal and subsequent simplification.
−14÷16
To divide fractions, multiply the first fraction by the reciprocal of the second. −14·61
Multiply. −64
Simplify. −32

Simplify each expression:

  1. ⓐ −34−16
  2. ⓑ −34·16
Solution
  1. ⓐ −1112
  2. ⓑ −18

Simplify each expression:

  1. ⓐ 56÷(−14)
  2. ⓑ 56−(−14)
Solution
  1. ⓐ −103
  2. ⓑ 1312
Simplify:
  1. ⓐ 5x6−310
  2. ⓑ 5x6·310
Solution

Solution

ⓐ The operation is subtraction. The fractions do not have a common denominator.

This table illustrates the step-by-step process of subtracting algebraic fractions by finding a common denominator.
5x6−310
Rewrite each fraction as an equivalent fraction with the LCD, 30. 5x·56·5−3·310·3
25x30−930
Subtract the numerators and place the difference over the common denominator. 25x−930

ⓑ The operation is multiplication; no need for a common denominator.

Step-by-step example demonstrating the multiplication and simplification of algebraic fractions.
5x6·310
To multiply fractions, multiply the numerators and multiply the denominators. 5x·36·10
Rewrite, showing common factors. 5·x·32·3·2·5
Remove common factors to simplify. x4
Simplify:
  1. ⓐ (27a−32)36
  2. ⓑ 2a3
Solution
  1. ⓐ (27a−32)36
  2. ⓑ 2a3
Simplify:
  1. ⓐ (24k+25)30
  2. ⓑ 24k5
Solution
  1. ⓐ (24k+25)30
  2. ⓑ 24k5

Use the Order of Operations to Simplify Complex Fractions

In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. We simplified complex fractions by rewriting them as division problems. For example,

3458=34÷58

Now we will look at complex fractions in which the numerator or denominator can be simplified. To follow the order of operations, we simplify the numerator and denominator separately first. Then we divide the numerator by the denominator.

Simplify complex fractions.

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

Simplify: (12)24+32.

Solution

Solution

Step-by-step simplification of the expression (1/2)^2 / (4 + 3^2), detailing each mathematical transformation to reach the final value of 1/52.
(12)24+32
Simplify the numerator. 144+32
Simplify the term with the exponent in the denominator. 144+9
Add the terms in the denominator. 1413
Divide the numerator by the denominator. 14÷13
Rewrite as multiplication by the reciprocal. 14·113
Multiply. 152

Simplify: (13)223+2.

Solution

190

Simplify: 1+42(14)2.

Solution

272

Simplify: 12+2334−16.

Solution

Solution

Step-by-step solution demonstrating the simplification of a complex fraction, detailing each operation and the resulting mathematical expression.
12+2334−16
Rewrite numerator with the LCD of 6 and denominator with LCD of 12. 36+46912−212
Add in the numerator. Subtract in the denominator. 76712
Divide the numerator by the denominator. 76÷712
Rewrite as multiplication by the reciprocal. 76·127
Rewrite, showing 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 also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.

Evaluate x+13 when
  1. ⓐ x=−13
  2. ⓑ x=−34.
Solution

Solution

ⓐ To evaluate x+13 when x=−13, substitute −13 for x in the expression.

x+13
The image displays the instruction 'Substitute -1/3 for x.' in a dark teal font, with the fraction '-1/3' highlighted in red, against a white background. The mathematical expression for negative one-third plus one-third, illustrating additive inverses that sum to zero.
Simplify. 0

ⓑ To evaluate x+13 when x=−34, we substitute −34 for x in the expression.

x+13
The text 'Substitute -3/4 for x.' is shown in a math problem, indicating an instruction to replace the variable 'x' with the fractional value -3/4. A mathematical expression showing the sum of negative three-fourths and one-third.
Rewrite as equivalent fractions with the LCD, 12. −3·34·3+1·43·4
Simplify the numerators and denominators. −912+412
Add. −512
Evaluate: x+34 when
  1. ⓐ x=−74
  2. ⓑ x=−54
Solution
  1. ⓐ −1
  2. ⓑ −12
Evaluate: y+12 when
  1. ⓐ y=23
  2. ⓑ y=−34
Solution
  1. ⓐ 76
  2. ⓑ −14

Evaluate y−56 when y=−23.

Solution

Solution

We substitute −23 for y in the expression.

y−56
The text 'Substitute -2/3 for y.' is displayed, instructing to replace the variable y with the fraction -2/3. A mathematical expression showing the subtraction of two fractions: negative two-thirds minus five-sixths. The first fraction's numerator and denominator are colored red.
Rewrite as equivalent fractions with the LCD, 6. −46−56
Subtract. −96
Simplify. −32

Evaluate: y−12 when y=−14.

Solution

−34

Evaluate: x−38 when x=−52.

Solution

−238

Evaluate 2x2y when x=14 and y=−23.

Solution

Solution

Substitute the values into the expression. In 2x2y, the exponent applies only to x.

The mathematical expression 2x^2y is shown in black text on a white background.
Substitute 1/4 for x and -2/3 for y. The image displays the mathematical expression 2(1/4)^2(-2/3), showing a multiplication of a whole number, a fraction squared, and a negative fraction. The numbers 1/4 are red, and -2/3 are blue.
Simplify exponents first. A mathematical expression showing the product of three terms: 2, the fraction 1/16, and the negative fraction -2/3. The terms are enclosed in parentheses, indicating multiplication.
Multiply. The product will be negative. A mathematical expression showing the multiplication of three fractions: negative two over one, one over sixteen, and two over three.
Simplify. A mathematical expression showing the fraction -4/48, presented in black text on a white background. The negative sign precedes the fraction bar, with '4' as the numerator and '48' as the denominator.
Remove the common factors. A mathematical expression showing a fraction with a negative sign, where the common factor '4' in the numerator and denominator has been crossed out as part of simplification, resulting in -(1*4)/(4*12).
Simplify. A mathematical expression displaying the fraction minus 1 over 12, written as -1/12.

Evaluate. 3ab2 when a=−23 and b=−12.

Solution

−12

Evaluate. 4c3d when c=−12 and d=−43.

Solution

23

Evaluate p+qr when p=−4,q=−2, and r=8.

Solution

Solution

We substitute the values into the expression and simplify.

p+qr
The image displays the text 'Substitute -4 for p, -2 for q and 8 for r.' against a white background. A mathematical expression showing the sum of negative four and negative two, all divided by eight.
Add in the numerator first. −68
Simplify. −34

Evaluate: a+bc when a=−8,b=−7, and c=6.

Solution

−52

Evaluate: x+yz when x=9,y=−18, and z=−6.

Solution

32

Key Concepts

  • Find the least common denominator (LCD) of two fractions.
    1. Factor each denominator into its primes.
    2. List the primes, matching primes in columns when possible.
    3. Bring down the columns.
    4. Multiply the factors. The product is the LCM of the denominators.
    5. The LCM of the denominators is the LCD of the fractions.
  • Equivalent Fractions Property
    • If a, b, and c are whole numbers where b≠0, c≠0 then
      ab = a⋅cb⋅c and a⋅cb⋅c=ab
  • Convert two fractions to equivalent fractions with their LCD as the common denominator.
    1. Find the LCD.
    2. For each fraction, determine the number needed to multiply the denominator to get the LCD.
    3. Use the Equivalent Fractions Property to multiply the numerator and denominator by the number from Step 2.
    4. Simplify the numerator and denominator.
  • Add or subtract fractions with different denominators.
    1. Find the LCD.
    2. Convert each fraction to an equivalent form with the LCD as the denominator.
    3. Add or subtract the fractions.
    4. Write the result in simplified form.
  • Summary of Fraction Operations
    • Fraction multiplication: Multiply the numerators and multiply the denominators.
      ab⋅cd=acbd
    • Fraction division: Multiply the first fraction by the reciprocal of the second.
      ab+cd=ab⋅dc
    • Fraction addition: Add the numerators and place the sum over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.
      ac+bc=a+bc
    • Fraction subtraction: Subtract the numerators and place the difference over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.
      ac-bc=a-bc
  • Simplify complex fractions.
    1. Simplify the numerator.
    2. Simplify the denominator.
    3. Divide the numerator by the denominator.
    4. Simplify if possible.

Practice Makes Perfect

Find the Least Common Denominator (LCD)

In the following exercises, find the least common denominator (LCD) for each set of fractions.

23 and 34

34 and 25

Solution

20

712 and 58

916 and 712

Solution

48

1330 and 2542

2330 and 548

Solution

240

2135 and 3956

1835 and 3349

Solution

245

23,16, and 34

23,14, and 35

Solution

60

Convert Fractions to Equivalent Fractions with the LCD

In the following exercises, convert to equivalent fractions using the LCD.

13 and 14, LCD =12

14 and 15, LCD =20

Solution

520,420

512 and 78, LCD =24

712 and 58, LCD =24

Solution

1424,1524

1316 and -1112, LCD =48

1116 and -512, LCD =48

Solution

3348,−2048

13,56, and 34, LCD =12

13,34, and 35, LCD =60

Solution

2060,4560,3660

Add and Subtract Fractions with Different Denominators

In the following exercises, add or subtract. Write the result in simplified form.

13+15

14+15

Solution

920

12+17

13+18

Solution

1124

13−(−19)

14−(−18)

Solution

38

15−(−110)

12−(−16)

Solution

23

23+34

34+25

Solution

2320

712+58

512+38

Solution

1924

712−916

716−512

Solution

148

1112−38

58−712

Solution

124

23−38

56−34

Solution

112

−1130+2740

−920+1730

Solution

760

−1330+2542

−2330+548

Solution

−5380

−3956−2235

−3349−1835

Solution

−291245

−23−(−34)

−34−(−45)

Solution

120

−916−(−45)

−720−(−58)

Solution

1140

1+78

1+56

Solution

116

1−59

1−310

Solution

710

x3+14

y2+23

Solution

3y+46

y4−35

x5−14

Solution

4x−520

Identify and Use Fraction Operations

In the following exercises, perform the indicated operations. Write your answers in simplified form.

  1. ⓐ 34+16
  2. ⓑ 34÷16
  1. ⓐ 23+16
  2. ⓑ 23÷16
Solution
  1. ⓐ 56
  2. ⓑ 4
  1. ⓐ -25−18
  2. ⓑ -25·18
  1. ⓐ -45−18
  2. ⓑ -45·18
Solution
  1. ⓐ −3740
  2. ⓑ −110
  1. ⓐ 5n6÷815
  2. ⓑ 5n6−815
  1. ⓐ 3a8÷712
  2. ⓑ 3a8−712
Solution
  1. ⓐ 9a14
  2. ⓑ 9a−1424
  1. ⓐ 910·(−11d12)
  2. ⓑ 910+(−11d12)
  1. ⓐ 415·(−5q9)
  2. ⓑ 415+(−5q9)
Solution
  1. ⓐ −4q27
  2. ⓑ 12−25q45

−38÷(−310)

−512÷(−59)

Solution

34

−38+512

−18+712

Solution

1124

56−19

59−16

Solution

718

38·(−1021)

712·(−835)

Solution

−215

−715−y4

−38−x11

Solution

−33−8x88

1112a·9a16

10y13·815y

Solution

1639

Use the Order of Operations to Simplify Complex Fractions

In the following exercises, simplify.

(15)22+32

(13)25+22

Solution

181

23+42(23)2

33−32(34)2

Solution

32

(35)2(37)2

(34)2(58)2

Solution

3625

213+15

514+13

Solution

607

23+1234−23

34+1256−23

Solution

152

78−2312+38

34−3514+25

Solution

313

Mixed Practice

In the following exercises, simplify.

12+23·512

13+25·34

Solution

1930

1−35÷110

1−56÷112

Solution

−9

23+16+34

23+14+35

Solution

9160

38−16+34

25+58−34

Solution

1140

12(920−415)

8(1516−56)

Solution

56

58+161924

16+3101430

Solution

1

(59+16)÷(23−12)

(34+16)÷(58−13)

Solution

227

In the following exercises, evaluate the given expression. Express your answers in simplified form, using improper fractions if necessary.

x+12 when
  1. ⓐ x=−18
  2. ⓑ x=−12
x+23 when
  1. ⓐ x=−16
  2. ⓑ x=−53
Solution
  1. ⓐ 12
  2. ⓑ −1
x+(−56) when
  1. ⓐ x=13
  2. ⓑ x=−16
x+(−1112) when
  1. ⓐ x=1112
  2. ⓑ x=34
Solution
  1. ⓐ 0
  2. ⓑ −16
x−25 when
  1. ⓐ x=35
  2. ⓑ x=−35
x−13 when
  1. ⓐ x=23
  2. ⓑ x=−23
Solution
  1. ⓐ 13
  2. ⓑ −1
710−w when
  1. ⓐ w=12
  2. ⓑ w=−12
512−w when
  1. ⓐ w=14
  2. ⓑ w=−14
Solution
  1. ⓐ 16
  2. ⓑ 23

4p2q when p=−12 and q=59

5m2n when m=−25 and n=13

Solution

415

2x2y3 when x=−23 and y=−12

8u2v3 when u=−34 and v=−12

Solution

−916

u+vw when u=−4,v=−8,w=2

m+np when m=−6,n=−2,p=4

Solution

−2

a+ba−b when a=−3,b=8

r−sr+s when r=10,s=−5

Solution

3

Everyday Math

Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs 316 yard of print fabric and 38 yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover?

Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs 114 cups of sugar for the chocolate chip cookies, and 118 cups for the oatmeal cookies How much sugar does she need altogether?

Solution

She needs 238 cups

Writing Exercises

Explain why it is necessary to have a common denominator to add or subtract fractions.

Explain how to find the LCD of two fractions.

Solution

Answers will vary.

Self Check

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

An empty self-evaluation chart for math students to gauge their understanding of fraction concepts, from basic addition to complex expressions, marked as '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?

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

Add and Subtract Mixed Numbers

Learning Objectives

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

  • Model addition of mixed numbers with a common denominator
  • Add mixed numbers with a common denominator
  • Model subtraction of mixed numbers
  • Subtract mixed numbers with a common denominator
  • Add and subtract mixed numbers with different denominators

Before you get started, take this readiness quiz.

Draw a model of the fraction 73.
If you missed this problem, review Example 6 in Visualize Fractions.

Solution

Three circles are shown. They are divided into 3 equal pieces. All 3 pieces are shaded on 2 of the circles and 1 piece is shaded on the third circle.

Change 114 to a mixed number.
If you missed this problem, review Example 9 in Visualize Fractions.

Solution

234

Change 312 to an improper fraction.
If you missed this problem, review Example 11 in Visualize Fractions.

Solution

72

Model Addition of Mixed Numbers with a Common Denominator

So far, we’ve added and subtracted proper and improper fractions, but not mixed numbers. Let’s begin by thinking about addition of mixed numbers using money.

If Ron has 1 dollar and 1 quarter, he has 114 dollars.

If Don has 2 dollars and 1 quarter, he has 214 dollars.

What if Ron and Don put their money together? They would have 3 dollars and 2 quarters. They add the dollars and add the quarters. This makes 324 dollars. Because two quarters is half a dollar, they would have 3 and a half dollars, or 312 dollars.

114+214________324=312

When you added the dollars and then added the quarters, you were adding the whole numbers and then adding the fractions.

114+214

We can use fraction circles to model this same example:

114+214
Start with 114. one whole and one 14 pieces An orange circle is positioned next to an orange quarter-circle, both outlined with a slightly darker hue on a white background. A numerical representation of the mixed fraction one and one-quarter, often written as 1 1/4, is displayed on a plain white background.
Add 214 more. two wholes and one 14 pieces An image displaying a plus sign, a horizontal line, two complete orange circles, and a single orange quarter circle, suggesting a mathematical operation or a conceptual representation of quantities. A mathematical expression showing the mixed number 2 1/4 preceded by a plus sign, positioned above a horizontal line, likely part of an addition problem.
The sum is: three wholes and two 14's Three light orange circles and a bisected orange semicircle, all outlined in red, arranged horizontally on a white background. The image displays a mathematical equation showing the mixed number 3 and 2/4, which simplifies to 3 and 1/2. The expression reads: three and two-fourths equals three and one-half.
Doing the Manipulative Mathematics activity "Model Mixed Number Addition/Subtraction" will help you develop a better understanding of adding and subtracting mixed numbers.

Model 213+123 and give the sum.

Solution

Solution

We will use fraction circles, whole circles for the whole numbers and 13 pieces for the fractions.

two wholes and one 13 Two reddish-orange circles and a quarter-circle shape are depicted on a white background. The image displays the mixed number 2 1/3, representing two whole units and one-third of a unit. This common mathematical notation combines an integer with a proper fraction.
plus one whole and two 13s A diagram shows a horizontal line, a plus sign, a whole peach-colored circle, and another peach-colored circle with one-third of its section removed, displaying two connected sectors. A mathematical expression showing '+ 1 and 2/3' with a line underneath, implying a calculation or sum is to be performed.
sum is three wholes and three 13s Four light red circles are shown in a row. The first three circles are solid and undivided, while the fourth circle is segmented into three equal parts, with lines meeting at its center. A mathematical equation displays '3 3/3 = 4' on a white background, demonstrating that the mixed number simplifies to 3 + 1, which equals 4.

This is the same as 4 wholes. So, 213+123=4.

Use a model to add the following. Draw a picture to illustrate your model.

125+335

Solution

5
The figure shows five circles. The last one is divided into five equal sections.

Use a model to add the following. Draw a picture to illustrate your model.

216+256

Solution

5
The figure shows five circles. The last one is divided into six equal sections.

Model 135+235 and give the sum as a mixed number.

Solution

Solution

We will use fraction circles, whole circles for the whole numbers and 15 pieces for the fractions.

one whole and three 15s A solid orange circle is positioned next to a segmented orange shape depicting three-quarters of a circle, with one section slightly detached. A mathematical expression displaying the mixed number 1 and 3/5, set against a plain white background.
plus two wholes and three 15s. Visualizing 'two and two-thirds' (2 2/3) with two full circles and one circle showing two out of three segments, positioned above a horizontal line next to a plus sign. A mathematical expression showing '+ 2 3/5' above an underscore, suggesting an addition problem or a step in arithmetic.
sum is three wholes and six 15s Three whole circles, followed by a circle divided into five equal parts, with one part detached, illustrating the concept of fractions or division into fifths. A mathematical equation displays the equality of mixed numbers: 3 6/5 = 4 1/5. This demonstrates how an improper fraction within a mixed number can be converted to simplify or change the whole number part.

Adding the whole circles and fifth pieces, we got a sum of 365. We can see that 65 is equivalent to 115, so we add that to the 3 to get 415.

Model, and give the sum as a mixed number. Draw a picture to illustrate your model.

256+156

Solution

423
The figure shows five circles. The last two are divided into six equal sections. Two sections of the last circle are white.

Model, and give the sum as a mixed number. Draw a picture to illustrate your model.

158+178

Solution

312
The figure shows four circles. The last two are divided into eight equal sections. Four sections of the last circle are white.

Add Mixed Numbers

Modeling with fraction circles helps illustrate the process for adding mixed numbers: We add the whole numbers and add the fractions, and then we simplify the result, if possible.

Add mixed numbers with a common denominator.

Step 1. Add the whole numbers.

Step 2. Add the fractions.

Step 3. Simplify, if possible.

Add: 349+229.

Solution

Solution

349+229
Add the whole numbers. A vertical addition problem shows two mixed numbers, 3 4/9 and 2 2/9, being added. The sum of the whole number parts, 5, is displayed below the line, with fractions yet to be combined.
Add the fractions. The addition of two mixed numbers is shown, where 3 and 4/9 plus 2 and 2/9 equals 5 and 6/9. The fractional components (4/9, 2/9, and 6/9) are highlighted in red.
Simplify the fraction. A vertical math problem showing the addition of mixed numbers: 3 4/9 + 2 2/9. The sum is initially 5 6/9, with the fraction highlighted in red, which simplifies to 5 2/3.

Find the sum: 447+127.

Solution

567

Find the sum: 2311+5611.

Solution

7911

In Example 3, the sum of the fractions was a proper fraction. Now we will work through an example where the sum is an improper fraction.

Find the sum: 959+579.

Solution

Solution

Step-by-step example demonstrating the addition of mixed numbers, including simplification of the result.
959+579
Add the whole numbers and then add the fractions.
959 +579_____ 14129
Rewrite 129 as a mixed number. 14+139
Add. 1539
Simplify. 1513

Find the sum: 878+758.

Solution

1612

Find the sum: 679+859.

Solution

1513

An alternate method for adding mixed numbers is to convert the mixed numbers to improper fractions and then add the improper fractions. This method is usually written horizontally.

Add by converting the mixed numbers to improper fractions: 378+438.

Solution

Solution

This table demonstrates the step-by-step process of adding two mixed numbers, including conversion to improper fractions and simplification.
378+438
Convert to improper fractions. 318+358
Add the fractions. 31+358
Simplify the numerator. 668
Rewrite as a mixed number. 828
Simplify the fraction. 814

Since the problem was given in mixed number form, we will write the sum as a mixed number.

Find the sum by converting the mixed numbers to improper fractions:

559+379.

Solution

913

Find the sum by converting the mixed numbers to improper fractions:

3710+2910.

Solution

635

Table 7 compares the two methods of addition, using the expression 325+645 as an example. Which way do you prefer?

Mixed Numbers Improper Fractions
325+6459659+659+1151015 325+645175+3455151015

Model Subtraction of Mixed Numbers

Let’s think of pizzas again to model subtraction of mixed numbers with a common denominator. Suppose you just baked a whole pizza and want to give your brother half of the pizza. What do you have to do to the pizza to give him half? You have to cut it into at least two pieces. Then you can give him half.

We will use fraction circles (pizzas!) to help us visualize the process.

Start with one whole.

A shaded circle is shown. Below it is a 1. There are arrows pointing to a shaded circle divided into 2 equal parts. Below it is 2 over 2. Next to this are two circles, each divided into 2 equal parts. The top circle has the right half shaded and the bottom circle has the left half shaded.

Algebraically, you would write:

On the left, it says 1 minus 1 half. There is an arrow pointing to 2 over 2 minus 1 over 2. There is another arrow pointing to 2 over 2 minus 1 over 2 equals 1 over 2.

Use a model to subtract: 1−13.

Solution

Solution

There is a table with five rows and three columns. The first column is not labeled. The second column is labeled “Model.” The third column is labeled “Math Notation.” In the first column, the first row says, “Rewrite vertically. Start with one whole.” The next row says, “Since one-third has denominator 3, cut the whole into 3 pieces. The 1 whole becomes 3 thirds.” The next row says, “Take away one-third.” The last row says, “There are two-thirds left.” In the “Model” column, there is a picture of a shaded circle. Below that is a picture of a shaded circle divided into 3 equal pieces. Below that is a picture of a circle divided into 3 equal pieces with 2 pieces shaded. In the “Math Notation” column, the first row shows 1 minus 1 third. The next row says 3 thirds minus 1 third. The last row says 3 thirds minus 1 third is 2 thirds.

Use a model to subtract: 1−14.

Solution

34
A series of three circles demonstrating fractions: a fully shaded circle, a circle divided into four shaded quadrants, and a circle with three out of four quadrants shaded.

Use a model to subtract: 1−15.

Solution

45
The image shows circles to demonstrate the subtraction problem.

What if we start with more than one whole? Let’s find out.

Use a model to subtract: 2−34.

Solution

Solution

There is a table with four rows and three columns. The first column is not labeled. The second column is labeled “Model.” The third column is labeled “Math Notation.” In the first column, the first row says, “Rewrite vertically. Start with two wholes.” The next row says, “Since three-fourths has denominator 4, cut one of the wholes into 4 pieces. You have one whole and 4 fourths.” The next row says, “Take away three-fourths.” The last row says, “There is 1 and 1 fourth left.” In the “Model” column, there is a picture of two shaded circles. Below that is a picture of two shaded circles. One of the circles is divided into 4 equal pieces. Below that is a picture of one full shaded circle and a circle divided into 4 equal pieces with 1 piece shaded. In the “Math Notation” column, the first row shows 2 minus 3 fourths. The next row says 1 and 4 fourths minus 3 fourths. The last row says 1 and 4 fourths minus 3 fourths equals 1 and 1 fourth.

Use a model to subtract: 2−15.

Solution

95
Six circles illustrating fractions. The first four are solid gray. The fifth circle is divided into five equal gray segments. The sixth shows four out of five segments gray and one white.

Use a model to subtract: 2−13.

Solution

53
A grid of six circles illustrating fractions. The top row shows whole circles. The bottom row progresses from a whole circle to one divided into thirds, ending with two-thirds shaded.

In the next example, we’ll subtract more than one whole.

Use a model to subtract: 2−125.

Solution

Solution

There is a table with five rows and three columns. The first column is not labeled. The second column is labeled “Model.” The third column is labeled “Math Notation.” In the first column, the first row says, “Rewrite vertically. Start with two wholes.” The next row says, “Since two-fifths has denominator 5, cut one of the wholes into 5 pieces. You have one whole and 5 fifths.” The next row says, “Take away 1 and two-fifths.” The last row says, “There is 3 fifths left.” In the “Model” column, there is a picture of two shaded circles. Below that is a picture of two shaded circles. One of the circles is divided into 5 equal pieces. Below that is a picture of one full unshaded circle and a circle divided into 5 equal pieces with 3 pieces shaded. In the “Math Notation” column, the first row shows 2 minus 1 and 2 fifths. The next row says 1 and 5 fifths minus 1 and 2 fifths. The last row says 1 and 5 fifths minus 1 and 2 fifths equals 3 fifths.

Use a model to subtract: 2−113.

Solution

23
A grid of six circles, illustrating various fractions including whole circles, 3/3 (a whole), and 2/3 of a circle, useful for teaching fundamental concepts of fractions and whole numbers.

Use a model to subtract: 2−114.

Solution

34
The image uses circles to illustrate the subtraction problem.

What if you start with a mixed number and need to subtract a fraction? Think about this situation: You need to put three quarters in a parking meter, but you have only a $1 bill and one quarter. What could you do? You could change the dollar bill into 4 quarters. The value of 4 quarters is the same as one dollar bill, but the 4 quarters are more useful for the parking meter. Now, instead of having a $1 bill and one quarter, you have 5 quarters and can put 3 quarters in the meter.

This models what happens when we subtract a fraction from a mixed number. We subtracted three quarters from one dollar and one quarter.

We can also model this using fraction circles, much like we did for addition of mixed numbers.

Use a model to subtract: 114−34

Solution

Solution

Rewrite vertically. Start with one whole and one fourth. Two reddish-orange geometric shapes: a large oval and a smaller quarter circle, side by side on a white background. A vertical math problem shows the subtraction of fractions. The top number is a mixed number, 1 and 1/4, colored red. Below it, the number 3/4 is shown, preceded by a minus sign. A horizontal line underneath indicates the operation is ready to be performed.
Since the fractions have denominator 4, cut the whole into 4 pieces.
You now have 44 and 14 which is 54.
An oval shape is divided into four equal quadrants, with a single, separate quadrant displayed to its right, illustrating a whole and one of its parts. A vertical math problem showing the subtraction of fractions: 5/4 minus 3/4. The 5/4 is in red text above, and the -3/4 is in black text below, with a line underneath indicating the operation.
Take away 34.
There is 12 left.
Visualizing fractions: a circle with two orange shaded quarters (half) and a separate, unshaded single quarter circle, demonstrating 2/4 and 1/4 parts of a whole. A mathematical equation shows the subtraction of two fractions with a common denominator: 5/4 minus 3/4 equals 2/4, which simplifies to 1/2. The 3/4 is highlighted in red.

Use a model to subtract. Draw a picture to illustrate your model.

113−23

Solution

23
A circle divided into three sections, two of which are shaded.

Use a model to subtract. Draw a picture to illustrate your model.

115−45

Solution

25
A circle divided into 5 sections, 2 of which are shaded.

Subtract Mixed Numbers with a Common Denominator

Now we will subtract mixed numbers without using a model. But it may help to picture the model in your mind as you read the steps.

    Subtract mixed numbers with common denominators.

  1. Rewrite the problem in vertical form.
  2. Compare the two fractions.
    • If the top fraction is larger than the bottom fraction, go to Step 3.
    • If not, in the top mixed number, take one whole and add it to the fraction part, making a mixed number with an improper fraction.
  3. Subtract the fractions.
  4. Subtract the whole numbers.
  5. Simplify, if possible.

Find the difference: 535−245.

Solution

Solution

A mathematical expression showing the subtraction of two mixed numbers: 5 and 3/5 minus 2 and 4/5.
Rewrite the problem in vertical form. A vertical subtraction problem showing the mixed number 5 and 3/5, with 2 and 4/5 written below it, indicating 5 3/5 minus 2 4/5.
Since 35 is less than 45, take 1 from the 5 and add it to the 35:(55+35=85) An image demonstrating the regrouping process in mixed number subtraction, converting 5 3/5 to 4 8/5 to enable subtraction when the minuend's fraction is smaller.
Subtract the fractions. A vertical subtraction problem of mixed numbers: 4 and 8/5 minus 2 and 4/5, with the result shown below the line as 4/5.
Subtract the whole parts.
The result is in simplest form.
A math problem showing the subtraction of two mixed numbers. 4 and 8/5 minus 2 and 4/5 results in 2 and 4/5. The whole number components of the original mixed numbers, 4 and 2, are colored red.

Since the problem was given with mixed numbers, we leave the result as mixed numbers.

Find the difference: 649−379.

Solution

223

Find the difference: 447−267.

Solution

157

Just as we did with addition, we could subtract mixed numbers by converting them first to improper fractions. We should write the answer in the form it was given, so if we are given mixed numbers to subtract we will write the answer as a mixed number.

Subtract mixed numbers with common denominators as improper fractions.

Step 1. Rewrite the mixed numbers as improper fractions.

Step 2. Subtract the numerators.

Step 3. Write the answer as a mixed number, simplifying the fraction part, if possible.

Find the difference by converting to improper fractions:

9611−71011.

Solution

Solution

A step-by-step example demonstrating the subtraction of mixed numbers, including conversion to improper fractions and rewriting the final answer.
9611−71011
Rewrite as improper fractions. 10511−8711
Subtract the numerators. 1811
Rewrite as a mixed number. 1711

Find the difference by converting the mixed numbers to improper fractions:

649−379.

Solution

223

Find the difference by converting the mixed numbers to improper fractions:

447−267.

Solution

157

Add and Subtract Mixed Numbers with Different Denominators

To add or subtract mixed numbers with different denominators, we first convert the fractions to equivalent fractions with the LCD. Then we can follow all the steps we used above for adding or subtracting fractions with like denominators.

Add: 212+523.

Solution

Solution

Since the denominators are different, we rewrite the fractions as equivalent fractions with the LCD, 6. Then we will add and simplify.

There are three vertical addition problems. The first shows 2 and 1 half plus 5 and 2 thirds. There is an arrow pointing to the next. This one shows 2 and 1 times a red 3 over 2 times a red 3, with an arrow pointing to the top red 3 that says, “Change into equivalent,” plus 5 and 2 times a red 2 over 3 times a red 2. There is an arrow pointing to the next. This one shows 2 and 3 sixths plus 5 and 4 sixths equals 7 and 7 sixths. Below are instructions to add and rewrite in simplest form. There is an arrow pointing to a red 8 and 1 sixth.

We write the answer as a mixed number because we were given mixed numbers in the problem.

Add: 156+434.

Solution

6712

Add: 345+812.

Solution

12310

Subtract: 434−278.

Solution

Solution

Since the denominators of the fractions are different, we will rewrite them as equivalent fractions with the LCD 8. Once in that form, we will subtract. But we will need to borrow 1 first.

There are four vertical subtraction problems. The first shows 4 and 3 fourths minus 2 and 7 eighths. There is an arrow pointing to the next. This shows 4 and 3 times a red 2 over 4 times a red 2, with an arrow above saying, “change into equivalent,” minus 2 and 7 eighths. There is an arrow pointing to the next. This shows 4 and 6 eighths minus 2 and 7 eighths. There is an arrow pointing to the next. It says to borrow 1 whole from the 4, since we cannot subtract 7 eighths from 6 eighths, and shows 3 and 14 eighths minus 2 and 7 eighths equals 1 and 7 eighths.

We were given mixed numbers, so we leave the answer as a mixed number.

Find the difference: 812−345.

Solution

4710

Find the difference: 434−156.

Solution

21112

Subtract: 3511−434.

Solution

Solution

We can see the answer will be negative since we are subtracting 4 from 3. Generally, when we know the answer will be negative it is easier to subtract with improper fractions rather than mixed numbers.

This table illustrates the step-by-step process of subtracting two mixed numbers, including conversion to improper fractions.
3511−434
Change to equivalent fractions with the LCD. 35·411·4−43·114·11

32044−43344
Rewrite as improper fractions. 15244−20944
Subtract. −5744
Rewrite as a mixed number. −11344

Subtract: 134−678.

Solution

−418

Subtract: 1037−2249.

Solution

−75763

ACCESS ADDITIONAL ONLINE RESOURCES

  • Adding Mixed Numbers
  • Subtracting Mixed Numbers

Key Concepts

  • Add mixed numbers with a common denominator.
    1. Add the whole numbers.
    2. Add the fractions.
    3. Simplify, if possible.
  • Subtract mixed numbers with common denominators.
    1. Rewrite the problem in vertical form.
    2. Compare the two fractions.
      If the top fraction is larger than the bottom fraction, go to Step 3.
      If not, in the top mixed number, take one whole and add it to the fraction part, making a mixed number with an improper fraction.
    3. Subtract the fractions.
    4. Subtract the whole numbers.
    5. Simplify, if possible.
  • Subtract mixed numbers with common denominators as improper fractions.
    1. Rewrite the mixed numbers as improper fractions.
    2. Subtract the numerators.
    3. Write the answer as a mixed number, simplifying the fraction part, if possible.

Practice Makes Perfect

Model Addition of Mixed Numbers

In the following exercises, use a model to find the sum. Draw a picture to illustrate your model.

115+315

213+113

Solution


Four circles are shown. The first three are shaded. The last circle is divided into 3 equal parts. 2 parts are shaded.
323

138+178

156+156

Solution


Four circles are shown. The first three are shaded. The last circle is divided into 3 equal parts. 2 parts are shaded.
323

Add Mixed Numbers with a Common Denominator

In the following exercises, add.

513+613

249+519

Solution

759

458+938

7910+3110

Solution

11

345+645

923+123

Solution

1113

6910+8310

849+289

Solution

1113

Model Subtraction of Mixed Numbers

In the following exercises, use a model to find the difference. Draw a picture to illustrate your model.

116−56

118−58

Solution


A circle is shown. It is divided into 8 equal pieces. 4 pieces are shaded.
12

Subtract Mixed Numbers with a Common Denominator

In the following exercises, find the difference.

278−138

2712−1512

Solution

116

81720−4920

191315−13715

Solution

625

837−447

529−349

Solution

179

258−178

2512−1712

Solution

56

Add and Subtract Mixed Numbers with Different Denominators

In the following exercises, write the sum or difference as a mixed number in simplified form.

314+613

216+534

Solution

71112

158+412

723+812

Solution

1616

9710−213

645−114

Solution

51120

223−312

278−413

Solution

−11124

Mixed Practice

In the following exercises, perform the indicated operation and write the result as a mixed number in simplified form.

258·134

123·416

Solution

61718

27+47

29+59

Solution

79

1512÷112

2310÷110

Solution

23

13512−9712

1558−678

Solution

834

59−49

1115−715

Solution

415

4−34

6−25

Solution

535

920÷34

724÷143

Solution

116

9611+71011

8513+4913

Solution

13113

325+534

256+415

Solution

7130

815·1019

512·89

Solution

1027

678−213

659−425

Solution

2745

529−445

438−323

Solution

1724

Everyday Math

Sewing Renata is sewing matching shirts for her husband and son. According to the patterns she will use, she needs 238 yards of fabric for her husband’s shirt and 118 yards of fabric for her son’s shirt. How much fabric does she need to make both shirts?

Sewing Pauline has 314 yards of fabric to make a jacket. The jacket uses 223 yards. How much fabric will she have left after making the jacket?

Solution

712 yards

Printing Nishant is printing invitations on his computer. The paper is 812 inches wide, and he sets the print area to have a 112-inch border on each side. How wide is the print area on the sheet of paper?

Framing a picture Tessa bought a picture frame for her son’s graduation picture. The picture is 8 inches wide. The picture frame is 258 inches wide on each side. How wide will the framed picture be?

Solution

1314 inches

Writing Exercises

Draw a diagram and use it to explain how to add 158+278.

Edgar will have to pay $3.75 in tolls to drive to the city.

ⓐ Explain how he can make change from a $10 bill before he leaves so that he has the exact amount he needs.

ⓑ How is Edgar’s situation similar to how you subtract 10−334?

Solution

Answers will vary.

Add 4512+378 twice, first by leaving them as mixed numbers and then by rewriting as improper fractions. Which method do you prefer, and why?

Subtract 378−4512 twice, first by leaving them as mixed numbers and then by rewriting as improper fractions. Which method do you prefer, and why?

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 rubric for students to gauge their understanding of adding and subtracting mixed numbers, indicating if they can perform tasks confidently, with some help, or don't get it.

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

Solve Equations with Fractions

Learning Objectives

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

  • Determine whether a fraction is a solution of an equation
  • Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality
  • Solve equations using the Multiplication Property of Equality
  • Translate sentences to equations and solve

Before you get started, take this readiness quiz. If you miss a problem, go back to the section listed and review the material.

Evaluate x+4 when x=−3
If you missed this problem, review Example 10 in Add Integers.

Solution

1

Solve: 2y−3=9.
If you missed this problem, review Example 2 in Solve Equations Using Integers; The Division Property of Equality.

Solution

y=6

Solve: y−3=−9
If you missed this problem, review Example 10 in Multiply and Divide Fractions.

Solution

-6

Determine Whether a Fraction is a Solution of an Equation

As we saw in Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. In those sections, we found whole number and integer solutions to equations. Now that we have worked with fractions, we are ready to find fraction solutions to equations.

The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.

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 each of the following is a solution of x−310=12.

  1. ⓐ x=1
  2. ⓑ x=45
  3. ⓒ x=−45
Solution

Solution

ⓐ
A mathematical equation shows 'x minus three tenths equals one half' on a white background. The equation is represented as x - 3/10 = 1/2.
The image displays the mathematical instruction 'Substitute 1 for x.' in a bold, teal-colored font, with the number '1' highlighted in red. Mathematical expression 1 - 3/10 ?= 1/2, questioning the equality. The number '1' is colored red.
Change to fractions with a LCD of 10. A mathematical equation shows '10/10 - 3/10 ? 5/10', where the question mark represents an unknown operator or symbol between the two sides of the equation.
Subtract. A mathematical expression states that 7/10 is not equal to 5/10. The numbers are written in black against a white background.

Since x=1 does not result in a true equation, 1 is not a solution to the equation.

ⓑ
The image displays the algebraic equation x - 3/10 = 1/2, where 'x' is an unknown variable and the fractions 3/10 and 1/2 are constants. This equation requires solving for the value of x.
The image shows the text 'Substitute 4/5 for x.' in a mathematical context, likely instructing to replace the variable 'x' with the fraction '4/5'. A mathematical equation asking whether 4/5 minus 3/10 equals 1/2. The fraction 4/5 is colored red.
The image displays a fractional subtraction problem: 8/10 - 3/10. A question mark over the equals sign asks if the result is 5/10, evaluating the truth of the statement.
Subtract. The image displays a mathematical equation: 5/10 = 5/10, followed by a checkmark, indicating that the equality is correct.

Since x=45 results in a true equation, 45 is a solution to the equation x−310=12.

ⓒ
A mathematical equation shows 'x - 3/10 = 1/2' on a white background, representing an algebraic problem to solve for the variable x.
The image shows the instruction 'Substitute -4/5 for x.' with the fraction -4/5 written in red. A mathematical expression shows the equation: negative four-fifths minus three-tenths equals, with a question mark over the equal sign, one-half. The fraction negative four-fifths is in red.
A mathematical equation shows '-8/10 - 3/10' on the left, an equals sign with a question mark above it in the middle, and '5/10' on the right. The first fraction's numerator and denominator are red.
Subtract. A mathematical expression showing that -11/10 is not equal to 5/10.

Since x=−45 does not result in a true equation, −45 is not a solution to the equation.

Determine whether each number is a solution of the given equation.

x−23=16:
  1. ⓐ x=1
  2. ⓑ x=56
  3. ⓒ x=−56
Solution
  1. ⓐ no
  2. ⓑ yes
  3. ⓒ no

Determine whether each number is a solution of the given equation.

y−14=38:
  1. ⓐ y=1
  2. ⓑ y=−58
  3. ⓒ y=58
Solution
  1. ⓐ no
  2. ⓑ no
  3. ⓒ yes

Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality

In Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, we solved equations using the Addition, Subtraction, and Division Properties of Equality. We will use these same properties to solve equations with fractions.

Addition, Subtraction, and Division Properties of Equality

For any numbers a,b, and c,

This table lists fundamental properties of equality including addition, subtraction, and division, showing their mathematical statements.
if a=b, then a+c=b+c. Addition Property of Equality
if a=b, then a−c=b−c. Subtraction Property of Equality
if a=b, then ac=bc,c≠0. Division Property of Equality

In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.

Solve: y+916=516.

Solution

Solution

A mathematical equation is displayed, reading 'y + 9/16 = 5/16'.
Subtract 916 from each side to undo the addition. An equation showing y + 9/16 - 9/16 = 5/16 - 9/16. The subtracted 9/16 terms are highlighted in red on both sides of the equality, suggesting they might be canceled out.
Simplify on each side of the equation. A mathematical equation shows 'y + 0 = -4/16' centered on a white background, representing a simplified algebraic expression where y is equal to -1/4.
Simplify the fraction. The mathematical equation y = -1/4 is displayed on a white background, representing a horizontal line at y equals negative one-fourth.
Check: A mathematical equation displays 'y + 9/16 = 5/16'.
Substitute y=−14. A mathematical equation shows -1/4 + 9/16 =? 5/16, representing the sum of two fractions with a question mark indicating a verification or unknown result.
Rewrite as fractions with the LCD. The mathematical expression -4/16 + 9/16 ?= 5/16, asking to verify the equality of the fractional sum.
Add. The image shows the fraction 5/16 equals 5/16, followed by a checkmark, indicating correctness or verification of the equality.

Since y=−14 makes y+916=516 a true statement, we know we have found the solution to this equation.

Solve: y+1112=512.

Solution

−12

Solve: y+815=415.

Solution

−415

We used the Subtraction Property of Equality in Example 2. Now we’ll use the Addition Property of Equality.

Solve: a−59=−89.

Solution

Solution

A mathematical equation is displayed on a white background. The equation reads as 'a - 5/9 = -8/9', where 'a' is a variable, and '5/9' and '-8/9' are fractions.
Add 59 from each side to undo the subtraction. The image shows a mathematical equation: a - 5/9 + 5/9 = -8/9 + 5/9. The fractions 5/9 on both sides of the equation are highlighted in red.
Simplify on each side of the equation. A mathematical equation shows 'a + 0 = -3/9' written in black text on a white background.
Simplify the fraction. The image displays a mathematical equation: a = -1/3. The variable 'a' is shown to be equal to negative one-third, presented in a clear, standard mathematical notation against a white background.
Check: A mathematical equation shows 'a minus five ninths equals negative eight ninths'.
Substitute a=−13. A mathematical equation with fractions: -1/3 (in red) - 5/9 with a question mark above the equals sign, followed by -8/9. It asks to verify the equality of the expression.
Change to common denominator. The equation -3/9 - 5/9 ?= -8/9, where a question mark above the equals sign prompts verification of the fractional subtraction.
Subtract. A mathematical equation displays -8/9 = -8/9, accompanied by a checkmark, confirming its correctness.

Since a=−13 makes the equation true, we know that a=−13 is the solution to the equation.

Solve: a−35=−85.

Solution

−1

Solve: n−37=−97.

Solution

−67

The next example may not seem to have a fraction, but let’s see what happens when we solve it.

Solve: 10q=44.

Solution

Solution

Illustrates the step-by-step process of solving the linear equation 10q=44 and verifying its solution.
10q=44
Divide both sides by 10 to undo the multiplication. 10q10=4410
Simplify. q=225
Check:
Substitute q=225 into the original equation. 10(225)=?44
Simplify. 102(225)=?44
Multiply. 44=44✓

The solution to the equation was the fraction 225. We leave it as an improper fraction.

Solve: 12u=−76.

Solution

−193

Solve: 8m=92.

Solution

232

Solve Equations with Fractions Using the Multiplication Property of Equality

Consider the equation x4=3. We want to know what number divided by 4 gives 3. So to “undo” the division, we will need to multiply by 4. The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.

The Multiplication Property of Equality

For any numbers a,b, and c,

ifa=b,thenac=bc.

If you multiply both sides of an equation by the same quantity, you still have equality.

Let’s use the Multiplication Property of Equality to solve the equation x7=−9.

Solve: x7=−9.

Solution

Solution

A mathematical equation is displayed on a white background, reading 'x/7 = -9'.
Use the Multiplication Property of Equality to multiply both sides by 7. This will isolate the variable. An algebraic step where both sides of the equation are multiplied by 7 to isolate 'x', shown as 7 * (x/7) = 7(-9).
Multiply. A mathematical equation is displayed on a white background: 7x/7 = -63.
Simplify. A mathematical equation displays 'x = -63' in black text against a white background.
The text reads 'Check.Substitute -63 for x for in the original equation.' with '-63' highlighted in red, indicating a step in solving an algebraic problem. A mathematical equation shows the fraction negative sixty-three over seven, followed by a question mark above an equals sign, and then negative nine. This setup asks if negative sixty-three divided by seven is equal to negative nine.
The equation is true. The mathematical equation -9 = -9 is shown, followed by a checkmark, indicating that the equality is correct.

Solve: f5=−25.

Solution

−125

Solve: h9=−27.

Solution

−243

Solve: p−8=−40.

Solution

Solution

Here, p is divided by −8. We must multiply by −8 to isolate p.

The mathematical equation p over -8 equals -40 is displayed on a white background.
Multiply both sides by −8 A mathematical equation shows '-8' multiplied by 'P divided by -8' on the left side, which equals '-8' multiplied by '-40' on the right side. The -8 coefficients are highlighted in red.
Multiply. A mathematical equation shows a fraction with '-8p' in the numerator and '-8' in the denominator, set equal to '320'.
Simplify. The image displays the mathematical equation p = 320 in black text against a white background.
Check:
Substitute p=320. A mathematical equation shows '320' in red, divided by '-8', with a question mark over an equals sign, followed by '-40'. It asks whether 320 divided by -8 is equal to -40.
The equation is true. A mathematical equation shows '-40 = -40' with a black checkmark symbol to its right, indicating that the equality is correct. The numbers and symbols are in a dark gray font against a white background.

Solve: c−7=−35.

Solution

245

Solve: x−11=−12.

Solution

132

Solve Equations with a Coefficient of −1

Look at the equation −y=15. Does it look as if y is already isolated? But there is a negative sign in front of y, so it is not isolated.

There are three different ways to isolate the variable in this type of equation. We will show all three ways in Example 7.

Solve: −y=15.

Solution
Solution

One way to solve the equation is to rewrite −y as −1y, and then use the Division Property of Equality to isolate y.

The image displays the equation -y = 15, representing a simple linear algebraic expression.
Rewrite −y as −1y. A mathematical equation is displayed, showing -1y = 15.
Divide both sides by −1. The equation shows -1y divided by -1 equals 15 divided by -1, demonstrating a step in solving for y by dividing both sides by -1.
Simplify each side. The image displays the equation y = -15, rendered in a clear, standard mathematical font on a white background.

Another way to solve this equation is to multiply both sides of the equation by −1.

The image shows a mathematical equation with a variable 'y'. The equation reads '-y = 15'.
Multiply both sides by −1. A mathematical equation shows '-1(-y) = -1(15)', where both sides are multiplied by -1. The -1 is highlighted in red on both sides.
Simplify each side. The image displays a mathematical equation in black text on a white background, stating 'y = -15'.

The third way to solve the equation is to read −y as “the opposite of y.” What number has 15 as its opposite? The opposite of 15 is −15. So y=−15.

For all three methods, we isolated y is isolated and solved the equation.

Check:

The image shows a mathematical equation with a variable 'y'. The equation reads '-y = 15'.
Substitute y=−15. A mathematical equation shows -(-15) ?= (15), asking if the negative of negative 15 is equal to 15, with the -15 in red text.
Simplify. The equation is true. The number 15 is equal to 15, confirmed by a checkmark indicating correctness or verification.

Solve: −y=48.

Solution

−48

Solve: −c=−23.

Solution

23

Solve Equations with a Fraction Coefficient

When we have an equation with a fraction coefficient we can use the Multiplication Property of Equality to make the coefficient equal to 1.

For example, in the equation:

34x=24

The coefficient of x is 34. To solve for x, we need its coefficient to be 1. Since the product of a number and its reciprocal is 1, our strategy here will be to isolate x by multiplying by the reciprocal of 34. We will do this in Example 8.

Solve: 34x=24.

Solution
Solution
A mathematical equation is displayed on a white background: 3/4x = 24. The equation represents a linear algebraic problem involving a fraction and a variable.
Multiply both sides by the reciprocal of the coefficient. An algebraic equation is shown: (4/3) multiplied by (3/4)x equals (4/3) multiplied by 24. The fraction 4/3 appears in red on both sides of the equation.
Simplify. A mathematical equation showing 1x equals the product of 4/3 and 24/1, which simplifies to 1x = 32. It demonstrates a step in solving for 'x' by multiplying fractions.
Multiply. The mathematical expression 'x = 32' is displayed on a white background.
Check: A mathematical equation is displayed on a white background, which reads '3/4x = 24'.
Substitute x=32. A mathematical equation asks if three-fourths multiplied by thirty-two equals twenty-four, with a question mark over the equals sign to indicate inquiry.
Rewrite 32 as a fraction. A mathematical equation shows (3/4) multiplied by (32/1) with a question mark over the equality sign, followed by the number 24, suggesting a check to see if the product equals 24.
Multiply. The equation is true. A mathematical statement showing 24 = 24, followed by a checkmark to indicate correctness or verification.

Notice that in the equation 34x=24, we could have divided both sides by 34 to get x by itself. Dividing is the same as multiplying by the reciprocal, so we would get the same result. But most people agree that multiplying by the reciprocal is easier.

Solve: 25n=14.

Solution

35

Solve: 56y=15.

Solution

18

Solve: −38w=72.

Solution
Solution

The coefficient is a negative fraction. Remember that a number and its reciprocal have the same sign, so the reciprocal of the coefficient must also be negative.

A mathematical equation showing negative three-eighths multiplied by 'w' equals 72.
Multiply both sides by the reciprocal of −38. The equation -8/3(-3/8w) = (-8/3)72 is displayed, illustrating a linear equation involving fractions and the variable 'w'.
Simplify; reciprocals multiply to one. The equation shows '1w = -8/3 ×72/1', representing a multiplication of two fractions, one negative and one positive, to solve for 'w'.
Multiply. A mathematical equation is displayed, showing the variable 'w' equal to the negative integer -192, presented in a clear, dark font against a stark white background.
Check: A mathematical equation displays '-3/8 w = 72' in black text against a white background.
Let w=−192. A mathematical expression showing negative three-eighths multiplied by negative one hundred ninety-two, followed by an equals sign with a question mark above it, and then the number seventy-two.
Multiply. It checks. The equation 72=72 is marked as correct with a checkmark.

Solve: −47a=52.

Solution

−91

Solve: −79w=84.

Solution

−108

Translate Sentences to Equations and Solve

Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.

Subtraction Property of Equality:
For any real numbers a, b, and c,

if a=b, then a−c=b−c.
Addition Property of Equality:
For any real numbers a, b, and c,

if a=b, then a+c=b+c.
Division Property of Equality:
For any numbers a, b, and c, where c≠0

if a=b, then ac=bc
Multiplication Property of Equality:
For any real numbers a, b, and c

if a=b, then ac=bc

When you add, subtract, multiply or divide the same quantity from both sides of an equation, you still have equality.

In the next few examples, we’ll translate sentences into equations and then solve the equations. It might be helpful to review the translation table in Evaluate, Simplify, and Translate Expressions.

Translate and solve: n divided by 6 is −24.

Solution

Solution

Translate. The image shows the phrase 'n divided by 6 is 24' with brackets indicating its parts. Below, the equation 'n/6 = -24' is displayed, translating the phrase but with a changed sign for the number 24.
Multiply both sides by 6. The equation 6 times n/6 equals 6 times -24, illustrating a step in solving for 'n'.
Simplify. The image displays a mathematical equation in a black serif font on a white background, which states 'n = -144'.
Check: Is −144 divided by 6 equal to −24?
Translate. A math problem displays the fraction -144/6, followed by a question mark above an equals sign, and then -24. It asks whether -144 divided by 6 is equal to -24.
Simplify. It checks. A mathematical equation displays '-24 = -24' followed by a checkmark, confirming the equality.

Translate and solve: n divided by 7 is equal to −21.

Solution

n7=−21;n=−147

Translate and solve: n divided by 8 is equal to −56.

Solution

n8=−56;n=−448

Translate and solve: The quotient of q and −5 is 70.

Solution

Solution

Translate. The image shows the verbal expression 'The quotient of q and -5 is 70' translated into the mathematical equation q/-5 = 70. Brackets link corresponding parts of the phrase and the equation.
Multiply both sides by −5. A mathematical equation shows -5 multiplied by q over -5, which equals -5 multiplied by 70.
Simplify. The mathematical equation 'q = -350' is displayed on a white background, representing a variable 'q' equal to the negative integer three hundred fifty.
Check: Is the quotient of −350 and −5 equal to 70?
Translate. A mathematical problem demonstrating the division of negative numbers, asking if -350 divided by -5 equals 70, symbolized by a question mark over an equals sign between the two values.
Simplify. It checks. The equation 70 = 70 is displayed, followed by a black checkmark, indicating correctness or verification.

Translate and solve: The quotient of q and −8 is 72.

Solution

q−8=72;q=−576

Translate and solve: The quotient of p and −9 is 81.

Solution

p−9=81;p=−729

Translate and solve: Two-thirds of f is 18.

Solution

Solution

Translate. A mathematical expression shows how the phrase 'Two-thirds of f is 18' translates into the algebraic equation '(2/3)f = 18,' with curly brackets indicating the corresponding parts.
Multiply both sides by 32. A math equation shows (3/2) * (2/3)f = (3/2) * 18, with the red fraction 3/2 applied to both sides, likely to solve for 'f' after simplification.
Simplify. The image displays the equation 'f = 27' in black text on a white background.
Check: Is two-thirds of 27 equal to 18?
Translate. A mathematical equation shows two-thirds multiplied by twenty-seven, followed by a question mark over an equals sign, then eighteen. The equation asks if 2/3(27) is equal to 18.
Simplify. It checks. The image displays the equation '18 = 18' followed by a checkmark, indicating that the mathematical statement is correct.

Translate and solve: Two-fifths of f is 16.

Solution

25f=16;f=40

Translate and solve: Three-fourths of f is 21.

Solution

34f=21;f=28

Translate and solve: The quotient of m and 56 is 34.

Solution

Solution

This table demonstrates the step-by-step process of solving and verifying an algebraic equation involving fractions for the variable 'm'.
The quotient of m and 56 is 34.
Translate. m56=34
Multiply both sides by 56 to isolate m. 56(m56)=56(34)
Simplify. m=5·36·4
Remove common factors and multiply. m=58
Check:
Is the quotient of 58 and 56 equal to 34? 5856=?34
Rewrite as division. 58÷56=?34
Multiply the first fraction by the reciprocal of the second. 58·65=?34
Simplify. 34=34✓

Our solution checks.

Translate and solve. The quotient of n and 23 is 512.

Solution

n23=512;n=518

Translate and solve The quotient of c and 38 is 49.

Solution

c38=49;c=16

Translate and solve: The sum of three-eighths and x is three and one-half.

Solution

Solution

Translate. This image translates the phrase 'The sum of three-eighths and x is three and one-half' into the algebraic equation 3/8 + x = 3 1/2, illustrating how to set up an equation from a word problem.
Use the Subtraction Property of Equality to subtract 38 from both sides. A mathematical equation is shown with the expression '3/8 + x - 3/8 = 3 1/2 - 3/8' on a white background, requiring the solving for the variable 'x'.
Combine like terms on the left side. A mathematical equation shows 'x = 3 1/2 - 3/8' centered on a white background, representing a mixed number subtraction problem.
Convert mixed number to improper fraction. A mathematical equation displays 'x = 7/2 - 3/8' centered on a white background.
Convert to equivalent fractions with LCD of 8. A mathematical equation shows x equals 28 over 8 minus 3 over 8, demonstrating subtraction of fractions with a common denominator.
Subtract. The equation x = 25/8 is displayed on a white background, representing a mathematical solution.
Write as a mixed number. A mathematical expression displays 'x = 3 1/8' on a white background.

We write the answer as a mixed number because the original problem used a mixed number.

Check:

Is the sum of three-eighths and 318 equal to three and one-half?

Step-by-step verification of a fractional addition problem.
38+318=?312
Add. 348=?312
Simplify. 312=312✓

The solution checks.

Translate and solve: The sum of five-eighths and x is one-fourth.

Solution

58+x=14;x=−38

Translate and solve: The difference of one-and-three-fourths and x is five-sixths.

Solution

134−x=56;x=1112

ACCESS ADDITIONAL ONLINE RESOURCES

  • Solve One Step Equations With Fractions
  • Solve One Step Equations With Fractions by Adding or Subtracting
  • Solve One Step Equations With Fraction by Multiplying

Key Concepts

  • 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.
  • Addition, Subtraction, and Division Properties of Equality
    • For any numbers a, b, and c,
      if a=b, then a+c=b+c. Addition Property of Equality
    • if a=b, then a-c=b-c. Subtraction Property of Equality
    • if a=b, then ac=bc, c≠0. Division Property of Equality
  • The Multiplication Property of Equality
    • For any numbers ab and c, a=b, then ac=bc.
    • If you multiply both sides of an equation by the same quantity, you still have equality.

Section Exercises

Practice Makes Perfect

Determine Whether a Fraction is a Solution of an Equation

In the following exercises, determine whether each number is a solution of the given equation.

x−25=110:
  1. ⓐ x=1
  2. ⓑ x=12
  3. ⓒ x=−12
y−13=512:
  1. ⓐ y=1
  2. ⓑ y=34
  3. ⓒ y=−34
Solution
  1. ⓐ no
  2. ⓑ yes
  3. ⓒ no
h+34=25:
  1. ⓐ h=1
  2. ⓑ h=720
  3. ⓒ h=−720
k+25=56:
  1. ⓐ k=1
  2. ⓑ k=1330
  3. ⓒ k=−1330
Solution
  1. ⓐ no
  2. ⓑ yes
  3. ⓒ no

Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality

In the following exercises, solve.

y+13=43

m+38=78

Solution

m=12

f+910=25

h+56=16

Solution

h=−23

a−58=−78

c−14=−54

Solution

c = −1

x−(−320)=−1120

z−(−512)=−712

Solution

z = −1

n−16=34

p−310=58

Solution

p=3740

s+(−12)=−89

k+(−13)=−45

Solution

k=−715

5j=17

7k=18

Solution

k=187

−4w=26

−9v=33

Solution

v=−113

Solve Equations with Fractions Using the Multiplication Property of Equality

In the following exercises, solve.

f4=−20

b3=−9

Solution

b = −27

y7=−21

x8=−32

Solution

x = −256

p−5=−40

q−4=−40

Solution

q = 160

r−12=−6

s−15=−3

Solution

s = 45

−x=23

−y=42

Solution

y = −42

−h=−512

−k=−1720

Solution

k=1720

45n=20

310p=30

Solution

p = 100

38q=−48

52m=−40

Solution

m = −16

−29a=16

−37b=9

Solution

b = −21

−611u=−24

−512v=−15

Solution

v = 36

Mixed Practice

In the following exercises, solve.

3x=0

8y=0

Solution

y = 0

4f=45

7g=79

Solution

g=19

p+23=112

q+56=112

Solution

q=−34

78m=110

14n=710

Solution

n=145

−25=x+34

−23=y+38

Solution

y=−2524

1120=-f

815=-d

Solution

d=−815

Translate Sentences to Equations and Solve

In the following exercises, translate to an algebraic equation and solve.

n divided by eight is −16.

n divided by six is −24.

Solution

n6=−24;n=−144

m divided by −9 is −7.

m divided by −7 is −8.

Solution

m−7=−8;m=56

The quotient of f and −3 is −18.

The quotient of f and −4 is −20.

Solution

f−4=−20;f=80

The quotient of g and twelve is 8.

The quotient of g and nine is 14.

Solution

g9=14;g=126

Three-fourths of q is 12.

Two-fifths of q is 20.

Solution

25q=20;q=50

Seven-tenths of p is −63.

Four-ninths of p is −28.

Solution

49p=−28;p=−63

m divided by 4 equals negative 6.

The quotient of h and 2 is 43.

Solution

h2=43;h=86

Three-fourths of z is 15.

The quotient of a and 23 is 34.

Solution

a23=34;a=12

The sum of five-sixths and x is 12.

The sum of three-fourths and x is 18.

Solution

34+x=18;x=−58

The difference of y and one-fourth is −18.

The difference of y and one-third is −16.

Solution

y−13=−16;y=16

Everyday Math

Shopping Teresa bought a pair of shoes on sale for $48. The sale price was 23 of the regular price. Find the regular price of the shoes by solving the equation 23p=48

Playhouse The table in a child’s playhouse is 35 of an adult-size table. The playhouse table is 18 inches high. Find the height of an adult-size table by solving the equation 35h=18.

Solution

30 inches

Writing Exercises

Example 6 describes three methods to solve the equation −y=15. Which method do you prefer? Why?

Richard thinks the solution to the equation 34x=24 is 16. Explain why Richard is wrong.

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 with 'I can...' statements related to solving equations with fractions, including determining solutions, using properties of equality, and translating sentences, with columns for confidence levels: Confidently, With some help, and No-I don't get it!

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Visualize Fractions

In the following exercises, name the fraction of each figure that is shaded.

A circle is shown. It is divided into 8 equal pieces. 5 pieces are shaded.
A square is shown. It is divided into 9 equal pieces. 5 pieces are shaded.
Solution

59

In the following exercises, name the improper fractions. Then write each improper fraction as a mixed number.

Two squares are shown. Both are divided into four equal pieces. The square on the left has all 4 pieces shaded. The square on the right has one piece shaded.
Two circles are shown. Both are divided into two equal pieces. The circle on the left has both pieces shaded. The circle on the right has one piece shaded.
Solution

32=112

In the following exercises, convert the improper fraction to a mixed number.

5815

6311

Solution

5811

In the following exercises, convert the mixed number to an improper fraction.

1214

945

Solution

495

Find three fractions equivalent to 25. Show your work, using figures or algebra.

Find three fractions equivalent to −43. Show your work, using figures or algebra.

Solution

Answers may vary.

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

58,43,334,4

14,−14,113,−113,72,−72

Solution


A number line is shown. Integers from negative 4 to 4 are labeled. Between negative 4 and negative 3, negative 7 halves is labeled and marked with a red dot. Between negative 2 and negative 1, negative 1 and 1 third is labeled and marked with a red dot. Between negative 1 and 0, negative 1 fourth is labeled and marked with a red dot. Between 0 and 1, 1 fourth is labeled and marked with a red dot. Between 1 and 2, 1 and 1 third is labeled and marked with a red dot. Between 3 and 4, 7 halves is labeled and marked with a red dot.

In the following exercises, order each pair of numbers, using < or >.

−1___−25

−212___−3

Solution

>

Multiply and Divide Fractions

In the following exercises, simplify.

−6384

−90120

Solution

−34

−14a14b

−8x8y

Solution

−xy

In the following exercises, multiply.

25·813

−13·127

Solution

−47

29·(−4532)

6m·411

Solution

24m11

−14(−32)

165·158

Solution

6

In the following exercises, find the reciprocal.

29

154

Solution

415

3

−14

Solution

−4

Fill in the chart.

Opposite Absolute Value Reciprocal
−513
310
94
−12

In the following exercises, divide.

23÷16

Solution

4

(−3x5)÷(−2y3)

45÷3

Solution

415

8÷83

518÷(−b9)

Solution

−52b

Multiply and Divide Mixed Numbers and Complex Fractions

In the following exercises, perform the indicated operation.

315·178

−5712·4411

Solution

−26811

8÷223

823÷1112

Solution

8

In the following exercises, translate the English phrase into an algebraic expression.

the quotient of 8 and y

the quotient of V and the difference of h and 6

Solution

Vh−6

In the following exercises, simplify the complex fraction

5845

89−4

Solution

−29

n438

−156−112

Solution

22

In the following exercises, simplify.

5+165

8·4−523·12

Solution

736

8·7+5(8−10)9·3−6·4

Add and Subtract Fractions with Common Denominators

In the following exercises, add.

38+28

Solution

58

45+15

25+15

Solution

35

1532+932

x10+710

Solution

x+710

In the following exercises, subtract.

811−611

1112−512

Solution

12

45−y5

−3130−730

Solution

−1915

32−(32)

1115−515−(−215)

Solution

815

Add and Subtract Fractions with Different Denominators

In the following exercises, find the least common denominator.

13 and 112

13 and 45

Solution

15

815 and 1120

34,16,and510

Solution

60

In the following exercises, change to equivalent fractions using the given LCD.

13 and 15, LCD =15

38 and 56, LCD =24

Solution

924 and 2024

−916 and 512, LCD =48

13,34 and 45, LCD =60

Solution

2060,4560 and 4860

In the following exercises, perform the indicated operations and simplify.

15+23

1112−23

Solution

14

−910−34

−1136−1120

Solution

−7790

−2225+940

y10−13

Solution

3y−1030

25+(−59)

411÷27d

Solution

14d11

25+(−3n8)(−29n)

(23)2(58)2

Solution

256225

(1112+38)÷(56−110)

In the following exercises, evaluate.

y−45 when
  1. ⓐ y=−45
  2. ⓑ y=14
Solution
  1. ⓐ −85
  2. ⓑ −1120

6mn2 when m=34 and n=−13

Add and Subtract Mixed Numbers

In the following exercises, perform the indicated operation.

413+913

Solution

1323

625+735

5811+2411

Solution

8111

358+378

91320−41120

Solution

5110

2310−1910

21112−1712

Solution

113

8611−2911

Solve Equations with Fractions

In the following exercises, determine whether the each number is a solution of the given equation.

x−12=16:

  1. ⓐ x=1
  2. ⓑ x=23
  3. ⓒ x=−13
Solution
  1. ⓐ no
  2. ⓑ yes
  3. ⓒ no

y+35=59:

  1. ⓐ y=12
  2. ⓑ y=5245
  3. ⓒ y=−245

In the following exercises, solve the equation.

n+911=411

Solution

n=−511

x−16=76

h−(−78)=−25

Solution

h=−5140

x5=−10

−z=23

Solution

z = −23

In the following exercises, translate and solve.

The sum of two-thirds and n is −35.

The difference of q and one-tenth is 12.

Solution

q−110=12;q=35

The quotient of p and −4 is −8.

Three-eighths of y is 24.

Solution

38y=24;y=64

Chapter Practice Test

Convert the improper fraction to a mixed number.

195

Convert the mixed number to an improper fraction.

327

Solution

237

Locate the numbers on a number line.

12,123,−234, and 94

In the following exercises, simplify.

520

Solution

14

18r27s

13·34

Solution

14

35·15

−36u(−49)

Solution

16u

−5712·4411

−56÷512

Solution

−2

711÷(−711)

9a10÷15a8

Solution

1225

−625÷4

(−1556)÷(−316)

Solution

5

−6611

p2q5

Solution

5p2q

−415−223

92−429−4

Solution

13

2d+9d

−313+(−413)

Solution

−713

−2225+940

25+(−75)

Solution

−1

−310+(−58)

−34÷x3

Solution

−94x

23−22(34)2

514+18956

Solution

3

Evaluate.

x+13 when
  1. ⓐ x=23
  2. ⓑ x=−56

In the following exercises, solve the equation.

y+35=75

Solution

y=45

a−310=−910

f+(−23)=512

Solution

f=1312

m−2=−16

−23c=18

Solution

c = −27

Translate and solve: The quotient of p and −4 is −8. Solve for p.

Introduction to Decimals

A gas station sign is shown. It lists unleaded as 3.999, mid-grade as 4.099, and premium as 4.199.
The price of a gallon of gasoline is written as a decimal number. (credit: Mark Turnauckus, Flickr)

Gasoline price changes all the time. They might go down for a period of time, but then they usually rise again. One thing that stays the same is that the price is not usually a whole number. Instead, it is shown using a decimal point to describe the cost in dollars and cents. We use decimal numbers all the time, especially when dealing with money. In this chapter, we will explore decimal numbers and how to perform operations using them.

Decimals

Learning Objectives

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

  • Name decimals
  • Write decimals
  • Convert decimals to fractions or mixed numbers
  • Locate decimals on the number line
  • Order decimals
  • Round decimals

Before you get started, take this readiness quiz.

Name the number 4,926,015 in words.
If you missed this problem, review Example 4 in Introduction to Whole Numbers.

Solution

Four million, nine hundred twenty-six thousand, fifteen

Round 748 to the nearest ten.
If you missed this problem, review Example 9 in Introduction to Whole Numbers.

Solution

750

Locate 310 on a number line.
If you missed this problem, review Example 16 in Visualize Fractions.

Solution

This image shows a number line from negative 5 to 5. A point is plotted at three tenths on the number line.

Name Decimals

You probably already know quite a bit about decimals based on your experience with money. Suppose you buy a sandwich and a bottle of water for lunch. If the sandwich costs $3.45, the bottle of water costs $1.25, and the total sales tax is $0.33, what is the total cost of your lunch?

A vertical addition problem is shown. The top line shows $3.45 for a sandwich, the next line shows $1.25 for water, and the last line shows $0.33 for tax. The total is shown to be $5.03.

The total is $5.03. Suppose you pay with a $5 bill and 3 pennies. Should you wait for change? No, $5 and 3 pennies is the same as $5.03.

Because 100 pennies=$1, each penny is worth 1100 of a dollar. We write the value of one penny as $0.01, since 0.01=1100.

Writing a number with a decimal is known as decimal notation. It is a way of showing parts of a whole when the whole is a power of ten. In other words, decimals are another way of writing fractions whose denominators are powers of ten. Just as the counting numbers are based on powers of ten, decimals are based on powers of ten. Table 1 shows the counting numbers.

Counting number Name
1 One
10=10 Ten
10·10=100 One hundred
10·10·10=1000 One thousand
10·10·10·10=10,000 Ten thousand

How are decimals related to fractions? Table 2 shows the relation.

Decimal Fraction Name
0.1 110 One tenth
0.01 1100 One hundredth
0.001 11,000 One thousandth
0.0001 110,000 One ten-thousandth

When we name a whole number, the name corresponds to the place value based on the powers of ten. In Whole Numbers, we learned to read 10,000 as ten thousand. Likewise, the names of the decimal places correspond to their fraction values. Notice how the place value names in Figure 1 relate to the names of the fractions from Table 2.

A chart is shown labeled “Place Value”. There are 12 columns. The columns are labeled, from left to right, Hundred thousands, Ten thousands, Thousands, Hundreds, Tens, Ones, Decimal Point, Tenths, Hundredths, Thousandths, Ten-thousandths, Hundred-thousandths.
This chart illustrates place values to the left and right of the decimal point.

Notice two important facts shown in Figure 1.

  • The “th” at the end of the name means the number is a fraction. “One thousand” is a number larger than one, but “one thousandth” is a number smaller than one.
  • The tenths place is the first place to the right of the decimal, but the tens place is two places to the left of the decimal.

Remember that $5.03 lunch? We read $5.03 as five dollars and three cents. Naming decimals (those that don’t represent money) is done in a similar way. We read the number 5.03 as five and three hundredths.

We sometimes need to translate a number written in decimal notation into words. As shown in Figure 2, we write the amount on a check in both words and numbers.

An image of a check is shown. The check is made out to Jane Doe. It shows the number $152.65 and says in words, “One hundred fifty two and 65 over 100 dollars.”
When we write a check, we write the amount as a decimal number as well as in words. The bank looks at the check to make sure both numbers match. This helps prevent errors.
This table provides a step-by-step guide on how to correctly name and pronounce a decimal number, using 15.68 as an example.
Let’s try naming a decimal, such as 15.68.
We start by naming the number to the left of the decimal. fifteen______
We use the word “and” to indicate the decimal point. fifteen and_____
Then we name the number to the right of the decimal point as if it were a whole number. fifteen and sixty-eight_____
Last, name the decimal place of the last digit. fifteen and sixty-eight hundredths

The number 15.68 is read fifteen and sixty-eight hundredths.

Name a decimal number.

  • Name the number to the left of the decimal point.
  • Write “and” for the decimal point.
  • Name the “number” part to the right of the decimal point as if it were a whole number.
  • Name the decimal place of the last digit.

Name each decimal: ⓐ 4.3 ⓑ 2.45 ⓒ 0.009 ⓓ −15.571.

Solution

Solution

Step-by-step guide demonstrating how to write the decimal number 4.3 in words, illustrating each component of its verbal representation.
ⓐ
4.3
Name the number to the left of the decimal point. four_____
Write "and" for the decimal point. four and_____
Name the number to the right of the decimal point as if it were a whole number. four and three_____
Name the decimal place of the last digit. four and three tenths
Illustrates the step-by-step process of writing the decimal number 2.45 in words.
ⓑ
2.45
Name the number to the left of the decimal point. two_____
Write "and" for the decimal point. two and_____
Name the number to the right of the decimal point as if it were a whole number. two and forty-five_____
Name the decimal place of the last digit. two and forty-five hundredths
Steps to name the decimal number 0.009, detailing how to identify its whole and fractional parts.
ⓒ
0.009
Name the number to the left of the decimal point. Zero is the number to the left of the decimal; it is not included in the name.
Name the number to the right of the decimal point as if it were a whole number. nine_____
Name the decimal place of the last digit. nine thousandths
Illustrates the step-by-step process of converting the decimal number -15.571 into its verbal form.
ⓓ
−15.571
Name the number to the left of the decimal point. negative fifteen
Write "and" for the decimal point. negative fifteen and_____
Name the number to the right of the decimal point as if it were a whole number. negative fifteen and five hundred seventy-one_____
Name the decimal place of the last digit. negative fifteen and five hundred seventy-one thousandths

Name each decimal:

ⓐ 6.7 ⓑ 19.58 ⓒ 0.018 ⓓ −2.053

Solution
  1. ⓐ six and seven tenths
  2. ⓑ nineteen and fifty-eight hundredths
  3. ⓒ eighteen thousandths
  4. ⓓ negative two and fifty-three thousandths

Name each decimal:

ⓐ 5.8 ⓑ 3.57 ⓒ 0.005 ⓓ −13.461

Solution
  1. ⓐ five and eight tenths
  2. ⓑ three and fifty-seven hundredths
  3. ⓒ five thousandths
  4. ⓓ negative thirteen and four hundred sixty-one thousandths

Write Decimals

Now we will translate the name of a decimal number into decimal notation. We will reverse the procedure we just used.

Let’s start by writing the number six and seventeen hundredths:

Step-by-step conversion of 'six and seventeen hundredths' to its decimal form (6.17).
six and seventeen hundredths
The word and tells us to place a decimal point. ___.___
The word before and is the whole number; write it to the left of the decimal point. 6._____
The decimal part is seventeen hundredths.
Mark two places to the right of the decimal point for hundredths.
6._ _
Write the numerals for seventeen in the places marked. 6.17

Write fourteen and thirty-seven hundredths as a decimal.

Solution

Solution

Steps for converting a number written in words to its decimal form, demonstrated with 'fourteen and thirty-seven hundredths'.
fourteen and thirty-seven hundredths
Place a decimal point under the word ‘and’. ______. _________
Translate the words before ‘and’ into the whole number and place it to the left of the decimal point. 14. _________
Mark two places to the right of the decimal point for “hundredths”. 14.__ __
Translate the words after “and” and write the number to the right of the decimal point. 14.37
Fourteen and thirty-seven hundredths is written 14.37.

Write as a decimal: thirteen and sixty-eight hundredths.

Solution

13.68

Write as a decimal: five and eight hundred ninety-four thousandths.

Solution

5.894

Write a decimal number from its name.

  1. Look for the word “and”—it locates the decimal point.
  2. Mark the number of decimal places needed to the right of the decimal point by noting the place value indicated by the last word.
    • Place a decimal point under the word “and.” Translate the words before “and” into the whole number and place it to the left of the decimal point.
    • If there is no “and,” write a “0” with a decimal point to its right.
  3. Translate the words after “and” into the number to the right of the decimal point. Write the number in the spaces—putting the final digit in the last place.
  4. Fill in zeros for place holders as needed.

The second bullet in Step 2 is needed for decimals that have no whole number part, like ‘nine thousandths’. We recognize them by the words that indicate the place value after the decimal – such as ‘tenths’ or ‘hundredths.’ Since there is no whole number, there is no ‘and.’ We start by placing a zero to the left of the decimal and continue by filling in the numbers to the right, as we did above.

Write twenty-four thousandths as a decimal.

Solution

Solution

twenty-four thousandths
Look for the word "and". There is no "and" so start with 0
0.
To the right of the decimal point, put three decimal places for thousandths. A math problem template showing '0.' followed by three blank spaces, labeled respectively 'tenths', 'hundredths', and 'thousandths', illustrating decimal place values.
Write the number 24 with the 4 in the thousandths place. A decimal number is shown as 0. _ (blank for tenths), 2 (hundredths), and 4 (thousandths). The place values 'tenths', 'hundredths', and 'thousandths' are explicitly labeled below their respective positions.
Put zeros as placeholders in the remaining decimal places. 0.024
So, twenty-four thousandths is written 0.024

Write as a decimal: fifty-eight thousandths.

Solution

0.058

Write as a decimal: sixty-seven thousandths.

Solution

0.067

Before we move on to our next objective, think about money again. We know that $1 is the same as $1.00. The way we write $1(or$1.00) depends on the context. In the same way, integers can be written as decimals with as many zeros as needed to the right of the decimal.

5=5.0−2=−2.05=5.00−2=−2.005=5.000−2=−2.000
and so on…

Convert Decimals to Fractions or Mixed Numbers

We often need to rewrite decimals as fractions or mixed numbers. Let’s go back to our lunch order to see how we can convert decimal numbers to fractions. We know that $5.03 means 5 dollars and 3 cents. Since there are 100 cents in one dollar, 3 cents means 3100 of a dollar, so 0.03=3100.

We convert decimals to fractions by identifying the place value of the 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

For our $5.03 lunch, we can write the decimal 5.03 as a mixed number.

5.03=53100

Notice that when the number to the left of the decimal is zero, we get a proper fraction. When the number to the left of the decimal is not zero, we get a mixed number.

Convert a decimal number to a fraction or mixed number.

  1. Look at the number to the left of the decimal.
    • If it is zero, the decimal converts to a proper fraction.
    • If it is not zero, the decimal converts to a mixed number.
      • Write the whole number.
  2. Determine the place value of the final digit.
  3. Write the fraction.
    • numerator—the ‘numbers’ to the right of the decimal point
    • denominator—the place value corresponding to the final digit
  4. Simplify the fraction, if possible.

Write each of the following decimal numbers as a fraction or a mixed number:

ⓐ 4.09 ⓑ 3.7 ⓒ −0.286

Solution

Solution

ⓐ
4.09
There is a 4 to the left of the decimal point.
Write "4" as the whole number part of the mixed number.
The number 4 is followed by a fraction with an empty square numerator and an empty square denominator, suggesting an incomplete mathematical expression or a prompt to fill in the blanks.
Determine the place value of the final digit. A partially obscured image displays a decimal number, with '4.' visible, followed by '0' and '9'. Below '0' is the word 'tenths' in light blue, and below '9' is 'hundredths' in light blue, indicating place values.
Write the fraction.
Write 9 in the numerator as it is the number to the right of the decimal point.
A mathematical expression showing a mixed number where the whole number is 4, the numerator of the fraction is 9, and the denominator is represented by an empty rectangle or box, indicating a missing value.
Write 100 in the denominator as the place value of the final digit, 9, is hundredth. A mixed number is displayed, consisting of the whole number 4, followed by the fraction 9/100, where 9 is the numerator and 100 is the denominator. The expression represents four and nine hundredths.
The fraction is in simplest form. The image shows the conversion of the decimal number 4.09 into a mixed fraction, expressed as 'So, 4.09 = 4 9/100' on a white background.

Did you notice that the number of zeros in the denominator is the same as the number of decimal places?

ⓑ
3.7
There is a 3 to the left of the decimal point.
Write "3" as the whole number part of the mixed number.
The number 3 is shown next to a horizontal fraction line, with an empty square box above the line and another empty square box below the line, against a white background.
Determine the place value of the final digit. The number '3. 7' is displayed above the word 'tenths', indicating the value 3 and 7 tenths.
Write the fraction.
Write 7 in the numerator as it is the number to the right of the decimal point.
A mathematical expression showing the number 3 next to a fraction bar with the number 7 above it and an empty square below it, representing an incomplete mixed number or fraction.
Write 10 in the denominator as the place value of the final digit, 7, is tenths. The mixed number 3 and 7/10 is displayed in a dark blue font against a white background.
The fraction is in simplest form. A mathematical equation showing the conversion of a decimal to a mixed number, specifically 'So, 3.7 = 3 7/10'.
ⓒ
−0.286
There is a 0 to the left of the decimal point.
Write a negative sign before the fraction.
A mathematical expression featuring a negative sign followed by a fraction, where both the numerator and denominator are depicted as empty square placeholders.
Determine the place value of the final digit and write it in the denominator. A decimal number -0.286 is shown with its place values labeled: 2 is in the tenths place, 8 is in the hundredths place, and 6 is in the thousandths place.
Write the fraction.
Write 286 in the numerator as it is the number to the right of the decimal point.
Write 1,000 in the denominator as the place value of the final digit, 6, is thousandths.
A negative fraction is displayed, with the number 286 in the numerator and 1000 in the denominator.
We remove a common factor of 2 to simplify the fraction. The image displays the negative fraction -143/500, a mathematical expression showing a division of 143 by 500 with a negative sign preceding it. It is presented on a plain white background.

Write as a fraction or mixed number. Simplify the answer if possible.

ⓐ 5.3 ⓑ 6.07 ⓒ −0.234

Solution
  1. ⓐ 5310
  2. ⓑ 67100
  3. ⓒ −117500

Write as a fraction or mixed number. Simplify the answer if possible.

ⓐ 8.7 ⓑ 1.03 ⓒ −0.024

Solution
  1. ⓐ 8710
  2. ⓑ 13100
  3. ⓒ −3125

Locate Decimals on the Number Line

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

Locate 0.4 on a number line.

Solution

Solution

The decimal 0.4 is equivalent to 410, so 0.4 is located between 0 and 1. On a number line, divide the interval between 0 and 1 into 10 equal parts and place marks to separate the parts.

Label the marks 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.
A number line is shown with 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, and 1.0 labeled. There is a red dot at 0.4.

Locate 0.6 on a number line.

Solution


This image shows a number line from 0.0 to 1.0 and segmented into tenths.  A point is plotted at 0.6 on the number line.

Locate 0.9 on a number line.

Solution


This image shows a number line from 0.0 to 1.0 and segmented into tenths.  A point is plotted at 0.9 on the number line.

Locate −0.74 on a number line.

Solution

Solution

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 multiples of -0.10 in the interval between 0 and −1 (−0.10, −0.20, etc.) and mark −0.74 between −0.70 and −0.80, a little closer to −0.70.
A number line is shown with negative 1.00, negative 0.90, negative 0.80, negative 0.70, negative 0.60, negative 0.50, negative 0.40, negative 0.30, negative 0.20, negative 0.10, and 0.00 labeled. There is a red dot between negative 0.80 and negative 0.70 labeled as negative 0.74.

Locate −0.63 on a number line.

Solution


This image shows a number line from -1.00 to 0.00 . A point is plotted at negative 0.63 on the number line.

Locate −0.25 on a number line.

Solution


This image shows a number line from -1.00 to 0.00 . A point is plotted at negative 0.25 on the number line.

Order Decimals

Which is larger, 0.04 or 0.40?

If you think of this as money, you know that $0.40 (forty cents) is greater than $0.04 (four cents). So,

0.40>0.04

In previous chapters, we used the number line to order numbers.

a<b‘ais less thanb’whenais to the left ofbon the number linea>b‘ais greater thanb’whenais to the right ofbon the number line

Where are 0.04 and 0.40 located on the number line?

A number line is shown with 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, and 1.0 labeled. There is a red dot between 0.0 and 0.1 labeled as 0.04. There is another red dot at 0.4.

We see that 0.40 is to the right of 0.04. So we know 0.40>0.04.

How does 0.31 compare to 0.308? This doesn’t translate into money to make the comparison easy. But if we convert 0.31 and 0.308 to fractions, we can tell which is larger.

0.31 0.308
Convert to fractions. 31100 3081000
We need a common denominator to compare them. A fraction with a numerator of 31 multiplied by 10 and a denominator of 100 multiplied by 10. The number 10 is highlighted in red in both the numerator and denominator. 3081000
3101000 3081000

Because 310>308, we know that 3101000>3081000. Therefore, 0.31>0.308.

Notice what we did in converting 0.31 to a fraction—we started with the fraction 31100 and ended with the equivalent fraction 3101000. Converting 3101000 back to a decimal gives 0.310. So 0.31 is equivalent to 0.310. Writing zeros at the end of a decimal does not change its value.

31100=3101000and0.31=0.310

If two decimals have the same value, they are said to be equivalent decimals.

0.31=0.310

We say 0.31 and 0.310 are equivalent decimals.

Equivalent Decimals

Two decimals are equivalent decimals if they convert to equivalent fractions.

Remember, writing zeros at the end of a decimal does not change its value.

Order decimals.

  1. Check to see if both numbers have the same number of decimal places. If not, write zeros at the end of the one with fewer digits to make them match.
  2. Compare the numbers to the right of the decimal point as if they were whole numbers.
  3. Order the numbers using the appropriate inequality sign.

Order the following decimals using <or>:

  1. ⓐ 0.64__0.6
  2. ⓑ 0.83__0.803
Solution

Solution

This table illustrates the step-by-step process of comparing decimal numbers, showing how to equalize decimal places and perform the comparison.
ⓐ
0.64__0.6
Check to see if both numbers have the same number of decimal places. They do not, so write one zero at the right of 0.6. 0.64__0.60
Compare the numbers to the right of the decimal point as if they were whole numbers. 64>60
Order the numbers using the appropriate inequality sign. 0.64>0.60

0.64>0.6
Illustrates the step-by-step process of comparing decimal numbers, using 0.83 and 0.803 as an example.
ⓑ
0.83__0.803
Check to see if both numbers have the same number of decimal places. They do not, so write one zero at the right of 0.83. 0.830__0.803
Compare the numbers to the right of the decimal point as if they were whole numbers. 830>803
Order the numbers using the appropriate inequality sign. 0.830>0.803

0.83>0.803

Order each of the following pairs of numbers, using <or>:

ⓐ 0.42__0.4 ⓑ 0.76__0.706

Solution
  1. ⓐ >
  2. ⓑ >

Order each of the following pairs of numbers, using <or>:

ⓐ 0.1__0.18 ⓑ 0.305__0.35

Solution
  1. ⓐ <
  2. ⓑ <

When we order negative decimals, it is important to remember how to order negative integers. Recall that larger numbers are to the right on the number line. For example, because −2 lies to the right of −3 on the number line, we know that −2>−3. Similarly, smaller numbers lie to the left on the number line. For example, because −9 lies to the left of −6 on the number line, we know that −9<−6.

A number line is shown with integers from negative 10 to 0. Blue dots are placed on negative nine and negative six. Red dots are placed at negative two and negative three.

If we zoomed in on the interval between 0 and −1, we would see in the same way that −0.2>−0.3and−0.9<−0.6.

Use <or> to order. −0.1__−0.8.

Solution

Solution

Illustrates the step-by-step process and reasoning for comparing two negative decimal numbers, -0.1 and -0.8.
−0.1__−0.8
Write the numbers one under the other, lining up the decimal points. −0.1

−0.8
They have the same number of digits.
Since −1>−8,−1 tenth is greater than −8 tenths. −0.1>−0.8

Order each of the following pairs of numbers, using <or>:

−0.3___−0.5

Solution

>

Order each of the following pairs of numbers, using <or>:

−0.6___−0.7

Solution

>

Round Decimals

In the United States, gasoline prices are usually written with the decimal part as thousandths of a dollar. For example, a gas station might post the price of unleaded gas at $3.279 per gallon. But if you were to buy exactly one gallon of gas at this price, you would pay $3.28, because the final price would be rounded to the nearest cent. In Whole Numbers, we saw that we round numbers to get an approximate value when the exact value is not needed. Suppose we wanted to round $2.72 to the nearest dollar. Is it closer to $2 or to $3? What if we wanted to round $2.72 to the nearest ten cents; is it closer to $2.70 or to $2.80? The number lines in Figure 3 can help us answer those questions.

In part a, a number line is shown with 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9 and 3. There is a dot between 2.7 and 2.8 labeled as 2.72.  In part b, a number line is shown with 2.70, 2.71, 2.72, 2.73, 2.74, 2.75, 2.76, 2.77, 2.78, 2.79, and 2.80. There is a dot at 2.72.
ⓐ We see that 2.72 is closer to 3 than to 2. So, 2.72 rounded to the nearest whole number is 3.
ⓑ We see that 2.72 is closer to 2.70 than 2.80. So we say that 2.72 rounded to the nearest tenth is 2.7.

Can we round decimals without number lines? Yes! We use a method based on the one we used to round whole numbers.

Round a decimal.

  1. Locate the given place value and mark it with an arrow.
  2. Underline the digit to the right of the given place value.
  3. Is this 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, removing all digits to the right of the given place value.

Round 18.379 to the nearest hundredth.

Solution

Solution

The number 18.379 is prominently displayed on a white background.
Locate the hundredths place and mark it with an arrow. The digit 7 occupies the hundredths place in the decimal number 18.379, as indicated by the arrow pointing from 'hundredths place' to the 7.
Underline the digit to the right of the 7. An arrow points from the text 'hundredths place' to the digit '7' in the number 18.379, indicating its position.
Because 9 is greater than or equal to 5, add 1 to the 7. An image illustrating a numerical operation, where the number 18.379 is shown with instructions to 'delete' the '.379' part and 'add 1' to the integer portion, effectively rounding up to 19.
Rewrite the number, deleting all digits to the right of the hundredths place. The numbers 18.38 are displayed in a dark teal font against a white background.
18.38 is 18.379 rounded to the nearest hundredth.

Round to the nearest hundredth: 1.047.

Solution

1.05

Round to the nearest hundredth: 9.173.

Solution

9.17

Round 18.379 to the nearest ⓐ tenth ⓑ whole number.

Solution

Solution

ⓐ Round 18.379 to the nearest tenth.
The number 18.379 is displayed in a dark teal font against a white background.
Locate the tenths place and mark it with an arrow. An arrow points from the text 'tenths place' to the digit '3' in the number '18.379', indicating the tenths place value in a decimal.
Underline the digit to the right of the tenths digit. An arrow points from 'tenths place' to the number 18.379, with the '3' underlined, illustrating the tenths place value in a decimal.
Because 7 is greater than or equal to 5, add 1 to the 3. Illustration of rounding 18.379 to the nearest whole number. Because the tenths digit '3' (indicated by 'add 1') is less than 5, the decimal part is 'delete'd, resulting in 18.
Rewrite the number, deleting all digits to the right of the tenths place. The number 18.4 is displayed in a teal or bluish-green color against a clean white background.
So, 18.379 rounded to the nearest tenth is 18.4.
ⓑ Round 18.379 to the nearest whole number.
A numerical value, '18.379', is displayed in a dark teal font against a plain white background.
Locate the ones place and mark it with an arrow. An arrow points from the text 'ones place' to the number '18.379', illustrating the concept of place value for the digit 8 in the ones place.
Underline the digit to the right of the ones place. An illustration of place value, with an arrow pointing from 'ones place' to the digit '8' in 18.379, demonstrating its position. The digit '3' is also underlined.
Since 3 is not greater than or equal to 5, do not add 1 to the 8. An image illustrating a numerical operation, showing the number 18.379 with instructions to 'delete' the .379 portion and 'do not add 1' (implying no rounding up) for the remaining 18.
Rewrite the number, deleting all digits to the right of the ones place. The number 18 is displayed in a dark blue-grey font on a clean white background, standing out with its simple yet clear presentation.
So 18.379 rounded to the nearest whole number is 18.

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

Solution
  1. ⓐ 6.58
  2. ⓑ 6.6
  3. ⓒ 7

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

Solution
  1. ⓐ 15.218
  2. ⓑ 15.22
  3. ⓒ 15.2

ACCESS ADDITIONAL ONLINE RESOURCES

  • Introduction to Decimal Notation
  • Write a Number in Decimal Notation from Words
  • Identify Decimals on the Number Line
  • Rounding Decimals
  • Writing a Decimal as a Simplified Fraction

Key Concepts

  • Name a decimal number.
    1. Name the number to the left of the decimal point.
    2. Write “and” for the decimal point.
    3. Name the “number” part to the right of the decimal point as if it were a whole number.
    4. Name the decimal place of the last digit.
  • Write a decimal number from its name.
    1. Look for the word “and”—it locates the decimal point.
      Place a decimal point under the word “and.” Translate the words before “and” into the whole number and place it to the left of the decimal point.
      If there is no “and,” write a “0” with a decimal point to its right.
    2. Mark the number of decimal places needed to the right of the decimal point by noting the place value indicated by the last word.
    3. Translate the words after “and” into the number to the right of the decimal point. Write the number in the spaces—putting the final digit in the last place.
    4. Fill in zeros for place holders as needed.
  • Convert a decimal number to a fraction or mixed number.
    1. Look at the number to the left of the decimal.
      If it is zero, the decimal converts to a proper fraction.
      If it is not zero, the decimal converts to a mixed number.
      Write the whole number.
    2. Determine the place value of the final digit.
    3. Write the fraction. numerator—the ‘numbers’ to the right of the decimal point denominator—the place value corresponding to the final digit
    4. Simplify the fraction, if possible.
  • Order decimals.
    1. Check to see if both numbers have the same number of decimal places. If not, write zeros at the end of the one with fewer digits to make them match.
    2. Compare the numbers to the right of the decimal point as if they were whole numbers.
    3. Order the numbers using the appropriate inequality sign.
  • Round a decimal.
    1. Locate the given place value and mark it with an arrow.
    2. Underline the digit to the right of the given place value.
    3. Is this 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, removing all digits to the right of the given place value.

Practice Makes Perfect

Name Decimals

In the following exercises, name each decimal.

5.5

Solution

five and five tenths

7.8

5.01

Solution

five and one hundredth

14.02

8.71

Solution

eight and seventy-one hundredths

2.64

0.002

Solution

two thousandths

0.005

0.381

Solution

three hundred eighty-one thousandths

0.479

−17.9

Solution

negative seventeen and nine tenths

−31.4

Write Decimals

In the following exercises, translate the name into a decimal number.

Eight and three hundredths

Solution

8.03

Nine and seven hundredths

Twenty-nine and eighty-one hundredths

Solution

29.81

Sixty-one and seventy-four hundredths

Seven tenths

Solution

0.7

Six tenths

One thousandth

Solution

0.001

Nine thousandths

Twenty-nine thousandths

Solution

0.029

Thirty-five thousandths

Negative eleven and nine ten-thousandths

Solution

−11.0009

Negative fifty-nine and two ten-thousandths

Thirteen and three hundred ninety-five ten thousandths

Solution

13.0395

Thirty and two hundred seventy-nine thousandths

Convert Decimals to Fractions or Mixed Numbers

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

1.99

Solution

199100

5.83

15.7

Solution

15710

18.1

0.239

Solution

2391000

0.373

0.13

Solution

13100

0.19

0.011

Solution

111000

0.049

−0.00007

Solution

−7100000

−0.00003

6.4

Solution

625

5.2

7.05

Solution

7120

9.04

4.006

Solution

43500

2.008

10.25

Solution

1014

12.75

1.324

Solution

181250

2.482

14.125

Solution

1418

20.375

Locate Decimals on the Number Line

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

0.8

Solution


There is a number line shown with integers from negative 4 to 4. There is a red dot between 0 and 1 labeled 0.8.

0.3

−0.2

Solution


There is a number line shown with integers from negative 4 to 4. There is a red dot between negative 1 and  0 labeled negative 0.2.

−0.9

3.1

Solution


This is an image of a number line. It spans from negative 5 on the left to 5 on the right. To the right of 0 are tick marks with the numbers 1, 2, 3, 4, 5 on the number line. To the left of the zero are tick marks with the numbers negative 1, negative 2, negative 3, negative 4, and negative 5. A point is plotted at 3.1.

2.7

−2.5

Solution


There is a number line shown with integers from negative 4 to 4. There is a red dot between negative 3 and negative 2 labeled negative 2.5.

−1.6

Order Decimals

In the following exercises, order each of the following pairs of numbers, using <or>.

0.9__0.6

Solution

>

0.7__0.8

0.37__0.63

Solution

<

0.86__0.69

0.6__0.59

Solution

>

0.27__0.3

0.91__0.901

Solution

>

0.415__0.41

−0.5__−0.3

Solution

<

−0.1_−0.4

−0.62_−0.619

Solution

<

−7.31_−7.3

Round Decimals

In the following exercises, round each number to the nearest tenth.

0.67

Solution

0.7

0.49

2.84

Solution

2.8

4.63

In the following exercises, round each number to the nearest hundredth.

0.845

Solution

0.85

0.761

5.7932

Solution

5.79

3.6284

0.299

Solution

0.30

0.697

4.098

Solution

4.10

7.096

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

5.781

Solution
  1. ⓐ 5.78
  2. ⓑ 5.8
  3. ⓒ 6

1.638

63.479

Solution
  1. ⓐ 63.48
  2. ⓑ 63.5
  3. ⓒ 63

84.281

Everyday Math

Salary Increase Danny got a raise and now makes $58,965.95 a year. Round this number to the nearest:

ⓐ dollar

ⓑ thousand dollars

ⓒ ten thousand dollars.

Solution
  1. ⓐ $58,966
  2. ⓑ $59,000
  3. ⓒ $60,000

New Car Purchase Selena’s new car cost $23,795.95. Round this number to the nearest:

ⓐ dollar

ⓑ thousand dollars

ⓒ ten thousand dollars.

Sales Tax Hyo Jin lives in San Diego. She bought a refrigerator for $1624.99 and when the clerk calculated the sales tax it came out to exactly $142.186625. Round the sales tax to the nearest ⓐ penny ⓑ dollar.

Solution
  1. ⓐ $142.19
  2. ⓑ $142

Sales Tax Jennifer bought a $1,038.99 dining room set for her home in Cincinnati. She calculated the sales tax to be exactly $67.53435. Round the sales tax to the nearest ⓐ penny ⓑ dollar.

Writing Exercises

How does your knowledge of money help you learn about decimals?

Solution

Answers will vary.

Explain how you write “three and nine hundredths” as a decimal.

Jim ran a 100-meter race in 12.32 seconds. Tim ran the same race in 12.3 seconds. Who had the faster time, Jim or Tim? How do you know?

Solution

Tim had the faster time. 12.3 is less than 12.32, so Tim had the faster time.

Gerry saw a sign advertising postcards marked for sale at “10for0.99¢.” What is wrong with the advertised price?

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 gauge their understanding of decimals, with categories: Confidently, With some help, and No-I don't get it! Tasks include naming, writing, converting, locating, ordering, and rounding decimals.

ⓑ 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.

equivalent decimals
Two decimals are equivalent decimals if they convert to equivalent fractions.

Decimal Operations

Learning Objectives

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

  • Add and subtract decimals
  • Multiply decimals
  • Divide decimals
  • Use decimals in money applications

Before you get started, take this readiness quiz.

Simplify 70100.
If you missed this problem, review Example 1 in Multiply and Divide Fractions.

Solution

710

Multiply 310·910.
If you missed this problem, review Example 7 in Multiply and Divide Fractions.

Solution

27100

Divide −36÷(−9).
If you missed this problem, review Example 3 in Multiply and Divide Integers.

Solution

4

Add and Subtract Decimals

Let’s take one more look at the lunch order from the start of Decimals, this time noticing how the numbers were added together.

A vertical addition problem is shown. The top line shows $3.45 for a sandwich, the next line shows $1.25 for water, and the last line shows $0.33 for tax. The total is shown to be $5.03.

All three items (sandwich, water, tax) were priced in dollars and cents, so we lined up the dollars under the dollars and the cents under the cents, with the decimal points lined up between them. Then we just added each column, as if we were adding whole numbers. By lining up decimals this way, we can add or subtract the corresponding place values just as we did with whole numbers.

Add or subtract decimals.

  1. Write the numbers vertically so the decimal points line up.
  2. Use zeros as place holders, as needed.
  3. Add or subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers.

Add: 3.7+12.4.

Solution

Solution

Step-by-step guide illustrating how to add decimal numbers, using 3.7 + 12.4 as an example.
3.7+12.4
Write the numbers vertically so the decimal points line up. 3.7 +12.4_____
Place holders are not needed since both numbers have the same number of decimal places.
Add the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers. 31.7 +12.4_____ 16.1

Add: 5.7+11.9.

Solution

17.6

Add: 18.32+14.79.

Solution

33.11

Add: 23.5+41.38.

Solution

Solution

23.5+41.38
Write the numbers vertically so the decimal points line up. A vertical addition problem with decimal numbers: 23.5 plus 41.38, with a horizontal line beneath them indicating the sum is to be calculated.
Place 0 as a place holder after the 5 in 23.5, so that both numbers have two decimal places. A vertical addition problem showing 23.50 plus 41.38. The trailing zero in 23.50 is highlighted in red, often to illustrate significant figures or decimal place alignment in math.
Add the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers. A vertical addition problem showing 23.50 plus 41.38 equals 64.88, demonstrating basic arithmetic with decimal numbers.

Add: 4.8+11.69.

Solution

16.49

Add: 5.123+18.47.

Solution

23.593

How much change would you get if you handed the cashier a $20 bill for a $14.65 purchase? We will show the steps to calculate this in the next example.

Subtract: 20−14.65.

Solution

Solution

20−14.65
Write the numbers vertically so the decimal points line up. Remember 20 is a whole number, so place the decimal point after the 0. A vertical subtraction problem showing the number 20. above 14.65, with a minus sign to the left of 14.65 and a horizontal line below it, indicating a calculation.
Place two zeros after the decimal point in 20, as place holders so that both numbers have two decimal places. Decimal subtraction problem setup: 20.00 - 14.65. The red '00' in 20.00 draws attention to borrowing digits during calculation.
Subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers. A vertical subtraction problem: 20.00 - 14.65 = 5.35, illustrating the borrowing process clearly from left to right with crossed-out digits and new values above them.

Subtract:

10−9.58.

Solution

0.42

Subtract:

50−37.42.

Solution

12.58

Subtract: 2.51−7.4.

Solution

Solution

If we subtract 7.4 from 2.51, the answer will be negative since 7.4>2.51. To subtract easily, we can subtract 2.51 from 7.4. Then we will place the negative sign in the result.
2.51−7.4
Write the numbers vertically so the decimal points line up. A vertical subtraction problem showing 7.4 minus 2.51, ready for calculation.
Place zero after the 4 in 7.4 as a place holder, so that both numbers have two decimal places. A vertical subtraction problem showing 7.40 minus 2.51, with the '0' in 7.40 highlighted in red to indicate the column where borrowing will begin for decimal subtraction.
Subtract and place the decimal in the answer. A vertical subtraction problem shows 7.40 minus 2.51, with the result being 4.89.
Remember that we are really subtracting 2.51−7.4 so the answer is negative. 2.51−7.4=−4.89

Subtract: 4.77−6.3.

Solution

−1.53

Subtract: 8.12−11.7.

Solution

−3.58

Multiply Decimals

Multiplying decimals is very much like multiplying whole numbers—we just have to determine where to place the decimal point. The procedure for multiplying decimals will make sense if we first review multiplying fractions.

Do you remember how to multiply fractions? To multiply fractions, you multiply the numerators and then multiply the denominators.

So let’s see what we would get as the product of decimals by converting them to fractions first. We will do two examples side-by-side in Table 5. Look for a pattern.

A B
(0.3)(0.7) (0.2)(0.46)
Convert to fractions. (310)(710) (210)(46100)
Multiply. 21100 921000
Convert back to decimals. 0.21 0.092

There is a pattern that we can use. In A, we multiplied two numbers that each had one decimal place, and the product had two decimal places. In B, we multiplied a number with one decimal place by a number with two decimal places, and the product had three decimal places.

How many decimal places would you expect for the product of (0.01)(0.004)? If you said “five”, you recognized the pattern. When we multiply two numbers with decimals, we count all the decimal places in the factors—in this case two plus three—to get the number of decimal places in the product—in this case five.

The top line says 0.01 times 0.004 equals 0.00004. Below the 0.01, it says 2 places. Below the 0.004, it says 3 places. Below the 0.00004, it says 5 places. The bottom line says 1 over 100 times 4 over 1000 equals 4 over 100,000.

Once we know how to determine the number of digits after the decimal point, we can multiply decimal numbers without converting them to fractions first. The number of decimal places in the product is the sum of the number of decimal places in the factors.

The rules for multiplying positive and negative numbers apply to decimals, too, of course.

Multiplying Two Numbers

When multiplying two numbers,
  • if their signs are the same, the product is positive.
  • if their signs are different, the product is negative.

When you multiply signed decimals, first determine the sign of the product and then multiply as if the numbers were both positive. Finally, write the product with the appropriate sign.

Multiply decimal numbers.

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

Multiply: (3.9)(4.075).

Solution

Solution

(3.9)(4.075)
Determine the sign of the product. The signs are the same. The product will be positive.
Write the numbers in vertical format, lining up the numbers on the right. A vertical multiplication problem shows 4.075 multiplied by 3.9, with a line underneath to indicate the calculation is set up to be performed.
Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points. A long multiplication problem is displayed, showing 4.075 multiplied by 3.9. The intermediate products are 36675 and 12225, summing to 158925, but the decimal point is not yet placed in the final answer.
Place the decimal point. 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 decimals: 4.075 by 3.9. It demonstrates summing the decimal places of the factors (3+1) to correctly place the decimal in the final product, yielding 15.8925.
The product is positive. (3.9)(4.075)=15.8925

Multiply: 4.5(6.107).

Solution

27.4815

Multiply: 10.79(8.12).

Solution

87.6148

Multiply: (−8.2)(5.19).

Solution

Solution

(−8.2)(5.19)
The signs are different. The product will be negative.
Write in vertical format, lining up the numbers on the right. 5.19 ×8.2_____
Multiply. 5.19 ×8.2_____ 1038 4152_____ 42558
The image shows how to determine the decimal point placement in the product of (-8.2) and (5.19). It states to 'Place the decimal point 3 places from the right,' indicating that -8.2 has 1 decimal place and 5.19 has 2 decimal places. 5.19 ×8.2_____ 1038 4152_____ 42.558
The product is negative. (−8.2)(5.19)=−42.558

Multiply: (4.63)(−2.9).

Solution

−13.427

Multiply: (−7.78)(4.9).

Solution

−38.122

In the next example, we’ll need to add several placeholder zeros to properly place the decimal point.

Multiply: (0.03)(0.045).

Solution

Solution

(0.03)(0.045)
The product is positive.
Write in vertical format, lining up the numbers on the right. A vertical multiplication problem showing 0.045 multiplied by 0.03, set up for manual calculation.
Multiply. A multiplication problem is displayed, showing 0.045 multiplied by 0.03, with a partial product of 135 below the line, indicating an intermediate step in calculating the product of these two decimal numbers.
Illustrates the rule for decimal placement in multiplication: sum the decimal places of the factors. For (0.03)(0.045), 2 + 3 = 5, so the product has 5 decimal places.
Add zeros as needed to get the 5 places.
An example of decimal multiplication: 0.045 multiplied by 0.03 equals 0.00135. The blue arrow illustrates counting decimal places to correctly position the decimal point in the product.
The product is positive. (0.03)(0.045)=0.00135

Multiply: (0.04)(0.087).

Solution

0.00348

Multiply: (0.09)(0.067).

Solution

0.00603

Multiply by Powers of 10

In many fields, especially in the sciences, it is common to multiply decimals by powers of 10. Let’s see what happens when we multiply 1.9436 by some powers of 10.

The top row says 1.9436 times 10, then 1.9436 times 100, then 1.9436 times 1000. Below each is a vertical multiplication problem. These show that 1.9436 times 10 is 19.4360, 1.9436 times 100 is 194.3600, and 1.9436 times 1000 is 1943.6000.

Look at the results without the final zeros. Do you notice a pattern?

1.9436(10)=19.4361.9436(100)=194.361.9436(1000)=1943.6

The number of places that the decimal point moved is the same as the number of zeros in the power of ten. Table 9 summarizes the results.

Multiply by Number of zeros Number of places decimal point moves
10 1 1 place to the right
100 2 2 places to the right
1,000 3 3 places to the right
10,000 4 4 places to the right

We can use this pattern as a shortcut to multiply by powers of ten instead of multiplying using the vertical format. We can count the zeros in the power of 10 and then move the decimal point that same of places to the right.

So, for example, to multiply 45.86 by 100, move the decimal point 2 places to the right.

45.86 times 100 is shown to equal 4586. There is an arrow from the decimal going over 2 places from after the 5 to after the 6.

Sometimes when we need to move the decimal point, there are not enough decimal places. In that case, we use zeros as placeholders. For example, let’s multiply 2.4 by 100. We need to move the decimal point 2 places to the right. Since there is only one digit to the right of the decimal point, we must write a 0 in the hundredths place.

2.4 times 100 is shown to equal 240. There is an arrow from the decimal going over 2 places from after the 2 to after the 0.

Multiply a decimal by a power of 10.

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

Multiply 5.63 by factors of ⓐ 10 ⓑ 100ⓒ 1000.

Solution
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.

ⓐ
56.3(10)
There is 1 zero in 10, so move the decimal point 1 place to the right. A blue arrow highlights the number 5.63.
56.3
ⓑ
5.63(100)
There are 2 zeros in 100, so move the decimal point 2 places to the right. The number 5.63 is displayed above a blue, wavy upward-pointing arrow, indicating a trend or value increase.
563
ⓒ
5.63(1000)
There are 3 zeros in 1000, so move the decimal point 3 places to the right. A blue squiggly arrow points upwards from the number 5.63, suggesting an increase or upward trend.
A zero must be added at the end. 5,630

Multiply 2.58 by factors of ⓐ 10 ⓑ 100 ⓒ 1000.

Solution
  1. ⓐ 25.8
  2. ⓑ 258
  3. ⓒ 2,580

Multiply 14.2 by factors of ⓐ 10 ⓑ 100 ⓒ 1000.

Solution
  1. ⓐ 142
  2. ⓑ 1,420
  3. ⓒ 14,200

Divide Decimals

Just as with multiplication, division of decimals is very much like dividing whole numbers. We just have to figure out where the decimal point must be placed.

To understand decimal division, let’s consider the multiplication problem

(0.2)(4)=0.8

Remember, a multiplication problem can be rephrased as a division problem. So we can write

0.8÷4=0.2

We can think of this as “If we divide 8 tenths into four groups, how many are in each group?” Figure 1 shows that there are four groups of two-tenths in eight-tenths. So 0.8÷4=0.2.

A number line is shown with 0, 0.2, 0.4, 0.6, 0.8, and 1. There are braces showing a distance of 0.2 between each adjacent set of 2 numbers.

Using long division notation, we would write

A division problem is shown. 0.8 is on the inside of the division sign, 4 is on the outside. Above the division sign is 0.2.

Notice that the decimal point in the quotient is directly above the decimal point in the dividend.

To divide a decimal by a whole number, we place the decimal point in the quotient above the decimal point in the dividend and then divide as usual. Sometimes we need to use extra zeros at the end of the dividend to keep dividing until there is no remainder.

Divide a decimal by a whole number.

  1. Write as long division, placing the decimal point in the quotient above the decimal point in the dividend.
  2. Divide as usual.

Divide: 0.12÷3.

Solution

Solution

0.12÷3
Write as long division, placing the decimal point in the quotient above the decimal point in the dividend. A long division problem showing 0.12 divided by 3, with a red dot above the first 1 indicating the decimal placement for the quotient.
Divide as usual. Since 3 does not go into 0 or 1 we use zeros as placeholders. Long division of 0.12 by 3, showing the quotient 0.04 and the step-by-step process with a final remainder of 0.
0.12÷3=0.04

Divide: 0.28÷4.

Solution

0.07

Divide: 0.56÷7.

Solution

0.08

In everyday life, we divide whole numbers into decimals—money—to find the price of one item. For example, suppose a case of 24 water bottles cost $3.99. To find the price per water bottle, we would divide $3.99 by 24, and round the answer to the nearest cent (hundredth).

Divide: $3.99÷24.

Solution

Solution

$3.99÷24
Place the decimal point in the quotient above the decimal point in the dividend. A long division problem showing 3.99 being divided by 24, with a decimal point placed above the '3' indicating the beginning of the quotient.
Divide as usual. When do we stop? Since this division involves money, we round it to the nearest cent (hundredth). To do this, we must carry the division to the thousandths place. A long division calculation showing 3.990 divided by 24, yielding a quotient of 0.166 with a remainder of 6.
Round to the nearest cent. $0.166≈$0.17
$3.99÷24≈$0.17

This means the price per bottle is 17 cents.

Divide: $6.99÷36.

Solution

$0.19

Divide: $4.99÷12.

Solution

$0.42

Divide a Decimal by Another Decimal

So far, we have divided a decimal by a whole number. What happens when we divide a decimal by another decimal? Let’s look at the same multiplication problem we looked at earlier, but in a different way.

(0.2)(4)=0.8

Remember, again, that a multiplication problem can be rephrased as a division problem. This time we ask, “How many times does 0.2 go into 0.8?” Because (0.2)(4)=0.8, we can say that 0.2 goes into 0.8 four times. This means that 0.8 divided by 0.2 is 4.

0.8÷0.2=4
A number line is shown with 0, 0.2, 0.4, 0.6, 0.8, and 1. There are braces showing a distance of 0.2 between each adjacent set of 2 numbers.

We would get the same answer, 4, if we divide 8 by 2, both whole numbers. Why is this so? Let’s think about the division problem as a fraction.

0.80.2(0.8)10(0.2)10824

We multiplied the numerator and denominator by 10 and ended up just dividing 8 by 2. To divide decimals, we multiply both the numerator and denominator by the same power of 10 to make the denominator a whole number. Because of the Equivalent Fractions Property, we haven’t changed the value of the fraction. The effect is to move the decimal points in the numerator and denominator the same number of places to the right.

We use the rules for dividing positive and negative numbers with decimals, too. 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.

It may help to review the vocabulary for division:

a divided by b is shown with a labeled as the dividend and b labeled as the divisor. Then a over b is shown with a labeled as the divided and b labeled as the divisor. Then a is shown inside a division problem with b on the outside with a labeled as the dividend and b labeled as the divisor.

Divide decimal numbers.

  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 to the right, writing 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: −2.89÷(3.4).

Solution
Solution
Determine the sign of the quotient. The quotient will be negative.
Make the divisor the 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 to the right. A long division problem with decimals, showing 2.89 being divided by 3.4.
Divide. Place the decimal point in the quotient above the decimal point in the dividend. Add zeros as needed until the remainder is zero. A long division calculation showing 28.90 divided by 34, yielding a quotient of 0.85 with a remainder of 0. The steps demonstrate the process of dividing the numbers.
Write the quotient with the appropriate sign. −2.89÷(3.4)=−0.85

Divide: −1.989÷5.1.

Solution

−0.39

Divide: −2.04÷5.1.

Solution

−0.4

Divide: −25.65÷(−0.06).

Solution
Solution
−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 blue arrows indicating the movement of the decimal point two places to the right in both the divisor and the dividend to simplify the division.
Divide.
Place the decimal point in the quotient above the decimal point in the dividend.
A long division calculation showing 2565 divided by 6, yielding a quotient of 427.5.
Write the quotient with the appropriate sign. −25.65÷(−0.06)=427.5

Divide: −23.492÷(−0.04).

Solution

587.3

Divide: −4.11÷(−0.12).

Solution

34.25

Now we will divide a whole number by a decimal number.

Divide: 4÷0.05.

Solution
Solution
4÷0.05
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, adding zeros as needed.
A long division problem is shown where 4.00 is divided by 0.05. Blue arrows beneath the numbers indicate the decimal point being shifted two places to the right in both the divisor and dividend.
Divide.
Place the decimal point in the quotient above the decimal point in the dividend.
A long division problem showing 400 divided by 5, resulting in a quotient of 80. The steps for calculating 40 divided by 5 (yielding 8) and then bringing down the zero, resulting in 0, are shown.
Write the quotient with the appropriate sign. 4÷0.05=80

We can relate this example to money. How many nickels are there in four dollars? Because 4÷0.05=80, there are 80 nickels in $4.

Divide: 6÷0.03.

Solution

200

Divide: 7÷0.02.

Solution

350

Use Decimals in Money Applications

We often apply decimals in real life, and most of the applications involving money. The Strategy for Applications we used in The Language of Algebra gives us a plan to follow to help find the answer. Take a moment to review that strategy now.

Strategy for Applications

  1. Identify what you are asked to find.
  2. Write a phrase that gives the information to find it.
  3. Translate the phrase to an expression.
  4. Simplify the expression.
  5. Answer the question with a complete sentence.

Paul received $50 for his birthday. He spent $31.64 on a video game. How much of Paul’s birthday money was left?

Solution

Solution

Steps for solving a word problem, showing the question, phrase, translation, simplification, and final answer.
What are you asked to find? How much did Paul have left?
Write a phrase. $50 less $31.64
Translate. 50−31.64
Simplify. 18.36
Write a sentence. Paul has $18.36 left.

Nicole earned $35 for babysitting her cousins, then went to the bookstore and spent $18.48 on books and coffee. How much of her babysitting money was left?

Solution

$16.52

Amber bought a pair of shoes for $24.75 and a purse for $36.90. The sales tax was $4.32. How much did Amber spend?

Solution

$65.97

Jessie put 8 gallons of gas in her car. One gallon of gas costs $3.529. How much does Jessie owe for the gas? (Round the answer to the nearest cent.)

Solution

Solution

A step-by-step solution demonstrating how to calculate a total gas cost from a word problem, including translation, simplification, and rounding.
What are you asked to find? How much did Jessie owe for all the gas?
Write a phrase. 8 times the cost of one gallon of gas
Translate. 8($3.529)
Simplify. $28.232
Round to the nearest cent. $28.23
Write a sentence. Jessie owes $28.23 for her gas purchase.

Hector put 13 gallons of gas into his car. One gallon of gas costs $3.175. How much did Hector owe for the gas? Round to the nearest cent.

Solution

$41.28

Christopher bought 5 pizzas for the team. Each pizza cost $9.75. How much did all the pizzas cost?

Solution

$48.75

Four friends went out for dinner. They shared a large pizza and a pitcher of soda. The total cost of their dinner was $31.76. If they divide the cost equally, how much should each friend pay?

Solution

Solution

Steps to solve a word problem involving equal division of a cost among friends.
What are you asked to find? How much should each friend pay?
Write a phrase. $31.76 divided equally among the four friends.
Translate to an expression. $31.76÷4
Simplify. $7.94
Write a sentence. Each friend should pay $7.94 for his share of the dinner.

Six friends went out for dinner. The total cost of their dinner was $92.82. If they divide the bill equally, how much should each friend pay?

Solution

$15.47

Chad worked 40 hours last week and his paycheck was $570. How much does he earn per hour?

Solution

$14.25

Be careful to follow the order of operations in the next example. Remember to multiply before you add.

Marla buys 6 bananas that cost $0.22 each and 4 oranges that cost $0.49 each. How much is the total cost of the fruit?

Solution

Solution

Step-by-step solution for calculating the total cost of fruit, detailing each stage from problem identification to the final answer.
What are you asked to find? How much is the total cost of the fruit?
Write a phrase. 6 times the cost of each banana plus 4 times the cost of each orange
Translate to an expression. 6($0.22)+4($0.49)
Simplify. $1.32+$1.96
Add. $3.28
Write a sentence. Marla's total cost for the fruit is $3.28.

Suzanne buys 3 cans of beans that cost $0.75 each and 6 cans of corn that cost $0.62 each. How much is the total cost of these groceries?

Solution

$5.97

Lydia bought movie tickets for the family. She bought two adult tickets for $9.50 each and four children’s tickets for $6.00 each. How much did the tickets cost Lydia in all?

Solution

$43.00

The Links to Literacy activity "Alexander Who Used to be Rich Last Sunday" will provide you with another view of the topics covered in this section.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Adding and Subtracting Decimals
  • Multiplying Decimals
  • Multiplying by Powers of Ten
  • Dividing Decimals
  • Dividing by Powers of Ten

Key Concepts

  • Add or subtract decimals.
    1. Write the numbers vertically so the decimal points line up.
    2. Use zeros as place holders, as needed.
    3. Add or subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers.
  • Multiply decimal numbers.
    1. Determine the sign of the product.
    2. Write the numbers in vertical format, lining up the numbers on the right.
    3. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
    4. Place the decimal point. The number of decimal places in the product is the sum of the number of decimal places in the factors. If needed, use zeros as placeholders.
    5. Write the product with the appropriate sign.
  • Multiply a decimal by a power of 10.
    1. Move the decimal point to the right the same number of places as the number of zeros in the power of 10.
    2. Write zeros at the end of the number as placeholders if needed.
  • Divide a decimal by a whole number.
    1. Write as long division, placing the decimal point in the quotient above the decimal point in the dividend.
    2. Divide as usual.
  • Divide decimal numbers.
    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 to the right, writing 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.
  • Strategy for Applications
    1. Identify what you are asked to find.
    2. Write a phrase that gives the information to find it.
    3. Translate the phrase to an expression.
    4. Simplify the expression.
    5. Answer the question with a complete sentence.

Practice Makes Perfect

Add and Subtract Decimals

In the following exercises, add or subtract.

16.92+7.56

Solution

24.48

18.37+9.36

256.37−85.49

Solution

170.88

248.25−91.29

21.76−30.99

Solution

−9.23

15.35−20.88

37.5+12.23

Solution

49.73

38.6+13.67

−16.53−24.38

Solution

−40.91

−19.47−32.58

−38.69+31.47

Solution

−7.22

−29.83+19.76

−4.2+(−9.3)

Solution

−13.5

−8.6+(−8.6)

100−64.2

Solution

35.8

100−65.83

72.5−100

Solution

−27.5

86.2−100

15+0.73

Solution

15.73

27+0.87

2.51+40

Solution

42.51

9.38+60

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 Decimals

In the following exercises, multiply.

(0.3)(0.4)

Solution

0.12

(0.6)(0.7)

(0.24)(0.6)

Solution

0.144

(0.81)(0.3)

(5.9)(7.12)

Solution

42.008

(2.3)(9.41)

(8.52)(3.14)

Solution

26.7528

(5.32)(4.86)

(−4.3)(2.71)

Solution

−11.653

(−8.5)(1.69)

(−5.18)(−65.23)

Solution

337.8914

(−9.16)(−68.34)

(0.09)(24.78)

Solution

2.2302

(0.04)(36.89)

(0.06)(21.75)

Solution

1.305

(0.08)(52.45)

(9.24)(10)

Solution

92.4

(6.531)(10)

(55.2)(1,000)

Solution

55,200

(99.4)(1,000)

Divide Decimals

In the following exercises, divide.

0.15÷5

Solution

0.03

0.27÷3

4.75÷25

Solution

0.19

12.04÷43

$8.49÷12

Solution

$0.71

$16.99÷9

$117.25÷48

Solution

$2.44

$109.24÷36

0.6÷0.2

Solution

3

0.8÷0.4

1.44÷(−0.3)

Solution

−4.8

1.25÷(−0.5)

−1.75÷(−0.05)

Solution

35

−1.15÷(−0.05)

5.2÷2.5

Solution

2.08

6.5÷3.25

12÷0.08

Solution

150

5÷0.04

11÷0.55

Solution

20

14÷0.35

Mixed Practice

In the following exercises, simplify.

6(12.4−9.2)

Solution

19.2

3(15.7−8.6)

24(0.5)+(0.3)2

Solution

12.09

35(0.2)+(0.9)2

1.15(26.83+1.61)

Solution

32.706

1.18(46.22+3.71)

$45+0.08($45)

Solution

$48.60

$63+0.18($63)

18÷(0.75+0.15)

Solution

20

27÷(0.55+0.35)

(1.43+0.27)÷(0.9−0.05)

Solution

2

(1.5−0.06)÷(0.12+0.24)

[$75.42+0.18($75.42)]÷5

Solution

$17.80

[$56.31+0.22($56.31)]÷4

Use Decimals in Money Applications

In the following exercises, use the strategy for applications to solve.

Spending money Brenda got $40 from the ATM. She spent $15.11 on a pair of earrings. How much money did she have left?

Solution

$24.89

Spending money Marissa found $20 in her pocket. She spent $4.82 on a smoothie. How much of the $20 did she have left?

Shopping Adam bought a t-shirt for $18.49 and a book for $8.92 The sales tax was $1.65. How much did Adam spend?

Solution

$29.06

Restaurant Roberto’s restaurant bill was $20.45 for the entrée and $3.15 for the drink. He left a $4.40 tip. How much did Roberto spend?

Coupon Emily bought a box of cereal that cost $4.29. She had a coupon for $0.55 off, and the store doubled the coupon. How much did she pay for the box of cereal?

Solution

$3.19

Coupon Diana bought a can of coffee that cost $7.99. She had a coupon for $0.75 off, and the store doubled the coupon. How much did she pay for the can of coffee?

Diet Leo took part in a diet program. He weighed 190 pounds at the start of the program. During the first week, he lost 4.3 pounds. During the second week, he had lost 2.8 pounds. The third week, he gained 0.7 pounds. The fourth week, he lost 1.9 pounds. What did Leo weigh at the end of the fourth week?

Solution

181.7 pounds

Snowpack On April 1, the snowpack at the ski resort was 4 meters deep, but the next few days were very warm. By April 5, the snow depth was 1.6 meters less. On April 8, it snowed and added 2.1 meters of snow. What was the total depth of the snow?

Coffee Noriko bought 4 coffees for herself and her co-workers. Each coffee was $3.75. How much did she pay for all the coffees?

Solution

$15.00

Subway Fare Arianna spends $4.50 per day on subway fare. Last week she rode the subway 6 days. How much did she spend for the subway fares?

Income Mayra earns $9.25 per hour. Last week she worked 32 hours. How much did she earn?

Solution

$296.00

Income Peter earns $8.75 per hour. Last week he worked 19 hours. How much did he earn?

Hourly Wage Alan got his first paycheck from his new job. He worked 30 hours and earned $382.50. How much does he earn per hour?

Solution

$12.75

Hourly Wage Maria got her first paycheck from her new job. She worked 25 hours and earned $362.50. How much does she earn per hour?

Restaurant Jeannette and her friends love to order mud pie at their favorite restaurant. They always share just one piece of pie among themselves. With tax and tip, the total cost is $6.00. How much does each girl pay if the total number sharing the mud pie is

ⓐ 2?

ⓑ 3?

ⓒ 4?

ⓓ 5?

ⓔ 6?

Solution
  1. ⓐ $3
  2. ⓑ $2
  3. ⓒ $1.50
  4. ⓓ $1.20
  5. ⓔ $1

Pizza Alex and his friends go out for pizza and video games once a week. They share the cost of a $15.60 pizza equally. How much does each person pay if the total number sharing the pizza is

ⓐ 2?

ⓑ 3?

ⓒ 4?

ⓓ 5?

ⓔ 6?

Fast Food At their favorite fast food restaurant, the Carlson family orders 4 burgers that cost $3.29 each and 2 orders of fries at $2.74 each. What is the total cost of the order?

Solution

$18.64

Home Goods Chelsea needs towels to take with her to college. She buys 2 bath towels that cost $9.99 each and 6 washcloths that cost $2.99 each. What is the total cost for the bath towels and washcloths?

Zoo The Lewis and Chousmith families are planning to go to the zoo together. Adult tickets cost $29.95 and children’s tickets cost $19.95. What will the total cost be for 4 adults and 7 children?

Solution

$259.45

Ice Skating Jasmine wants to have her birthday party at the local ice skating rink. It will cost $8.25 per child and $12.95 per adult. What will the total cost be for 12 children and 3 adults?

Everyday Math

Paycheck Annie has two jobs. She gets paid $14.04 per hour for tutoring at City College and $8.75 per hour at a coffee shop. Last week she tutored for 8 hours and worked at the coffee shop for 15 hours.

ⓐ How much did she earn?

ⓑ If she had worked all 23 hours as a tutor instead of working both jobs, how much more would she have earned?

Solution
  1. ⓐ $243.57
  2. ⓑ $79.35

Paycheck Jake has two jobs. He gets paid $7.95 per hour at the college cafeteria and $20.25 at the art gallery. Last week he worked 12 hours at the cafeteria and 5 hours at the art gallery.

ⓐ How much did he earn?

ⓑ If he had worked all 17 hours at the art gallery instead of working both jobs, how much more would he have earned?

Writing Exercises

At the 2010 winter Olympics, two skiers took the silver and bronze medals in the Men's Super-G ski event. Miller's time was 1 minute 30.62 seconds and Weibrecht's time was 1 minute 30.65 seconds. Find the difference in their times and then write the name of that decimal.

Solution

The difference: 0.03 seconds. Three hundredths of a second.

Find the quotient of 0.12÷0.04 and explain in words all the steps taken.

Self Check

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

A self-assessment table for decimal skills, allowing users to rate their confidence in adding, subtracting, multiplying, dividing, and applying decimals in money.

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

Decimals and Fractions

Learning Objectives

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

  • Convert fractions to decimals
  • Order decimals and fractions
  • Simplify expressions using the order of operations
  • Find the circumference and area of circles

Before you get started, take this readiness quiz.

Divide: 0.24÷8.
If you missed this problem, review Example 9 in Decimal Operations.

Solution

0.03

Order 0.64__0.6 using < or >.
If you missed this problem, review Example 7 in Decimals.

Solution

>

Order −0.2__−0.1 using < or >.
If you missed this problem, review Example 8 in Decimals.

Solution

<

Convert Fractions to Decimals

In Decimals, we learned to convert decimals to fractions. Now we will do the reverse—convert fractions to decimals. Remember that the fraction bar indicates division. So 45 can be written 4÷5 or 54. This means that we can convert a fraction to a decimal by treating it as a division problem.

Convert a Fraction to a Decimal

To convert a fraction to a decimal, divide the numerator of the fraction by the denominator of the fraction.

Write the fraction 34 as a decimal.

Solution

Solution

Steps to convert the fraction 3/4 to the decimal 0.75 using long division, with textual explanations and visual examples.
A fraction bar means division, so we can write the fraction 34 using division. A division problem is shown. 3 is on the inside of the division sign and 4 is on the outside.
Divide. A division problem is shown. 3.00 is on the inside of the division sign and 4 is on the outside. Below the 3.00 is a 28 with a line below it. Below the line is a 20. Below the 20 is another 20 with a line below it. Below the line is a 0. Above the division sign is 0.75.
So the fraction 34 is equal to 0.75.

Write the fraction as a decimal: 14.

Solution

0.25

Write the fraction as a decimal: 38.

Solution

0.375

Write the fraction −72 as a decimal.

Solution

Solution

Steps and example for converting a negative fraction (-7/2) to its decimal equivalent (-3.5).
The value of this fraction is negative. After dividing, the value of the decimal will be negative. We do the division ignoring the sign, and then write the negative sign in the answer. −72
Divide 7 by 2. A division problem is shown. 7.0 is on the inside of the division sign and 2 is on the outside. Below the 7 is a 6 with a line below it. Below the line is a 10. Below the 10 is another 10 with a line below it. Below the line is a 0. 3.5 is written above the division sign.
So, −72=−3.5.

Write the fraction as a decimal: −94.

Solution

−2.25

Write the fraction as a decimal: −112.

Solution

−5.5

Repeating Decimals

So far, in all the examples converting fractions to decimals the division resulted in a remainder of zero. This is not always the case. Let’s see what happens when we convert the fraction 43 to a decimal. First, notice that 43 is an improper fraction. Its value is greater than 1. The equivalent decimal will also be greater than 1.

We divide 4 by 3.

A division problem is shown. 4.000 is on the inside of the division sign and 3 is on the outside. Below the 4 is a 3 with a line below it. Below the line is a 10. Below the 10 is a 9 with a line below it. Below the line is another 10, followed by another 9 with a line, followed by another 10, followed by another 9 with a line, followed by a 1. Above the division sign is 1.333...

No matter how many more zeros we write, there will always be a remainder of 1, and the threes in the quotient will go on forever. The number 1.333… is called a repeating decimal. Remember that the “…” means that the pattern repeats.

Repeating Decimal

A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly.

How do you know how many ‘repeats’ to write? Instead of writing 1.333… we use a shorthand notation by placing a line over the digits that repeat. The repeating decimal 1.333… is written 1.3–. The line above the 3 tells you that the 3 repeats endlessly. So 1.333…=1.3–

For other decimals, two or more digits might repeat. Table 3 shows some more examples of repeating decimals.

1.333…=1.3– 3 is the repeating digit
4.1666…=4.16– 6 is the repeating digit
4.161616…=4.16— 16 is the repeating block
0.271271271…=0.271––– 271 is the repeating block

Write 4322 as a decimal.

Solution
Solution

Divide 43 by 22.
A division problem is shown. 43.00000 is on the inside of the division sign and 22 is on the outside. Below the 43 is a 22 with a line below it. Below the line is a 210 with a 198 with a line below it. Below the line is a 120 with 110 and a line below it. Below the line is 100 with 88 and a line below it. Below the line is 120 with 110 and a line below it. Below the line is 100 with 88 and a line below it. Below the line is an ellipses. There are arrows pointing to the 120s saying 120 repeats. There are arrows pointing to the 100s saying 100 repeats. There are arrows pointing to the 88s saying, in red, “The pattern repeats, so the numbers in the quotient will repeat as well.” The quotient is shown above the division sign. It is 1.95454.

Notice that the differences of 120 and 100 repeat, so there is a repeat in the digits of the quotient; 54 will repeat endlessly. The first decimal place in the quotient, 9, is not part of the pattern. So,

4322=1.954—

Write as a decimal: 2711.

Solution

2.45—

Write as a decimal: 5122.

Solution

2.318—

It is useful to convert between fractions and decimals when we need to add or subtract numbers in different forms. To add a fraction and a decimal, for example, we would need to either convert the fraction to a decimal or the decimal to a fraction.

Simplify: 78+6.4.

Solution
Solution
78+6.4
Change 78 to a decimal. Long division calculation demonstrating 7 divided by 8 equals 0.875, showing each step of the process with a final remainder of 0. 0.875+6.4
Add. 7.275

Simplify: 38+4.9.

Solution

5.275

Simplify: 5.7+1320.

Solution

6.35

Order Decimals and Fractions

In Decimals, we compared two decimals and determined which was larger. To compare a decimal to a fraction, we will first convert the fraction to a decimal and then compare the decimals.

Order 38__0.4 using < or >.

Solution

Solution

Step-by-step comparison of a fraction (3/8) and a decimal (0.4), showing the conversion of the fraction to a decimal.
38__0.4
Convert 38 to a decimal. 0.375__0.4
Compare 0.375 to 0.4 0.375<0.4
Rewrite with the original fraction. 38<0.4

Order each of the following pairs of numbers, using < or >.

1720__0.82

Solution

>

Order each of the following pairs of numbers, using < or >.

34__0.785

Solution

<

When ordering negative numbers, remember that larger numbers are to the right on the number line and any positive number is greater than any negative number.

Order −0.5___−34 using < or >.

Solution

Solution

Steps demonstrating how to compare a decimal and a fraction by converting the fraction to its decimal equivalent.
−0.5___−34
Convert −34 to a decimal. −0.5___−0.75
Compare −0.5 to −0.75. −0.5>−0.75
Rewrite the inequality with the original fraction. −0.5>−34

Order each of the following pairs of numbers, using < or >:

−58__−0.58

Solution

<

Order each of the following pairs of numbers, using < or >:

−0.53__−1120

Solution

>

Write the numbers 1320,0.61,1116 in order from smallest to largest.

Solution

Solution

This table illustrates the step-by-step process of ordering a mixed set of fractions and decimals from smallest to largest by converting fractions to decimals.
1320,0.61,1116
Convert the fractions to decimals. 0.65,0.61,0.6875
Write the smallest decimal number first. 0.61,____,_____
Write the next larger decimal number in the middle place. 0.61,0.65,_____
Write the last decimal number (the larger) in the third place. 0.61,0.65,0.6875
Rewrite the list with the original fractions. 0.61,1320,1116

Write each set of numbers in order from smallest to largest: 78,45,0.82.

Solution

45,0.82,78

Write each set of numbers in order from smallest to largest: 0.835,1316,34.

Solution

34,1316,0.835

Simplify Expressions Using the Order of Operations

The order of operations introduced in Use the Language of Algebra also applies to decimals. Do you remember what the phrase “Please excuse my dear Aunt Sally” stands for?

Simplify the expressions:

  1. ⓐ 7(18.3−21.7)
  2. ⓑ 23(8.3−3.8)
Solution

Solution

Step-by-step simplification of the mathematical expression 7(18.3-21.7).
ⓐ
7(18.3−21.7)
Simplify inside parentheses. 7(−3.4)
Multiply. −23.8
Step-by-step simplification of the mathematical expression (2/3)(8.3 - 3.8).
ⓑ
23(8.3−3.8)
Simplify inside parentheses. 23(4.5)
Write 4.5 as a fraction. 23(4.51)
Multiply. 93
Simplify. 3

Simplify: ⓐ 8(14.6−37.5) ⓑ 35(9.6−2.1).

Solution
  1. ⓐ −183.2
  2. ⓑ 4.5

Simplify: ⓐ 25(25.69−56.74) ⓑ 27(11.9−4.2).

Solution
  1. ⓐ −776.25
  2. ⓑ 2.2

Simplify each expression:

  1. ⓐ 6÷0.6+(0.2)4−(0.1)2
  2. ⓑ (110)2+(3.5)(0.9)
Solution

Solution

Step-by-step simplification of a mathematical expression, demonstrating the order of operations.
ⓐ
6÷0.6+(0.2)4−(0.1)2
Simplify exponents. 6÷0.6+(0.2)4−0.01
Divide. 10+(0.2)4−0.01
Multiply. 10+0.8−0.01
Add. 10.8−0.01
Subtract. 10.79
Simplification steps for the expression (1/10)^2 + (3.5)(0.9), detailing each calculation leading to the final result of 3.16.
ⓑ
(110)2+(3.5)(0.9)
Simplify exponents. 1100+(3.5)(0.9)
Multiply. 1100+3.15
Convert 1100 to a decimal. 0.01+3.15
Add. 3.16

Simplify: 9÷0.9+(0.4)3−(0.2)2.

Solution

11.16

Simplify: (12)2+(0.3)(4.2).

Solution

1.51

Find the Circumference and Area of Circles

The properties of circles have been studied for over 2,000 years. All circles have exactly the same shape, but their sizes are affected by the length of the radius, a line segment from the center to any point on the circle. A line segment that passes through a circle’s center connecting two points on the circle is called a diameter. The diameter is twice as long as the radius. See Figure 1.

The size of a circle can be measured in two ways. The distance around a circle is called its circumference.

A circle is shown. A dotted line running through the widest portion of the circle is labeled as a diameter. A dotted line from the center of the circle to a point on the circle is labeled as a radius. Along the edge of the circle is the circumference.

Archimedes discovered that for circles of all different sizes, dividing the circumference by the diameter always gives the same number. The value of this number is pi, symbolized by Greek letter π (pronounced pie). However, the exact value of π cannot be calculated since the decimal never ends or repeats (we will learn more about numbers like this in The Properties of Real Numbers.)

Doing the Manipulative Mathematics activity Pi Lab will help you develop a better understanding of pi.

If we want the exact circumference or area of a circle, we leave the symbol π in the answer. We can get an approximate answer by substituting 3.14 as the value of π. We use the symbol ≈ to show that the result is approximate, not exact.

Properties of Circles

A circle is shown. A line runs through the widest portion of the circle. There is a red dot at the center of the circle. The half of the line from the center of the circle to a point on the right of the circle is labeled with an r. The half of the line from the center of the circle to a point on the left of the circle is also labeled with an r. The two sections labeled r have a brace drawn underneath showing that the entire segment is labeled d.
ris the length of the radius.dis the length of the diameter.
The circumference is2πr.C=2πrThe area isπr2.A=πr2

Since the diameter is twice the radius, another way to find the circumference is to use the formula C=πd.

Suppose we want to find the exact area of a circle of radius 10 inches. To calculate the area, we would evaluate the formula for the area when r=10 inches and leave the answer in terms of π.

A=πr2A=π(102)A=π·100

We write π after the 100. So the exact value of the area is A=100π square inches.

To approximate the area, we would substitute π≈3.14.

A = 100 π ≈ 100 · 3.14 ≈ 314 square inches

Remember to use square units, such as square inches, when you calculate the area.

A circle has radius 10 centimeters. Approximate its ⓐ circumference and ⓑ area.

Solution

Solution

This table illustrates the step-by-step calculation of a circle's circumference given a radius of 10 units.
ⓐ Find the circumference when r=10.
Write the formula for circumference. C=2πr
Substitute 3.14 for π and 10 for ,r. C≈2(3.14)(10)
Multiply. C≈62.8centimeters
Steps to calculate the area of a circle with a radius of 10, using the formula A = πr² and π ≈ 3.14.
ⓑ Find the area when r=10.
Write the formula for area. A=πr2
Substitute 3.14 for π and 10 for r. A≈(3.14)(10)2
Multiply. A≈314square centimeters

A circle has radius 50 inches. Approximate its ⓐ circumference and ⓑ area.

Solution
  1. ⓐ 314 in.
  2. ⓑ 7850 sq. in.

A circle has radius 100 feet. Approximate its ⓐ circumference and ⓑ area.

Solution
  1. ⓐ 628 ft.
  2. ⓑ 31,400 sq. ft.

A circle has radius 42.5 centimeters. Approximate its ⓐ circumference and ⓑ area.

Solution

Solution

Steps to calculate the circumference of a circle given its radius.
ⓐ Find the circumference when r=42.5.
Write the formula for circumference. C=2πr
Substitute 3.14 for π and 42.5 for r C≈2(3.14)(42.5)
Multiply. C≈266.9centimeters
This table demonstrates the step-by-step calculation of the area of a circle with a radius of 42.5 units, showing the formula, substitution, and final result.
ⓑ Find the area when r=42.5.
Write the formula for area. A=πr2
Substitute 3.14 for π and 42.5 for r. A≈(3.14)(42.5)2
Multiply. A≈5671.625square centimeters

A circle has radius 51.8 centimeters. Approximate its ⓐ circumference and ⓑ area.

Solution
  1. ⓐ 325.304 cm
  2. ⓑ 8425.3736 sq. cm

A circle has radius 26.4 meters. Approximate its ⓐ circumference and ⓑ area.

Solution
  1. ⓐ 165.792 m
  2. ⓑ 2188.4544 sq. m

Approximate π with a Fraction

Convert the fraction 227 to a decimal. If you use your calculator, the decimal number will fill up the display and show 3.14285714. But if we round that number to two decimal places, we get 3.14, the decimal approximation of π. When we have a circle with radius given as a fraction, we can substitute 227 for π instead of 3.14. And, since 227 is also an approximation of π, we will use the ≈ symbol to show we have an approximate value.

A circle has radius 1415 meter. Approximate its ⓐ circumference and ⓑ area.

Solution
Solution
Step-by-step calculation of a circle's circumference, given a radius of 14/15.
ⓐ Find the circumference when r=1415.
Write the formula for circumference. C=2πr
Substitute 227 for π and 1415 for r. C≈2(227)(1415)
Multiply. C≈8815meters
Calculates a circle's area step-by-step, substituting a fractional radius and an approximate value for pi to find the final area.
ⓑ Find the area when r=1415.
Write the formula for area. A=πr2
Substitute 227 for π and 1415 for r. A≈(227)(1415)2
Multiply. A≈616225square meters

A circle has radius 521 meters. Approximate its ⓐ circumference and ⓑ area.

Solution
  1. ⓐ 220147m
  2. ⓑ 5503087sq. m

A circle has radius 1033 inches. Approximate its ⓐ circumference and ⓑ area.

Solution
  1. ⓐ 4021in.
  2. ⓑ 200693sq.in.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Converting a Fraction to a Decimal - Part 2
  • Convert a Fraction to a Decimal (repeating)
  • Compare Fractions and Decimals using Inequality Symbols
  • Determine the Area of a Circle
  • Determine the Circumference of a Circle

Key Concepts

  • Convert a Fraction to a Decimal To convert a fraction to a decimal, divide the numerator of the fraction by the denominator of the fraction.
  • Properties of Circles A labeled circle displaying its radius (r), diameter (d), and the central point, illustrating the relation d = 2r. r is the length of the radius
    d is the length of the diameter
    The circumference is 2πr. C=2πr
    The area is πr2. A=πr2

Practice Makes Perfect

Convert Fractions to Decimals

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

25

Solution

0.4

45

−38

Solution

−0.375

−58

1720

Solution

0.85

1320

114

Solution

2.75

174

−31025

Solution

−12.4

−28425

59

Solution

0.5–

29

1511

Solution

1.36—

1811

15111

Solution

0.135—

25111

In the following exercises, simplify the expression.

12+6.5

Solution

7

14+10.75

2.4+58

Solution

3.025

3.9+920

9.73+1720

Solution

10.58

6.29+2140

Order Decimals and Fractions

In the following exercises, order each pair of numbers, using < or >.

18___0.8

Solution

<

14___0.4

25___0.25

Solution

>

35___0.35

0.725___34

Solution

<

0.92___78

0.66___23

Solution

<

0.83___56

−0.75___−45

Solution

>

−0.44___−920

−34___−0.925

Solution

>

−23___−0.632

In the following exercises, write each set of numbers in order from least to greatest.

35,916,0.55

Solution

0.55,916,35

38,720,0.36

0.702,1320,58

Solution

58,1320,0.702

0.15,316,15

−0.3,−13,−720

Solution

−720,−13,−0.3

−0.2,−320,−16

−34,−79,−0.7

Solution

−79,−34,−0.7

−89,−45,−0.9

Simplify Expressions Using the Order of Operations

In the following exercises, simplify.

10(25.1−43.8)

Solution

−187

30(18.1−32.5)

62(9.75−4.99)

Solution

295.12

42(8.45−5.97)

34(12.4−4.2)

Solution

6.15

45(8.6+3.9)

512(30.58+17.9)

Solution

20.2

916(21.96−9.8)

10÷0.1+(1.8)4−(0.3)2

Solution

107.11

5÷0.5+(3.9)6−(0.7)2

(37.1+52.7)÷(12.5÷62.5)

Solution

449

(11.4+16.2)÷(18÷60)

(15)2+(1.4)(6.5)

Solution

9.14

(12)2+(2.1)(8.3)

−910·815+0.25

Solution

−0.23

−38·1415+0.72

Mixed Practice

In the following exercises, simplify. Give the answer as a decimal.

314−6.5

Solution

−3.25

525−8.75

10.86÷23

Solution

16.29

5.79÷34

78(103.48)+112(361)

Solution

632.045

516(117.6)+213(699)

3.6(98−2.72)

Solution

−5.742

5.1(125−3.91)

Find the Circumference and Area of Circles

In the following exercises, approximate the ⓐ circumference and ⓑ area of each circle. If measurements are given in fractions, leave answers in fraction form.

radius=5 in.

Solution
  1. ⓐ 31.4 in
  2. ⓑ 78.5 sq.in.

radius=20 in.

radius=9 ft.

Solution
  1. ⓐ 56.52.ft.
  2. ⓑ 254.34 sq.ft.

radius=4 ft.

radius=46 cm

Solution
  1. ⓐ 288.88 cm
  2. ⓑ 6644.24 sq.cm

radius=38 cm

radius=18.6 m

Solution
  1. ⓐ 116.808 m
  2. ⓑ 1086.3144 sq.m

radius=57.3 m

radius=710mile

Solution
  1. ⓐ 225mile
  2. ⓑ 7750sq.mile

radius=711mile

radius=38yard

Solution
  1. ⓐ 3314yard
  2. ⓑ 99224sq.yard

radius=512yard

diameter=56m

Solution
  1. ⓐ 5521m
  2. ⓑ 275504sq.m

diameter=34m

Everyday Math

Kelly wants to buy a pair of boots that are on sale for 23 of the original price. The original price of the boots is $84.99. What is the sale price of the shoes?

Solution

$56.66

An architect is planning to put a circular mosaic in the entry of a new building. The mosaic will be in the shape of a circle with radius of 6 feet. How many square feet of tile will be needed for the mosaic? (Round your answer up to the next whole number.)

Writing Exercises

Is it easier for you to convert a decimal to a fraction or a fraction to a decimal? Explain.

Solution

Answers will vary.

Describe a situation in your life in which you might need to find the area or circumference of a circle.

Self Check

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

A self-assessment chart for math skills, including converting fractions to decimals, ordering numbers, simplifying expressions, and finding circle area and circumference, with columns for confidence levels.

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

circumference of a circle
The distance around a circle is called its circumference.
diameter of a circle
A diameter of a circle is a line segment that passes through a circle’s center connecting two points on the circle.
radius of a circle
A radius of a circle is a line segment from the center to any point on the circle.
repeating decimal
A repeating decimal is a decimal in which the last digit or group of digits repeats endlessly.

Solve Equations with Decimals

Learning Objectives

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

  • Determine whether a decimal is a solution of an equation
  • Solve equations with decimals
  • Translate to an equation and solve

Before you get started, take this readiness quiz.

Evaluate x+23whenx=−14.
If you missed this problem, review Example 15 in Add and Subtract Fractions with Different Denominators.

Solution

512

Evaluate 15−y when y=−5.
If you missed this problem, review Example 12 in Subtract Integers.

Solution

20

Solve n−7=42.
If you missed this problem, review Example 5 in Solve Equations with Fractions.

Solution

−294

Determine Whether a Decimal is a Solution of an Equation

Solving equations with decimals is important in our everyday lives because money is usually written with decimals. When applications involve money, such as shopping for yourself, making your family’s budget, or planning for the future of your business, you’ll be solving equations with decimals.

Now that we’ve worked with decimals, we are ready to find solutions to equations involving decimals. The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, a fraction, or a decimal. We’ll list these steps here again for easy reference.

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 so, the number is a solution.
    • If not, the number is not a solution.

Determine whether each of the following is a solution of x−0.7=1.5:

ⓐ x=1ⓑ x=−0.8ⓒ x=2.2

Solution

Solution

ⓐ
The image shows the mathematical equation x - 0.7 = 1.5, presented in a clear, sans-serif font. The equation is horizontally centered against a white background.
The text 'Substitute 1 for x.' is shown in a dark teal color with the number 1 highlighted in red. A mathematical expression shows '1 - 0.7' on the left side, followed by an equals sign with a question mark above it, and '1.5' on the right side. The number '1' is highlighted in red.
Subtract. A white background displays the mathematical expression '0.3 ≠ 1.5' in black text, indicating that 0.3 is not equal to 1.5.

Since x=1 does not result in a true equation, 1 is not a solution to the equation.

ⓑ
A mathematical equation is displayed, showing 'x - 0.7 = 1.5' in black text against a white background.
Substitute -0.8 for x. A mathematical expression displays -0.8 (in red) - 0.7, followed by an equals sign with a question mark on top, and then 1.5, posing whether the two sides are equal.
Subtract. A mathematical expression showing that -1.5 is not equal to 1.5, displayed in black text on a white background.

Since x=−0.8 does not result in a true equation, −0.8 is not a solution to the equation.

ⓒ
A mathematical equation is displayed on a white background, reading 'x - 0.7 = 1.5'.
The text 'Substitute 2.2 for x.' is shown, with the number '2.2' highlighted in red. The expression 2.2 - 0.7 ?= 1.5, testing whether the subtraction of decimals results in 1.5. The number 2.2 is highlighted in red.
Subtract. The equation 1.5 = 1.5 is shown with a checkmark, indicating it is correct or verified.

Since x=2.2 results in a true equation, 2.2 is a solution to the equation.

Determine whether each value is a solution of the given equation.

x−0.6=1.3:ⓐ x=0.7ⓑ x=1.9ⓒ x=−0.7

Solution
  1. ⓐ no
  2. ⓑ yes
  3. ⓒ no

Determine whether each value is a solution of the given equation.

y−0.4=1.7:ⓐ y=2.1ⓑ y=1.3ⓒ −1.3

Solution
  1. ⓐ yes
  2. ⓑ no
  3. ⓒ no

Solve Equations with Decimals

In previous chapters, we solved equations using the Properties of Equality. We will use these same properties to solve equations with decimals.

Properties of Equality

Subtraction Property of Equality
For any numbers a,b,andc,
If a=b, then a−c=b−c.
Addition Property of Equality
For any numbers a,b,andc,
If a=b, then a+c=b+c.
The Division Property of Equality
For any numbers a,b,andc,andc≠0
If a=b, then ac=bc
The Multiplication Property of Equality
For any numbers a,b,andc,
If a=b, then ac=bc

When you add, subtract, multiply or divide the same quantity from both sides of an equation, you still have equality.

Solve: y+2.3=−4.7.

Solution

Solution

We will use the Subtraction Property of Equality to isolate the variable.
A mathematical equation is displayed: y + 2.3 = -4.7. This is a linear equation with one variable 'y' that requires solving for 'y' by isolating it on one side of the equation.
The text reads 'Subtract 2.3 from each side, to undo the addition.' A mathematical equation illustrating the process of solving for 'y' by subtracting 2.3 from both sides: y + 2.3 - 2.3 = -4.7 - 2.3, with the subtracted 2.3 highlighted in red.
Simplify. The image displays the equation y = -7 in black text on a white background, representing a horizontal line in a coordinate system.
Check: A mathematical equation displayed on a white background, which reads 'y + 2.3 = -4.7'.
The image shows the text 'Substitute y = -7.' in a mathematical context, indicating an instruction to replace the variable 'y' with the value -7. A mathematical equation checks if -7 + 2.3 equals -4.7, which is true.
Simplify. A mathematical expression shows '-4.7 = -4.7' followed by a checkmark, indicating the equality is correct.

Since y=−7 makes y+2.3=−4.7 a true statement, we know we have found a solution to this equation.

Solve: y+2.7=−5.3.

Solution

y = −8

Solve: y+3.6=−4.8.

Solution

y = −8.4

Solve: a−4.75=−1.39.

Solution

Solution

We will use the Addition Property of Equality.
A mathematical equation is displayed, showing 'a - 4.75 = -1.39' against a white background.
Add 4.75 to each side, to undo the subtraction. A mathematical equation showing the addition of 4.75 to both sides of an equation to isolate the variable 'a'. The numbers added are highlighted in red.
Simplify. The image displays a mathematical equation in black text on a white background, which reads 'a = 3.36'.
Check: A mathematical equation is displayed with the variable 'a' minus 4.75, which equals -1.39.
The text 'Substitute A mathematical equation, '3.36 - 4.75 ?= -1.39', queries if the subtraction on the left side equals the negative value on the right. The first number, 3.36, is highlighted in red.
A mathematical equation shows '-1.39 = -1.39' followed by a checkmark, indicating the equality is verified and correct.

Since the result is a true statement, a=3.36 is a solution to the equation.

Solve: a−3.93=−2.86.

Solution

a = 1.07

Solve: n−3.47=−2.64.

Solution

n = 0.83

Solve: −4.8=0.8n.

Solution

Solution

We will use the Division Property of Equality.

Use the Properties of Equality to find a value for n.
A mathematical equation shows '-4.8 = 0.8n' against a white background.
We must divide both sides by 0.8 to isolate n. The equation -4.8/0.8 = 0.8n/0.8, demonstrating division by 0.8 on both sides to solve for 'n', with the divisor highlighted in red.
Simplify. -6 = n, an algebraic equation where the variable n is equal to negative six.
Check: An image showing the algebraic equation -4.8 = 0.8n, which involves a negative decimal, an equals sign, another decimal, and the variable 'n'.
The text reads 'Substitute n = -6.' on a white background. The word 'Substitute' is in dark teal, 'n = ' is in dark teal, and '-6.' is in red. A mathematical equation asks whether -4.8 is equal to 0.8 multiplied by -6, with the -6 highlighted in red. The equality holds true as 0.8 * -6 also equals -4.8.
The image displays the equation '-4.8 = -4.8' followed by a black checkmark, indicating that the equality is confirmed or correct.

Since n=−6 makes −4.8=0.8n a true statement, we know we have a solution.

Solve: −8.4=0.7b.

Solution

b = −12

Solve: −5.6=0.7c.

Solution

c = −8

Solve: p−1.8=−6.5.

Solution

Solution

We will use the Multiplication Property of Equality.
The image displays the algebraic equation p divided by -1.8 equals -6.5, presented in a black font against a white background.
Here, p is divided by −1.8. We must multiply by −1.8 to isolate p The mathematical equation -1.8(p/-1.8) = -1.8(-6.5) is shown.
Multiply. The variable 'p' is assigned the numerical value of 11.7, displayed in a simple mathematical expression.
Check: A mathematical equation shows p divided by -1.8, which equals -6.5.
The text reads 'Substitute p = 11.7.', instructing to replace the variable p with the numerical value 11.7 in a mathematical context. The number 11.7 is highlighted in red. A mathematical equation questions if the division of 11.7 by -1.8 is equal to -6.5.
The equation -6.5 = -6.5 is displayed, followed by a checkmark, indicating the equality is correct.

A solution to p−1.8=−6.5 is p=11.7.

Solve: c−2.6=−4.5.

Solution

c = 11.7

Solve: b−1.2=−5.4.

Solution

b = 6.48

Translate to an Equation and Solve

Now that we have solved equations with decimals, we are ready to translate word sentences to equations and solve. Remember to look for words and phrases that indicate the operations to use.

Translate and solve: The difference of n and 4.3 is 2.1.

Solution

Solution

Translate. Demonstration of translating the word phrase 'The difference of n and 4.3 is 2.1' into the mathematical equation 'n - 4.3 = 2.1', using brackets for visual mapping.
Add 4.3 to both sides of the equation. Algebraic solution step: Adding 4.3 to both sides of the equation n - 4.3 = 2.1 to solve for n. The red numbers highlight the operation.
Simplify. A mathematical expression on a white background displays 'n = 6.4' in bold, black text.
Check: Is the difference of n and 4.3 equal to 2.1?
Let n=6.4: Is the difference of 6.4 and 4.3 equal to 2.1?
Translate. A mathematical expression displaying '6.4 - 4.3' with a small '2' superscript on '4.3', followed by a symbol resembling an equality sign also with a small '2' superscript, then '2.1'.
Simplify. The equation 2.1 = 2.1 is shown with a checkmark, indicating that the mathematical statement is correct.

Translate and solve: The difference of y and 4.9 is 2.8.

Solution

y − 4.9 = 2.8; y = 7.7

Translate and solve: The difference of z and 5.7 is 3.4.

Solution

z − 5.7 = 3.4; z = 9.1

Translate and solve: The product of −3.1 and x is 5.27.

Solution

Solution

Translate. An image translates 'The product of 3.1 and x is 5.27' into the algebraic equation '-3.1x = 5.27', showing an unexpected negative sign in the term -3.1x compared to the verbal statement.
Divide both sides by −3.1. A step in solving an algebraic equation, showing both sides of the equation -3.1x = 5.27 being divided by -3.1, highlighting the division operation to isolate x.
Simplify. A mathematical expression 'x = -1.7' is displayed on a plain white background.
Check: Is the product of −3.1 and x equal to 5.27?
Let x=−1.7: Is the product of −3.1 and −1.7 equal to 5.27?
Translate. A mathematical problem displaying the equation -3.1(-1.7)?=5.27, which asks to determine if -3.1 multiplied by -1.7 indeed equals 5.27.
Simplify. The equation 5.27 = 5.27 with a checkmark, indicating correctness.

Translate and solve: The product of −4.3 and x is 12.04.

Solution

−4.3x = 12.04; x = −2.8

Translate and solve: The product of −3.1 and m is 26.66.

Solution

−3.1m = 26.66; m = −8.6

Translate and solve: The quotient of p and −2.4 is 6.5.

Solution

Solution

Translate. This image translates the verbal statement 'The quotient of p and -2.4 is 6.5' into the mathematical equation p / -2.4 = 6.5, visually connecting each part of the phrase to its symbolic representation.
Multiply both sides by −2.4. A mathematical equation shows '-2.4(P/-2.4) = -2.4(6.5)'. The number -2.4 is highlighted in red, indicating multiplication by -2.4 on both sides of the equation.
Simplify. The image displays a mathematical equation: p = -15.6, rendered in a black serif font on a plain white background.
Check: Is the quotient of p and −2.4 equal to 6.5?
Let p=−15.6: Is the quotient of −15.6 and −2.4 equal to 6.5?
Translate. A mathematical equation questions whether -15.6 divided by -2.4 is approximately equal to 6.5.
Simplify. The equation 6.5 = 6.5 is shown with a checkmark, indicating that the equality is correct or verified.

Translate and solve: The quotient of q and −3.4 is 4.5.

Solution

q−3.4=4.5;q=−15.3

Translate and solve: The quotient of r and −2.6 is 2.5.

Solution

r−2.6=2.5;r=−6.5

Translate and solve: The sum of n and 2.9 is 1.7.

Solution

Solution

Translate. An image explaining how to translate a word problem into an algebraic equation. It shows 'The sum of n and 2.9 is 1.7.' translated to 'n + 2.9 = 1.7', illustrating the breakdown of the sentence into its mathematical components.
Subtract 2.9 from each side. An algebraic equation showing the step of subtracting 2.9 from both sides: n + 2.9 - 2.9 = 1.7 - 2.9, with the subtracted 2.9 highlighted in red.
Simplify. The image displays the mathematical equation 'n = -1.2' in black text on a plain white background, occupying the top right portion of the frame.
Check: Is the sum n and 2.9 equal to 1.7?
Let n=−1.2: Is the sum −1.2 and 2.9 equal to 1.7?
Translate. A mathematical equation reads '-1.2 + 2.9 =? 1.7', showing the addition of two decimal numbers with a question mark next to the equals sign, suggesting a verification or problem to solve.
Simplify. The image shows the mathematical equality '1.7 = 1.7' followed by a checkmark, indicating the correctness of the statement.

Translate and solve: The sum of j and 3.8 is 2.6.

Solution

j + 3.8 = 2.6; j = −1.2

Translate and solve: The sum of k and 4.7 is 0.3.

Solution

k + 4.7 = 0.3; k = −4.4

ACCESS ADDITIONAL ONLINE RESOURCES

  • Solving One Step Equations Involving Decimals
  • Solve a One Step Equation With Decimals by Adding and Subtracting
  • Solve a One Step Equation With Decimals by Multiplying
  • Solve a One Step Equation With Decimals by Dividing

Key Concepts

  • Determine whether a number is a solution to an equation.
    • Substitute the number for the variable in the equation.
    • Simplify the expressions on both sides of the equation.
    • Determine whether the resulting equation is true.
      If so, the number is a solution.
      If not, the number is not a solution.
  • Properties of Equality
Summary of four fundamental properties of equality: Subtraction, Addition, Division, and Multiplication, demonstrating how these operations maintain balance in equations.
Subtraction Property of Equality Addition Property of Equality
For any numbers a, b, and c,
Ifa=b thena−c=b−c
For any numbers a, b, and c,
Ifa=b thena+c=b+c
Division of Property of Equality Multiplication Property of Equality
For any numbers a, b, and c≠0,
Ifa=b thenac=bc
For any numbers a, b, and c,
Ifa=b thena⋅c=b⋅c

Practice Makes Perfect

Determine Whether a Decimal is a Solution of an Equation

In the following exercises, determine whether each number is a solution of the given equation.

x−0.8=2.3
ⓐ x=2ⓑ x=−1.5ⓒ x=3.1

Solution
  1. ⓐ no
  2. ⓑ no
  3. ⓒ yes

y+0.6=−3.4
ⓐ y=−4ⓑ y=−2.8ⓒ y=2.6

h1.5=−4.3
ⓐ h=6.45ⓑ h=−6.45ⓒ h=−2.1

Solution
  1. ⓐ no
  2. ⓑ yes
  3. ⓒ no

0.75k=−3.6
ⓐ k=−0.48ⓑ k=−4.8ⓒ k=−2.7

Solve Equations with Decimals

In the following exercises, solve the equation.

y+2.9=5.7

Solution

y = 2.8

m+4.6=6.5

f+3.45=2.6

Solution

f = −0.85

h+4.37=3.5

a+6.2=−1.7

Solution

a = −7.9

b+5.8=−2.3

c+1.15=−3.5

Solution

c = −4.65

d+2.35=−4.8

n−2.6=1.8

Solution

n = 4.4

p−3.6=1.7

x−0.4=−3.9

Solution

x = −3.5

y−0.6=−4.5

j−1.82=−6.5

Solution

j = −4.68

k−3.19=−4.6

m−0.25=−1.67

Solution

m = −1.42

q−0.47=−1.53

0.5x=3.5

Solution

x = 7

0.4p=9.2

−1.7c=8.5

Solution

c = −5

−2.9x=5.8

−1.4p=−4.2

Solution

p = 3

−2.8m=−8.4

−120=1.5q

Solution

q = −80

−75=1.5y

0.24x=4.8

Solution

x = 20

0.18n=5.4

−3.4z=−9.18

Solution

z = 2.7

−2.7u=−9.72

a0.4=−20

Solution

a = −8

b0.3=−9

x0.7=−0.4

Solution

x = −0.28

y0.8=−0.7

p−5=−1.65

Solution

p = 8.25

q−4=−5.92

r−1.2=−6

Solution

r = 7.2

s−1.5=−3

Mixed Practice

In the following exercises, solve the equation. Then check your solution.

x−5=−11

Solution

x = −6

−25=x+34

p+8=−2

Solution

p = −10

p+23=112

−4.2m=−33.6

Solution

m = 8

q+9.5=−14

q+56=112

Solution

q=−34

8.615=−d

78m=110

Solution

m=435

j−6.2=−3

−23=y+38

Solution

y=−2524

s−1.75=−3.2

1120=−f

Solution

f=−1120

−3.6b=2.52

−4.2a=3.36

Solution

a = −0.8

−9.1n=−63.7

r−1.25=−2.7

Solution

r = −1.45

14n=710

h−3=−8

Solution

h = 24

y−7.82=−16

Translate to an Equation and Solve

In the following exercises, translate and solve.

The difference of n and 1.9 is 3.4.

Solution

n−1.9=3.4;5.3

The difference n and 1.5 is 0.8.

The product of −6.2 and x is −4.96.

Solution

−6.2x = −4.96; 0.8

The product of −4.6 and x is −3.22.

The quotient of y and −1.7 is −5.

Solution

y−1.7=−5;8.5

The quotient of z and −3.6 is 3.

The sum of n and −7.3 is 2.4.

Solution

n + (−7.3) = 2.4; 9.7

The sum of n and −5.1 is 3.8.

Everyday Math

Shawn bought a pair of shoes on sale for $78. Solve the equation 0.75p=78 to find the original price of the shoes, p.

Solution

$104

Mary bought a new refrigerator. The total price including sales tax was $1,350. Find the retail price, r, of the refrigerator before tax by solving the equation 1.08r=1,350.

Writing Exercises

Think about solving the equation 1.2y=60, but do not actually solve it. Do you think the solution should be greater than 60 or less than 60? Explain your reasoning. Then solve the equation to see if your thinking was correct.

Solution

Answers will vary.

Think about solving the equation 0.8x=200, but do not actually solve it. Do you think the solution should be greater than 200 or less than 200? Explain your reasoning. Then solve the equation to see if your thinking was correct.

Self Check

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

A self-assessment chart for math skills related to decimals and equations. It lists three skills: determining if a decimal is a solution, solving equations with decimals, and translating to an equation and solving.

ⓑ 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?

Averages and Probability

Learning Objectives

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

  • Calculate the mean of a set of numbers
  • Find the median of a set of numbers
  • Find the mode of a set of numbers
  • Apply the basic definition of probability

Before you get started, take this readiness quiz.

Simplify: 4+9+23.
If you missed this problem, review Example 12 in Multiply and Divide Mixed Numbers and Complex Fractions.

Solution

5

Simplify: 4(8)+6(3).
If you missed this problem, review Example 8 in Use the Language of Algebra.

Solution

50

Convert 52 to a decimal.
If you missed this problem, review Example 1 in Decimals and Fractions.

Solution

2.5

One application of decimals that arises often is finding the average of a set of numbers. What do you think of when you hear the word average? Is it your grade point average, the average rent for an apartment in your city, the batting average of a player on your favorite baseball team? The average is a typical value in a set of numerical data. Calculating an average sometimes involves working with decimal numbers. In this section, we will look at three different ways to calculate an average.

Calculate the Mean of a Set of Numbers

The mean is often called the arithmetic average. It is computed by dividing the sum of the values by the number of values. Students want to know the mean of their test scores. Climatologists report that the mean temperature has, or has not, changed. City planners are interested in the mean household size.

Suppose Ethan’s first three test scores were 85,88,and94. To find the mean score, he would add them and divide by 3.

85+88+943267389

His mean test score is 89 points.

The Mean

The mean of a set of n numbers is the arithmetic average of the numbers.

mean=sum of values in data setn

Calculate the mean of a set of numbers.

  1. Write the formula for the mean
    mean=sum of values in data setn
  2. Find the sum of all the values in the set. Write the sum in the numerator.
  3. Count the number, n, of values in the set. Write this number in the denominator.
  4. Simplify the fraction.
  5. Check to see that the mean is reasonable. It should be greater than the least number and less than the greatest number in the set.

Find the mean of the numbers 8,12,15,9,and6.

Solution

Solution

Detailed steps and mathematical expressions for calculating the mean of a numerical data set.
Write the formula for the mean: mean=sum of all the numbersn
Write the sum of the numbers in the numerator. mean=8+12+15+9+6n
Count how many numbers are in the set. There are 5 numbers in the set, so n=5. mean=8+12+15+9+65
Add the numbers in the numerator. mean=505
Then divide. mean=10
Check to see that the mean is 'typical': 10 is neither less than 6 nor greater than 15. The mean is 10.

Find the mean of the numbers: 8,9,7,12,10,5.

Solution

8.5

Find the mean of the numbers: 9,13,11,7,5.

Solution

9

The ages of the members of a family who got together for a birthday celebration were 16,26,53,56,65,70,93,and97 years. Find the mean age.

Solution

Solution

This table provides a step-by-step example of how to calculate the arithmetic mean of a set of numbers.
Write the formula for the mean: mean=sum of all the numbersn
Write the sum of the numbers in the numerator. mean=16+26+53+56+65+70+93+97n
Count how many numbers are in the set. Call this n and write it in the denominator. mean=16+26+53+56+65+70+93+978
Simplify the fraction. mean=4768
mean=59.5

Is 59.5 ‘typical’? Yes, it is neither less than 16 nor greater than 97. The mean age is 59.5 years.

The ages of the four students in Ben’s carpool are 25,18,21,and22. Find the mean age of the students.

Solution

21.5 years

Yen counted the number of emails she received last week. The numbers were 4,9,15,12,10,12,and8. Find the mean number of emails.

Solution

10

Did you notice that in the last example, while all the numbers were whole numbers, the mean was 59.5, a number with one decimal place? It is customary to report the mean to one more decimal place than the original numbers. In the next example, all the numbers represent money, and it will make sense to report the mean in dollars and cents.

For the past four months, Daisy’s cell phone bills were $42.75,$50.12,$41.54,$48.15. Find the mean cost of Daisy’s cell phone bills.

Solution

Solution

Step-by-step calculation of the mean, illustrating each procedure with its corresponding mathematical expression.
Write the formula for the mean. mean=sum of all the numbersn
Count how many numbers are in the set. Call this n and write it in the denominator. mean=sum of all the numbers4
Write the sum of all the numbers in the numerator. mean=42.75+50.12+41.54+48.154
Simplify the fraction. mean=182.564
mean=45.64

Does $45.64 seem ‘typical’ of this set of numbers? Yes, it is neither less than $41.54 nor greater than $50.12.

The mean cost of her cell phone bill was $45.64

Last week Ray recorded how much he spent for lunch each workday. He spent $6.50,$7.25,$4.90,$5.30,and$12.00. Find the mean of how much he spent each day.

Solution

$7.19

Lisa has kept the receipts from the past four trips to the gas station. The receipts show the following amounts: $34.87,$42.31,$38.04,and$43.26. Find the mean.

Solution

$39.62

Find the Median of a Set of Numbers

When Ann, Bianca, Dora, Eve, and Francine sing together on stage, they line up in order of their heights. Their heights, in inches, are shown in Table 4.

Ann Bianca Dora Eve Francine
59 60 65 68 70

Dora is in the middle of the group. Her height, 65″, is the median of the girls’ heights. Half of the heights are less than or equal to Dora’s height, and half are greater than or equal. The median is the middle value.

The numbers 59, 60, 65, 68, and 70 are listed. 59 and 60 have a brace beneath them and in red are labeled “2 below.” 68 and 70 have a brace beneath them and in red are labeled “2 above.” 65 has an arrow pointing to it and is labeled as the median.

Median

The median of a set of data values is the middle value.

  • Half the data values are less than or equal to the median.
  • Half the data values are greater than or equal to the median.

What if Carmen, the pianist, joins the singing group on stage? Carmen is 62 inches tall, so she fits in the height order between Bianca and Dora. Now the data set looks like this:

59,60,62,65,68,70

There is no single middle value. The heights of the six girls can be divided into two equal parts.

The numbers 59, 60, and 62 are listed, followed by a blank space, then 65, 68, and 70.

Statisticians have agreed that in cases like this the median is the mean of the two values closest to the middle. So the median is the mean of 62and65,62+652. The median height is 63.5 inches.

The numbers 9, 11, 12, 13, 15, 18, and 19 are listed. 9, 11, and 12 have a brace beneath them and are labeled “3 below.” 15, 18, and 19 have a brace beneath them and are labeled “3 above.” 13 has an arrow pointing to it and is labeled as the median.

Notice that when the number of girls was 5, the median was the third height, but when the number of girls was 6, the median was the mean of the third and fourth heights. In general, when the number of values is odd, the median will be the one value in the middle, but when the number is even, the median is the mean of the two middle values.

Find the median of a set of numbers.

  1. List the numbers from smallest to largest.
  2. Count how many numbers are in the set. Call this n.
  3. Is n odd or even?
    • If n is an odd number, the median is the middle value.
    • If n is an even number, the median is the mean of the two middle values.

Find the median of 12,13,19,9,11,15,and18.

Solution

Solution

List the numbers in order from smallest to largest. 9, 11, 12, 13, 15, 18, 19
Count how many numbers are in the set. Call this n. n=7
Is n odd or even? odd
The median is the middle value. The image shows the median, 13, as the central value in the ordered dataset (9, 11, 12, 13, 15, 18, 19), with an equal count of 3 values below and 3 values above it.
The middle is the number in the 4th position. So the median of the data is 13.

Find the median of the data set: 43,38,51,40,46.

Solution

43

Find the median of the data set: 15,35,20,45,50,25,30.

Solution

30

Kristen received the following scores on her weekly math quizzes:

83,79,85,86,92,100,76,90,88,and64. Find her median score.

Solution

Solution

Find the median of 83, 79, 85, 86, 92, 100, 76, 90, 88, and 64.
List the numbers in order from smallest to largest. 64, 76, 79, 83, 85, 86, 88, 90, 92, 100
Count the number of data values in the set. Call this n. n=10
Is n odd or even? even
The median is the mean of the two middle values, the 5th and 6th numbers. An ordered set of 10 numbers (64, 76, 79, 83, 85, 86, 88, 90, 92, 100) is shown, divided into two groups of 5, with the middle numbers 85 and 86 highlighted to illustrate finding the median.
Find the mean of 85 and 86. mean=85+862
mean=85.5
Kristen's median score is 85.5.

Find the median of the data set: 8,7,5,10,9,12.

Solution

8.5

Find the median of the data set: 21,25,19,17,22,18,20,24.

Solution

20.5

Identify the Mode of a Set of Numbers

The average is one number in a set of numbers that is somehow typical of the whole set of numbers. The mean and median are both often called the average. Yes, it can be confusing when the word average refers to two different numbers, the mean and the median! In fact, there is a third number that is also an average. This average is the mode. The mode of a set of numbers is the number that occurs the most. The frequency, is the number of times a number occurs. So the mode of a set of numbers is the number with the highest frequency.

Mode

The mode of a set of numbers is the number with the highest frequency.

Suppose Jolene kept track of the number of miles she ran since the start of the month, as shown in Figure 1.

An image of a calendar is shown. On Thursday the first, labeled New Year's Day, is written 2 mi. On Saturday the third is written 15 mi. On the 4th, 8 mi. On the 6th, 3 mi. On the 7th, 8 mi. On the 9th, 5 mi. On the 10th, 8 mi.

If we list the numbers in order it is easier to identify the one with the highest frequency.

2,3,5,8,8,8,15

Jolene ran 8 miles three times, and every other distance is listed only once. So the mode of the data is 8 miles.

Identify the mode of a set of numbers.

  1. List the data values in numerical order.
  2. Count the number of times each value appears.
  3. The mode is the value with the highest frequency.

The ages of students in a college math class are listed below. Identify the mode.

18,18,18,18,19,19,19,20,20,20,20,20,20,20,21,21,22,22,22,22,22,23,24,24,25,29,30,40,44

Solution

Solution

The ages are already listed in order. We will make a table of frequencies to help identify the age with the highest frequency.

A table is shown with 2 rows. The first row is labeled 'Age' and lists the values: 18, 19, 20, 21, 22, 23, 24, 25, 29, 30, 40, and 44. The second row is labeled 'Frequency' and lists the values: 4, 3, 7, 2, 5, 1, 2, 1, 1, 1, 1, and 1.

Now look for the highest frequency. The highest frequency is 7, which corresponds to the age 20. So the mode of the ages in this class is 20 years.

The number of sick days employees used last year: 3,6,2,3,7,5,6,2,4,2. Identify the mode.

Solution

2

The number of handbags owned by women in a book club: 5,6,3,1,5,8,1,5,8,5. Identify the mode.

Solution

5

The data lists the heights (in inches) of students in a statistics class. Identify the mode.

56 61 63 64 65 66 67 67
60 62 63 64 65 66 67 70
60 63 63 64 66 66 67 74
61 63 64 65 66 67 67
Solution

Solution

List each number with its frequency.

A table is shown with 2 rows. The first row is labeled “Number” and lists the values: 56, 60, 61, 62, 63, 64, 65, 66, 67, 70, and 74. The second row is labeled “Frequency” and lists the values: 1, 2, 2, 1, 5, 4, 3, 5, 6, 1, and 1.

Now look for the highest frequency. The highest frequency is 6, which corresponds to the height 67 inches. So the mode of this set of heights is 67 inches.

The ages of the students in a statistics class are listed here: 19, 20, 23, 23, 38, 21, 19, 21, 19, 21, 20, 43, 20, 23, 17, 21, 21, 20, 29, 18, 28. What is the mode?

Solution

21

Students listed the number of members in their household as follows: 6, 2, 5, 6, 3, 7, 5, 6, 5, 3, 4, 4, 5, 7, 6, 4, 5, 2, 1, 5. What is the mode?

Solution

5

Some data sets do not have a mode because no value appears more than any other. And some data sets have more than one mode. In a given set, if two or more data values have the same highest frequency, we say they are all modes.

Use the Basic Definition of Probability

The probability of an event tells us how likely that event is to occur. We usually write probabilities as fractions or decimals.

For example, picture a fruit bowl that contains five pieces of fruit - three bananas and two apples.

If you want to choose one piece of fruit to eat for a snack and don’t care what it is, there is a 35 probability you will choose a banana, because there are three bananas out of the total of five pieces of fruit. The probability of an event is the number of favorable outcomes divided by the total number of outcomes.

Two equations are shown. The top equation says the probability of an event equals the number of favorable outcomes over the total number of outcomes. The bottom equation says the probability of choosing a banana equals 3 over 5. There is a blue arrow pointing to the 3 with the text, 'There are 3 bananas.' There is a blue arrow pointing to the 5 with the text, 'There are 5 pieces of fruit.'

Probability

The probability of an event is the number of favorable outcomes divided by the total number of outcomes possible.

Probability=number of favorable outcomestotal number of outcomes

Converting the fraction 35 to a decimal, we would say there is a 0.6 probability of choosing a banana.

Probability of choosing a banana=35Probability of choosing a banana=0.6

This basic definition of probability assumes that all the outcomes are equally likely to occur. If you study probabilities in a later math class, you’ll learn about several other ways to calculate probabilities.

The ski club is holding a raffle to raise money. They sold 100 tickets. All of the tickets are placed in a jar. One ticket will be pulled out of the jar at random, and the winner will receive a prize. Cherie bought one raffle ticket.

ⓐ Find the probability she will win the prize.

ⓑ Convert the fraction to a decimal.

Solution

Solution

This table outlines the step-by-step process to calculate the probability of Cherie winning a prize, applying the definition of probability.
ⓐ
What are you asked to find? The probability Cherie wins the prize.
What is the number of favorable outcomes? 1, because Cherie has 1 ticket.
Use the definition of probability. Probability of an event=number of favorable outcomestotal number of outcomes
Substitute into the numerator and denominator. Probability Cherie wins=1100
Illustrates converting a probability from fractional form (1/100) to its decimal equivalent (0.01).
ⓑ
Convert the fraction to a decimal.
Write the probability as a fraction. Probability=1100
Convert the fraction to a decimal. Probability=0.01

Ignaly is attending a fashion show where the guests are seated at tables of ten. One guest from each table will be selected at random to receive a door prize. ⓐ Find the probability Ignaly will win the door prize for her table. ⓑ Convert the fraction to a decimal.

Solution
  1. ⓐ 110
  2. ⓑ 0.1

Hoang is among 20 people available to sit on a jury. One person will be chosen at random from the 20. ⓐ Find the probability Hoang will be chosen. ⓑ Convert the fraction to a decimal.

Solution
  1. ⓐ 120
  2. ⓑ 0.05

Three women and five men interviewed for a job. One of the candidates will be offered the job.

ⓐ Find the probability the job is offered to a woman.

ⓑ Convert the fraction to a decimal.

Solution

Solution

This table outlines the steps involved in calculating the probability of a specific event, using the example of a woman being offered a job.
ⓐ
What are you asked to find? The probability the job is offered to a woman.
What is the number of favorable outcomes? 3, because there are three women.
What are the total number of outcomes? 8, because 8 people interviewed.
Use the definition of probability. Probability of an event=number of favorable outcomestotal number of outcomes
Substitute into the numerator and denominator. Probability=38
This table illustrates the step-by-step conversion of a probability expressed as a fraction to its decimal form.
ⓑ
Convert the fraction to a decimal.
Write the probability as a fraction. Probability=38
Convert the fraction to a decimal. Probability=0.375

A bowl of Halloween candy contains 5 chocolate candies and 3 lemon candies. Tanya will choose one piece of candy at random. ⓐ Find the probability Tanya will choose a chocolate candy. ⓑ Convert the fraction to a decimal.

Solution
  1. ⓐ 58
  2. ⓑ 0.625

Dan has 2 pairs of black socks and 6 pairs of blue socks. He will choose one pair at random to wear tomorrow. ⓐ Find the probability Dan will choose a pair of black socks ⓑ Convert the fraction to a decimal.

Solution
  1. ⓐ 28
  2. ⓑ 0.25

ACCESS ADDITIONAL ONLINE RESOURCES

  • Mean, Median, and Mode
  • Find the Mean of a Data Set
  • Find the Median of a Data Set
  • Find the Mode of a Data Set

Key Concepts

  • Calculate the mean of a set of numbers.
    1. Write the formula for the mean mean=sum of values in data setn
    2. Find the sum of all the values in the set. Write the sum in the numerator.
    3. Count the number, n, of values in the set. Write this number in the denominator.
    4. Simplify the fraction.
    5. Check to see that the mean is reasonable. It should be greater than the least number and less than the greatest number in the set.
  • Find the median of a set of numbers.
    1. List the numbers from least to greatest.
    2. Count how many numbers are in the set. Call this n.
    3. Is n odd or even?
      If n is an odd number, the median is the middle value.
      If n is an even number, the median is the mean of the two middle values
  • Identify the mode of a set of numbers.
    1. List the data values in numerical order.
    2. Count the number of times each value appears.
    3. The mode is the value with the highest frequency.

Practice Makes Perfect

Calculate the Mean of a Set of Numbers

In the following exercises, find the mean.

3, 8, 2, 2, 5

Solution

4

6, 1, 9, 3, 4, 7

65, 13, 48, 32, 19, 33

Solution

35

34, 45, 29, 61, and 41

202, 241, 265, 274

Solution

245.5

525, 532, 558, 574

12.45, 12.99, 10.50, 11.25, 9.99, 12.72

Solution

11.65

28.8, 32.9, 32.5, 27.9, 30.4, 32.5, 31.6, 32.7

Four girls leaving a mall were asked how much money they had just spent. The amounts were $0, $14.95, $35.25, and $25.16. Find the mean amount of money spent.

Solution

$18.84

Juan bought 5 shirts to wear to his new job. The costs of the shirts were $32.95, $38.50, $30.00, $17.45, and $24.25. Find the mean cost.

The number of minutes it took Jim to ride his bike to school for each of the past six days was 21, 18, 16, 19, 24, and 19. Find the mean number of minutes.

Solution

19.5 minutes

Norris bought six books for his classes this semester. The costs of the books were $74.28, $120.95, $52.40, $10.59, $35.89, and $59.24. Find the mean cost.

The top eight hitters in a softball league have batting averages of .373, .360, .321, .321, .320, .312, .311, and .311. Find the mean of the batting averages. Round your answer to the nearest thousandth.

Solution

0.329

The monthly snowfall at a ski resort over a six-month period was 60.3, 79.7, 50.9, 28.0, 47.4, and 46.1 inches. Find the mean snowfall.

Find the Median of a Set of Numbers

In the following exercises, find the median.

24, 19, 18, 29, 21

Solution

21

48, 51, 46, 42, 50

65, 56, 35, 34, 44, 39, 55, 52, 45

Solution

45

121, 115, 135, 109, 136, 147, 127, 119, 110

4, 8, 1, 5, 14, 3, 1, 12

Solution

4.5

3, 9, 2, 6, 20, 3, 3, 10

99.2, 101.9, 98.6, 99.5, 100.8, 99.8

Solution

99.65

28.8, 32.9, 32.5, 27.9, 30.4, 32.5, 31.6, 32.7

Last week Ray recorded how much he spent for lunch each workday. He spent $6.50, $7.25, $4.90, $5.30, and $12.00. Find the median.

Solution

$6.50

Michaela is in charge of 6 two-year olds at a daycare center. Their ages, in months, are 25, 24, 28, 32, 29, and 31. Find the median age.

Brian is teaching a swim class for 6 three-year olds. Their ages, in months, are 38,41,45,36,40,and42. Find the median age.

Solution

40.5 months

Sal recorded the amount he spent for gas each week for the past 8 weeks. The amounts were $38.65, $32.18, $40.23, $51.50, $43.68, $30.96, $41.37, and $44.72. Find the median amount.

Identify the Mode of a Set of Numbers

In the following exercises, identify the mode.

2, 5, 1, 5, 2, 1, 2, 3, 2, 3, 1

Solution

2

8, 5, 1 , 3, 7, 1 , 1, 7 , 1, 8 , 7

18, 22, 17, 20, 19, 20, 22, 19, 29, 18, 23, 25, 22, 24, 23, 22, 18, 20, 22, 20

Solution

22

42, 28, 32, 35, 24, 32, 48, 32, 32, 24, 35, 28, 30, 35, 45, 32, 28, 32, 42, 42, 30

The number of children per house on one block: 1, 4, 2, 3, 3, 2, 6, 2, 4 , 2, 0, 3, 0.

Solution

2 children

The number of movies watched each month last year: 2, 0, 3, 0, 0, 8, 6, 5, 0, 1, 2, 3.

The number of units being taken by students in one class: 12, 5, 11, 10, 10, 11, 5, 11, 11, 11, 10, 12.

Solution

11 units

The number of hours of sleep per night for the past two weeks: 8, 5 , 7, 8, 8, 6 , 6, 6, 6, 9, 7, 8, 8, 8.

Use the Basic Definition of Probability

In the following exercises, express the probability as both a fraction and a decimal. (Round to three decimal places, if necessary.)

Josue is in a book club with 20 members. One member is chosen at random each month to select the next month’s book. Find the probability that Josue will be chosen next month.

Solution

120,0.05

Jessica is one of eight kindergarten teachers at Mandela Elementary School. One of the kindergarten teachers will be selected at random to attend a summer workshop. Find the probability that Jessica will be selected.

There are 24 people who work in Dane’s department. Next week, one person will be selected at random to bring in doughnuts. Find the probability that Dane will be selected. Round your answer to the nearest thousandth.

Solution

124,0.0416–≈0.042

Monica has two strawberry yogurts and six banana yogurts in her refrigerator. She will choose one yogurt at random to take to work. Find the probability Monica will choose a strawberry yogurt.

Michel has four rock CDs and six country CDs in his car. He will pick one CD to play on his way to work. Find the probability Michel will pick a rock CD.

Solution

25,0.4

Noah is planning his summer camping trip. He can’t decide among six campgrounds at the beach and twelve campgrounds in the mountains, so he will choose one campground at random. Find the probability that Noah will choose a campground at the beach.

Donovan is considering transferring to a 4-year college. He is considering 10 out-of state colleges and 4 colleges in his state. He will choose one college at random to visit during spring break. Find the probability that Donovan will choose an out-of-state college.

Solution

57,0.714285———≈0.714

There are 258,890,850 number combinations possible in the Mega Millions lottery. One winning jackpot ticket will be chosen at random. Brent chooses his favorite number combination and buys one ticket. Find the probability Brent will win the jackpot. Round the decimal to the first digit that is not zero, then write the name of the decimal.

Everyday Math

Joaquin gets paid every Friday. His paychecks for the past 8 Fridays were $315, $236.25, $236.25, $236.25,$315, $315, $236.25, $393.75. Find the ⓐ mean, ⓑ median, and ⓒ mode.

Solution
  1. ⓐ $285.47
  2. ⓑ $275.63
  3. ⓒ $236.25

The cash register receipts each day last week at a coffee shop were $1,845, $1,520, $1,438, $1,682, $1,850, $2,721, $2,539. Find the ⓐ mean, ⓑ median, and ⓒ mode.

Writing Exercises

Explain in your own words the difference between the mean, median, and mode of a set of numbers.

Solution

Answers will vary.

Make an example of probability that relates to your life. Write your answer as a fraction and explain what the numerator and denominator represent.

Self Check

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

A self-assessment table for statistical skills, including calculating mean, median, mode, and using probability, with options to mark confidence levels: 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?

mean
The mean of a set of n numbers is the arithmetic average of the numbers. The formula is mean=sum of values in data setn
median
The median of a set of data values is the middle value.
  • Half the data values are less than or equal to the median.
  • Half the data values are greater than or equal to the median.
mode
The mode of a set of numbers is the number with the highest frequency.

Ratios and Rate

Learning Objectives

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

  • Write a ratio as a fraction
  • Write a rate as a fraction
  • Find unit rates
  • Find unit price
  • Translate phrases to expressions with fractions

Before you get started, take this readiness quiz.

Simplify: 1624.
If you missed this problem, review Example 1 in Multiply and Divide Fractions.

Solution

23

Divide: 2.76÷11.5.
If you missed this problem, review Example 9 in Decimal Operations.

Solution

0.24

Simplify: 112234.
If you missed this problem, review Example 7 in Multiply and Divide Mixed Numbers and Complex Fractions.

Solution

611

Write a Ratio as a Fraction

When you apply for a mortgage, the loan officer will compare your total debt to your total income to decide if you qualify for the loan. This comparison is called the debt-to-income ratio. A ratio compares two quantities that are measured with the same unit. If we compare a and b, the ratio is written as atob,ab,ora:b.

Ratios

A ratio compares two numbers or two quantities that are measured with the same unit. The ratio of a to b is written atob,ab,ora:b.

In this section, we will use the fraction notation. When a ratio is written in fraction form, the fraction should be simplified. If it is an improper fraction, we do not change it to a mixed number. Because a ratio compares two quantities, we would leave a ratio as 41 instead of simplifying it to 4 so that we can see the two parts of the ratio.

Write each ratio as a fraction: ⓐ 15to27ⓑ 45to18.

Solution

Solution

This table illustrates the step-by-step process of converting a given ratio into a simplified fraction.
ⓐ
15 to 27
Write as a fraction with the first number in the numerator and the second in the denominator. 1527
Simplify the fraction. 59
This table demonstrates converting and simplifying the ratio 45 to 18 into a fraction, showing the steps involved.
ⓑ
45 to 18
Write as a fraction with the first number in the numerator and the second in the denominator. 4518
Simplify. 52

We leave the ratio in ⓑ as an improper fraction.

Write each ratio as a fraction: ⓐ 21to56ⓑ 48to32.

Solution
  1. ⓐ 38
  2. ⓑ 32

Write each ratio as a fraction: ⓐ 27to72ⓑ 51to34.

Solution
  1. ⓐ 38
  2. ⓑ 32

Ratios Involving Decimals

We will often work with ratios of decimals, especially when we have ratios involving money. In these cases, we can eliminate the decimals by using the Equivalent Fractions Property to convert the ratio to a fraction with whole numbers in the numerator and denominator.

For example, consider the ratio 0.8to0.05. We can write it as a fraction with decimals and then multiply the numerator and denominator by 100 to eliminate the decimals.

A fraction is shown with 0.8 in the numerator and 0.05 in the denominator. Below it is the same fraction with both the numerator and denominator multiplied by 100. Below that is a fraction with 80 in the numerator and 5 in the denominator.

Do you see a shortcut to find the equivalent fraction? Notice that 0.8=810 and 0.05=5100. The least common denominator of 810 and 5100 is 100. By multiplying the numerator and denominator of 0.80.05 by 100, we ‘moved’ the decimal two places to the right to get the equivalent fraction with no decimals. Now that we understand the math behind the process, we can find the fraction with no decimals like this:

Illustrates dividing decimals by converting to whole numbers and simplifying the resulting fraction.
The top line says 0.80 over 0.05. There are blue arrows moving the decimal points over 2 places to the right.
"Move" the decimal 2 places. 805
Simplify. 161

You do not have to write out every step when you multiply the numerator and denominator by powers of ten. As long as you move both decimal places the same number of places, the ratio will remain the same.

Write each ratio as a fraction of whole numbers:

  1. ⓐ 4.8to11.2
  2. ⓑ 2.7to0.54
Solution
Solution
This table illustrates the step-by-step process of converting the decimal ratio 4.8 to 11.2 into its simplest fractional form, resulting in 3/7.
ⓐ 4.8 to 11.2
Write as a fraction. 4.811.2
Rewrite as an equivalent fraction without decimals, by moving both decimal points 1 place to the right. 48112
Simplify. 37

So 4.8to11.2 is equivalent to 37.

Steps to simplify a fraction with decimals, demonstrating how to clear decimals and reduce the fraction to its simplest form.
ⓑ
The numerator has one decimal place and the denominator has 2. To clear both decimals we need to move the decimal 2 places to the right.
2.7to0.54
Write as a fraction. 2.70.54
Move both decimals right two places. 27054
Simplify. 51

So 2.7to0.54 is equivalent to 51.

Write each ratio as a fraction: ⓐ 4.6to11.5ⓑ 2.3to0.69.

Solution
  1. ⓐ 25
  2. ⓑ 103

Write each ratio as a fraction: ⓐ 3.4to15.3ⓑ 3.4to0.68.

Solution
  1. ⓐ 29
  2. ⓑ 51

Some ratios compare two mixed numbers. Remember that to divide mixed numbers, you first rewrite them as improper fractions.

Write the ratio of 114to238 as a fraction.

Solution
Solution
This table illustrates the step-by-step process of simplifying a ratio of two mixed numbers into a single simplified fraction.
114to238
Write as a fraction. 114238
Convert the numerator and denominator to improper fractions. 54198
Rewrite as a division of fractions. 54÷198
Invert the divisor and multiply. 54·819
Simplify. 1019

Write each ratio as a fraction: 134to258.

Solution

23

Write each ratio as a fraction: 118to234.

Solution

922

Applications of Ratios

One real-world application of ratios that affects many people involves measuring cholesterol in blood. The ratio of total cholesterol to HDL cholesterol is one way doctors assess a person's overall health. A ratio of less than 5 to 1 is considered good.

Hector's total cholesterol is 249 mg/dl and his HDL cholesterol is 39 mg/dl. ⓐ Find the ratio of his total cholesterol to his HDL cholesterol. ⓑ Assuming that a ratio less than 5 to 1 is considered good, what would you suggest to Hector?

Solution
Solution

ⓐ First, write the words that express the ratio. We want to know the ratio of Hector's total cholesterol to his HDL cholesterol.

Step-by-step process for expressing and simplifying a ratio of cholesterol values as a fraction.
Write as a fraction. total cholesterolHDL cholesterol
Substitute the values. 24939
Simplify. 8313

ⓑ Is Hector's cholesterol ratio ok? If we divide 83 by 13 we obtain approximately 6.4, so 8313≈6.41. Hector's cholesterol ratio is high! Hector should either lower his total cholesterol or raise his HDL cholesterol.

Find the patient's ratio of total cholesterol to HDL cholesterol using the given information.

Total cholesterol is 185 mg/dL and HDL cholesterol is 40 mg/dL.

Solution

378

Find the patient’s ratio of total cholesterol to HDL cholesterol using the given information.

Total cholesterol is 204 mg/dL and HDL cholesterol is 38 mg/dL.

Solution

10219

Ratios of Two Measurements in Different Units

To find the ratio of two measurements, we must make sure the quantities have been measured with the same unit. If the measurements are not in the same units, we must first convert them to the same units.

We know that to simplify a fraction, we divide out common factors. Similarly in a ratio of measurements, we divide out the common unit.

The Americans with Disabilities Act (ADA) Guidelines for wheel chair ramps require a maximum vertical rise of 1 inch for every 1 foot of horizontal run. What is the ratio of the rise to the run?

Solution
Solution

In a ratio, the measurements must be in the same units. We can change feet to inches, or inches to feet. It is usually easier to convert to the smaller unit, since this avoids introducing more fractions into the problem.

Write the words that express the ratio.

Step-by-step calculation of the rise to run ratio, demonstrating unit conversion and simplification.
Ratio of the rise to the run
Write the ratio as a fraction. riserun
Substitute in the given values. 1 inch1 foot
Convert 1 foot to inches. 1 inch12 inches
Simplify, dividing out common factors and units. 112

So the ratio of rise to run is 1 to 12. This means that the ramp should rise 1 inch for every 12 inches of horizontal run to comply with the guidelines.

Find the ratio of the first length to the second length: 32 inches to 1 foot.

Solution

83

Find the ratio of the first length to the second length: 1 foot to 54 inches.

Solution

29

Write a Rate as a Fraction

Frequently we want to compare two different types of measurements, such as miles to gallons. To make this comparison, we use a rate. Examples of rates are 120 miles in 2 hours, 160 words in 4 minutes, and $5 dollars per 64 ounces.

Rate

A rate compares two quantities of different units. A rate is usually written as a fraction.

When writing a fraction as a rate, we put the first given amount with its units in the numerator and the second amount with its units in the denominator. When rates are simplified, the units remain in the numerator and denominator.

Bob drove his car 525 miles in 9 hours. Write this rate as a fraction.

Solution

Solution

Demonstrates converting a rate into a fraction and simplifying it.
525 miles in 9 hours
Write as a fraction, with 525 miles in the numerator and 9 hours in the denominator. 525 miles9 hours
175 miles3 hours

So 525 miles in 9 hours is equivalent to 175 miles3 hours.

Write the rate as a fraction: 492 miles in 8 hours.

Solution

123 miles2 hours

Write the rate as a fraction: 242 miles in 6 hours.

Solution

121 miles3 hours

Find Unit Rates

In the last example, we calculated that Bob was driving at a rate of 175 miles3 hours. This tells us that every three hours, Bob will travel 175 miles. This is correct, but not very useful. We usually want the rate to reflect the number of miles in one hour. A rate that has a denominator of 1 unit is referred to as a unit rate.

Unit Rate

A unit rate is a rate with denominator of 1 unit.

Unit rates are very common in our lives. For example, when we say that we are driving at a speed of 68 miles per hour we mean that we travel 68 miles in 1 hour. We would write this rate as 68 miles/hour (read 68 miles per hour). The common abbreviation for this is 68 mph. Note that when no number is written before a unit, it is assumed to be 1.

So 68 miles/hour really means 68 miles/1 hour.

Two rates we often use when driving can be written in different forms, as shown:

Example Rate Write Abbreviate Read
68 miles in 1 hour 68 miles1 hour 68 miles/hour 68 mph 68 miles per hour
36 miles to 1 gallon 36 miles1 gallon 36 miles/gallon 36 mpg 36 miles per gallon

Another example of unit rate that you may already know about is hourly pay rate. It is usually expressed as the amount of money earned for one hour of work. For example, if you are paid $12.50 for each hour you work, you could write that your hourly (unit) pay rate is $12.50/hour (read $12.50 per hour.)

To convert a rate to a unit rate, we divide the numerator by the denominator. This gives us a denominator of 1.

Anita was paid $384 last week for working 32 hours. What is Anita’s hourly pay rate?

Solution

Solution

Steps demonstrating how to calculate an hourly rate from a given total earning and hours worked.
Start with a rate of dollars to hours. Then divide. $384 last week for 32 hours
Write as a rate. $38432 hours
Divide the numerator by the denominator. $121 hour
Rewrite as a rate. $12/hour

Anita’s hourly pay rate is $12 per hour.

Find the unit rate: $630 for 35 hours.

Solution

$18.00/hour

Find the unit rate: $684 for 36 hours.

Solution

$19.00/hour

Sven drives his car 455 miles, using 14 gallons of gasoline. How many miles per gallon does his car get?

Solution

Solution

Start with a rate of miles to gallons. Then divide.

This table illustrates the steps to convert a given ratio of miles to gallons into its equivalent unit rate.
455 miles to 14 gallons of gas
Write as a rate. 455 miles14 gallons
Divide 455 by 14 to get the unit rate. 32.5 miles1 gallon

Sven’s car gets 32.5 miles/gallon, or 32.5 mpg.

Find the unit rate: 423 miles to 18 gallons of gas.

Solution

23.5 mpg

Find the unit rate: 406 miles to 14.5 gallons of gas.

Solution

28 mpg

Find Unit Price

Sometimes we buy common household items ‘in bulk’, where several items are packaged together and sold for one price. To compare the prices of different sized packages, we need to find the unit price. To find the unit price, divide the total price by the number of items. A unit price is a unit rate for one item.

Unit price

A unit price is a unit rate that gives the price of one item.

The grocery store charges $3.99 for a case of 24 bottles of water. What is the unit price?

Solution

Solution

What are we asked to find? We are asked to find the unit price, which is the price per bottle.

Steps for calculating unit price per bottle, from initial rate to rounded final amount.
Write as a rate. $3.9924 bottles
Divide to find the unit price. $0.166251 bottle
Round the result to the nearest penny. $0.171 bottle

The unit price is approximately $0.17 per bottle. Each bottle costs about $0.17.

Find the unit price. Round your answer to the nearest cent if necessary.

24-pack of juice boxes for $6.99

Solution

$0.29/box

Find the unit price. Round your answer to the nearest cent if necessary.

24-pack of bottles of ice tea for $12.72

Solution

$0.53/bottle

Unit prices are very useful if you comparison shop. The better buy is the item with the lower unit price. Most grocery stores list the unit price of each item on the shelves.

Paul is shopping for laundry detergent. At the grocery store, the liquid detergent is priced at $14.99 for 64 loads of laundry and the same brand of powder detergent is priced at $15.99 for 80 loads.

Which detergent has the lowest cost per load?

Solution

Solution

To compare the prices, we first find the unit price for each type of detergent.
Liquid Powder
Write as a rate. $14.9964 loads $15.9980 loads
Find the unit price. $0.234…1 load $0.199…1 load
Round to the nearest cent. $0.23/load(23 cents per load.) $0.20/load(20 cents per load)

Now we compare the unit prices. The unit price of the liquid detergent is about $0.23 per load and the unit price of the powder detergent is about $0.20 per load. The powder is the better buy.

Find each unit price and then determine the better buy. Round to the nearest cent if necessary.

Brand A Storage Bags, $4.59 for 40 count, or Brand B Storage Bags, $3.99 for 30 count

Solution

Brand A costs $0.11 per bag. Brand B costs $0.13 per bag. Brand A is the better buy.

Find each unit price and then determine the better buy. Round to the nearest cent if necessary.

Brand C Chicken Noodle Soup, $1.89 for 26 ounces, or Brand D Chicken Noodle Soup, $0.95 for 10.75 ounces

Solution

Brand C costs $0.07 per ounce. Brand D costs $0.09 per ounce. Brand C is the better buy.

Notice in Example 10 that we rounded the unit price to the nearest cent. Sometimes we may need to carry the division to one more place to see the difference between the unit prices.

Translate Phrases to Expressions with Fractions

Have you noticed that the examples in this section used the comparison words ratio of, to, per, in, for, on, and from? When you translate phrases that include these words, you should think either ratio or rate. If the units measure the same quantity (length, time, etc.), you have a ratio. If the units are different, you have a rate. In both cases, you write a fraction.

Translate the word phrase into an algebraic expression:

  1. ⓐ 427 miles per h hours
  2. ⓑ x students to 3 teachers
  3. ⓒ y dollars for 18 hours
Solution

Solution

This table demonstrates how to convert a verbal rate description into its mathematical fractional representation.
ⓐ
427 miles perhhours
Write as a rate. 427 mileshhours
This table demonstrates how to convert a verbal ratio, such as 'x students to 3 teachers', into a fractional rate expression.
ⓑ
xstudents to 3 teachers
Write as a rate. xstudents3 teachers
Demonstrates converting 'y dollars for 18 hours' into a mathematical rate.
ⓒ
ydollars for 18 hours
Write as a rate. $y18 hours

Translate the word phrase into an algebraic expression.

ⓐ 689 miles per h hours ⓑ y parents to 22 students ⓒ d dollars for 9 minutes

Solution
  1. ⓐ 689 mi/h hours
  2. ⓑ y parents/22 students
  3. ⓒ $d/9 min

Translate the word phrase into an algebraic expression.

ⓐ m miles per 9 hours ⓑ x students to 8 buses ⓒ y dollars for 40 hours

Solution
  1. ⓐ m mi/9 h
  2. ⓑ x students/8 buses
  3. ⓒ $y/40 h

ACCESS ADDITIONAL ONLINE RESOURCES

  • Ratios
  • Write Ratios as a Simplified Fractions Involving Decimals and Fractions
  • Write a Ratio as a Simplified Fraction
  • Rates and Unit Rates
  • Unit Rate for Cell Phone Plan

Practice Makes Perfect

Write a Ratio as a Fraction

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

20 to 36

Solution

59

20 to 32

42 to 48

Solution

78

45 to 54

49 to 21

Solution

73

56 to 16

84 to 36

Solution

73

6.4 to 0.8

0.56 to 2.8

Solution

15

1.26 to 4.2

123 to 256

Solution

1017

134 to 258

416 to 313

Solution

54

535 to 335

$18 to $63

Solution

27

$16 to $72

$1.21 to $0.44

Solution

114

$1.38 to $0.69

28 ounces to 84 ounces

Solution

13

32 ounces to 128 ounces

12 feet to 46 feet

Solution

623

15 feet to 57 feet

246 milligrams to 45 milligrams

Solution

8215

304 milligrams to 48 milligrams

total cholesterol of 175 to HDL cholesterol of 45

Solution

359

total cholesterol of 215 to HDL cholesterol of 55

27 inches to 1 foot

Solution

94

28 inches to 1 foot

Write a Rate as a Fraction

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

140 calories per 12 ounces

Solution

35 calories3 ounces

180 calories per 16 ounces

8.2 pounds per 3 square inches

Solution

41 lbs15 sq. in.

9.5 pounds per 4 square inches

488 miles in 7 hours

Solution

488 miles7 hours

527 miles in 9 hours

$595 for 40 hours

Solution

$1198 hours

$798 for 40 hours

Find Unit Rates

In the following exercises, find the unit rate. Round to two decimal places, if necessary.

140 calories per 12 ounces

Solution

11.67 calories/ounce

180 calories per 16 ounces

8.2 pounds per 3 square inches

Solution

2.73 lbs./sq. in.

9.5 pounds per 4 square inches

488 miles in 7 hours

Solution

69.71 mph

527 miles in 9 hours

$595 for 40 hours

Solution

$14.88/hour

$798 for 40 hours

576 miles on 18 gallons of gas

Solution

32 mpg

435 miles on 15 gallons of gas

43 pounds in 16 weeks

Solution

2.69 lbs./week

57 pounds in 24 weeks

46 beats in 0.5 minute

Solution

92 beats/minute

54 beats in 0.5 minute

The bindery at a printing plant assembles 96,000 magazines in 12 hours. How many magazines are assembled in one hour?

Solution

8,000

The pressroom at a printing plant prints 540,000 sections in 12 hours. How many sections are printed per hour?

Find Unit Price

In the following exercises, find the unit price. Round to the nearest cent.

Soap bars at 8 for $8.69

Solution

$1.09/bar

Soap bars at 4 for $3.39

Women’s sports socks at 6 pairs for $7.99

Solution

$1.33/pair

Men’s dress socks at 3 pairs for $8.49

Snack packs of cookies at 12 for $5.79

Solution

$0.48/pack

Granola bars at 5 for $3.69

CD-RW discs at 25 for $14.99

Solution

$0.60/disc

CDs at 50 for $4.49

The grocery store has a special on macaroni and cheese. The price is $3.87 for 3 boxes. How much does each box cost?

Solution

$1.29/box

The pet store has a special on cat food. The price is $4.32 for 12 cans. How much does each can cost?

In the following exercises, find each unit price and then identify the better buy. Round to three decimal places.

Mouthwash, 50.7-ounce size for $6.99 or 33.8-ounce size for $4.79

Solution

The 50.7-ounce size costs $0.138 per ounce. The 33.8-ounce size costs $0.142 per ounce. The 50.7-ounce size is the better buy.

Toothpaste, 6 ounce size for $3.19 or 7.8-ounce size for $5.19

Breakfast cereal, 18 ounces for $3.99 or 14 ounces for $3.29

Solution

The 18-ounce size costs $0.222 per ounce. The 14-ounce size costs $0.235 per ounce. The 18-ounce size is a better buy.

Breakfast Cereal, 10.7 ounces for $2.69 or 14.8 ounces for $3.69

Ketchup, 40-ounce regular bottle for $2.99 or 64-ounce squeeze bottle for $4.39

Solution

The regular bottle costs $0.075 per ounce. The squeeze bottle costs $0.069 per ounce. The squeeze bottle is a better buy.

Mayonnaise 15-ounce regular bottle for $3.49 or 22-ounce squeeze bottle for $4.99

Cheese $6.49 for 1 lb. block or $3.39 for 12 lb. block

Solution

The half-pound block costs $6.78/lb, so the 1-lb. block is a better buy.

Candy $10.99 for a 1 lb. bag or $2.89 for 14 lb. of loose candy

Translate Phrases to Expressions with Fractions

In the following exercises, translate the English phrase into an algebraic expression.

793 miles per p hours

Solution

793 milesphours

78 feet per r seconds

$3 for 0.5 lbs.

Solution

$30.5 lbs.

j beats in 0.5 minutes

105 calories in x ounces

Solution

105 caloriesxounces

400 minutes for m dollars

the ratio of y and 5x

Solution

y5x

the ratio of 12x and y

Everyday Math

One elementary school in Ohio has 684 students and 45 teachers. Write the student-to-teacher ratio as a unit rate.

Solution

15.2 students per teacher

The average American produces about 1,600 pounds of paper trash per year (365 days). How many pounds of paper trash does the average American produce each day? (Round to the nearest tenth of a pound.)

A popular fast food burger weighs 7.5 ounces and contains 540 calories, 29 grams of fat, 43 grams of carbohydrates, and 25 grams of protein. Find the unit rate of ⓐ calories per ounce ⓑ grams of fat per ounce ⓒ grams of carbohydrates per ounce ⓓ grams of protein per ounce. Round to two decimal places.

Solution
  1. ⓐ 72 calories/ounce
  2. ⓑ 3.87 grams of fat/ounce
  3. ⓒ 5.73 grams carbs/ounce
  4. ⓓ 3.33 grams protein/ounce

A 16-ounce chocolate mocha coffee with whipped cream contains 470 calories, 18 grams of fat, 63 grams of carbohydrates, and 15 grams of protein. Find the unit rate of ⓐ calories per ounce ⓑ grams of fat per ounce ⓒ grams of carbohydrates per ounce ⓓ grams of protein per ounce.

Writing Exercises

Would you prefer the ratio of your income to your friend’s income to be 3/1 or 1/3? Explain your reasoning.

Solution

Answers will vary.

The parking lot at the airport charges $0.75 for every 15 minutes. ⓐ How much does it cost to park for 1 hour? ⓑ Explain how you got your answer to part ⓐ. Was your reasoning based on the unit cost or did you use another method?

Kathryn ate a 4-ounce cup of frozen yogurt and then went for a swim. The frozen yogurt had 115 calories. Swimming burns 422 calories per hour. For how many minutes should Kathryn swim to burn off the calories in the frozen yogurt? Explain your reasoning.

Solution

Kathryn should swim for approximately 16.35 minutes. Explanations will vary.

Mollie had a 16-ounce cappuccino at her neighborhood coffee shop. The cappuccino had 110 calories. If Mollie walks for one hour, she burns 246 calories. For how many minutes must Mollie walk to burn off the calories in the cappuccino? Explain your reasoning.

Self Check

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

Self-assessment checklist for students to evaluate their understanding of ratios, rates, unit prices, and translating phrases to fractions, categorized by confidence level.

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

ratio
A ratio compares two numbers or two quantities that are measured with the same unit. The ratio of a to b is written a to b, ab, or a:b.
rate
A rate compares two quantities of different units. A rate is usually written as a fraction.
unit rate
A unit rate is a rate with denominator of 1 unit.
unit price
A unit price is a unit rate that gives the price of one item.

Simplify and Use Square Roots

Learning Objectives

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

  • Simplify expressions with square roots
  • Estimate square roots
  • Approximate square roots
  • Simplify variable expressions with square roots
  • Use square roots in applications

Before you get started, take this readiness quiz.

Simplify: (−9)2.
If you missed this problem, review Example 6 in Multiply and Divide Integers.

Solution

81

Round 3.846 to the nearest hundredth.
If you missed this problem, review Example 9 in Decimals.

Solution

3.85

Evaluate 12d for d=80.
If you missed this problem, review Example 2 in Evaluate, Simplify, and Translate Expressions.

Solution

960

Simplify Expressions with Square Roots

To start this section, we need to review some important vocabulary and notation.

Remember that when a number n is multiplied by itself, we can write this as n2, which we read aloud as “nsquared.” For example, 82 is read as “8squared.”

We call 64 the square of 8 because 82=64. Similarly, 121 is the square of 11, because 112=121.

Square of a Number

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

Modeling Squares

Do you know why we use the word square? If we construct a square with three tiles on each side, the total number of tiles would be nine.

A square is shown with 3 tiles on each side. There are a total of 9 tiles in the square.

This is why we say that the square of three is nine.

32=9

The number 9 is called a perfect square because it is the square of a whole number.

Doing the Manipulative Mathematics activity Square Numbers will help you develop a better understanding of perfect square numbers

The chart shows the squares of the counting numbers 1 through 15. You can refer to it to help you identify the perfect squares.

A table with two columns is shown. The first column is labeled “Number” and has the values: n, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15. The second column is labeled “Square” and has the values: n squared, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, and 225.

Perfect Squares

A perfect square is the square of a whole number.

What happens when you square a negative number?

(−8)2=(−8)(−8)=64

When we multiply two negative numbers, the product is always positive. So, the square of a negative number is always positive.

The chart shows the squares of the negative integers from −1 to −15.

A table is shown with 2 columns. The first column is labeled “Number” and contains the values: n, negative 1, negative 2, negative 3, negative 4, negative 5, negative 6, negative 7, negative 8, negative 9, negative 10, negative 11, negative 12, negative 13, negative 14, and negative 15. The next column is labeled “Square” and contains the values: n squared, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, and 225.

Did you notice that these squares are the same as the squares of the positive numbers?

Square Roots

Sometimes we will need to look at the relationship between numbers and their squares in reverse. Because 102=100, we say 100 is the square of 10. We can also say that 10 is a square root of 100.

Square Root of a Number

A number whose square is m is called a square root of m.

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.

What if we only want the positive square root of a positive number? The radical sign, 0, stands for the positive square root. The positive square root is also called the principal square root.

Square Root Notation

m is read as “the square root of m.”

Ifm=n2,thenm=nforn≥0.

A picture of an m inside a square root sign is shown. The sign is labeled as a radical sign and the m is labeled as the radicand.

We can also use the radical sign for the square root of zero. Because 02=0,0=0. Notice that zero has only one square root.

The chart shows the square roots of the first 15 perfect square numbers.

A table is shown with 2 columns. The first column contains the values: square root of 1, square root of 4, square root of 9, square root of 16, square root of 25, square root of 36, square root of 49, square root of 64, square root of 81, square root of 100, square root of 121, square root of 144, square root of 169, square root of 196, and square root of 225. The second column contains the values: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15.

Simplify: ⓐ 25ⓑ 121.

Solution
Solution
This table illustrates the process of finding the square root of 25, presenting the expression, its result, and the underlying mathematical justification.
ⓐ
25
Since 52=25 5
Demonstrates finding the square root of 121, presenting the mathematical justification and the resulting value.
ⓑ
121
Since 112=121 11

Simplify: ⓐ 36ⓑ 169.

Solution
  1. ⓐ 6
  2. ⓑ 13

Simplify: ⓐ 16ⓑ 196.

Solution
  1. ⓐ 4
  2. ⓑ 14

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.

Simplify. ⓐ −9ⓑ −144.

Solution
Solution
Simplification of -√9, emphasizing the negative sign's position outside the radical.
ⓐ
−9
The negative is in front of the radical sign. −3
This table demonstrates the evaluation of the expression -√144, explaining that the negative sign is applied after calculating the square root.
ⓑ
−144
The negative is in front of the radical sign. −12

Simplify: ⓐ −4ⓑ −225.

Solution
  1. ⓐ −2
  2. ⓑ −15

Simplify: ⓐ −81ⓑ −64.

Solution
  1. ⓐ −9
  2. ⓑ −8

Square Root of a Negative Number

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 have a square that is −25. Why? Any positive number squared is positive, and any negative number squared is also positive. In the next chapter we will see that all the numbers we work with are called the real numbers. So we say there is no real number equal to −25. If we are asked to find the square root of any negative number, we say that the solution is not a real number.

Simplify: ⓐ −169ⓑ −121.

Solution
Solution

ⓐ There is no real number whose square is −169. Therefore, −169 is not a real number.

ⓑ The negative is in front of the radical sign, so we find the opposite of the square root of 121.

Demonstrates evaluation of a negative square root, highlighting the sign's placement and showing the final result.
−121
The negative is in front of the radical. −11

Simplify: ⓐ −196ⓑ −81.

Solution
  1. ⓐ not a real number
  2. ⓑ −9

Simplify: ⓐ −49ⓑ −121.

Solution
  1. ⓐ not a real number
  2. ⓑ −11

Square Roots and the Order of Operations

When using the order of operations to simplify an expression that has square roots, we treat the radical sign as a grouping symbol. We simplify any expressions under the radical sign before performing other operations.

Simplify: ⓐ 25+144ⓑ 25+144.

Solution
Solution
Step-by-step solution demonstrating the order of operations for evaluating the expression 25 + 144.
ⓐ Use the order of operations.
25+144
Simplify each radical. 5+12
Add. 17
Step-by-step simplification of a square root expression using the order of operations.
ⓑ Use the order of operations.
25+144
Add under the radical sign. 169
Simplify. 13

Simplify: ⓐ 9+16ⓑ 9+16.

Solution
  1. ⓐ 7
  2. ⓑ 5

Simplify: ⓐ 64+225ⓑ 64+225.

Solution
  1. ⓐ 17
  2. ⓑ 23

Notice the different answers in parts ⓐ and ⓑ of Example 4. It is important to follow the order of operations correctly. In ⓐ , we took each square root first and then added them. In ⓑ , we added under the radical sign first and then found the square root.

Estimate Square Roots

So far we have only worked with square roots of perfect squares. The square roots of other numbers are not whole numbers.

A table is shown with 2 columns. The first column is labeled “Number” and contains the values: 4, 5, 6, 7, 8, 9. The second column is labeled “Square root” and contains the values: square root of 4 equals 2, square root of 5, square root of 6, square root of 7, square root of 8, square root of 9 equals 3.

We might conclude that the square roots of numbers between 4 and 9 will be between 2 and 3, and they will not be whole numbers. Based on the pattern in the table above, we could say that 5 is between 2 and 3. Using inequality symbols, we write

2<5<3

Estimate 60 between two consecutive whole numbers.

Solution

Solution

Think of the perfect squares closest to 60. Make a small table of these perfect squares and their squares roots.

A table is shown with 2 columns. The first column is labeled “Number” and contains the values: 36, 49, 64, and 81. There is a balloon coming out of the table between 49 and 64 that says 60. The second column is labeled “Square root” and contains the values: 6, 7, 8, and 9. There is a balloon coming out of the table between 7 and 8 that says square root of 60.
Illustrates bounding a number by consecutive perfect squares and its square root by their respective roots.
Locate 60 between two consecutive perfect squares. 49<60<64
60is between their square roots. 7<60<8

Estimate 38 between two consecutive whole numbers.

Solution

6<38<7

Estimate 84 between two consecutive whole numbers.

Solution

9<84<10

Approximate Square Roots with a Calculator

There are mathematical methods to approximate square roots, but it is much more convenient to use a calculator to find square roots. Find the 0 or x key on your calculator. You will need to use this key to approximate square roots. When you use your calculator to find the square root of a number that is not a perfect square, the answer that you see is not the exact number. It is an approximation, 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. Using it to find the square root of 5 will give 2.236067977. This is the approximate square root of 5. When we report the answer, we should use the “approximately equal to” sign instead of an equal sign.

5≈2.236067978

You will seldom use this many digits for applications in algebra. So, if you wanted to round 5 to two decimal places, you would write

5≈2.24

How do we know these values are approximations and not the exact values? Look at what happens when we square them.

2.2360679782=5.0000000022.242=5.0176

The squares are close, but not exactly equal, to 5.

Round 17 to two decimal places using a calculator.

Solution

Solution

Demonstrates the step-by-step calculation and rounding of the square root of 17, showing the initial expression, calculator result, and final approximation.
17
Use the calculator square root key. 4.123105626
Round to two decimal places. 4.12
17≈4.12

Round 11 to two decimal places.

Solution

≈ 3.32

Round 13 to two decimal places.

Solution

≈ 3.61

Simplify Variable Expressions with Square Roots

Expressions with square root that we have looked at so far have not had any variables. What happens when we have to find a square root of a variable expression?

Consider 9x2, where x≥0. Can you think of an expression whose square is 9x2?

(?)2=9x2(3x)2=9x2so9x2=3x

When we use a variable in a square root expression, for our work, we will assume that the variable represents a non-negative number. In every example and exercise that follows, each variable in a square root expression is greater than or equal to zero.

Simplify: x2.

Solution

Solution

Think about what we would have to square to get x2. Algebraically, (?)2=x2

Illustrates the algebraic simplification of `sqrt(x^2)` to `x`, based on the property `(x)^2=x^2`.
x2
Since (x)2=x2 x

Simplify: y2.

Solution

y

Simplify: m2.

Solution

m

Simplify: 16x2.

Solution

Solution

Illustrates the simplification of the square root of 16x squared, showing the original expression, reasoning, and final simplified form.
16x2
Since(4x)2=16x2 4x

Simplify: 64x2.

Solution

8x

Simplify: 169y2.

Solution

13y

Simplify: −81y2.

Solution

Solution

This table illustrates the simplification of the mathematical expression -sqrt(81y^2) to -9y, including the reasoning.
−81y2
Since(9y)2=81y2 −9y

Simplify: −121y2.

Solution

−11y

Simplify: −100p2.

Solution

−10p

Simplify: 36x2y2.

Solution

Solution

Illustrates the simplification of a square root expression along with its underlying reason.
36x2y2
Since(6xy)2=36x2y2 6xy

Simplify: 100a2b2.

Solution

10ab

Simplify: 225m2n2.

Solution

15mn

Use Square Roots in Applications

As you progress through your college courses, you’ll encounter several applications of square roots. Once again, if we use our strategy for applications, it will give us a plan for finding the answer!

Use a strategy for applications with square roots.

  1. Identify what you are asked to find.
  2. Write a phrase that gives the information to find it.
  3. Translate the phrase to an expression.
  4. Simplify the expression.
  5. Write a complete sentence that answers the question.

Square Roots and Area

We have solved applications with area before. If we were given the length of the sides of a square, we could find its area by squaring the length of its sides. Now we can find the length of the sides of a square if we are given the area, by finding the square root of the area.

If the area of the square is A square units, the length of a side is A units. See Table 14.

Area (square units) Length of side (units)
9 9=3
144 144=12
A A

Mike and Lychelle want to make a square patio. They have enough concrete for an area of 200 square feet. To the nearest tenth of a foot, how long can a side of their square patio be?

Solution
Solution

We know the area of the square is 200 square feet and want to find the length of the side. If the area of the square is A square units, the length of a side is A units.

Step-by-step guide demonstrating how to calculate the side length of a square patio from its area, using square roots and rounding to one decimal place.
What are you asked to find? The length of each side of a square patio
Write a phrase. The length of a side
Translate to an expression. A
Evaluate A when A=200. 200
Use your calculator. 14.142135...
Round to one decimal place. 14.1 feet
Write a sentence. Each side of the patio should be 14.1 feet.

Katie wants to plant a square lawn in her front yard. She has enough sod to cover an area of 370 square feet. To the nearest tenth of a foot, how long can a side of her square lawn be?

Solution

19.2 feet

Sergio wants to make a square mosaic as an inlay for a table he is building. He has enough tile to cover an area of 2704 square centimeters. How long can a side of his mosaic be?

Solution

52 centimeters

Square Roots and Gravity

Another application of square roots involves gravity. 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 evaluating the expression 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 evaluating 644.

Steps to simplify the mathematical expression sqrt(64)/4.
644
Take the square root of 64. 84
Simplify the fraction. 2

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Christy dropped her sunglasses from a bridge 400 feet above a river. How many seconds does it take for the sunglasses to reach the river?

Solution
Solution
This table outlines the step-by-step solution to a problem involving calculating the time for sunglasses to fall into a river, from problem statement to final numerical answer.
What are you asked to find? The number of seconds it takes for the sunglasses to reach the river
Write a phrase. The time it will take to reach the river
Translate to an expression. h4
Evaluate h4 when h=400. 4004
Find the square root of 400. 204
Simplify. 5
Write a sentence. It will take 5 seconds for the sunglasses to reach the river.

A helicopter drops a rescue package from a height of 1296 feet. How many seconds does it take for the package to reach the ground?

Solution

9 seconds

A window washer drops a squeegee from a platform 196 feet above the sidewalk. How many seconds does it take for the squeegee to reach the sidewalk?

Solution

3.5 seconds

Square Roots and Accident Investigations

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. According to some formulas, if the length of the skid marks is d feet, then the speed of the car can be found by evaluating 24d.

After a car accident, the skid marks for one car measured 190 feet. To the nearest tenth, what was the speed of the car (in mph) before the brakes were applied?

Solution
Solution
This table outlines the step-by-step process of calculating a car's speed using a square root formula and a given distance, resulting in a numerical approximation.
What are you asked to find? The speed of the car before the brakes were applied
Write a phrase. The speed of the car
Translate to an expression. 24d
Evaluate24dwhend=190. 24·190
Multiply. 4,560
Use your calculator. 67.527772...
Round to tenths. 67.5
Write a sentence. The speed of the car was approximately 67.5 miles per hour.

An accident investigator measured the skid marks of a car and found their length was 76 feet. To the nearest tenth, what was the speed of the car before the brakes were applied?

Solution

42.7 mph

The skid marks of a vehicle involved in an accident were 122 feet long. To the nearest tenth, how fast had the vehicle been going before the brakes were applied?

Solution

54.1 mph

The Links to Literacy activity "Sea Squares" will provide you with another view of the topics covered in this section.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Introduction to Square Roots
  • Estimating Square Roots with a Calculator

Key Concepts

  • Square Root Notation m is read ‘the square root of m’
    If m=n2, then m=n, for n≥0. This image labels the parts of a radical: the radical sign (square root symbol) and the radicand (the number or expression 'm' inside it).
  • Use a strategy for applications with square roots.
    • Identify what you are asked to find.
    • Write a phrase that gives the information to find it.
    • Translate the phrase to an expression.
    • Simplify the expression.
    • Write a complete sentence that answers the question.

Section Exercises

Practice Makes Perfect

Simplify Expressions with Square Roots

In the following exercises, simplify.

36

Solution

6

4

64

Solution

8

144

−4

Solution

−2

−100

−1

Solution

−1

−121

−121

Solution

not a real number

−36

−9

Solution

not a real number

−49

9+16

Solution

5

25+144

9+16

Solution

7

25+144

Estimate Square Roots

In the following exercises, estimate each square root between two consecutive whole numbers.

70

Solution

8<70<9

55

200

Solution

14<200<15

172

Approximate Square Roots with a Calculator

In the following exercises, use a calculator to approximate each square root and round to two decimal places.

19

Solution

4.36

21

53

Solution

7.28

47

Simplify Variable Expressions with Square Roots

In the following exercises, simplify. (Assume all variables are greater than or equal to zero.)

y2

Solution

y

b2

49x2

Solution

7x

100y2

−64a2

Solution

−8a

−25x2

144x2y2

Solution

12xy

196a2b2

Use Square Roots in Applications

In the following exercises, solve. Round 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. How long can a side of his garden be?

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. How long can a side of his patio be?

Gravity An airplane dropped a flare from a height of 1,024 feet above a lake. How many seconds did it take for the flare to reach the water?

Solution

8 seconds

Gravity A hang glider dropped his cell phone from a height of 350 feet. How many seconds did it take for the cell phone to reach the ground?

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4,000 feet above the Colorado River. How many seconds did it take for the hammer to reach the river?

Solution

15.8 seconds

Accident investigation The skid marks from a car involved in an accident measured 54 feet. What was the speed of the car before the brakes were applied?


Accident investigation The skid marks from a car involved in an accident measured 216 feet. What was the speed of the car before the brakes were applied?

Solution

72 mph

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. What was the speed of the vehicle before the brakes were applied?

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 117 feet. What was the speed of the vehicle before the brakes were applied?

Solution

53.0 mph

Everyday Math

Decorating Denise wants to install a square accent of designer tiles in her new shower. She can afford to buy 625 square centimeters of the designer tiles. How long can a side of the accent be?

Decorating Morris wants to have a square mosaic inlaid in his new patio. His budget allows for 2,025 tiles. Each tile is square with an area of one square inch. How long can a side of the mosaic be?

Solution

45 inches

Writing Exercises

Why is there no real number equal to −64?

What is the difference between 92 and 9?

Solution

Answers will vary. 92 reads: “nine squared” and means nine times itself. The expression 9 reads: “the square root of nine” which gives us the number such that if it were multiplied by itself would give you the number inside of the square root.

Self Check

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

A self-assessment table for square root skills, with columns for 'Confidently,' 'With some help,' and 'No-I don't get it!' for topics like simplifying, estimating, approximating, and applications.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Decimals

Name Decimals

In the following exercises, name each decimal.

0.8

0.375

Solution

three hundred seventy-five thousandths

0.007

5.24

Solution

five and twenty-four hundredths

−12.5632

−4.09

Solution

negative four and nine hundredths

Write Decimals

In the following exercises, write as a decimal.

three tenths

nine hundredths

Solution

0.09

twenty-seven hundredths

ten and thirty-five thousandths

Solution

10.035

negative twenty and three tenths

negative five hundredths

Solution

−0.05

Convert Decimals to Fractions or Mixed Numbers

In the following exercises, convert each decimal to a fraction. Simplify the answer if possible.

0.43

0.825

Solution

3340

9.7

3.64

Solution

31625

Locate Decimals on the Number Line

ⓐ 0.6

ⓑ −0.9

ⓒ 2.2

ⓓ −1.3

Order Decimals

In the following exercises, order each of the following pairs of numbers, using < or >.

0.6___0.8

Solution

<

0.2___0.15

0.803____0.83

Solution

<

−0.56____−0.562

Round Decimals

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

12.529

Solution
  1. ⓐ 12.53
  2. ⓑ 12.5
  3. ⓒ 13

4.8447

5.897

Solution
  1. ⓐ 5.90
  2. ⓑ 5.9
  3. ⓒ 6

Decimal Operations

Add and Subtract Decimals

In the following exercises, add or subtract.

5.75+8.46

32.89−8.22

Solution

24.67

24−19.31

10.2+14.631

Solution

24.831

−6.4+(−2.9)

1.83−4.2

Solution

−2.37

Multiply Decimals

In the following exercises, multiply.

(0.3)(0.7)

(−6.4)(0.25)

Solution

−1.6

(−3.35)(−12.7)

(15.4)(1000)

Solution

15,400

Divide Decimals

In the following exercises, divide.

0.48÷6

4.32÷24

Solution

0.18

$6.29÷12

(−0.8)÷(−0.2)

Solution

4

1.65÷0.15

9÷0.045

Solution

200

Use Decimals in Money Applications

In the following exercises, use the strategy for applications to solve.

Miranda got $40 from her ATM. She spent $9.32 on lunch and $16.99 on a book. How much money did she have left? Round to the nearest cent if necessary.

Jessie put 8 gallons of gas in her car. One gallon of gas costs $3.528. How much did Jessie owe for all the gas?

Solution

$28.22

A pack of 16 water bottles cost $6.72. How much did each bottle cost?

Alice bought a roll of paper towels that cost $2.49. She had a coupon for $0.35 off, and the store doubled the coupon. How much did Alice pay for the paper towels?

Solution

$1.79

Decimals and Fractions

Convert Fractions to Decimals

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

35

78

Solution

0.875

−1920

−214

Solution

−5.25

13

611

Solution

0.54—

Order Decimals and Fractions

In the following exercises, order each pair of numbers, using < or >.

12___0.2

35___0.

Solution

>

−78___−0.84

−512___−0.42

Solution

>

0.625___1320

0.33___516

Solution

>

In the following exercises, write each set of numbers in order from least to greatest.

23,1720,0.65

79,0.75,1115

Solution

1115,0.75,79

Simplify Expressions Using the Order of Operations

In the following exercises, simplify

4(10.3−5.8)

34(15.44−7.4)

Solution

6.03

30÷(0.45+0.15)

1.6+38

Solution

1.975

52(0.5)+(0.4)2

−25·910+0.14

Solution

−0.22

Find the Circumference and Area of Circles

In the following exercises, approximate the ⓐ circumference and ⓑ area of each circle.

radius=6 in.

radius=3.5 ft.

Solution
  1. ⓐ 21.98 ft.
  2. ⓑ 38.465 sq.ft.

radius=733m

diameter=11 cm

Solution
  1. ⓐ 34.54 cm
  2. ⓑ 94.985 sq.cm

Solve Equations with Decimals

Determine Whether a Decimal is a Solution of an Equation

In the following exercises, determine whether the each number is a solution of the given equation.

x−0.4=2.1
ⓐ x=1.7 ⓑ x=2.5

y+3.2=−1.5
ⓐ y=1.7ⓑ y=−4.7

Solution
  1. ⓐ no
  2. ⓑ yes

u2.5=−12.5
ⓐ u=−5ⓑ u=−31.25

0.45v=−40.5
ⓐ v=−18.225ⓑ v=−90

Solution
  1. ⓐ no
  2. ⓑ yes

Solve Equations with Decimals

In the following exercises, solve.

m+3.8=7.5

h+5.91=2.4

Solution

h = −3.51

a+2.26=−1.1

p−4.3=−1.65

Solution

p = 2.65

x−0.24=−8.6

j−7.42=−3.7

Solution

j = 3.72

0.6p=13.2

−8.6x=34.4

Solution

x = −4

−22.32=−2.4z

a0.3=−24

Solution

a = −7.2

p−7=−4.2

s−2.5=−10

Solution

s = 25

Translate to an Equation and Solve

In the following exercises, translate and solve.

The difference of n and 15.2 is 4.4.

The product of −5.9 and x is −3.54.

Solution

−5.9x = −3.54; x = 0.6

The quotient of y and −1.8 is −9.

The sum of m and (−4.03) is 6.8.

Solution

m + (−4.03) = 6.8; m = 10.83

Averages and Probability

Find the Mean of a Set of Numbers

In the following exercises, find the mean of the numbers.

2,4,1,0,1,and1

$270, $310.50, $243.75, and $252.15

Solution

$269.10

Each workday last week, Yoshie kept track of the number of minutes she had to wait for the bus. She waited 3,0,8,1,and8 minutes. Find the mean.

In the last three months, Raul’s water bills were $31.45,$48.76,and$42.60. Find the mean.

Solution

$40.94

Find the Median of a Set of Numbers

In the following exercises, find the median.

41, 45, 32, 60, 58

25, 23, 24, 26, 29, 19, 18, 32

Solution

24.5

The ages of the eight men in Jerry’s model train club are 52,63,45,51,55,75,60,and59. Find the median age.

The number of clients at Miranda’s beauty salon each weekday last week were 18,7,12,16,and20. Find the median number of clients.

Solution

16 clients

Find the Mode of a Set of Numbers

In the following exercises, identify the mode of the numbers.

6, 4, 4,5, 6,6, 4, 4, 4, 3, 5

The number of siblings of a group of students: 2, 0, 3, 2, 4, 1, 6, 5, 4, 1, 2, 3

Solution

2

Use the Basic Definition of Probability

In the following exercises, solve. (Round decimals to three places.)

The Sustainability Club sells 200 tickets to a raffle, and Albert buys one ticket. One ticket will be selected at random to win the grand prize. Find the probability Albert will win the grand prize. Express your answer as a fraction and as a decimal.

Luc has to read 3 novels and 12 short stories for his literature class. The professor will choose one reading at random for the final exam. Find the probability that the professor will choose a novel for the final exam. Express your answer as a fraction and as a decimal.

Solution

15;0.2

Ratios and Rate

Write a Ratio as a Fraction

In the following exercises, write each ratio as a fraction. Simplify the answer if possible.

28 to 40

56 to 32

Solution

74

3.5 to 0.5

1.2 to 1.8

Solution

23

134to158

213to514

Solution

49

64 ounces to 30 ounces

28 inches to 3 feet

Solution

79

Write a Rate as a Fraction

In the following exercises, write each rate as a fraction. Simplify the answer if possible.

180 calories per 8 ounces

90 pounds per 7.5 square inches

Solution

12pounds1square inch

126 miles in 4 hours

$612.50 for 35 hours

Solution

$352hours

Find Unit Rates

In the following exercises, find the unit rate.

180 calories per 8 ounces

90 pounds per 7.5 square inches

Solution

12 pounds/sq.in.

126 miles in 4 hours

$612.50 for 35 hours

Solution

$17.50/hour

Find Unit Price

In the following exercises, find the unit price.

t-shirts: 3 for $8.97

Highlighters: 6 for $2.52

Solution

$0.42

An office supply store sells a box of pens for $11. The box contains 12 pens. How much does each pen cost?

Anna bought a pack of 8 kitchen towels for $13.20. How much did each towel cost? Round to the nearest cent if necessary.

Solution

$1.65

In the following exercises, find each unit price and then determine the better buy.

Shampoo: 12 ounces for $4.29 or 22 ounces for $7.29?

Vitamins: 60 tablets for $6.49 or 100 for $11.99?

Solution

$0.11, $0.12; 60 tablets for $6.49

Translate Phrases to Expressions with Fractions

In the following exercises, translate the English phrase into an algebraic expression.

535 miles per hhours

a adults to 45 children

Solution

aadults45children

the ratio of 4y and the difference of x and 10

the ratio of 19 and the sum of 3 and n

Solution

193+n

Simplify and Use Square Roots

Simplify Expressions with Square Roots

In the following exercises, simplify.

64

144

Solution

12

−25

−81

Solution

−9

−9

−36

Solution

not a real number

64+225

64+225

Solution

17

Estimate Square Roots

In the following exercises, estimate each square root between two consecutive whole numbers.

28

155

Solution

12<155<13

Approximate Square Roots

In the following exercises, approximate each square root and round to two decimal places.

15

57

Solution

7.55

Simplify Variable Expressions with Square Roots

In the following exercises, simplify. (Assume all variables are greater than or equal to zero.)

q2

64b2

Solution

8b

−121a2

225m2n2

Solution

15mn

−100q2

49y2

Solution

7y

4a2b2

121c2d2

Solution

11cd

Use Square Roots in Applications

In the following exercises, solve. Round to one decimal place.

Art Diego has 225 square inch tiles. He wants to use them to make a square mosaic. How long can each side of the mosaic be?

Landscaping Janet wants to plant a square flower garden in her yard. She has enough topsoil to cover an area of 30 square feet. How long can a side of the flower garden be?

Solution

5.5 feet

Gravity A hiker dropped a granola bar from a lookout spot 576 feet above a valley. How long did it take the granola bar to reach the valley floor?

Accident investigation The skid marks of a car involved in an accident were 216 feet. How fast had the car been going before applying the brakes?

Solution

72 mph

Chapter Practice Test

Write six and thirty-four thousandths as a decimal.

Write 1.73 as a fraction.

Solution

173100

Write 58 as a decimal.

Round 16.749 to the nearest ⓐ tenth ⓑ hundredth ⓒ whole number

Solution
  1. ⓐ 16.7
  2. ⓑ 16.75
  3. ⓒ 17

Write the numbers 45,−0.1,0.804,29,−7.4,0.21 in order from smallest to largest.

In the following exercises, simplify each expression.

15.4+3.02

Solution

18.42

20−5.71

(0.64)(0.3)

Solution

0.192

(−4.2)(100)

0.96÷(−12)

Solution

−0.08

−5÷0.025

−0.6÷(−0.3)

Solution

2

(0.7)2

24÷(0.1+0.02)

Solution

200

4(10.3−5.8)

1.6+38

Solution

1.975

23(14.65−4.6)

In the following exercises, solve.

m+3.7=2.5

Solution

−1.2

h0.5=4.38

−6.5y=−57.2

Solution

8.8

1.94=a−2.6

Three friends went out to dinner and agreed to split the bill evenly. The bill was $79.35. How much should each person pay?

Solution

$26.45

A circle has radius 12. Find the ⓐ circumference and ⓑ area. [Use3.14forπ.]

The ages, in months, of 10 children in a preschool class are:
55, 55, 50, 51, 52, 50, 53, 51, 55, 49
Find the ⓐ mean ⓑ median ⓒ mode

Solution
  1. ⓐ 52.1
  2. ⓑ 51.5
  3. ⓒ 55

Of the 16 nurses in Doreen’s department, 12 are women and 4 are men. One of the nurses will be assigned at random to work an extra shift next week. ⓐ Find the probability a woman nurse will be assigned the extra shift. ⓑ Convert the fraction to a decimal.

Find each unit price and then the better buy.
Laundry detergent: 64 ounces for $10.99 or 48 ounces for $8.49

Solution

The unit prices are $0.172 per ounce for 64 ounces, and $0.177 per ounce for 48 ounces; 64 ounces is the better buy.

In the following exercises, simplify.

36+64

144n2

Solution

12n

Estimate 54 to between two whole numbers.

Yanet wants a square patio in her backyard. She has 225 square feet of tile. How long can a side of the patio be?

Solution

15 feet

Introduction to Percents

This image shows a sign at a bank indicating annual percentage rates for CDs. The top line says “Earn big with a locked-in rate”. The second line reads 2-year CD with one percent annual percentage rate; the third line reads 3-year CD with a 1.5 percent  annual percentage rate, and the fourth line reads 5-year CD with a 2 percent annual percentage rate.
Banks provide money for savings and charge money for loans. The interest on savings and loans is usually given as a percent. (credit: Mike Mozart, Flickr)

When you deposit money in a savings account at a bank, it earns additional money. Figuring out how your money will grow involves understanding and applying concepts of percents. In this chapter, we will find out what percents are and how we can use them to solve problems.

Understand Percent

Learning Objectives

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

  • Use the definition of percent
  • Convert percents to fractions and decimals
  • Convert decimals and fractions to percents

Before you get started, take this readiness quiz.

Translate “the ratio of 33 to 5” into an algebraic expression.
If you missed this problem, review Table 13 in Evaluate, Simplify, and Translate Expressions.

Solution

335

Write 35 as a decimal.
If you missed this problem, review Example 1 in Decimals and Fractions.

Solution

0.6

Write 0.62 as a fraction.
If you missed this problem, review Example 4 in Decimals.

Solution

3150

Use the Definition of Percent

How many cents are in one dollar? There are 100 cents in a dollar. How many years are in a century? There are 100 years in a century. Does this give you a clue about what the word “percent” means? It is really two words, “per cent,” and means per one hundred. A percent is a ratio whose denominator is 100. We use the percent symbol %, to show percent.

Percent

A percent is a ratio whose denominator is 100.

According to data from the American Association of Community Colleges (2015), about 57% of community college students are female. This means 57 out of every 100 community college students are female, as Figure 1 shows. Out of the 100 squares on the grid, 57 are shaded, which we write as the ratio 57100.

The figure shows a hundred flat with 57 units shaded.
Among every 100 community college students, 57 are female.

Similarly, 25% means a ratio of 25100,3% means a ratio of 3100 and 100% means a ratio of 100100. In words, "one hundred percent" means the total 100% is 100100, and since 100100=1, we see that 100% means 1 whole.

According to the Public Policy Institute of California (2010),44% of parents of public school children would like their youngest child to earn a graduate degree. Write this percent as a ratio.

Solution

Solution

This table illustrates the step-by-step conversion of a percentage (44%) into a ratio.
The amount we want to convert is 44%. 44%
Write the percent as a ratio. Remember that percent means per 100. 44100

Write the percent as a ratio.

According to a survey, 89% of college students have a smartphone.

Solution

89100

Write the percent as a ratio.

A study found that 72% of U.S. teens send text messages regularly.

Solution

72100

In 2007, according to a U.S. Department of Education report, 21 out of every 100 first-time freshmen college students at 4-year public institutions took at least one remedial course. Write this as a ratio and then as a percent.

Solution

Solution

Steps demonstrating the conversion of a quantity out of 100 into its ratio and percentage forms.
The amount we want to convert is 21 out of 100. 21 out of 100
Write as a ratio. 21100
Convert the 21 per 100 to percent. 21%

Write as a ratio and then as a percent: The American Association of Community Colleges reported that 62 out of 100 full-time community college students balance their studies with full-time or part time employment.

Solution

62100,62%

Write as a ratio and then as a percent: In response to a student survey, 41 out of 100 Santa Ana College students expressed a goal of earning an Associate's degree or transferring to a four-year college.

Solution

41100,41%

Convert Percents to Fractions and Decimals

Since percents are ratios, they can easily be expressed as fractions. Remember that percent means per 100, so the denominator of the fraction is 100.

Convert a percent to a fraction.

  1. Write the percent as a ratio with the denominator 100.
  2. Simplify the fraction if possible.
Convert each percent to a fraction:
  1. ⓐ 36%
  2. ⓑ 125%
Solution

Solution

This table demonstrates the step-by-step conversion of a percentage (36%) into its simplified fraction form.
ⓐ
36%
Write as a ratio with denominator 100. 36100
Simplify. 925
Steps demonstrating the conversion of a percentage into a simplified fraction.
ⓑ
125%
Write as a ratio with denominator 100. 125100
Simplify. 54
Convert each percent to a fraction:
  1. ⓐ 48%
  2. ⓑ 110%
Solution
  1. ⓐ 1225
  2. ⓑ 1110
Convert each percent to a fraction:
  1. ⓐ 64%
  2. ⓑ 150%
Solution
  1. ⓐ 1625
  2. ⓑ 32

The previous example shows that a percent can be greater than 1. We saw that 125% means 125100, or 54. These are improper fractions, and their values are greater than one.

Convert each percent to a fraction:
  1. ⓐ 24.5%
  2. ⓑ 3313%
Solution

Solution

This table illustrates the step-by-step process of converting 24.5% into its simplest fractional form, demonstrating each mathematical operation.
ⓐ
24.5%
Write as a ratio with denominator 100. 24.5100
Clear the decimal by multiplying numerator and denominator by 10. 24.5(10)100(10)
Multiply. 2451000
Rewrite showing common factors. 5·495·200
Simplify. 49200
Steps to convert the percentage 33 1/3% to its equivalent fraction 1/3, detailing each mathematical transformation.
ⓑ
3313%
Write as a ratio with denominator 100. 3313100
Write the numerator as an improper fraction. 1003100
Rewrite as fraction division, replacing 100 with 1001. 1003÷1001
Multiply by the reciprocal. 1003⋅1100
Simplify. 13
Convert each percent to a fraction:
  1. ⓐ 64.4%
  2. ⓑ 6623%
Solution
  1. ⓐ 161250
  2. ⓑ 23
Convert each percent to a fraction:
  1. ⓐ 42.5%
  2. ⓑ 834%
Solution
  1. ⓐ 1740
  2. ⓑ 780

In Decimals, we learned how to convert fractions to decimals. To convert a percent to a decimal, we first convert it to a fraction and then change the fraction to a decimal.

Convert a percent to a decimal.

  1. Write the percent as a ratio with the denominator 100.
  2. Convert the fraction to a decimal by dividing the numerator by the denominator.
Convert each percent to a decimal:
  1. ⓐ 6%
  2. ⓑ 78%
Solution

Solution

Because we want to change to a decimal, we will leave the fractions with denominator 100 instead of removing common factors.

This table illustrates the step-by-step conversion of 6% into its equivalent fractional (6/100) and decimal (0.06) representations.
ⓐ
6%
Write as a ratio with denominator 100. 6100
Change the fraction to a decimal by dividing the numerator by the denominator. 0.06
Steps to convert a percentage (78%) into its equivalent ratio and decimal forms.
ⓑ
78%
Write as a ratio with denominator 100. 78100
Change the fraction to a decimal by dividing the numerator by the denominator. 0.78
Convert each percent to a decimal:
  1. ⓐ 9%
  2. ⓑ 87%
Solution
  1. ⓐ 0.09
  2. ⓑ 0.87
Convert each percent to a decimal:
  1. ⓐ 3%
  2. ⓑ 91%
Solution
  1. ⓐ 0.03
  2. ⓑ 0.91
Convert each percent to a decimal:
  1. ⓐ 135%
  2. ⓑ 12.5%
Solution

Solution

Steps for converting a percentage to its fraction and decimal representations.
ⓐ
135%
Write as a ratio with denominator 100. 135100
Change the fraction to a decimal by dividing the numerator by the denominator. 1.35
Step-by-step conversion of 12.5% from percentage to its decimal equivalent.
ⓑ
12.5%
Write as a ratio with denominator 100. 12.5100
Change the fraction to a decimal by dividing the numerator by the denominator. 0.125
Convert each percent to a decimal:
  1. ⓐ 115%
  2. ⓑ 23.5%
Solution
  1. ⓐ 1.15
  2. ⓑ 0.235
Convert each percent to a decimal:
  1. ⓐ 123%
  2. ⓑ 16.8%
Solution
  1. ⓐ 1.23
  2. ⓑ 0.168

Let's summarize the results from the previous examples in Table 11, and look for a pattern we could use to quickly convert a percent number to a decimal number.

Percent Decimal
6% 0.06
78% 0.78
135% 1.35
12.5% 0.125

Do you see the pattern?

To convert a percent number to a decimal number, we move the decimal point two places to the left and remove the % sign. (Sometimes the decimal point does not appear in the percent number, but just like we can think of the integer 6 as 6.0, we can think of 6% as 6.0%.) Notice that we may need to add zeros in front of the number when moving the decimal to the left.

Figure 2 uses the percents in Table 11 and shows visually how to convert them to decimals by moving the decimal point two places to the left.

The figures shows two columns and five rows . The  first row is a header row and it labels each column “Percent” and “Decimal”. Under the “Percent” column are the values: 6%, 78%, 135%, 12.5%. Under the “Decimal” column are the values: 0.06, 0.78, 1.35, 0.125. There are two jumps for each percent to show how to convert it to a decimal.

Among a group of business leaders, 77% believe that poor math and science education in the U.S. will lead to higher unemployment rates.

Convert the percent to: ⓐ a fraction ⓑ a decimal

Solution

Solution

This table illustrates the conversion of 77% into its equivalent ratio form, 77/100, by expressing it with a denominator of 100.
ⓐ
77%
Write as a ratio with denominator 100. 77100
Demonstrates converting the fraction 77/100 to the decimal 0.77 by dividing the numerator by the denominator.
ⓑ
77100
Change the fraction to a decimal by dividing the numerator by the denominator. 0.77

Convert the percent to: ⓐ a fraction and ⓑ a decimal

Twitter's share of web traffic jumped 24% when one celebrity tweeted live on air.

Solution
  1. ⓐ 625
  2. ⓑ 0.24

Convert the percent to: ⓐ a fraction and ⓑ a decimal

The U.S. Census estimated that in 2013,44% of the population of Boston age 25 or older have a bachelor's or higher degrees.

Solution
  1. ⓐ 1125
  2. ⓑ 0.44
There are four suits of cards in a deck of cards—hearts, diamonds, clubs, and spades. The probability of randomly choosing a heart from a shuffled deck of cards is 25%. Convert the percent to:
  1. ⓐ a fraction
  2. ⓑ a decimal
The figure shows someone holding a deck of cards.
(credit: Riles32807, Wikimedia Commons)
Solution

Solution

Steps to convert a percentage (25%) into a simplified fraction (1/4).
ⓐ
25%
Write as a ratio with denominator 100. 25100
Simplify. 14
This table illustrates the process of converting the fraction 1/4 to its decimal equivalent, 0.25, by dividing the numerator by the denominator.
ⓑ 14
Change the fraction to a decimal by dividing the numerator by the denominator. 0.25

Convert the percent to: ⓐ a fraction, and ⓑ a decimal

The probability that it will rain Monday is 30%.

Solution
  1. ⓐ 310
  2. ⓑ 0.3

Convert the percent to: ⓐ a fraction, and ⓑ a decimal

The probability of getting heads three times when tossing a coin three times is 12.5%.

Solution
  1. ⓐ 18
  2. ⓑ 0.125

Convert Decimals and Fractions to Percents

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.

Convert a decimal to a percent.

  1. Write the decimal as a fraction.
  2. If the denominator of the fraction is not 100, rewrite it as an equivalent fraction with denominator 100.
  3. Write this ratio as a percent.

Convert each decimal to a percent: ⓐ 0.05 ⓑ 0.83

Solution

Solution

Demonstrates converting the decimal 0.05 into its equivalent fraction (5/100) and percentage (5%) forms with step-by-step instructions.
ⓐ
0.05
Write as a fraction. The denominator is 100. 5100
Write this ratio as a percent. 5%
Steps showing the conversion of the decimal 0.83 to its equivalent fraction (83/100) and percentage (83%).
ⓑ
0.83
The denominator is 100. 83100
Write this ratio as a percent. 83%

Convert each decimal to a percent: ⓐ 0.01 ⓑ 0.17.

Solution
  1. ⓐ 1%
  2. ⓑ 17%

Convert each decimal to a percent: ⓐ 0.04 ⓑ 0.41

Solution
  1. ⓐ 4%
  2. ⓑ 41%

To convert a mixed number to a percent, we first write it as an improper fraction.

Convert each decimal to a percent: ⓐ 1.05 ⓑ 0.075

Solution

Solution

Illustrates the step-by-step conversion of the decimal 1.05 into a mixed fraction, improper fraction, and a percentage.
ⓐ
1.05
Write as a fraction. 15100
Write as an improper fraction. The denominator is 100. 105100
Write this ratio as a percent. 105%

Notice that since 1.05>1, the result is more than 100%.

Steps to convert a decimal (0.075) into a fraction and then a percentage.
ⓑ
0.075
Write as a fraction. The denominator is 1,000. 751,000
Divide the numerator and denominator by 10, so that the denominator is 100. 7.5100
Write this ratio as a percent. 7.5%

Convert each decimal to a percent: ⓐ 1.75 ⓑ 0.0825

Solution
  1. ⓐ 175%
  2. ⓑ 8.25%

Convert each decimal to a percent: ⓐ 2.25 ⓑ 0.0925

Solution
  1. ⓐ 225%
  2. ⓑ 9.25%

Let's summarize the results from the previous examples in Table 20 so we can look for a pattern.

Decimal Percent
0.05 5%
0.83 83%
1.05 105%
0.075 7.5%

Do you see 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 4 uses the decimal numbers in Table 20 and shows visually to convert them to percents by moving the decimal point two places to the right and then writing the % sign.

The figure shows two columns and five rows. The  first row is a header row and it labels each column “Decimal” and “Percent”. Under the “Decimal” column are the values: 0.05, 0.83, 1.05, 0.075, 0.3. Under the “Percent” column are the values: 5%, 83%, 105%, 7.5%, 30%. There are two jumps for each decimal to show how to convert it to a percent.

In Decimals, we learned how to convert fractions to decimals. Now we also know how to change decimals to percents. So to convert a fraction to a percent, we first change it to a decimal and then convert that decimal to a percent.

Convert a fraction to a percent.

  1. Convert the fraction to a decimal.
  2. Convert the decimal to a percent.

Convert each fraction or mixed number to a percent: ⓐ 34 ⓑ 118 ⓒ 215

Solution

Solution

To convert a fraction to a decimal, divide the numerator by the denominator.

ⓐ
Change to a decimal. 34
Write as a percent by moving the decimal two places. The decimal number 0.75 is shown, with two blue arrows below it, likely illustrating movement of the decimal point or place value shifts in a mathematical context.
75%
ⓑ
Change to a decimal. 118
Write as a percent by moving the decimal two places. The number 1.375 is shown with light blue curved arrows beneath the decimal point, highlighting the digits '3' and '75' to emphasize the decimal places.
137.5%
ⓒ
Write as an improper fraction. 215
Change to a decimal. 115
Write as a percent. The number 2.20 is displayed with two blue arcs, visually representing the shift of the decimal point two places to the right, a common step in percentage conversions.
220%

Notice that we needed to add zeros at the end of the number when moving the decimal two places to the right.

Convert each fraction or mixed number to a percent: ⓐ 58 ⓑ 114 ⓒ 325

Solution
  1. ⓐ 62.5%
  2. ⓑ 275%
  3. ⓒ 340%

Convert each fraction or mixed number to a percent: ⓐ 78 ⓑ 94 ⓒ 135

Solution
  1. ⓐ 87.5%
  2. ⓑ 225%
  3. ⓒ 160%

Sometimes when changing a fraction to a decimal, the division continues for many decimal places and we will round off the quotient. The number of decimal places we round to will depend on the situation. If the decimal involves money, we round to the hundredths place. For most other cases in this book we will round the number to the nearest thousandth, so the percent will be rounded to the nearest tenth.

Convert 57 to a percent.

Solution

Solution

To change a fraction to a decimal, we divide the numerator by the denominator.

Conversions of a fraction (5/7) to its decimal (rounded to thousandths) and percentage forms.
57
Change to a decimal—rounding to the nearest thousandth. 0.714
Write as a percent. 71.4%

Convert the fraction to a percent: 37

Solution

42.9%

Convert the fraction to a percent: 47

Solution

57.1%

When we first looked at fractions and decimals, we saw that some fractions converted to a repeating decimal. For example, when we converted the fraction 43 to a decimal, we wrote the answer as 1.3¯. We will use this same notation, as well as fraction notation, when we convert fractions to percents in the next example.

An article in a medical journal claimed that approximately 13 of American adults are obese. Convert the fraction 13 to a percent.

Solution

Solution

13
Change to a decimal. Long division showing 1 divided by 3 equals the repeating decimal 0.333... This illustrates how a fraction translates to an infinite decimal.
Write as a repeating decimal. 0.333…
Write as a percent. 3313%

We could also write the percent as 33.3_%.

Convert the fraction to a percent:

According to the U.S. Census Bureau, about 19 of United States housing units have just 1 bedroom.

Solution

11.1–%,or1119%

Convert the fraction to a percent:

According to the U.S. Census Bureau, about 16 of Colorado residents speak a language other than English at home.

Solution

16.6–%,or1623%

Key Concepts

  • Convert a percent to a fraction.
    1. Write the percent as a ratio with the denominator 100.
    2. Simplify the fraction if possible.
  • Convert a percent to a decimal.
    1. Write the percent as a ratio with the denominator 100.
    2. Convert the fraction to a decimal by dividing the numerator by the denominator.
  • Convert a decimal to a percent.
    1. Write the decimal as a fraction.
    2. If the denominator of the fraction is not 100, rewrite it as an equivalent fraction with denominator 100.
    3. Write this ratio as a percent.
  • Convert a fraction to a percent.
    1. Convert the fraction to a decimal.
    2. Convert the decimal to a percent.

Practice Makes Perfect

Use the Definition of Percents

In the following exercises, write each percent as a ratio.

In 2014, the unemployment rate for those with only a high school degree was 6.0%.

Solution

6100

In 2015, among the unemployed, 29% were long-term unemployed.

The unemployment rate for those with Bachelor's degrees was 3.2% in 2014.

Solution

321000

The unemployment rate in Michigan in 2014 was 7.3%.

In the following exercises, write as
  1. ⓐ a ratio and
  2. ⓑ a percent

57 out of 100 nursing candidates received their degree at a community college.

Solution
  1. ⓐ 57100
  2. ⓑ 57%

80 out of 100 firefighters and law enforcement officers were educated at a community college.

42 out of 100 first-time freshmen students attend a community college.

Solution
  1. ⓐ 42100
  2. ⓑ 42%

71 out of 100 full-time community college faculty have a master's degree.

Convert Percents to Fractions and Decimals

In the following exercises, convert each percent to a fraction and simplify all fractions.

4%

Solution

125

8%

17%

Solution

17100

19%

52%

Solution

1325

78%

125%

Solution

54

135%

37.5%

Solution

38

42.5%

18.4%

Solution

23125

46.4%

912%

Solution

19200

812%

513%

Solution

475

623%

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

5%

Solution

0.05

9%

1%

Solution

0.01

2%

63%

Solution

0.63

71%

40%

Solution

0.4

50%

115%

Solution

1.15

125%

150%

Solution

1.5

250%

21.4%

Solution

0.214

39.3%

7.8%

Solution

0.078

6.4%

In the following exercises, convert each percent to
  1. ⓐ a simplified fraction and
  2. ⓑ a decimal

In 2010,1.5% of home sales had owner financing. (Source: Bloomberg Businessweek, 5/23–29/2011)

Solution
  1. ⓐ 3200
  2. ⓑ 0.015

In 2000,4.2% of the United States population was of Asian descent. (Source: www.census.gov)

According to government data, in 2013 the number of cell phones in India was 70.23% of the population.

Solution
  1. ⓐ 702310,000
  2. ⓑ 0.7023

According to the U.S. Census Bureau, among Americans age 25 or older who had doctorate degrees in 2014,37.1% are women.

A couple plans to have two children. The probability they will have two girls is 25%.

Solution
  1. ⓐ 14
  2. ⓑ 0.25

Javier will choose one digit at random from 0 through 9. The probability he will choose 3 is 10%.

According to the local weather report, the probability of thunderstorms in New York City on July 15 is 60%.

Solution
  1. ⓐ 35
  2. ⓑ 0.6

A club sells 50 tickets to a raffle. Osbaldo bought one ticket. The probability he will win the raffle is 2%.

Convert Decimals and Fractions to Percents

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

0.01

Solution

1%

0.03

0.18

Solution

18%

0.15

1.35

Solution

135%

1.56

3

Solution

300%

4

0.009

Solution

0.9%

0.008

0.0875

Solution

8.75%

0.0625

1.5

Solution

150%

2.2

2.254

Solution

225.4%

2.317

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

14

Solution

25%

15

38

Solution

37.5%

58

74

Solution

175%

98

645

Solution

680%

514

512

Solution

4123%or41.6–%

1112

223

Solution

266.6–%

123

37

Solution

42.9%

67

59

Solution

55.5–%

49

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

14 of washing machines needed repair.

Solution

25%

15 of dishwashers needed repair.

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

According to the National Center for Health Statistics, in 2012,720 of American adults were obese.

Solution

35%

The U.S. Census Bureau estimated that in 2013,85% of Americans lived in the same house as they did 1 year before.

In the following exercises, complete the table.

Fraction Decimal Percent
12
0.45
18%
13
0.008
2
Solution
Fraction Decimal Percent
12 0.5 50%
920 0.45 45%
950 0.18 18%
13 0.3 33.3%
1125 0.008 0.8%
2 2.0 200%
Fraction Decimal Percent
14
0.65
22%
23
0.004
3

Everyday Math

Sales tax Felipa says she has an easy way to estimate the sales tax when she makes a purchase. The sales tax in her city is 9.05%. She knows this is a little less than 10%.

  1. ⓐ Convert 10% to a fraction.
  2. ⓑ Use your answer from ⓐ to estimate the sales tax Felipa would pay on a $95 dress.
Solution
  1. ⓐ 110
  2. ⓑ approximately $9.50

Savings Ryan has 25% of each paycheck automatically deposited in his savings account.

  1. ⓐ Write 25% as a fraction.
  2. ⓑ Use your answer from ⓐ to find the amount that goes to savings from Ryan's $2,400 paycheck.

Amelio is shopping for textbooks online. He found three sellers that are offering a book he needs for the same price, including shipping. To decide which seller to buy from he is comparing their customer satisfaction ratings. The ratings are given in the chart.

Use the chart to answer the following questions

Seller Rating
A 4/5
B 3.5/4
C 85%

Write seller C’s rating as a fraction and a decimal.

Solution

1720;0.85

Write seller B’s rating as a percent and a decimal.

Write seller A’s rating as a percent and a decimal.

Solution

80%; 0.8

Which seller should Amelio buy from and why?

Writing Exercises

Convert 25%,50%,75%,and100% to fractions. Do you notice a pattern? Explain what the pattern is.

Solution

14,12,34,1.

Convert 110,210,310,410,510,610,710,810, and 910 to percents. Do you notice a pattern? Explain what the pattern is.

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.

Solution

The Szetos sold their home for five times what they paid 30 years ago.

According to cnn.com, cell phone use in 2008 was 600% of what it had been in 2001. Explain what 600% means in this context.

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 evaluate their understanding of percentages, including defining percent and converting between percents, fractions, and decimals.

ⓑ 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.

percent
A perfecnt is a ratio whose denominator is 100.

Solve General Applications of Percent

Learning Objectives

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

  • Translate and solve basic percent equations
  • Solve applications of percent
  • Find percent increase and percent decrease

Before you get started, take this readiness quiz.

Translate and solve: 34 of x is 24.
If you missed this problem, review Example 11 in Solve Equations with Fractions.

Solution

32

Simplify: (4.5)(2.38).
If you missed this problem, review Example 5 in Decimal Operations.

Solution

10.71

Solve: 3.5=0.7n.
If you missed this problem, review Example 4 in Solve Equations with Decimals.

Solution

5

Translate and Solve Basic Percent Equations

We will solve percent equations by using the methods we used to solve equations with fractions or decimals. In the past, you may have solved percent problems by setting them up as proportions. That was the best method available when you did not have the tools of algebra. Now as a prealgebra student, you can translate word sentences into algebraic equations, and then solve the equations.

We'll look at a common application of percent—tips to a server at a restaurant—to see how to set up a basic percent application.

When Aolani and her friends ate dinner at a restaurant, the bill came to $80. They wanted to leave a 20% tip. What amount would the tip be?

To solve this, we want to find what amount is 20% of $80. The $80 is called the base. The amount of the tip would be 0.20(80), or $16 See Figure 1. To find the amount of the tip, we multiplied the percent by the base.

The figure shows a customer copy of a restaurant receipt with the amount of the bill, $80, and the amount of the tip, $16. There is a group of bills totaling $16.
A 20% tip for an $80 restaurant bill comes out to $16.

In the next examples, we will find the amount. We must be sure to change the given percent to a decimal when we translate the words into an equation.

What number is 35% of 90?

Solution

Solution

Translate into algebra. Let n=the number. A visual representation of translating the word problem 'What number is 35% of 90?' into the algebraic equation 'n = 0.35 * 90'.
Multiply. A mathematical equation is displayed on a white background, showing the variable 'n' equals '31.5'.
31.5 is 35% of 90

What number is 45% of 80?

Solution

36

What number is 55% of 60?

Solution

33

125% of 28 is what number?

Solution

Solution

Translate into algebra. Let a=the number. An image illustrating the conversion of the word problem '125% of 28 is what number?' into the algebraic equation '1.25 * 28 = a', with visual cues mapping words to their mathematical forms.
Multiply. A simple mathematical equation shows '35 = a' with the number 35 equal to the variable 'a'.
125% of 28 is 35.

Remember that a percent over 100 is a number greater than 1. We found that 125% of 28 is 35, which is greater than 28.

150% of 78 is what number?

Solution

117

175% of 72 is what number?

Solution

126

In the next examples, we are asked to find the base.

Translate and solve: 36 is 75% of what number?

Solution

Solution

Translate. Let b= the number. A diagram illustrates how to translate the word problem '36 is 75% of what number?' into the mathematical equation '36 = 0.75 . b' by breaking down each part of the sentence.
Divide both sides by 0.75. A mathematical equation is displayed where both sides of an equality are divided by 0.75 to solve for the variable 'b', represented as '36 / 0.75 = 0.75b / 0.75'.
Simplify. The image displays two lines of text: '48 = b' and '36 is 75% of 48', illustrating a mathematical relationship between numbers and a percentage.

17 is 25% of what number?

Solution

68

40 is 62.5% of what number?

Solution

64

6.5% of what number is $1.17?

Solution

Solution

Translate. Let b= the number. Translating the percentage word problem '6.5% of what number is $1.17?' into the equation '0.065 * b = 1.17', showing term-by-term correspondence.
Divide both sides by 0.065. A mathematical equation shows '0.065n over 0.065 equals 1.17 over 0.065' on a white background, demonstrating the step of dividing both sides of an equation by 0.065 to solve for 'n'.
Simplify. The image displays two lines of text: 'n = 18' at the top right, and '6.5% of $18 is $1.17.' at the bottom left, illustrating a percentage calculation.

7.5% of what number is $1.95?

Solution

$26

8.5% of what number is $3.06?

Solution

$36

In the next examples, we will solve for the percent.

What percent of 36 is 9?

Solution

Solution

Translate into algebra. Let p= the percent. A visual guide translating the phrase 'What percent of 36 is 9?' into the algebraic equation 'p . 36 = 9', showing the correspondence between words and mathematical symbols.
Divide by 36. A mathematical equation shows '36p over 36 equals 9 over 36'.
Simplify. The image displays a mathematical equation on a white background, stating 'p = 1/4'. The letter 'p' is shown in italics, followed by an equals sign, and then the fraction one over four.
Convert to decimal form. The image displays the equation 'p = 0.25' in a clear, sans-serif font against a plain white background, indicating a probability or a numerical value.
Convert to percent. The image displays the mathematical statement '25% of 36 is 9.' and also shows 'p = 25%,' illustrating a percentage calculation and its variable representation.

What percent of 76 is 57?

Solution

75%

What percent of 120 is 96?

Solution

80%

144 is what percent of 96?

Solution

Solution

Translate into algebra. Let p= the percent. Translating the phrase '144 is what percent of 96?' into an algebraic equation '144 = p . 96' to solve for the unknown percentage 'p'.
Divide by 96. A mathematical equation shows '144 divided by 96 equals 96p divided by 96' on a white background.
Simplify. The equation 1.5 = p is displayed in black text on a white background, representing a simple algebraic statement or a given value for the variable 'p'.
Convert to percent. Two lines of text display mathematical concepts: the first line shows '150% = p', and the second line states '144 is 150% of 96.'

110 is what percent of 88?

Solution

125%

126 is what percent of 72?

Solution

175%

Solve Applications of Percent

Many applications of percent occur in our daily lives, such as tips, sales tax, discount, and interest. To solve these applications we'll translate to a basic percent equation, just like those we solved in the previous examples in this section. Once you translate the sentence into a percent equation, you know how to solve it.

We will update the strategy we used in our earlier applications to include equations now. Notice that we will translate a sentence into an equation.

Solve an application

  1. Identify what you are asked to find and choose a variable to represent it.
  2. Write a sentence that gives the information to find it.
  3. Translate the sentence into an equation.
  4. Solve the equation using good algebra techniques.
  5. Check the answer in the problem and make sure it makes sense.
  6. Write a complete sentence that answers the question.

Now that we have the 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'll solve involve everyday situations, you can rely on your own experience.

Dezohn and his girlfriend enjoyed a dinner at a restaurant, and the bill was $68.50. They want to leave an 18% tip. If the tip will be 18% of the total bill, how much should the tip be?

Solution

Solution

What are you asked to find? the amount of the tip
Choose a variable to represent it. Let t= amount of tip.
Write a sentence that give the information to find it. The tip is 18% of the total bill.
Translate the sentence into an equation. Algebraic equation representing a calculation: The variable "t" equals zero point one eight times sixty eight dollars and fifty cents. This shows how to convert eighteen percent to a decimal and interpret "of" as multiplication.
Multiply. The image shows a mathematical expression
Check. Is this answer reasonable?
If we approximate the bill to $70 and the percent to 20%, we would have a tip of $14.
So a tip of $12.33 seems reasonable.
Write a complete sentence that answers the question. The couple should leave a tip of $12.33.

Cierra and her sister enjoyed a special dinner in a restaurant, and the bill was $81.50. If she wants to leave 18% of the total bill as her tip, how much should she leave?

Solution

$14.67

Kimngoc had lunch at her favorite restaurant. She wants to leave 15% of the total bill as her tip. If her bill was $14.40, how much will she leave for the tip?

Solution

$2.16

The label on Masao's breakfast cereal said that one serving of cereal provides 85 milligrams (mg) of potassium, which is 2% of the recommended daily amount. What is the total recommended daily amount of potassium?

The figures shows the nutrition facts for cereal.
Solution

Solution

What are you asked to find? the total amount of potassium recommended
Choose a variable to represent it. Let a= total amount of potassium.
Write a sentence that gives the information to find it. 85 mg is 2% of the total amount.
Translate the sentence into an equation. Translating the word problem '85 mg is 2% of a?' into the algebraic equation '85 = 0.02 * a'.
Divide both sides by 0.02. A mathematical equation shows both sides being divided by 0.02: 85/0.02 = 0.02a/0.02.
Simplify. The image displays a mathematical equation: 4,250 equals the variable 'a'.
Check: Is this answer reasonable?
Yes. 2% is a small percent and 85 is a small part of 4,250.
Write a complete sentence that answers the question. The amount of potassium that is recommended is 4250 mg.

One serving of wheat square cereal has 7 grams of fiber, which is 29% of the recommended daily amount. What is the total recommended daily amount of fiber?

Solution

24.1 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

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?

Solution

Solution

What are you asked to find? the percent of the total calories from fat
Choose a variable to represent it. Let p= percent from fat.
Write a sentence that gives the information to find it. What percent of 480 is 240?
Translate the sentence into an equation. Visual representation of translating 'What percent of 480 is 240?' into the algebraic equation 'p * 480 = 240', showing how words map to mathematical symbols.
Divide both sides by 480. A mathematical equation shows p multiplied by 480 and divided by 480, which is equal to 240 divided by 480.
Simplify. The mathematical expression p = 0.5 is displayed, suggesting a probability or a specific value for the variable p.
Convert to percent form. The image displays the mathematical expression 'p = 50%' in black text against a plain white background, indicating a probability or percentage value.
Check. Is this answer reasonable?
Yes. 240 is half of 480, so 50% makes sense.
Write a complete sentence that answers the question. Of the total calories in each brownie, 50% is fat.

Veronica is planning to make muffins from a mix. The package says each muffin will be 230 calories and 60 calories will be from fat. What percent of the total calories is from fat? (Round to the nearest whole percent.)

Solution

26%

The brownie mix Ricardo plans to use says that each brownie will be 190 calories, and 70 calories are from fat. What percent of the total calories are from fat?

Solution

37%

Find Percent Increase and Percent Decrease

People in the media often talk about how much an amount has increased or decreased over a certain period of time. They usually express this increase or decrease as a percent.

To find the percent increase, first we find the amount of increase, which is the difference between the new amount and the original amount. Then we find what percent the amount of increase is of the original amount.

Find Percent Increase.

Step 1. Find the amount of increase.

  • increase=new amount−original amount

Step 2. Find the percent increase as a percent of the original amount.

In 2011, the California governor proposed raising community college fees from $26 per unit to $36 per unit. Find the percent increase. (Round to the nearest tenth of a percent.)

Solution

Solution

What are you asked to find? the percent increase
Choose a variable to represent it. Let p= percent.
Find the amount of increase. A mathematical equation illustrating the calculation of an increase: 36 (new amount) - 26 (original amount) = 10 (increase).
Find the percent increase. The increase is what percent of the original amount?
Translate to an equation. A mathematical problem asks '10 is what percent of 26?' above its algebraic translation: '10 = p . 26'. Each part of the word problem is shown with a bracket above its corresponding mathematical symbol below.
Divide both sides by 26. An algebraic equation showing 10 divided by 26 equals 26p divided by 26, presented in fraction form on a white background.
Round to the nearest thousandth. The equation 0.385 = p is displayed in black text on a white background.
Convert to percent form. A mathematical equation displays '38.5% = p' in a simple, clear font on a white background. This equation assigns the percentage value of 38.5 to the variable 'p', indicating a direct equivalency.
Write a complete sentence. The new fees represent a 38.5% increase over the old fees.

In 2011, the IRS increased the deductible mileage cost to 55.5 cents from 51 cents. Find the percent increase. (Round to the nearest tenth of a percent.)

Solution

8.8%

In 1995, the standard bus fare in Chicago was $1.50. In 2008, the standard bus fare was $2.25. Find the percent increase. (Round to the nearest tenth of a percent.)

Solution

50%

Finding the percent decrease is very similar to finding the percent increase, but now the amount of decrease is the difference between the original amount and the final amount. Then we find what percent the amount of decrease is of the original amount.

Find percent decrease.

  1. Find the amount of decrease.
    • decrease=original amount−new amount
  2. Find the percent decrease as a percent of the original amount.

The average price of a gallon of gas in one city in June 2014 was $3.71. The average price in that city in July was $3.64. Find the percent decrease.

Solution

Solution

What are you asked to find? the percent decrease
Choose a variable to represent it. Let p= percent.
Find the amount of decrease. An image displays the subtraction 3.71 - 3.64 = 0.07. The number 3.71 is labeled 'original amount', 3.64 is labeled 'new amount', and 0.07 is labeled 'decrease'.
Find the percent of decrease. The decrease is what percent of the original amount?
Translate to an equation. A visual representation of translating the English phrase '0.07 is what percent of 3.71?' into the algebraic equation '0.07 = p * 3.71' for solving percentages.
Divide both sides by 3.71. A mathematical equation shows both sides being divided by 3.71: 0.07 / 3.71 = 3.71p / 3.71.
Round to the nearest thousandth. A mathematical equation displays '0.019 = p' in black text against a plain white background.
Convert to percent form. The image displays a mathematical equation: 1.9% = p. The equation shows the percentage 1.9% equated to the variable p, suggesting a conversion or a given value for p.
Write a complete sentence. The price of gas decreased 1.9%.

The population of one city was about 672,000 in 2010. The population of the city is projected to be about 630,000 in 2020. Find the percent decrease. (Round to the nearest tenth of a percent.)

Solution

6.3%

Last year Sheila's salary was $42,000. Because of furlough days, this year her salary was $37,800. Find the percent decrease. (Round to the nearest tenth of a percent.)

Solution

10%

ACCESS ADDITIONAL ONLINE RESOURCES

  • Percent Increase and Percent Decrease Visualization

Key Concepts

  • Solve an application.
    1. Identify what you are asked to find and choose a variable to represent it.
    2. Write a sentence that gives the information to find it.
    3. Translate the sentence into an equation.
    4. Solve the equation using good algebra techniques.
    5. Write a complete sentence that answers the question.
    6. Check the answer in the problem and make sure it makes sense.
  • Find percent increase.
    1. Find the amount of increase:
      increase=new amount−original amount
    2. Find the percent increase as a percent of the original amount.
  • Find percent decrease.
    1. Find the amount of decrease.
      decrease=original amount−new amount
    2. Find the percent decrease as a percent of the original amount.

Practice Makes Perfect

Translate and Solve Basic Percent Equations

In the following exercises, translate and solve.

What number is 45% of 120?

Solution

54

What number is 65% of 100?

What number is 24% of 112?

Solution

26.88

What number is 36% of 124?

250% of 65 is what number?

Solution

162.5

150% of 90 is what number?

800% of 2,250 is what number?

Solution

18,000

600% of 1,740 is what number?

28 is 25% of what number?

Solution

112

36 is 25% of what number?

81 is 75% of what number?

Solution

108

93 is 75% of what number?

8.2% of what number is $2.87?

Solution

$35

6.4% of what number is $2.88?

11.5% of what number is $108.10?

Solution

$940

12.3% of what number is $92.25?

What percent of 260 is 78?

Solution

30%

What percent of 215 is 86?

What percent of 1,500 is 540?

Solution

36%

What percent of 1,800 is 846?

30 is what percent of 20?

Solution

150%

50 is what percent of 40?

840 is what percent of 480?

Solution

175%

790 is what percent of 395?

Solve Applications of Percents

In the following exercises, solve the applications of percents.

Geneva treated her parents to dinner at their favorite restaurant. The bill was $74.25. She 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 the bill was $90.50. They want to leave 18% of the total bill as a tip. How much should the tip be?

Trong has 12% of each paycheck automatically deposited to his savings account. His last paycheck was $2,165. How much money was deposited to Trong's savings account?

Solution

$259.80

Cherise deposits 8% of each paycheck into her retirement account. Her last paycheck was $1,485. How much did Cherise deposit into her retirement account?

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 grams

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 2,070 milligrams (mg) of sodium, which is 86% of the recommended daily amount. What is the total recommended daily amount of sodium?

Solution

2,407 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?

Find Percent Increase and Percent Decrease

In the following exercises, find the percent increase or percent decrease.

Tamanika got a raise in her hourly pay, from $15.50 to $17.55. Find the percent increase.

Solution

13.2%

Ayodele got a raise in her hourly pay, from $24.50 to $25.48. Find the percent increase.

Annual student fees at the University of California rose from about $4,000 in 2000 to about $9,000 in 2014. Find the percent increase.

Solution

125%

The price of a share of one stock rose from $12.50 to $50. Find the percent increase.

According to Time magazine (7/19/2011) annual global seafood consumption rose from 22 pounds per person in 1960 to 38 pounds per person today. Find the percent increase. (Round to the nearest tenth of a percent.)

Solution

72.7%

In one month, the median home price in the Northeast rose from $225,400 to $241,500. Find the percent increase. (Round to the nearest tenth of a percent.)

A grocery store reduced the price of a loaf of bread from $2.80 to $2.73. Find the percent decrease.

Solution

2.5%

The price of a share of one stock fell from $8.75 to $8.54. Find the percent decrease.

Hernando's salary was $49,500 last year. This year his salary was cut to $44,055. Find the percent decrease.

Solution

11%

From 2000 to 2010, the population of Detroit fell from about 951,000 to about 714,000. Find the percent decrease. (Round to the nearest tenth of a percent.)

In one month, the median home price in the West fell from $203,400 to $192,300. Find the percent decrease. (Round to the nearest tenth of a percent.)

Solution

5.5%

Sales of video games and consoles fell from $1,150 million to $1,030 million in one year. Find the percent decrease. (Round to the nearest tenth of a percent.)

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%

Late Fees Alison was late paying her credit card bill of $249. She was charged a 5% late fee. What was the amount of the late fee?

Writing Exercises

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.

Solution

The original number should be greater than 44.80% is less than 100%, so when 80% is converted to a decimal and multiplied to the base in the percent equation, the resulting amount of 44 is less. 44 is only the larger number in cases where the percent is greater than 100%.

Without solving the problem “What is 20% of 300?” think about what the solution might be. Should it be a number that is greater than 300 or less than 300? 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

Alex should have packed half as many shorts and twice as many shirts.

Because of road construction in one city, commuters were advised to plan their Monday morning commute to 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 objectives of this section.

A self-assessment table titled 'I can...' asks users to rate their ability to 'translate and solve basic percent equations,' 'solve applications of percent,' and 'find percent increase and percent decrease' with options '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?

percent increase
The percent increase is the percent the amount of increase is of the original amount.
percent decrease
The percent decrease is the percent the amount of decrease is of the original amount.

Solve Sales Tax, Commission, and Discount Applications

Learning Objectives

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

  • Solve sales tax applications
  • Solve commission applications
  • Solve discount applications
  • Solve mark-up applications

Before you get started, take this readiness quiz.

Solve 0.0875(720) through multiplication.
If you missed this problem, review Example 7 in Decimal Operations.

Solution

63

Solve 12.96÷0.04 through division.
If you missed this problem, review Example 12 in Decimal Operations.

Solution

324

Solve Sales Tax Applications

Sales tax and commissions are applications of percent in our everyday lives. To solve these applications, we will follow the same strategy we used in the section on decimal operations. We show it again here for easy reference.

Solve an application
  1. Identify what you are asked to find and choose a variable to represent it.
  2. Write a sentence that gives the information to find it.
  3. Translate the sentence into an equation.
  4. Solve the equation using good algebra techniques.
  5. Check the answer in the problem and make sure it makes sense.
  6. Write a complete sentence that answers the question.

Remember that whatever the application, once we write the sentence with the given information (Step 2), we can translate it to a percent equation and then solve it.

Do you pay a tax when you shop in your city or state? In many parts of the United States, sales tax is added to the purchase price of an item. See Figure 1. The sales tax is determined by computing a percent of the purchase price.

To find the sales tax multiply the purchase price by the sales tax rate. Remember to convert the sales tax rate from a percent to a decimal number. Once the sales tax is calculated, it is added to the purchase price. The result is the total cost—this is what the customer pays.

The figure shows a restaurant check with sales tax
The sales tax is calculated as a percent of the purchase price.

Sales Tax

The sales tax is a percent of the purchase price.

Sales Tax=Tax Rate·Purchase PriceTotal Cost=Purchase Price+Sales Tax
Cathy bought a bicycle in Washington, where the sales tax rate was 6.5% of the purchase price. What was
  1. ⓐ the sales tax and
  2. ⓑ the total cost of a bicycle if the purchase price of the bicycle was $392?
Solution

Solution

ⓐ
Identify what you are asked to find. What is the sales tax?
Choose a variable to represent it. Let t= sales tax.
Write a sentence that gives the information to find it. The sales tax is 6.5% of the purchase price.
Translate into an equation. (Remember to change the percent to a decimal). A visual breakdown of a sales tax problem, showing how 'The sales tax is 6.5% of the $392 purchase price' translates to the equation 't = 0.065 . 392'.
Simplify. The image displays the equation 't = 25.48' in a clear, central font against a white background.
Check: Is this answer reasonable?
Yes, because the sales tax amount is less than 10% of the purchase price.
Write a complete sentence that answers the question. The sales tax is $25.48.
This table outlines the systematic steps for solving a word problem, demonstrating how to identify, set up, calculate, and verify the total cost of a bicycle with sales tax.
ⓑ
Identify what you are asked to find. What is the total cost of the bicycle?
Choose a variable to represent it. Let c= total cost of bicycle.
Write a sentence that gives the information to find it. The total cost is the purchase price plus the sales tax.
Translate into an equation. A diagram shows the translation of the phrase 'The total cost is $392 plus $25.48' into the equation 'c = 392 + 25.48', highlighting how words map to mathematical symbols like 'is' to '=' and 'plus' to '+'.
Simplify. A white background displays a mathematical equation in black text, stating 'c = 417.48'.
Check: Is this answer reasonable?
Yes, because the total cost is a little more than the purchase price.
Write a complete sentence that answers the question. The total cost of the bicycle is $417.48.

Find ⓐ the sales tax and ⓑ the total cost: Alexandra bought a television set for $724 in Boston, where the sales tax rate was 6.25% of the purchase price.

Solution
  1. ⓐ $45.25
  2. ⓑ $769.25

Find ⓐ the sales tax and ⓑ the total cost: Kim bought a winter coat for $250 in St. Louis, where the sales tax rate was 8.2% of the purchase price.

Solution
  1. ⓐ $20.50
  2. ⓑ $270.50

Evelyn bought a new smartphone for $499 plus tax. She was surprised when she got the receipt and saw that the tax was $42.42. What was the sales tax rate for this purchase?

Solution

Solution

Identify what you are asked to find. What is the sales tax rate?
Choose a variable to represent it. Let r= sales tax.
Write a sentence that gives the information to find it. What percent of the price is the sales tax?
Translate into an equation. A word problem demonstrating how to translate a question about tax percentage into an algebraic equation: 'What percent of $499 is $42.42?' translates to 'r * 499 = 42.42'.
Divide. An algebraic equation showing 499r divided by 499 equals 42.42 divided by 499, illustrating the process of isolating the variable 'r' by dividing both sides by 499.
Simplify. The image displays the equation 'r = 0.085' centered on a plain white background, rendered in a standard black font.
Check. Is this answer reasonable?
Yes, because 8.5% is close to 10%.
10% of $500 is $50, which is close to $42.42.
Write a complete sentence that answers the question. The sales tax rate is 8.5%.

Diego bought a new car for $26,525. He was surprised that the dealer then added $2,387.25. What was the sales tax rate for this purchase?

Solution

9%

What is the sales tax rate if a $7,594 purchase will have $569.55 of sales tax added to it?

Solution

7.5%

Solve Commission Applications

Sales people often receive a commission, or percent of total sales, for their sales. Their income may be just the commission they earn, or it may be their commission added to their hourly wages or salary. The commission they earn is calculated as a certain percent of the price of each item they sell. That percent is called the rate of commission.

Commission

A commission is a percentage of total sales as determined by the rate of commission.

commission=rate of commission·total sales

To find the commission on a sale, multiply the rate of commission by the total sales. Just as we did for computing sales tax, remember to first convert the rate of commission from a percent to a decimal.

Helene is a realtor. She receives 3% commission when she sells a house. How much commission will she receive for selling a house that costs $260,000?

Solution

Solution

Identify what you are asked to find. What is the commission?
Choose a variable to represent it. Let c= the commission.
Write a sentence that gives the information to find it. The commission is 3% of the price.
Translate into an equation. This image illustrates how to translate the word problem 'The commission is 3% of the $260,000 price' into the mathematical equation: c = 0.03 . 260,000, showing each component's conversion.
Simplify. A mathematical expression 'c = 7800' is shown in black text on a white background.
Check. Is this answer reasonable?
Yes. 1% of $260,000 is $2,600, and $7,800 is three times $2,600.
Write a complete sentence that answers the question. Helene will receive a commission of $7,800.

Bob is a travel agent. He receives 7% commission when he books a cruise for a customer. How much commission will he receive for booking a $3,900 cruise?

Solution

$273

Fernando receives 18% commission when he makes a computer sale. How much commission will he receive for selling a computer for $2,190?

Solution

$394.20

Rikki earned $87 commission when she sold a $1,450 stove. What rate of commission did she get?

Solution

Solution

Identify what you are asked to find. What is the rate of commission?
Choose a variable to represent it. Let r= the rate of commission.
Write a sentence that gives the information to find it. The commission is what percent of the sale?
Translate into an equation. An illustration converting the question 'The $87 commission is what percent of the $1450 sale?' into the equation '87 = r * 1450' to solve for the unknown percentage 'r'.
Divide. A mathematical equation is displayed, showing the fraction 87/1450 equal to 1450r/1450, illustrating a step in solving for 'r' where both sides of the equation are divided by 1450.
Simplify. A mathematical equation shows '0.06 = r' in black text against a plain white background.
Change to percent form. The image displays the mathematical expression 'r = 6%' in a black serif font against a plain white background. This notation typically represents an interest rate or percentage value.
Check if this answer is reasonable.
Yes. A 10% commission would have been $145.
The 6% commission, $87, is a little more than half of that.
Write a complete sentence that answers the question. The commission was 6% of the price of the stove.

Homer received $1,140 commission when he sold a car for $28,500. What rate of commission did he get?

Solution

4%

Bernice earned $451 commission when she sold an $8,200 living room set. What rate of commission did she get?

Solution

5.5%

Solve Discount Applications

Applications of discount are very common in retail settings Figure 2. When you buy an item on sale, the original price of the item has been reduced 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. We summarize the discount model in the box below.

The figure shows a sale sign with a discount rate
Applications of discounts are common in everyday life. (credit: Charleston's TheDigitel, Flickr)

Discount

An amount of discount is a percent off the original price.

amount of discount=discount rate·original pricesale price=original price−discount

The sale price should always be less than the original price. In some cases, the amount of discount is a fixed dollar amount. Then we just find the sale price by subtracting the amount of discount from the original price.

Jason bought a pair of sunglasses that were on sale for $10 off. The original price of the sunglasses was $39. What was the sale price of the sunglasses?

Solution

Solution

Identify what you are asked to find. What is the sale price?
Choose a variable to represent it. Let s= the sale price.
Write a sentence that gives the information to find it. The sale price is the original price minus the discount.
Translate into an equation. A visual breakdown of how to formulate an equation for a sale price (s), showing s = 39 - 10, where 39 is the original price and 10 is the discount.
Simplify. The image displays the equation s = 29, written in black text on a white background.
Check if this answer is reasonable.
Yes. The sale price, $29, is less than the original price, $39.
Write a complete sentence that answers the question. The sale price of the sunglasses was $29.

Marta bought a dishwasher that was on sale for $75 off. The original price of the dishwasher was $525. What was the sale price of the dishwasher?

Solution

$450

Orlando bought a pair of shoes that was on sale for $30 off. The original price of the shoes was $112. What was the sale price of the shoes?

Solution

$82

In Example 5, the amount of discount was a set amount, $10. In Example 6 the discount is given as a percent of the original price.

Elise bought a dress that was discounted 35% off of the original price of $140. What was ⓐ the amount of discount and ⓑ the sale price of the dress?

Solution

Solution

ⓐ Before beginning, you may find it helpful to organize the information in a list.
Original price = $140
Discount rate = 35%
Amount of discount = ?

Identify what you are asked to find. What is the amount of discount?
Choose a variable to represent it. Let d= the amount of discount.
Write a sentence that gives the information to find it. The discount is 35% of the original price.
Translate into an equation. The image shows the breakdown of a discount calculation: 'The discount is 35% of the $140 original price', which translates to 's = 0.35 . 140'.
Simplify. The image displays the mathematical expression 'd=49' centered on a white background.
Check if this answer is reasonable.
Yes. A $49 discount is reasonable for a $140 dress.
Write a complete sentence that answers the question. The amount of discount was $49.

ⓑ
Original price = $140
Amount of discount = $49
Sale price = ?

Identify what you are asked to find. What is the sale price of the dress?
Choose a variable to represent it. Let s= the sale price.
Write a sentence that gives the information to find it. The sale price is the original price minus the discount.
Translate into an equation. Sale price calculation example. The sentence "The sale price is the one hundred forty dollars minus the forty-nine dollar discount" translates to the equation s equals one hundred forty minus forty-nine.
Simplify. The equation 's=91' is displayed on a white background, suggesting a mathematical or data-related context. The text is clear and centrally located.
Check if this answer is reasonable.
Yes. The sale price, $91, is less than the original price, $140.
Write a complete sentence that answers the question. The sale price of the dress was $91.

Find ⓐ the amount of discount and ⓑ the sale price: Sergio bought a belt that was discounted 40% from an original price of $29.

Solution
  1. ⓐ $11.60
  2. ⓑ $17.40

Find ⓐ the amount of discount and ⓑ the sale price: Oscar bought a barbecue grill that was discounted 65% from an original price of $395.

Solution
  1. ⓐ $256.75
  2. ⓑ $138.25

There may be times when you buy something on sale and want to know the discount rate. The next example will show this case.

Jeannette bought a swimsuit at a sale price of $13.95. The original price of the swimsuit was $31. Find the ⓐ amount of discount and ⓑ discount rate.

Solution

Solution

ⓐ Before beginning, you may find it helpful to organize the information in a list.
Original price = $31
Amount of discount = ?
Sale price = $13.95

Identify what you are asked to find. What is the amount of discount?
Choose a variable to represent it. Let d= the amount of discount.
Write a sentence that gives the information to find it. The discount is the original price minus the sale price.
Translate into an equation. Translating a discount word problem into an equation: 'The discount is the $31 original price minus the $13.95 sale price' becomes 'd = 31 - 13.95'.
Simplify. The mathematical equation 'd = 17.05' is displayed in black text on a plain white background.
Check if this answer is reasonable.
Yes. The $17.05 discount is less than the original price.
Write a complete sentence that answers the question. The amount of discount was $17.05.

ⓑ Before beginning, you may find it helpful to organize the information in a list.
Original price = $31
Amount of discount = $17.05
Discount rate = ?

Identify what you are asked to find. What is the discount rate?
Choose a variable to represent it. Let r= the discount rate.
Write a sentence that gives the information to find it. The discount is what percent of the original price?
Translate into an equation. An image demonstrating how to translate the word problem 'The discount of $17.05 is what percent of the $31 original price' into a mathematical equation. It shows 'The discount of $17.05' represented as 'd', 'is' as '=', 'what percent' as 'r', 'of' as a multiplication dot, and 'the $31 original price' as '31', forming the equation d = r . 31.
Divide. An algebra problem presenting the equation 17.05/31 = r(31)/31.
Simplify. A close-up view of the number 0.55 equals r, presented as a black text on a white background.
Check if this answer is reasonable.
The rate of discount was a little more than 50% and the amount of discount is a little more than half of $31.
Write a complete sentence that answers the question. The rate of discount was 55%.

Find ⓐ the amount of discount and ⓑ the discount rate: Lena bought a kitchen table at the sale price of $375.20. The original price of the table was $560.

Solution
  1. ⓐ $184.80
  2. ⓑ 33%

Find ⓐ the amount of discount and ⓑ the discount rate: Nick bought a multi-room air conditioner at a sale price of $340. The original price of the air conditioner was $400.

Solution
  1. ⓐ $60
  2. ⓑ 15%

Solve Mark-up Applications

Applications of mark-up are very common in retail settings. The price a retailer pays for an item is called the wholesale price. The retailer then adds a mark-up to the wholesale price to get the list price, the price he sells the item for. The mark-up is usually calculated as a percent of the wholesale price. The percent is called the mark-up rate. To determine the amount of mark-up, multiply the mark-up rate by the wholesale price. We summarize the mark-up model in the box below.

Mark-up

The mark-up is the amount added to the wholesale price.

amount of mark-up=mark-up rate·wholesale pricelist price=wholesale price+mark up

The list price should always be more than the wholesale price.

Adam's art gallery bought a photograph at the wholesale price of $250. Adam marked the price up 40%. Find the ⓐ amount of mark-up and ⓑ the list price of the photograph.

Solution

Solution

ⓐ
Identify what you are asked to find. What is the amount of mark-up?
Choose a variable to represent it. Let m= the amount of each mark-up.
Write a sentence that gives the information to find it. The mark-up is 40% of the wholesale price.
Translate into an equation. The image illustrates the calculation for a mark-up (m), showing it is 40% of the $250 wholesale price, translating to m = 0.40 * 250.
Simplify. The image displays a mathematical expression 'm = 100' in black text against a white background, representing a variable 'm' assigned a value of 100.
Check if this answer is reasonable.
Yes. The markup rate is less than 50% and $100 is less than half of $250.
Write a complete sentence that answers the question. The mark-up on the photograph was $100.
ⓑ
Identify what you are asked to find. What is the list price?
Choose a variable to represent it. Let p= the list price.
Write a sentence that gives the information to find it. The list price is the wholesale price plus the mark-up.
Translate into an equation. This image illustrates how to calculate the list price (p) by adding the wholesale price ($250) and the mark-up ($100), forming the equation p = 250 + 100.
Simplify. The text 'p = 350' is displayed on a white background, suggesting a mathematical or scientific context.
Check if this answer is reasonable.
Yes. The list price, $350, is more than the wholesale price, $250.
Write a complete sentence that answers the question. The list price of the photograph was $350.

Jim's music store bought a guitar at wholesale price $1,200. Jim marked the price up 50%. Find the ⓐ amount of mark-up and ⓑ the list price.

Solution
  1. ⓐ $600
  2. ⓑ $1,800

The Auto Resale Store bought Pablo's Toyota for $8,500. They marked the price up 35%. Find the ⓐ amount of mark-up and ⓑ the list price.

Solution
  1. ⓐ $2,975
  2. ⓑ $11,475

Key Concepts

  • Sales Tax The sales tax is a percent of the purchase price.
    • sales tax=tax rate⋅purchase price
    • total cost=purchase price+sales tax
  • Commission A commission is a percentage of total sales as determined by the rate of commission.
    • commission=rate of commission⋅original price
  • Discount An amount of discount is a percent off the original price, determined by the discount rate.
    • amount of discount=discount rate⋅original price
    • sale price=original price –discount
  • Mark-up The mark-up is the amount added to the wholesale price, determined by the mark-up rate.
    • amount of mark-up=mark-up rate wholesale price
    • list price=wholesale price+mark up

Practice Makes Perfect

Solve Sales Tax Applications

In the following exercises, find ⓐ the sales tax and ⓑ the total cost.

The cost of a pair of boots was $84. The sales tax rate is 5% of the purchase price.

Solution
  1. ⓐ $4.20
  2. ⓑ $88.20

The cost of a refrigerator was $1,242. The sales tax rate is 8% of the purchase price.

The cost of a microwave oven was $129. The sales tax rate is 7.5% of the purchase price.

Solution
  1. ⓐ $9.68
  2. ⓑ $138.68

The cost of a tablet computer is $350. The sales tax rate is 8.5% of the purchase price.

The cost of a file cabinet is $250. The sales tax rate is 6.85% of the purchase price.

Solution
  1. ⓐ $17.13
  2. ⓑ $267.13

The cost of a luggage set $400. The sales tax rate is 5.75% of the purchase price.

The cost of a 6-drawer dresser $1,199. The sales tax rate is 5.125% of the purchase price.

Solution
  1. ⓐ $61.45
  2. ⓑ $1,260.45

The cost of a sofa is $1,350. The sales tax rate is 4.225% of the purchase price.

In the following exercises, find the sales tax rate.

Shawna bought a mixer for $300. The sales tax on the purchase was $19.50.

Solution

6.5%

Orphia bought a coffee table for $400. The sales tax on the purchase was $38.

Bopha bought a bedroom set for $3,600. The sales tax on the purchase was $246.60.

Solution

6.85%

Ruth bought a washer and dryer set for $2,100. The sales tax on the purchase was $152.25.

Solve Commission Applications

In the following exercises, find the commission.

Christopher sold his dinette set for $225 through an online site, which charged him 9% of the selling price as commission. How much was the commission?

Solution

$20.25

Michele rented a booth at a craft fair, which charged her 8% commission on her sales. One day her total sales were $193. How much was the commission?

Farrah works in a jewelry store and receives 12% commission when she makes a sale. How much commission will she receive for selling a $8,125 ring?

Solution

$975

Jamal works at a car dealership and receives 9% commission when he sells a car. How much commission will he receive for selling a $32,575 car?

Hector receives 17.5% commission when he sells an insurance policy. How much commission will he receive for selling a policy for $4,910?

Solution

$859.25

Denise receives 10.5% commission when she books a tour at the travel agency. How much commission will she receive for booking a tour with total cost $7,420?

In the following exercises, find the rate of commission.

Dontay is a realtor and earned $11,250 commission on the sale of a $375,000 house. What is his rate of commission?

Solution

3%

Nevaeh is a cruise specialist and earned $364 commission after booking a cruise that cost $5,200. What is her rate of commission?

As a waitress, Emily earned $420 in tips on sales of $2,625 last Saturday night. What was her rate of commission?

Solution

16%

Alejandra earned $1,393.74 commission on weekly sales of $15,486 as a salesperson at the computer store. What is her rate of commission?

Maureen earned $7,052.50 commission when she sold a $45,500 car. What was the rate of commission?

Solution

15.5%

Lucas earned $4,487.50 commission when he brought a $35,900 job to his office. What was the rate of commission?

Solve Discount Applications

In the following exercises, find the sale price.

Perla bought a cellphone that was on sale for $50 off. The original price of the cellphone was $189.

Solution

$139

Sophie saw a dress she liked on sale for $15 off. The original price of the dress was $96.

Solution

$81

Rick wants to buy a tool set with original price $165. Next week the tool set will be on sale for $40 off.

Solution

$125

Angelo's store is having a sale on TV sets. One set, with an original price of $859, is selling for $125 off.

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
  1. ⓐ $26.97
  2. ⓑ $17.98

Errol bought a skateboard helmet on sale at 40% off. The original price was $49.95.

Kathy wants to buy a camera that lists for $389. The camera is on sale with a 33% discount.

Solution
  1. ⓐ $128.37
  2. ⓑ $260.63

Colleen bought a suit that was discounted 25% from an original price of $245.

Erys bought a treadmill on sale at 35% off. The original price was $949.95.

Solution
  1. ⓐ $332.48
  2. ⓑ $617.47

Jay bought a guitar on sale at 45% off. The original price was $514.75.

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
  1. ⓐ $576
  2. ⓑ 30%

Hiroshi bought a lawnmower at the sale price of $240. The original price of the lawnmower is $300.

Patty bought a baby stroller on sale for $301.75. The original price of the stroller was $355.

Solution
  1. ⓐ $53.25
  2. ⓑ 15%

Bill found a book he wanted on sale for $20.80. The original price of the book was $32.

Nikki bought a patio set on sale for $480. The original price was $850.

Solution
  1. ⓐ $370
  2. ⓑ 43.5%

Stella bought a dinette set on sale for $725. The original price was $1,299.

Solve Mark-up Applications

In the following exercises, find ⓐ the amount of the mark-up and ⓑ the list price.

Daria bought a bracelet at wholesale cost $16 to sell in her handicraft store. She marked the price up 45%.

Solution
  1. ⓐ $7.20
  2. ⓑ $23.20

Regina bought a handmade quilt at wholesale cost $120 to sell in her quilt store. She marked the price up 55%.

Tom paid $0.60 a pound for tomatoes to sell at his produce store. He added a 33% mark-up.

Solution
  1. ⓐ $0.20
  2. ⓑ $0.80

Flora paid her supplier $0.74 a stem for roses to sell at her flower shop. She added an 85% mark-up.

Alan bought a used bicycle for $115. After re-conditioning it, he added 225% mark-up and then advertised it for sale.

Solution
  1. ⓐ $258.75
  2. ⓑ $373.75

Michael bought a classic car for $8,500. He restored it, then added 150% mark-up before advertising it for sale.

Everyday Math

Coupons Yvonne can use two coupons for the same purchase at her favorite department store. One coupon gives her $20 off and the other gives her 25% off. She wants to buy a bedspread that sells for $195.

  1. ⓐ Calculate the discount price if Yvonne uses the $20 coupon first and then takes 25% off.
  2. ⓑ Calculate the discount price if Yvonne uses the 25% off coupon first and then uses the $20 coupon.
  3. ⓒ In which order should Yvonne use the coupons?
Solution
  1. ⓐ $131.25
  2. ⓑ $126.25
  3. ⓒ 25% off first, then $20 off

Cash Back Jason can buy a bag of dog food for $35 at two different stores. One store offers 6% cash back on the purchase plus $5 off his next purchase. The other store offers 20% cash back.

  1. ⓐ Calculate the total savings from the first store, including the savings on the next purchase.
  2. ⓑ Calculate the total savings from the second store.
  3. ⓒ Which store should Jason buy the dog food from? Why?

Writing Exercises

Priam bought a jacket that was on sale for 40% off. The original price of the jacket was $150. While the sales clerk figured the price by calculating the amount of discount and then subtracting that amount from $150, Priam found the price faster by calculating 60% of $150.

  1. ⓐ Explain why Priam was correct.
  2. ⓑ Will Priam's method work for any original price?
Solution
  1. ⓐ Priam is correct. The original price is 100%. Since the discount rate was 40%, the sale price was 60% of the original price.
  2. ⓑ Yes.

Roxy bought a scarf on sale for 50% off. The original price of the scarf was $32.90. Roxy claimed that the price she paid for the scarf was the same as the amount she saved. Was Roxy correct? Explain.

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, asking users to rate their ability to solve sales tax, commission, discount, and mark-up applications as '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?

commission
A commission is a percentage of total sales as determined by the rate of commission.
discount
An amount of discount is a percent off the original price, determined by the discount rate.
mark-up
The mark-up is the amount added to the wholesale price, determined by the mark-up rate.
sales tax
The sales tax is a percent of the purchase price.

Solve Simple Interest Applications

Learning Objectives

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

  • Use the simple interest formula
  • Solve simple interest applications

Before you get started, take this readiness quiz.

Solve 0.6y=45.
If you missed this problem, review Example 4 in Solve Equations with Decimals.

Solution

75

Solve n1.45=4.6.
If you missed this problem, review Example 5 in Solve Equations with Decimals.

Solution

6.67

Use the Simple Interest Formula

Do you know that banks pay you to let them keep your money? The money you put in the bank is called the principal, P, and the bank pays you interest, I. 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 for time, t, represents the number of years the money is left in the account.

Simple Interest

If an amount of money, P, the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is

I=Prt

where

I=interestP=principalr=ratet=time

Interest earned according to this formula is called simple interest.

The formula we use to calculate simple interest is I=Prt. To use the simple interest 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 by listing all four variables and filling in the given information.

Find the simple interest earned after 3 years on $500 at an interest rate of 6%.

Solution

Solution

Organize the given information in a list.

I=?P=$500r=6%t=3 years

We will use the simple interest formula to find the interest.

This table illustrates the step-by-step calculation of simple interest (I=Prt), showing formula application, simplification, and a reasonableness check for the final interest amount.
Write the formula. I=Prt
Substitute the given information. Remember to write the percent in decimal form. I=(500)(0.06)(3)
Simplify. I=90
Check your answer. Is $90 a reasonable interest earned on $500 in 3 years?
In 3 years the money earned 18%. If we rounded to 20%, the interest would have been 500(0.20) or $100. Yes, $90 is reasonable.
Write a complete sentence that answers the question. The simple interest is $90.

Find the simple interest earned after 4 years on $800 at an interest rate of 5%.

Solution

$160

Find the simple interest earned after 2 years on $700 at an interest rate of 4%.

Solution

$56

In the next example, we will use the simple interest formula to find the principal.

Find the principal invested if $178 interest was earned in 2 years at an interest rate of 4%.

Solution

Solution

Organize the given information in a list.

I=$178P=?r=4%t=2 years

We will use the simple interest formula to find the principal.

Detailed steps to calculate the principal (P) using the simple interest formula I=Prt, including substitution, simplification, and verification.
Write the formula. I=Prt
Substitute the given information. 178=P(0.04)(2)
Divide. 1780.08=0.08P0.08
Simplify. 2,225=P
Check your answer. Is it reasonable that $2,225 would earn $178 in 2 years?
I=Prt
178=?2,225(0.04)(2)
178=178✓
Write a complete sentence that answers the question. The principal is $2,225.

Find the principal invested if $495 interest was earned in 3 years at an interest rate of 6%.

Solution

$2,750

Find the principal invested if $1,246 interest was earned in 5 years at an interest rate of 7%.

Solution

$3,560

Now we will solve for the rate of interest.

Find the rate if a principal of $8,200 earned $3,772 interest in 4 years.

Solution

Solution

Organize the given information.

I=$3,772P=$8,200r=?t=4 years

We will use the simple interest formula to find the rate.

Step-by-step calculation of the interest rate 'r' using the simple interest formula I=Prt, including solution verification.
Write the formula. I=Prt
Substitute the given information. 3,772=8,200r(4)
Multiply. 3,772=32,800r
Divide. 3,77232,800=32,800r32,800
Simplify. 0.115=r
Write as a percent. 11.5%=r
Check your answer. Is 11.5% a reasonable rate if $3,772 was earned in 4 years?
I=Prt
3,772=?8,200(0.115)(4)
3,772=3,772✓
Write a complete sentence that answers the question. The rate was 11.5%.

Find the rate if a principal of $5,000 earned $1,350 interest in 6 years.

Solution

4.5%

Find the rate if a principal of $9,000 earned $1,755 interest in 3 years.

Solution

6.5%

Solve Simple Interest Applications

Applications with simple interest usually involve either investing money or borrowing money. To solve these applications, we continue to use the same strategy for applications that we have used earlier in this chapter. The only difference is that in place of translating to get an equation, we can use the simple interest formula.

We will start by solving a simple interest application to find the interest.

Nathaly deposited $12,500 in her bank account where it will earn 4% interest. How much interest will Nathaly earn in 5 years?

Solution

Solution

We are asked to find the Interest, I.

Organize the given information in a list.

I=?P=$12,500r=4%t=5 years

Step-by-step calculation of simple interest, demonstrating the application of the I=Prt formula to find the interest on $12,500 at 4% for 5 years.
Write the formula. I=Prt
Substitute the given information. I=(12,500)(0.04)(5)
Simplify. I=2,500
Check your answer. Is $2,500 a reasonable interest on $12,500 over 5 years?
At 4% interest per year, in 5 years the interest would be 20% of the principal. Is 20% of $12,500 equal to $2,500? Yes.
Write a complete sentence that answers the question. The interest is $2,500.

Areli invested a principal of $950 in her bank account with interest rate 3%. How much interest did she earn in 5 years?

Solution

$142.50

Susana invested a principal of $36,000 in her bank account with interest rate 6.5%. How much interest did she earn in 3 years?

Solution

$7,020

There may be times when you know the amount of interest earned on a given principal over a certain length of time, but you don't know the rate. For instance, this might happen when family members lend or borrow money among themselves instead of dealing with a bank. In the next example, we'll show how to solve for the rate.

Loren lent his brother $3,000 to help him buy a car. In 4 years his brother paid him back the $3,000 plus $660 in interest. What was the rate of interest?

Solution

Solution

We are asked to find the rate of interest, r.

Organize the given information.

I=660P=$3,000r=?t=4 years

Step-by-step calculation to determine the interest rate using the simple interest formula (I=Prt) and verification of the result.
Write the formula. I=Prt
Substitute the given information. 660=(3,000)r(4)
Multiply. 660=(12,000)r
Divide. 66012,000=(12,000)r12,000
Simplify. 0.055=r
Change to percent form. 5.5%=r
Check your answer. Is 5.5% a reasonable interest rate to pay your brother?
I=Prt
660=?(3,000)(0.055)(4)
660=660✓
Write a complete sentence that answers the question. The rate of interest was 5.5%.

Jim lent his sister $5,000 to help her buy a house. In 3 years, she paid him the $5,000, plus $900 interest. What was the rate of interest?

Solution

6%

Hang borrowed $7,500 from her parents to pay her tuition. In 5 years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of interest?

Solution

4%

There may be times when you take a loan for a large purchase and the amount of the principal is not clear. This might happen, for instance, in making a car purchase when the dealer adds the cost of a warranty to the price of the car. In the next example, we will solve a simple interest application for the principal.

Eduardo noticed that his new car loan papers stated that with an interest rate of 7.5%, he would pay $6,596.25 in interest over 5 years. How much did he borrow to pay for his car?

Solution

Solution

We are asked to find the principal, P.

Organize the given information.

I=6,596.25P=?r=7.5%t=5 years

Step-by-step calculation of the principal amount (P) using the simple interest formula I=Prt, demonstrating formula application, computation, and verification.
Write the formula. I=Prt
Substitute the given information. 6,596.25=P(0.075)(5)
Multiply. 6,596.25=0.375P
Divide. 6,596.250.375=0.375P0.375
Simplify. 17,590=P
Check your answer. Is $17,590 a reasonable amount to borrow to buy a car?
I=Prt
6,596.25=?(17,590)(0.075)(5)
6,596.25=6,596.25✓
Write a complete sentence that answers the question. The amount borrowed was $17,590.

Sean's new car loan statement said he would pay $4,866.25 in interest from an interest rate of 8.5% over 5 years. How much did he borrow to buy his new car?

Solution

$11,450

In 5 years, Gloria's bank account earned $2,400 interest at 5%. How much had she deposited in the account?

Solution

$9,600

In the simple interest formula, the rate of interest is given as an annual rate, the rate for one year. So the units of time must be in years. If the time is given in months, we convert it to years.

Caroline got $900 as graduation gifts and invested it in a 10-month certificate of deposit that earned 2.1% interest. How much interest did this investment earn?

Solution

Solution

We are asked to find the interest, I.

Organize the given information.

I=?P=$900r=2.1%t=10 months

Step-by-step calculation of simple interest, including formula application, computation, and result verification.
Write the formula. I=Prt
Substitute the given information, converting 10 months to 1012 of a year. I=$900(0.021)(1012)
Multiply. I=15.75
Check your answer. Is $15.75 a reasonable amount of interest?
If Caroline had invested the $900 for a full year at 2% interest, the amount of interest would have been $18. Yes, $15.75 is reasonable.
Write a complete sentence that answers the question. The interest earned was $15.75.

Adriana invested $4,500 for 8 months in an account that paid 1.9% interest. How much interest did she earn?

Solution

$57.00

Milton invested $2,460 for 20 months in an account that paid 3.5% interest How much interest did he earn?

Solution

$143.50

Key Concepts

  • Simple interest
    • If an amount of money, P, the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is I=Prt
    • Interest earned according to this formula is called simple interest.

Practice Makes Perfect

Use the Simple Interest Formula

In the following exercises, use the simple interest formula to fill in the missing information.

Interest Principal Rate Time (years)
$1200 3% 5
Solution

$180

Interest Principal Rate Time (years)
$1500 2% 4
Interest Principal Rate Time (years)
$4410 4.5% 7
Solution

$14,000

Interest Principal Rate Time (years)
$2112 3.2% 6
Interest Principal Rate Time (years)
$577.08 $4580 2
Solution

6.3%

Interest Principal Rate Time (years)
$528.12 $3260 3

In the following exercises, solve the problem using the simple interest formula.

Find the simple interest earned after 5 years on $600 at an interest rate of 3%.

Solution

$90

Find the simple interest earned after 4 years on $900 at an interest rate of 6%.

Find the simple interest earned after 2 years on $8,950 at an interest rate of 3.24%.

Solution

$579.96

Find the simple interest earned after 3 years on $6,510 at an interest rate of 2.85%.

Find the simple interest earned after 8 years on $15,500 at an interest rate of 11.425%.

Solution

$14,167

Find the simple interest earned after 6 years on $23,900 at an interest rate of 12.175%.

Find the principal invested if $656 interest was earned in 5 years at an interest rate of 4%.

Solution

$3,280

Find the principal invested if $177 interest was earned in 2 years at an interest rate of 3%.

Find the principal invested if $70.95 interest was earned in 3 years at an interest rate of 2.75%.

Solution

$860

Find the principal invested if $636.84 interest was earned in 6 years at an interest rate of 4.35%.

Find the principal invested if $15,222.57 interest was earned in 6 years at an interest rate of 10.28%.

Solution

$24,679.91

Find the principal invested if $10,953.70 interest was earned in 5 years at an interest rate of 11.04%.

Find the rate if a principal of $5,400 earned $432 interest in 2 years.

Solution

4%

Find the rate if a principal of $2,600 earned $468 interest in 6 years.

Find the rate if a principal of $11,000 earned $1,815 interest in 3 years.

Solution

5.5%

Find the rate if a principal of $8,500 earned $3,230 interest in 4 years.

Solve Simple Interest Applications

In the following exercises, solve the problem using the simple interest formula.

Casey deposited $1,450 in a bank account with interest rate 4%. How much interest was earned in 2 years?

Solution

$116

Terrence deposited $5,720 in a bank account with interest rate 6%. How much interest was earned in 4 years?

Robin deposited $31,000 in a bank account with interest rate 5.2%. How much interest was earned in 3 years?

Solution

$4,836

Carleen deposited $16,400 in a bank account with interest rate 3.9%. How much interest was earned in 8 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 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 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 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 interest?

In 10 years, a bank account that paid 5.25% earned $18,375 interest. What was the principal of the account?

Solution

$35,000

In 25 years, a bond that paid 4.75% earned $2,375 interest. What was the principal of the bond?

Joshua's computer loan statement said he would pay $1,244.34 in interest for a 3 year loan at 12.4%. How much did Joshua borrow to buy the computer?

Solution

$3,345

Margaret's car loan statement said she would pay $7,683.20 in interest for a 5 year loan at 9.8%. How much did Margaret borrow to buy the car?

Caitlin invested $8,200 in an 18-month certificate of deposit paying 2.7% interest. How much interest did she earn form this investment?

Solution

$332.10

Diego invested $6,100 in a 9-month certificate of deposit paying 1.8% interest. How much interest did he earn form this investment?

Airin borrowed $3,900 from her parents for the down payment on a car and promised to pay them back in 15 months at a 4% rate of interest. How much interest did she owe her parents?

Solution

$195.00

Yuta borrowed $840 from his brother to pay for his textbooks and promised to pay him back in 5 months at a 6% rate of interest. How much interest did Yuta owe his brother?

Everyday Math

Interest on savings Find the interest rate your local bank pays on savings accounts.

  1. ⓐ What is the interest rate?
  2. ⓑ Calculate the amount of interest you would earn on a principal of $8,000 for 5 years.
Solution

Answers will vary.

Interest on a loan Find the interest rate your local bank charges for a car loan.

  1. ⓐ What is the interest rate?
  2. ⓑ Calculate the amount of interest you would pay on a loan of $8,000 for 5 years.

Writing Exercises

Why do banks pay interest on money deposited in savings accounts?

Solution

Answers will vary.

Why do banks charge interest for lending money?

Self Check

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

A self-assessment table for simple interest, asking users to rate their ability to use the formula and solve applications with options: 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?

simple interest
If an amount of money, P, the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is I=Prt. Interest earned according to this formula is called simple interest.

Solve Proportions and their Applications

Learning Objectives

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

  • Use the definition of proportion
  • Solve proportions
  • Solve applications using proportions
  • Write percent equations as proportions
  • Translate and solve percent proportions

Before you get started, take this readiness quiz.

Simplify: 134.
If you missed this problem, review Example 8 in Multiply and Divide Mixed Numbers and Complex Fractions.

Solution

112

Solve: x4=20.
If you missed this problem, review Example 5 in Solve Equations with Fractions.

Solution

80

Write as a rate: Sale rode his bike 24 miles in 2 hours.
If you missed this problem, review Example 6 in Ratios and Rate.

Solution

24 miles2 hours

Use the Definition of Proportion

In the section on Ratios and Rates we saw some ways they are used in our daily lives. When two ratios or rates 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 states two ratios or rates are equal. 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”.

If we compare quantities with units, we have to be sure we are comparing them in the right order. For example, in the proportion 20 students1 teacher=60 students3 teachers we compare the number of students to the number of teachers. We put students in the numerators and teachers in the denominators.

Write each sentence as a proportion:
  1. ⓐ 3 is to 7 as 15 is to 35.
  2. ⓑ 5 hits in 8 at bats is the same as 30 hits in 48 at-bats.
  3. ⓒ $1.50 for 6 ounces is equivalent to $2.25 for 9 ounces.
Solution

Solution

Example illustrating how to convert a verbal proportion statement into its equivalent mathematical expression.
ⓐ
3 is to 7 as 15 is to 35.
Write as a proportion. 37=1535
Comparison of two equivalent batting ratios (hits to at-bats) demonstrated through fractions and proportions.
ⓑ
5 hits in 8 at-bats is the same as 30 hits in 48 at-bats.
Write each fraction to compare hits to at-bats. hitsat-bats=hitsat-bats
Write as a proportion. 58=3048
Illustrates the steps to set up and verify equivalent ratios comparing dollars to ounces, demonstrating a proportion calculation.
ⓒ
$1.50 for 6 ounces is equivalent to $2.25 for 9 ounces.
Write each fraction to compare dollars to ounces. $ounces=$ounces
Write as a proportion. 1.506=2.259
Write each sentence as a proportion:
  1. ⓐ 5 is to 9 as 20 is to 36.
  2. ⓑ 7 hits in 11 at-bats is the same as 28 hits in 44 at-bats.
  3. ⓒ $2.50 for 8 ounces is equivalent to $3.75 for 12 ounces.
Solution
  1. ⓐ 59=2036
  2. ⓑ 711=2844
  3. ⓒ 2.508=3.7512
Write each sentence as a proportion:
  1. ⓐ 6 is to 7 as 36 is to 42.
  2. ⓑ 8 adults for 36 children is the same as 12 adults for 54 children.
  3. ⓒ $3.75 for 6 ounces is equivalent to $2.50 for 4 ounces.
Solution
  1. ⓐ 67=3642
  2. ⓑ 836=1254
  3. ⓒ 3.756=2.504

Look at the proportions 12=48 and 23=69. From our work with equivalent fractions we know these equations are true. But how do we know if an equation is a proportion with equivalent fractions if it contains fractions with larger numbers?

To determine if a proportion is true, we find the cross products of each proportion. To find the cross products, we multiply each denominator with the opposite numerator (diagonally across the equal sign). The results are called a cross product because of the cross formed. If, and only if, the given proportion is true, that is, the two sides are equal, then the cross products of a proportion will be equal.

The figure shows cross multiplication of two proportions. There is the proportion 1 is to 2 as 4 is to 8. Arrows are shown diagonally across the equal sign to show cross products. The equations formed by cross multiplying are 8 · 1 = 8 and 2 · 4 = 8. There is the proportion 2 is to 3 as 6 is to 9. Arrows are shown diagonally across the equal sign to show cross products. The equations formed by cross multiplying are 9 · 2 = 18 and 3 · 6 = 18.

Cross Products of a Proportion

For any proportion of the form ab=cd, where b≠0,d≠0, its cross products are equal.

The image demonstrates cross products of a proportion with numerator a multiplying with denominator d, and denominator b multiplying with numerator c.

Cross products can be used to test whether a proportion is true. To test whether an equation makes a proportion, we find the cross products. If they are both equal, we have a proportion.

Determine whether each equation is a proportion:
  1. ⓐ 49=1228
  2. ⓑ 17.537.5=715
Solution

Solution

To determine if the equation is a proportion, we find the cross products. If they are equal, the equation is a proportion.

ⓐ
A mathematical equation showing the fractions 4/9 equals 12/28. Both fractions are written in black against a plain white background.
Find the cross products. 28⋅4=1129⋅12=108
The image illustrates an inequality between two fractions, 4/9 and 12/28, using cross-multiplication with blue arrows to show that 4 multiplied by 28 is not equal to 9 multiplied by 12.

Since the cross products are not equal, 28·4≠9·12, the equation is not a proportion.

ⓑ
A mathematical equation displays the fraction 17.5/37.5 simplified to 7/15, demonstrating equivalence between the two ratios.
Find the cross products. 15⋅17.5=262.537.5⋅7=262.5
An equation showing 17.5 divided by 37.5 is equal to 7 divided by 15, with blue arrows illustrating the concept of cross-multiplication.

Since the cross products are equal, 15·17.5=37.5·7, the equation is a proportion.

Determine whether each equation is a proportion:
  1. ⓐ 79=5472
  2. ⓑ 24.545.5=713
Solution
  1. ⓐ no
  2. ⓑ yes
Determine whether each equation is a proportion:
  1. ⓐ 89=5673
  2. ⓑ 28.552.5=815
Solution
  1. ⓐ no
  2. ⓑ no

Solve Proportions

To solve a proportion containing a variable, we remember that the proportion is an equation. All of the techniques we have used so far to solve equations still apply. In the next example, we will solve a proportion by multiplying by the Least Common Denominator (LCD) using the Multiplication Property of Equality.

Solve: x63=47.

Solution

Solution

A mathematical equation is displayed, showing a fraction on the left side, x over 63, which is equal to the fraction 4 over 7 on the right side.
To isolate x, multiply both sides by the LCD, 63. A mathematical equation shows 63 multiplied by the fraction x over 63, which equals 63 multiplied by the fraction 4 over 7. The number 63 is highlighted in red on both sides of the equation.
Simplify. A mathematical equation displays x = (9 * 7 * 4) / 7.
Divide the common factors. A mathematical equation displays 'x = 36' in a black serif font on a plain white background, centered within the frame.
Check: To check our answer, we substitute into the original proportion.
A mathematical equation shows x divided by 63 is equal to 4 divided by 7.
The image shows the text 'Substitute x = 36' in blue and red font against a white background. A mathematical equation displays the fraction 36/63, a question mark, and the fraction 4/7, inviting a comparison or to determine if the fractions are equivalent. The number 36 is highlighted in red.
Show common factors. A mathematical expression asks if the fraction (4 times 9) divided by (7 times 9) is equal to 4/7. This illustrates the principle of simplifying fractions by canceling common factors.
Simplify. The equation 4/7 = 4/7 with a checkmark, confirming its truth.

Solve the proportion: n84=1112.

Solution

77

Solve the proportion: y96=1312.

Solution

104

When the variable is in a denominator, we’ll use the fact that the cross products of a proportion are equal to solve the proportions.

We can find the cross products of the proportion and then set them equal. Then we solve the resulting equation using our familiar techniques.

Solve: 144a=94.

Solution

Solution

Notice that the variable is in the denominator, so we will solve by finding the cross products and setting them equal.

A mathematical equation illustrating cross-multiplication for solving proportions: 144 over a equals 9 over 4, with arrows indicating the cross-multiplication step.
Find the cross products and set them equal. A mathematical equation shows '4 * 144 = a * 9'.
Simplify. A mathematical equation displays '576 = 9a' in black text on a white background, representing a single linear equation with one unknown variable 'a'.
Divide both sides by 9. A mathematical equation shows a fraction 576 over 9 equal to a fraction 9a over 9.
Simplify. The mathematical equation '64 = a' is displayed in black font on a white background, indicating that the value of 'a' is 64.
Check your answer.
A mathematical equation shows '144 divided by a equals 9 divided by 4' against a white background.
The image shows the text 'Substitute a = 64' in a dark grey font, with the number '64' highlighted in red. A mathematical equation shows two fractions, 144/64 and 9/4, with a question mark and an equals sign between them, asking if they are equivalent. The denominator 64 is in red.
Show common factors.. A mathematical equation questions if (9 multiplied by 16) divided by (4 multiplied by 16) is equal to 9 divided by 4, illustrating the concept of simplifying fractions by cancelling common factors.
Simplify. Mathematical equation 9/4 = 9/4 with a checkmark.

Another method to solve this would be to multiply both sides by the LCD, 4a. Try it and verify that you get the same solution.

Solve the proportion: 91b=75.

Solution

65

Solve the proportion: 39c=138.

Solution

24

Solve: 5291=−4y.

Solution

Solution

Find the cross products and set them equal. The equation 52/91 = -4/y, with visual cues for cross-multiplication to solve for y.
A mathematical equation is displayed, showing 'y * 52 = 91(-4)'. The equation involves multiplication and a negative number, where 'y' is the unknown variable.
Simplify. A mathematical equation is displayed, showing '52y = -364' in black text against a white background, representing an algebraic problem.
Divide both sides by 52. The step of dividing both sides of 52y = -364 by 52 to find the value of y.
Simplify. The image displays a mathematical equation in black text on a white background, stating 'y = -7'.
Check:
A mathematical equation displays the fraction 52 over 91, which is set equal to the fraction -4 over y.
The text reads 'Substitute y = -7', with the word 'Substitute' and 'y = ' in dark gray, and '-7' in red. The text is rendered against a light background. A mathematical expression displaying two fractions, 52/91 and -4/-7, separated by an equals sign with a question mark above it, indicating a query about their equivalence.
Show common factors. A mathematical equation asking if (13 * 4) / (13 * 4) is equal to -4 / -7. The left side simplifies to 1, while the right side simplifies to 4/7, indicating they are not equal.
Simplify. The equation 4/7 = 4/7 is displayed with a checkmark, indicating its correctness.

Solve the proportion: 8498=−6x.

Solution

−7

Solve the proportion: −7y=105135.

Solution

−9

Solve Applications Using Proportions

The strategy for solving applications that we have used earlier in this chapter, also works for proportions, since proportions are equations. When we set up the proportion, we must make sure the units are correct—the units in the numerators match and the units in the denominators match.

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

Solution

This table outlines the sequential steps for calculating acetaminophen dosage using proportions, from problem identification to final solution.
Identify what you are asked to find. How many ml of acetaminophen the doctor will prescribe
Choose a variable to represent it. 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. A simple mathematical identity: ml/pounds = ml/pounds.
Substitute given values—be careful of the units. A mathematical equation showing the fraction 5/25 equal to the fraction a/80.
Multiply both sides by 80. A mathematical equation shows '80 multiplied by the fraction 5 over 25' on the left side, which is set equal to '80 multiplied by the fraction a over 80' on the right side.
Multiply and show common factors. A mathematical equation illustrating fraction simplification: (16 x 5 x 5) / (5 x 5) = 80a / 80.
Simplify. A mathematical equation displays the number 16 equal to the variable 'a', written as '16 = a'.
Check if the answer is reasonable.
Yes. Since 80 is about 3 times 25, the medicine should be about 3 times 5.
Write a complete sentence. The pediatrician would prescribe 16 ml of acetaminophen to Zoe.

You could also solve this proportion by setting the cross products equal.

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

12 ml

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

180 mg

One brand of microwave popcorn has 120 calories per serving. A whole bag of this popcorn has 3.5 servings. How many calories are in a whole bag of this microwave popcorn?

Solution

Solution

Outlines systematic steps to solve a word problem using proportions, illustrated by calculating total calories in a bag of microwave popcorn.
Identify what you are asked to find. How many calories are in a whole bag of microwave popcorn?
Choose a variable to represent it. Let c= number of calories.
Write a sentence that gives the information to find it. If there are 120 calories per serving, how many calories are in a whole bag with 3.5 servings?
Translate into a proportion. A mathematical equation shows that 'calories / serving' equals 'calories / serving,' representing an identity or a tautology in unit measurement.
Substitute given values. A mathematical equation shows a fraction 120 over 1 equal to the fraction c over 3.5. This represents a proportion where 120 divided by 1 is equal to c divided by 3.5.
Multiply both sides by 3.5. An algebraic equation: (3.5)(120/1) = (3.5)(c/3.5), demonstrating the multiplication of a decimal by a fraction on both sides.
Multiply. The image displays the mathematical equation '420 = c' in a simple, clear font against a white background.
Check if the answer is reasonable.
Yes. Since 3.5 is between 3 and 4, the total calories should be between 360 (3⋅120) and 480 (4⋅120).
Write a complete sentence. The whole bag of microwave popcorn has 420 calories.

Marissa loves the Caramel Macchiato at the coffee shop. The 16 oz. medium size has 240 calories. How many calories will she get if she drinks the large 20 oz. size?

Solution

300

Yaneli loves Starburst candies, but wants to keep her snacks to 100 calories. If the candies have 160 calories for 8 pieces, how many pieces can she have in her snack?

Solution

5 pieces

Josiah went to Mexico for spring break and changed $325 dollars into Mexican pesos. At that time, the exchange rate had $1 U.S. is equal to 12.54 Mexican pesos. How many Mexican pesos did he get for his trip?

Solution

Solution

Step-by-step process for solving a currency conversion problem using proportions, detailing each stage from problem identification to final answer.
Identify what you are asked to find. How many Mexican pesos did Josiah get?
Choose a variable to represent it. Let p= number of pesos.
Write a sentence that gives the information to find it. If $1 U.S. is equal to 12.54 Mexican pesos, then $325 is how many pesos?
Translate into a proportion. A diagram showing the expression "dollars per peso equals dollars per peso." The image illustrates that the ratio of dollars to pesos remains constant.
Substitute given values. A mathematical equation is displayed on a white background, showing 1 divided by 12.54 equals 325 divided by p.
The variable is in the denominator, so find the cross products and set them equal. The image displays the mathematical equation p * 1 = 12.54(325).
Simplify. A mathematical equation displays 'c = 4,075.5' in black font against a white background.
Check if the answer is reasonable.
Yes, $100 would be $1,254 pesos. $325 is a little more than 3 times this amount.
Write a complete sentence. Josiah has 4075.5 pesos for his spring break trip.

Yurianna is going to Europe and wants to change $800 dollars into Euros. At the current exchange rate, $1 US is equal to 0.738 Euro. How many Euros will she have for her trip?

Solution

590 Euros

Corey and Nicole are traveling to Japan and need to exchange $600 into Japanese yen. If each dollar is 94.1 yen, how many yen will they get?

Solution

56,460 yen

Write Percent Equations As Proportions

Previously, we solved percent equations by applying the properties of equality we have used to solve equations throughout this text. Some people prefer to solve percent equations by using the proportion method. The proportion method for solving percent problems involves a percent proportion. A percent proportion is an equation where a percent is equal to an equivalent ratio.

For example, 60%=60100 and we can simplify 60100=35. Since the equation 60100=35 shows a percent equal to an equivalent ratio, we call it a percent proportion. Using the vocabulary we used earlier:

amountbase=percent100
35=60100

Percent Proportion

The amount is to the base as the percent is to 100.

amountbase=percent100

If we restate the problem in the words of a proportion, it may be easier to set up the proportion:

The amount is to the base as the percent is to one hundred.

We could also say:

The amount out of the base is the same as the percent out of one hundred.

First we will practice translating into a percent proportion. Later, we’ll solve the proportion.

Translate to a proportion. What number is 75% of 90?

Solution

Solution

If you look for the word "of", it may help you identify the base.
Identify the parts of the percent proportion. An image explaining parts of a percentage problem: 'What number is 75% of 90?'. 'What number' is the amount, '75%' is the percent, and '90' is the base. 'Of' is highlighted in red.
Restate as a proportion. A math problem asks: 'What number out of 90 is the same as 75 out of 100?' The number '90' is highlighted in red.
Set up the proportion. Let n=number. n90=75100

Translate to a proportion: What number is 60% of 105?

Solution

n105=60100

Translate to a proportion: What number is 40% of 85?

Solution

n85=40100

Translate to a proportion. 19 is 25% of what number?

Solution

Solution

Identify the parts of the percent proportion. A breakdown of a percentage word problem: '19 is 25% of what number?', labeling '19' as amount, '25%' as percent, and 'what number' as base.
Restate as a proportion. A mathematical word problem asks: '19 out of what number is the same as 25 out of 100?' The word 'of' in the first phrase is highlighted in red.
Set up the proportion. Let n=number. 19n=25100

Translate to a proportion: 36 is 25% of what number?

Solution

36n=25100

Translate to a proportion: 27 is 36% of what number?

Solution

27n=36100

Translate to a proportion. What percent of 27 is 9?

Solution

Solution

Identify the parts of the percent proportion. A math problem asks 'What percent of 27 is 9?' with 'What percent' labeled as 'percent', '27' as 'base', and '9' as 'amount', illustrating how to identify parts of a percentage equation.
Restate as a proportion. A math question asks: '9 out of 27 is the same as what number out of 100?' The word 'of' between 'out' and '27' is highlighted in red, while the rest of the text is in a dark teal color.
Set up the proportion. Let p=percent. 927=p100

Translate to a proportion: What percent of 52 is 39?

Solution

n100=3952

Translate to a proportion: What percent of 92 is 23?

Solution

n100=2392

Translate and Solve Percent Proportions

Now that we have written percent equations as proportions, we are ready to solve the equations.

Translate and solve using proportions: What number is 45% of 80?

Solution

Solution

Identify the parts of the percent proportion. A math problem asks 'What number is 45% of 80?' with 'What number' labeled as amount, '45%' as percent, and '80' as base, illustrating how to identify parts of a percentage problem.
Restate as a proportion. A math problem asks: 'What number out of 80 is the same as 45 out of 100?'
Set up the proportion. Let n= number. A mathematical equation displays the fraction n over 80 set equal to the fraction 45 over 100, which is commonly used to represent a proportion or percentage problem.
Find the cross products and set them equal. A mathematical equation shows '100 multiplied by n equals 80 multiplied by 45'.
Simplify. A mathematical equation is displayed, reading '100n = 3,600'. The numbers and variable 'n' are in a bold, dark font against a white background.
Divide both sides by 100. A mathematical equation showing 100n divided by 100 equals 3,600 divided by 100. This step demonstrates dividing both sides of an equation by 100 to solve for 'n'.
Simplify. The image displays the equation n = 36 in bold, black mathematical text centered against a plain white background. The variable 'n' is followed by an equals sign and the number '36', indicating a simple numerical assignment.
Check if the answer is reasonable.
Yes. 45 is a little less than half of 100 and 36 is a little less than half 80.
Write a complete sentence that answers the question. 36 is 45% of 80.

Translate and solve using proportions: What number is 65% of 40?

Solution

n40=65100;n=26

Translate and solve using proportions: What number is 85% of 40?

Solution

n40=85100;n=34

In the next example, the percent is more than 100, which is more than one whole. So the unknown number will be more than the base.

Translate and solve using proportions: 125% of 25 is what number?

Solution

Solution

Identify the parts of the percent proportion. A math problem asking '125% is 25 of what number?', with '125%' labeled as 'percent', '25' as 'base', and 'what number?' as 'amount', demonstrating a percentage calculation.
Restate as a proportion. A mathematical word problem asks: 'What number out of 25 is the same as 125 out of 100?'
Set up the proportion. Let n= number. A mathematical equation showing the proportion n/25 = 125/100, where 'n' is an unknown variable. The equation is presented in a clear, digital format on a white background.
Find the cross products and set them equal. A mathematical equation shows '100 multiplied by n equals 25 multiplied by 125' in black text on a white background.
Simplify. The image displays the algebraic equation 100n = 3,125.
Divide both sides by 100. A mathematical equation shows '100n over 100 equals 3,125 over 100,' demonstrating how to solve for 'n' by dividing both sides by 100, which will result in n = 3,125.
Simplify. A mathematical expression 'n = 31.25' is displayed in a serif font on a white background.
Check if the answer is reasonable.
Yes. 125 is more than 100 and 31.25 is more than 25.
Write a complete sentence that answers the question. 125% of 25 is 31.25.

Translate and solve using proportions: 125% of 64 is what number?

Solution

n64=125100;n=80

Translate and solve using proportions: 175% of 84 is what number?

Solution

n84=175100;n=147

Percents with decimals and money are also used in proportions.

Translate and solve: 6.5% of what number is $1.56?

Solution

Solution

Step-by-step guide demonstrating how to solve a percentage problem using proportions, illustrating the procedure and mathematical solution for 6.5% of $24.
Identify the parts of the percent proportion. A percentage problem breaking down '6.5% of what number is $1.56?' into its components: percent, base, and amount.
Restate as a proportion. $1.56 out of what number is the same as 6.5 out of 100?
Set up the proportion. Letn= number. A mathematical equation shows a proportion: 1.56 divided by n is equal to 6.5 divided by 100.
Find the cross products and set them equal. A mathematical equation is displayed, reading 100 multiplied by 1.56 equals 'n' multiplied by 6.5.
Simplify. An image displays the mathematical equation '156 = 6.5n' on a white background. The numbers and symbols are rendered in a clear, dark font.
Divide both sides by 6.5 to isolate the variable. An equation showing 156 divided by 6.5 equals 6.5n divided by 6.5, illustrating a step in solving for 'n'.
Simplify. A mathematical equation displays '24 = n' in a black font against a white background.
Check if the answer is reasonable.
Yes. 6.5% is a small amount and $1.56 is much less than $24.
Write a complete sentence that answers the question. 6.5% of $24 is $1.56.

Translate and solve using proportions: 8.5% of what number is $3.23?

Solution

3.23n=8.5100;n=38

Translate and solve using proportions: 7.25% of what number is $4.64?

Solution

4.64n=7.25100;n=64

Translate and solve using proportions: What percent of 72 is 9?

Solution

Solution

A step-by-step guide demonstrating how to solve for a percentage in a proportion problem.
Identify the parts of the percent proportion. A percentage problem asks, 'What percent of 72 is 9?' It labels 'What percent' as 'percent', '72' as 'base', and '9' as 'amount', with 'of' highlighted in red to show its role.
Restate as a proportion. A mathematical problem asking to determine what number out of 100 is equivalent to 9 out of 72.
Set up the proportion. Let n= number. A mathematical equation shows the fraction 9/72 equal to the fraction n/100, which can be used to solve for the variable 'n' in a proportion or percentage problem.
Find the cross products and set them equal. A mathematical equation is displayed, showing '72 times n equals 100 times 9' in a bold, black font on a white background.
Simplify. A mathematical equation is displayed, showing '72n = 900' against a white background.
Divide both sides by 72. A mathematical equation is displayed, showing '72n over 72 equals 900 over 72' with fractions represented by horizontal lines.
Simplify. The mathematical equation n = 12.5 is displayed, showing the variable 'n' assigned a numerical value of twelve and a half.
Check if the answer is reasonable.
Yes. 9 is 18 of 72 and 18 is 12.5%.
Write a complete sentence that answers the question. 12.5% of 72 is 9.

Translate and solve using proportions: What percent of 72 is 27?

Solution

2772=n100;n=37.5%

Translate and solve using proportions: What percent of 92 is 23?

Solution

2392=n100;n=25%

Key Concepts

  • Proportion
    • A proportion is an equation of the form ab=cd, where b≠0, d≠0.The proportion states two ratios or rates are equal. The proportion is read “a is to b, as c is to d”.
  • Cross Products of a Proportion
    • For any proportion of the form ab=cd, where b≠0, its cross products are equal: a⋅d=b⋅c.
  • Percent Proportion
    • The amount is to the base as the percent is to 100. amountbase=percent100

Section Exercises

Practice Makes Perfect

Use the Definition of Proportion

In the following exercises, write each sentence as a proportion.

4 is to 15 as 36 is to 135.

Solution

415=36135

7 is to 9 as 35 is to 45.

12 is to 5 as 96 is to 40.

Solution

125=9640

15 is to 8 as 75 is to 40.

5 wins in 7 games is the same as 115 wins in 161 games.

Solution

57=115161

4 wins in 9 games is the same as 36 wins in 81 games.

8 campers to 1 counselor is the same as 48 campers to 6 counselors.

Solution

81=486

6 campers to 1 counselor is the same as 48 campers to 8 counselors.

$9.36 for 18 ounces is the same as $2.60 for 5 ounces.

Solution

9.3618=2.605

$3.92 for 8 ounces is the same as $1.47 for 3 ounces.

$18.04 for 11 pounds is the same as $4.92 for 3 pounds.

Solution

18.0411=4.923

$12.42 for 27 pounds is the same as $5.52 for 12 pounds.

In the following exercises, determine whether each equation is a proportion.

715=56120

Solution

yes

512=45108

116=2116

Solution

no

94=3934

1218=4.997.56

Solution

no

916=2.163.89

13.58.5=31.0519.55

Solution

yes

10.18.4=3.032.52

Solve Proportions

In the following exercises, solve each proportion.

x56=78

Solution

x = 49

n91=813

4963=z9

Solution

z = 7

5672=y9

5a=65117

Solution

a = 9

4b=64144

98154=−7p

Solution

p = −11

72156=−6q

a−8=−4248

Solution

a = 7

b−7=−3042

2.63.9=c3

Solution

c = 2

2.73.6=d4

2.7j=0.90.2

Solution

j = 0.6

2.8k=2.11.5

121=m8

Solution

m = 4

133=9n

Solve Applications Using Proportions

In the following exercises, solve the proportion problem.

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

Brianna, who weighs 6 kg, just received her shots and needs a pain killer. The pain killer is prescribed for children at 15 milligrams (mg) for every 1 kilogram (kg) of the child’s weight. How many milligrams will the doctor prescribe?

At the gym, Carol takes her pulse for 10 sec and counts 19 beats. How many beats per minute is this? Has Carol met her target heart rate of 140 beats per minute?

Solution

114 beats/minute. Carol has not met her target heart rate.

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?

A new energy drink advertises 106 calories for 8 ounces. How many calories are in 12 ounces of the drink?

Solution

159 cal

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?

Karen eats 12 cup of oatmeal that counts for 2 points on her weight loss program. Her husband, Joe, can have 3 points of oatmeal for breakfast. How much oatmeal can he have?

Solution

34cup

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?

Janice is traveling to Canada and will change $250 US dollars into Canadian dollars. At the current exchange rate, $1 US is equal to $1.01 Canadian. How many Canadian dollars will she get for her trip?

Solution

$252.50

Todd 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?

Steve changed $600 into 480 Euros. How many Euros did he receive per US dollar?

Solution

0.8 Euros

Martha changed $350 US into 385 Australian dollars. How many Australian dollars did she receive per US dollar?

At the laundromat, Lucy changed $12.00 into quarters. How many quarters did she get?

Solution

48 quarters

When she arrived at a casino, Gerty changed $20 into nickels. How many nickels did she get?

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?

Solution

19 gallons, $58.71

Danny wants to drive to Phoenix to see his grandfather. Phoenix is 370 miles from Danny’s home and his car gets 18.5 miles per gallon. How many gallons of gas will Danny need to get to and from Phoenix? If gas is $3.19 per gallon, what is the total cost for the gas to drive to see his grandfather?

Hugh leaves early one morning to drive from his home in Chicago to go to Mount Rushmore, 812 miles away. After 3 hours, he has gone 190 miles. At that rate, how long will the whole drive take?

Solution

12.8 hours

Kelly leaves her home in Seattle to drive to Spokane, a distance of 280 miles. After 2 hours, she has gone 152 miles. At that rate, how long will the whole drive take?

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

April wants to paint the exterior of her house. One gallon of paint covers about 350 square feet, and the exterior of the house measures approximately 2000 square feet. How many gallons of paint will she have to buy?

Write Percent Equations as Proportions

In the following exercises, translate to a proportion.

What number is 35% of 250?

Solution

n250=35100

What number is 75% of 920?

What number is 110% of 47?

Solution

n47=110100

What number is 150% of 64?

45 is 30% of what number?

Solution

45n=30100

25 is 80% of what number?

90 is 150% of what number?

Solution

90n=150100

77 is 110% of what number?

What percent of 85 is 17?

Solution

1785=p100

What percent of 92 is 46?

What percent of 260 is 340?

Solution

340260=p100

What percent of 180 is 220?

Translate and Solve Percent Proportions

In the following exercises, translate and solve using proportions.

What number is 65% of 180?

Solution

n180=65100; 117

What number is 55% of 300?

18% of 92 is what number?

Solution

n92=18100; 16.56

22% of 74 is what number?

175% of 26 is what number?

Solution

n26=175100; 45.5

250% of 61 is what number?

What is 300% of 488?

Solution

n488=300100; 1464

What is 500% of 315?

17% of what number is $7.65?

Solution

7.65n=17100; 45

19% of what number is $6.46?

$13.53 is 8.25% of what number?

Solution

13.53n=8.25100; 164

$18.12 is 7.55% of what number?

What percent of 56 is 14?

Solution

1456=p100; 25%

What percent of 80 is 28?

What percent of 96 is 12?

Solution

1296=p100; 12.5%

What percent of 120 is 27?

Everyday Math

Mixing a concentrate Sam bought a large bottle of concentrated cleaning solution at the warehouse store. He must mix the concentrate with water to make a solution for washing his windows. The directions tell him to mix 3 ounces of concentrate with 5 ounces of water. If he puts 12 ounces of concentrate in a bucket, how many ounces of water should he add? How many ounces of the solution will he have altogether?

Solution

He must add 20 oz of water to obtain a final solution of 32 oz.

Mixing a concentrate Travis is going to wash his car. The directions on the bottle of car wash concentrate say to mix 2 ounces of concentrate with 15 ounces of water. If Travis puts 6 ounces of concentrate in a bucket, how much water must he mix with the concentrate?

Writing Exercises

To solve “what number is 45% of 350” do you prefer to use an equation like you did in the section on Decimal Operations or a proportion like you did in this section? Explain your reason.

Solution

Answers will vary.

To solve “what percent of 125 is 25” do you prefer to use an equation like you did in the section on Decimal Operations or a proportion like you did in this section? Explain your reason.

Self Check

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

A self-assessment checklist for students to rate their understanding of proportions and percentages, with options: Confidently, With some help, or No-I don't get it!

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Understand Percent

In the following exercises, write each percent as a ratio.

32% admission rate for the university

Solution

32100

53.3% rate of college students with student loans

In the following exercises, write as a ratio and then as a percent.

13 out of 100 architects are women.

Solution

13100,13%

9 out of every 100 nurses are men.

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

48%

Solution

1225

175%

64.1%

Solution

6411000

814%

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

6%

Solution

0.06

23%

128%

Solution

1.28

4.9%

In the following exercises, convert each percent to ⓐ a simplified fraction and ⓑ a decimal.

In 2012,13.5% of the United States population was age 65 or over. (Source: www.census.gov)

Solution
  1. ⓐ 27200
  2. ⓑ 0.135

In 2012,6.5% of the United States population was under 5 years old. (Source: www.census.gov)

When a die is tossed, the probability it will land with an even number of dots on the top side is 50%.

Solution
  1. ⓐ 12
  2. ⓑ 0.5

A couple plans to have three children. The probability they will all be girls is 12.5%.

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

0.04

Solution

4%

0.15

2.82

Solution

282%

3

0.003

Solution

0.3%

1.395

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

34

Solution

75%

115

358

Solution

362.5%

29

According to the Centers for Disease Control, 25 of adults do not take a vitamin or supplement.

Solution

40%

According to the Centers for Disease Control, among adults who do take a vitamin or supplement, 34 take a multivitamin.

In the following exercises, translate and solve.

What number is 46% of 350?

Solution

161

120% of 55 is what number?

84 is 35% of what number?

Solution

240

15 is 8% of what number?

200% of what number is 50?

Solution

25

7.9% of what number is $4.74?

What percent of 120 is 81.6?

Solution

68%

What percent of 340 is 595?

Solve General Applications of Percents

In the following exercises, solve.

When Aurelio and his family ate dinner at a restaurant, the bill was $83.50. Aurelio wants to leave 20% of the total bill as a tip. How much should the tip be?

Solution

$16.70

One granola bar has 2 grams of fiber, which is 8% of the recommended daily amount. What is the total recommended daily amount of fiber?

The nutrition label on a package of granola bars says that each granola bar has 190 calories, and 54 calories are from fat. What percent of the total calories is from fat?

Solution

28.4%

Elsa gets paid $4,600 per month. Her car payment is $253. What percent of her monthly pay goes to her car payment?

In the following exercises, solve.

Jorge got a raise in his hourly pay, from $19.00 to $19.76. Find the percent increase.

Solution

4%

Last year Bernard bought a new car for $30,000. This year the car is worth $24,000. Find the percent decrease.

Solve Sales Tax, Commission, and Discount Applications

In the following exercises, find ⓐ the sales tax ⓑ the total cost.

The cost of a lawn mower was $750. The sales tax rate is 6% of the purchase price.

Solution
  1. ⓐ $45
  2. ⓑ $795

The cost of a water heater is $577. The sales tax rate is 8.75% of the purchase price.

In the following exercises, find the sales tax rate.

Andy bought a piano for $4,600. The sales tax on the purchase was $333.50.

Solution

7.25%

Nahomi bought a purse for $200. The sales tax on the purchase was $16.75.

In the following exercises, find the commission.

Ginny is a realtor. She receives 3% commission when she sells a house. How much commission will she receive for selling a house for $380,000?

Solution

$11,400

Jackson receives 16.5% commission when he sells a dinette set. How much commission will he receive for selling a dinette set for $895?

In the following exercises, find the rate of commission.

Ruben received $675 commission when he sold a $4,500 painting at the art gallery where he works. What was the rate of commission?

Solution

15%

Tori received $80.75 for selling a $950 membership at her gym. What was her rate of commission?

In the following exercises, find the sale price.

Aya bought a pair of shoes that was on sale for $30 off. The original price of the shoes was $75.

Solution

$45

Takwanna saw a cookware set she liked on sale for $145 off. The original price of the cookware was $312.

In the following exercises, find ⓐ the amount of discount and ⓑ the sale price.

Nga bought a microwave for her office. The microwave was discounted 30% from an original price of $84.90.

Solution
  1. ⓐ $25.47
  2. ⓑ $59.43

Jarrett bought a tie that was discounted 65% from an original price of $45.

In the following exercises, find ⓐ the amount of discount ⓑ the discount rate. (Round to the nearest tenth of a percent if needed.)

Hilda bought a bedspread on sale for $37. The original price of the bedspread was $50.

Solution
  1. ⓐ $13
  2. ⓑ 26%

Tyler bought a phone on sale for $49.99. The original price of the phone was $79.99.

In the following exercises, find
  1. ⓐ the amount of the mark-up
  2. ⓑ the list price

Manny paid $0.80 a pound for apples. He added 60% mark-up before selling them at his produce stand. What price did he charge for the apples?

Solution
  1. ⓐ $0.48
  2. ⓑ $1.28

It cost Noelle $17.40 for the materials she used to make a purse. She added a 325% mark-up before selling it at her friend’s store. What price did she ask for the purse?

Solve Simple Interest Applications

In the following exercises, solve the simple interest problem.

Find the simple interest earned after 4 years on $2,250 invested at an interest rate of 5%.

Solution

$450

Find the simple interest earned after 7 years on $12,000 invested at an interest rate of 8.5%.

Find the principal invested if $660 interest was earned in 5 years at an interest rate of 3%.

Solution

$4400

Find the interest rate if $2,898 interest was earned from a principal of $23,000 invested for 3 years.

Kazuo deposited $10,000 in a bank account with interest rate 4.5%. How much interest was earned in 2 years?

Solution

$900

Brent invested $23,000 in a friend’s business. In 5 years the friend paid him the $23,000 plus $9,200 interest. What was the rate of interest?

Fresia lent her son $5,000 for college expenses. Three years later he repaid her the $5,000 plus $375 interest. What was the rate of interest?

Solution

2.5%

In 6 years, a bond that paid 5.5% earned $594 interest. What was the principal of the bond?

Solve Proportions and their Applications

In the following exercises, write each sentence as a proportion.

3 is to 8 as 12 is to 32.

Solution

38=1232

95 miles to 3 gallons is the same as 475 miles to 15 gallons.

1 teacher to 18 students is the same as 23 teachers to 414 students.

Solution

118=23414

$7.35 for 15 ounces is the same as $2.94 for 6 ounces.

In the following exercises, determine whether each equation is a proportion.

513=3078

Solution

yes

167=4823

1218=6.9910.99

Solution

no

11.69.2=37.1229.44

In the following exercises, solve each proportion.

x36=59

Solution

20

7a=−684

1.21.8=d6

Solution

4

122=m20

In the following exercises, solve the proportion problem.

The children’s dosage of acetaminophen is 5 milliliters (ml) for every 25 pounds of a child’s weight. How many milliliters of acetaminophen will be prescribed for a 60 pound child?

Solution

12 ml

After a workout, Dennis takes his pulse for 10 sec and counts 21 beats. How many beats per minute is this?

An 8 ounce serving of ice cream has 272 calories. If Lavonne eats 10 ounces of ice cream, how many calories does she get?

Solution

340 calories

Alma is going to Europe and wants to exchange $1,200 into Euros. If each dollar is 0.75 Euros, how many Euros will Alma get?

Zack wants to drive from Omaha to Denver, a distance of 494 miles. If his car gets 38 miles to the gallon, how many gallons of gas will Zack need to get to Denver?

Solution

13 gallons

Teresa is planning a party for 100 people. Each gallon of punch will serve 18 people. How many gallons of punch will she need?

In the following exercises, translate to a proportion.

What number is 62% of 395?

Solution

n395=62100

42 is 70% of what number?

What percent of 1,000 is 15?

Solution

151000=p100

What percent of 140 is 210?

In the following exercises, translate and solve using proportions.

What number is 85% of 900?

Solution

n900=85100,765

6% of what number is $24?

$3.51 is 4.5% of what number?

Solution

3.51n=4.5100,$78

What percent of 3,100 is 930?

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

24%

Solution

0.24,625

5%

350%

Solution

3.5,72

In the following exercises, convert each fraction to a percent. (Round to 3 decimal places if needed.)

78

13

Solution

33.3¯%or3313%

1112

In the following exercises, solve the percent problem.

65 is what percent of 260?

Solution

25%

What number is 27% of 3,000?

150% of what number is 60?

Solution

40

Yuki’s monthly paycheck is $3,825. She pays $918 for rent. What percent of her paycheck goes to rent?

The total number of vehicles on one freeway dropped from 84,000 to 74,000. Find the percent decrease (round to the nearest tenth of a percent).

Solution

11.9%

Kyle bought a bicycle in Denver where the sales tax was 7.72% of the purchase price. The purchase price of the bicycle was $600. What was the total cost?

Mara received $31.80 commission when she sold a $795 suit. What was her rate of commission?

Solution

4%

Kiyoshi bought a television set on sale for $899. The original price was $1,200. Find:
  1. ⓐ the amount of discount
  2. ⓑ the discount rate (round to the nearest tenth of a percent)

Oxana bought a dresser at a garage sale for $20. She refinished it, then added a 250% markup before advertising it for sale. What price did she ask for the dresser?

Solution

$70

Find the simple interest earned after 5 years on $3000 invested at an interest rate of 4.2%.

Brenda borrowed $400 from her brother. Two years later, she repaid the $400 plus $50 interest. What was the rate of interest?

Solution

6.25%

Write as a proportion: 4 gallons to 144 miles is the same as 10 gallons to 360 miles.

Solve for a: 12a=−1565

Solution

−52

Vin read 10 pages of a book in 12 minutes. At that rate, how long will it take him to read 35 pages?

proportion
A proportion is an equation of the form ab=cd, where b≠0, d≠0.The proportion states two ratios or rates are equal. The proportion is read “a is to b, as c is to d”.

Introduction to the Properties of Real Numbers

A photo of an Eskimo family, mother, father and child, dressed in fur coats.
Quiltmakers know that by rearranging the same basic blocks the resulting quilts can look very different. What happens when we rearrange the numbers in an expression? Does the resulting value change? We will answer these questions in this chapter as we will learn about the properties of numbers. (credit: Hans, Public Domain)

A quilt is formed by sewing many different pieces of fabric together. The pieces can vary in color, size, and shape. The combinations of different kinds of pieces provide for an endless possibility of patterns. Much like the pieces of fabric, mathematicians distinguish among different types of numbers. The kinds of numbers in an expression provide for an endless possibility of outcomes. We have already described counting numbers, whole numbers, and integers. In this chapter, we will learn about other types of numbers and their properties.

Rational and Irrational Numbers

Learning Objectives

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

  • Identify rational numbers and irrational numbers
  • Classify different types of real numbers

Before you get started, take this readiness quiz.

Write 3.19 as an improper fraction.
If you missed this problem, review Example 4 in Decimals.

Solution

319100

Write 511 as a decimal.
If you missed this problem, review Example 3 in Decimals and Fractions.

Solution

0.45¯

Simplify: 144.
If you missed this problem, review Example 1 in Simplify and Use Square Roots.

Solution

12

Identify Rational Numbers and Irrational Numbers

Congratulations! You have completed the first six chapters of this book! It's time to take stock of what you have done so far in this course and think about what is ahead. You have learned how to add, subtract, multiply, and divide whole numbers, fractions, integers, and decimals. You have become familiar with the language and symbols of algebra, and have simplified and evaluated algebraic expressions. You have solved many different types of applications. You have established a good solid foundation that you need so you can be successful in algebra.

In this chapter, we'll make sure your skills are firmly set. We'll take another look at the kinds of numbers we have worked with in all previous chapters. We'll work with properties of numbers that will help you improve your number sense. And we'll practice using them in ways that we'll use when we solve equations and complete other procedures in algebra.

We have already described numbers as counting numbers, whole numbers, and integers. Do you remember what the difference is among these types of numbers?

counting numbers 1,2,3,4…
whole numbers 0,1,2,3,4…
integers …−3,−2,−1,0,1,2,3,4…

Rational Numbers

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

Rational Numbers

A rational number is a number that can be written in the form pq, where p and q are integers and q≠0.

All fractions, both positive and negative, are rational numbers. A few examples are

45,−78,134,and−203

Each numerator and each denominator is an integer.

We need to look at all the numbers we have used so far and verify that they are rational. The definition of rational numbers tells us that all fractions are rational. We will now look at the counting numbers, whole numbers, integers, and decimals to make sure they are rational.

Are integers rational numbers? To decide if an integer is a rational number, we try to write it as a ratio of two integers. An easy way to do this is to write it as a fraction with denominator one.

3=31−8=−810=01

Since any integer can be written as the ratio of two integers, all integers are rational numbers. Remember that all the counting numbers and all the whole numbers are also integers, and so they, too, are rational.

What about decimals? Are they rational? Let's look at a few to see if we can write each of them as the ratio of two integers. We've already seen that integers are rational numbers. The integer −8 could be written as the decimal −8.0. So, clearly, some decimals are rational.

Think about the decimal 7.3. Can we write it as a ratio of two integers? Because 7.3 means 7310, we can write it as an improper fraction, 7310. So 7.3 is the ratio of the integers 73 and 10. It is a rational number.

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.

Write each as the ratio of two integers: ⓐ −15ⓑ 6.81ⓒ −367.

Solution
Solution
This table demonstrates how to write an integer as a fraction with a denominator of 1.
ⓐ
−15
Write the integer as a fraction with denominator 1. −151
This table demonstrates the conversion of a decimal number to a mixed number and then to an improper fraction.
ⓑ
6.81
Write the decimal as a mixed number. 681100
Then convert it to an improper fraction. 681100
This table demonstrates the conversion of the mixed number -3 6/7 to its equivalent improper fraction, -27/7, detailing the instruction and the resulting mathematical value.
ⓒ
−367
Convert the mixed number to an improper fraction. −277

Write each as the ratio of two integers: ⓐ −24ⓑ 3.57.

Solution
  1. ⓐ −241
  2. ⓑ 357100

Write each as the ratio of two integers: ⓐ −19 ⓑ 8.41.

Solution
  1. ⓐ −191
  2. ⓑ 841100

Let's look at the decimal form of the numbers we know are rational. We have seen that every integer is a rational number, since a=a1 for any integer, a. We can also change any integer to a decimal by adding a decimal point and a zero.

Integer−2,−1,0,1,2,3Decimal−2.0,−1.0,0.0,1.0,2.0,3.0These decimal numbers stop.

We have also seen that every fraction is a rational number. Look at the decimal form of the fractions we just considered.

Ratio of Integers 45, −78, 134, −203 Decimal Forms 0.8, −0.875, 3.25, −6.666… These decimals either stop or repeat. −6.66—

What do these examples tell you? Every rational number can be written both as a ratio of integers and as a decimal that either stops or repeats. The table below shows the numbers we looked at expressed as a ratio of integers and as a decimal.

Rational Numbers
Fractions Integers
Number 45,−78,134,−203 −2,−1,0,1,2,3
Ratio of Integer 45,−78,134,−203 −21,−11,01,11,21,31
Decimal number 0.8,−0.875,3.25,−6.6–, −2.0,−1.0,0.0,1.0,2.0,3.0

Irrational Numbers

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.

π=3.141592654.......

Similarly, the decimal representations of square roots of whole numbers that are not perfect squares never stop and never repeat. For example,

5=2.236067978.....

A decimal that does not stop and does not repeat cannot be written as the ratio of integers. We call this kind of number 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.

If the decimal form of a number

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

Identify each of the following as rational or irrational:

  1. ⓐ 0.583–
  2. ⓑ 0.475
  3. ⓒ 3.605551275…
Solution
Solution

ⓐ 0.583–
The bar above the 3 indicates that it repeats. Therefore, 0.583– is a repeating decimal, and is therefore a rational number.

ⓑ 0.475
This decimal stops after the 5, so it is a rational number.

ⓒ 3.605551275…
The ellipsis (…) means that this number does not stop. There is no repeating pattern of digits. Since the number doesn't stop and doesn't repeat, it is irrational.

Identify each of the following as rational or irrational:

ⓐ 0.29ⓑ 0.816–ⓒ 2.515115111…

Solution
  1. ⓐ rational
  2. ⓑ rational
  3. ⓒ irrational

Identify each of the following as rational or irrational:

ⓐ 0.23–ⓑ 0.125ⓒ 0.418302…

Solution
  1. ⓐ rational
  2. ⓑ rational
  3. ⓒ irrational

Let's think about square roots now. Square roots of perfect squares are always whole numbers, so they are rational. But the decimal forms of square roots of numbers that are not perfect squares never stop and never repeat, so these square roots are irrational.

Identify each of the following as rational or irrational:

  1. ⓐ 36
  2. ⓑ 44

Solution
Solution

ⓐ The number 36 is a perfect square, since 62=36. So 36=6. Therefore 36 is rational.

ⓑ Remember that 62=36 and 72=49, so 44 is not a perfect square.

This means 44 is irrational.

Identify each of the following as rational or irrational:

  1. ⓐ 81
  2. ⓑ 17
Solution
  1. ⓐ rational
  2. ⓑ irrational

Identify each of the following as rational or irrational:

  1. ⓐ 116

  2. ⓑ 121

Solution
  1. ⓐ irrational
  2. ⓑ rational

Classify Real Numbers

We have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. Irrational numbers are a separate category of their own. When we put together the rational numbers and the irrational numbers, we get the set of real numbers.

Figure 1 illustrates how the number sets are related.

The image shows a large rectangle labeled “Real Numbers”. The rectangle is split in half vertically. The right half is labeled “Irrational Numbers”. The left half is labeled “Rational Numbers” and contains three concentric rectangles. The outer most rectangle is labeled “Integers”, the next rectangle is “Whole Numbers” and the inner most rectangle is “Natural Numbers”.
This diagram illustrates the relationships between the different types of real numbers.

Real Numbers

Real numbers are numbers that are either rational or irrational.

Does the term “real numbers” seem strange to you? Are there any numbers that are not “real”, and, if so, what could they be? For centuries, the only numbers people knew about were what we now call the real numbers. Then mathematicians discovered the set of imaginary numbers. You won't encounter imaginary numbers in this course, but you will later on in your studies of algebra.

Determine whether each of the numbers in the following list is a ⓐ whole number, ⓑ integer, ⓒ rational number, ⓓ irrational number, and ⓔ real number.

−7,145,8,5,5.9,−64
Solution

Solution

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

ⓑ The integers are the whole numbers, their opposites, and 0. From the given numbers, −7 and 8 are integers. Also, notice that 64 is the square of 8 so −64=−8. So the integers are −7,8,−64.

ⓒ Since all integers are rational, the numbers −7,8,and−64 are also rational. Rational numbers also include fractions and decimals that terminate or repeat, so 145and5.9 are rational.

ⓓ The number 5 is not a perfect square, so 5 is irrational.

ⓔ All of the numbers listed are real.

We'll summarize the results in a table.

Number Whole Integer Rational Irrational Real
−7 ✓ ✓ ✓
145 ✓ ✓
8 ✓ ✓ ✓ ✓
5 ✓ ✓
5.9 ✓ ✓
−64 ✓ ✓ ✓

Determine whether each number is a ⓐ whole number, ⓑ integer, ⓒ rational number, ⓓ irrational number, and ⓔ real number: −3,−2,0.3–,95,4,49.

Solution


The table has seven rows and six columns. The first row is a header row that labels each column. The first column is labeled “Number”, the second column “Whole”, the third “Integer”, the fourth “Rational” the fifth “Irrational” and the sixth “Real”. Each row has a number in the “Number” column then an x in each column that corresponds to the type of number it is. The second row has the number negative 3 in the “Number” column and an x marked in the “Integer”, “Rational” and “Real” columns. The third row has the number negative square root of 2 in the “Number” column and an x marked in the “Irrational” and “Real” columns. The fourth row has the number 0.3 repeating in the “Number” column and an x marked in the “Rational” and “Real” columns. The fifth row has the number  square root of negative 49 in the “Number” column with no other columns marked. The sixth row has the number 4 in the “Number” column and an x marked in the “Whole”, “Integer”, “Rational” and “Real” columns.  The last row has the number 9 fifths in the “Number” column and an x marked in the “Rational” and “Real” columns.

Determine whether each number is a ⓐ whole number, ⓑ integer, ⓒ rational number, ⓓ irrational number, and ⓔ real number: −25,−38,−1,6,121,2.041975…

Solution


The table has seven rows and six columns. The first row is a header row that labels each column. The first column is labeled “Number”, the second column “Whole”, the third “Integer”, the fourth “Rational” the fifth “Irrational” and the sixth “Real”. Each row has a number in the “Number” column then an x in each column that corresponds to the type of number it is. The second row has the number negative square root of 25 in the “Number” column and an x marked in the “Integer”, “Rational” and “Real” columns. The third row has the number negative 3 eights in the “Number” column and an x marked in the “Rational” and “Real” columns. The fourth row has the number negative 1 in the “Number” column and an x marked in the “Integer”, “Rational” and “Real” columns. The fifth row has the number 6 in the “Number” column and an x marked in the “Whole”, “Integer”, “Rational” and “Real” columns. The sixth row has the number square root of 121 in the “Number” column and an x marked in the “Whole”, “Integer”, “Rational” and “Real” columns. The last row has the number 2.041975 followed by an ellipsis in the “Number” column and an x marked in the “Irrational” and “Real” columns.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Sets of Real Numbers
  • Real Numbers

Key Concepts

  • Real numbers
    • The image shows a large rectangle labeled “Real Numbers”. The rectangle is split in half vertically. The right half is labeled “Irrational Numbers”. The left half is labeled “Rational Numbers” and contains three concentric rectangles. The outer most rectangle is labeled “Integers”, the next rectangle is “Whole Numbers” and the inner most rectangle is “Natural Numbers”.

Practice Makes Perfect

Rational Numbers

In the following exercises, write as the ratio of two integers.

  1. ⓐ 5
  2. ⓑ 3.19
Solution
  1. ⓐ 51
  2. ⓑ 319100
  1. ⓐ 8
  2. ⓑ −1.61
  1. ⓐ −12
  2. ⓑ 9.279
Solution
  1. ⓐ −121
  2. ⓑ 92791000
  1. ⓐ −16
  2. ⓑ 4.399

In the following exercises, determine which of the given numbers are rational and which are irrational.

0.75, 0.223–, 1.39174…

Solution

Rational: 0.75,0.223–. Irrational: 1.39174…

0.36, 0.94729…, 2.528–

0.45—, 1.919293…, 3.59

Solution

Rational: 0.45—, 3.59. Irrational: 1.919293…

0.13–,0.42982…, 1.875

In the following exercises, identify whether each number is rational or irrational.

  1. ⓐ 25
  2. ⓑ 30
Solution
  1. ⓐ rational
  2. ⓑ irrational
  1. ⓐ 44
  2. ⓑ 49
  1. ⓐ 164
  2. ⓑ 169
Solution
  1. ⓐ irrational
  2. ⓑ rational
  1. ⓐ 225
  2. ⓑ 216

Classifying Real Numbers

In the following exercises, determine whether each number is whole, integer, rational, irrational, and real.

−8, 0,1.95286...., 125, 36, 9

Solution


The table has seven rows and six columns. The first row is a header row that labels each column. The first column is labeled “Number”, the second column “Whole”, the third “Integer”, the fourth “Rational” the fifth “Irrational” and the sixth “Real”. Each row has a number in the “Number” column then an x in each column that corresponds to the type of number it is. The second row has the number negative 8 in the “Number” column and an x marked in the “Integer”, “Rational” and “Real” columns. The third row has the number 0 in the “Number” column and an x marked in the “Whole”, “Integer”, “Rational” and “Real” columns. The fourth row has the number 1.95286 followed by and ellipsis in the “Number” column and an x marked in the “Irrational” and “Real” columns. The fifth row has the number 12 fifths in the “Number” column and an x marked in the “Rational” and “Real” columns. The sixth row has the number square root of 36 in the “Number” column and an x marked in the “Whole”, “Integer”, “Rational” and “Real” columns. The last row has the number 9 in the “Number” column and an x marked in the “Whole”, “Integer”, “Rational” and “Real” columns.

−9, −349, −9, 0.409—,116, 7

−100, −7, −83, −1, 0.77, 314

Solution


The table has seven rows and six columns. The first row is a header row that labels each column. The first column is labeled “Number”, the second column “Whole”, the third “Integer”, the fourth “Rational” the fifth “Irrational” and the sixth “Real”. Each row has a number in the “Number” column then an x in each column that corresponds to the type of number it is. The second row has the number negative 100 in the “Number” column and an x marked in the “Integer”, “Rational” and “Real” columns. The third row has the number negative 7 in the “Number” column and an x marked in the “Integer”, “Rational” and “Real” columns. The fourth row has the number negative 8 thirds in the “Number” column and an x marked in the “Rational” and “Real” columns. The fifth row has the number negative 1 in the “Number” column and an x marked in the “Integer”, “Rational” and “Real” columns. The sixth row has the number 0.77 in the “Number” column and an x marked in the “Rational” and “Real” columns. The last row has the number 3 and 1 quarter in the “Number” column and an x marked in the “Rational” and “Real” columns.

Everyday Math

Field trip All the 5th graders at Lincoln Elementary School will go on a field trip to the science museum. Counting all the children, teachers, and chaperones, there will be 147 people. Each bus holds 44 people.

ⓐ How many buses will be needed?

ⓑ Why must the answer be a whole number?

ⓒ Why shouldn't you round the answer the usual way?

Child care Serena wants to open a licensed child care center. Her state requires that there be no more than 12 children for each teacher. She would like her child care center to serve 40 children.

  1. ⓐ How many teachers will be needed?

  2. ⓑ Why must the answer be a whole number?

  3. ⓒ Why shouldn't you round the answer the usual way?

Solution
  1. ⓐ 4
  2. ⓑ Teachers cannot be divided
  3. ⓒ It would result in a lower number.

Writing Exercises

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

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

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 table on number classification, asking students to rate their confidence in identifying rational/irrational numbers and classifying real numbers as '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.

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.
Rational number
A rational number is a number that can be written in 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.

Commutative and Associative Properties

Learning Objectives

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

  • Use the commutative and associative properties
  • Evaluate expressions using the commutative and associative properties
  • Simplify expressions using the commutative and associative properties

Before you get started, take this readiness quiz.

Simplify: 7y+2+y+13.
If you missed this problem, review Example 10 in Evaluate, Simplify, and Translate Expressions.

Solution

8y+15

Multiply: 23·18.
If you missed this problem, review Example 10 in Multiply and Divide Fractions.

Solution

12

Find the opposite of 15.
If you missed this problem, review Example 3 in Introduction to Integers.

Solution

−15

In the next few sections, we will take a look at the properties of real numbers. Many of these properties will describe things you already know, but it will help to give names to the properties and define them formally. This way we’ll be able to refer to them and use them as we solve equations in the next chapter.

Use the Commutative and Associative Properties

Think about adding two numbers, such as 5 and 3.

5+33+588

The results are the same. 5+3=3+5

Notice, the order in which we add does not matter. The same is true when multiplying 5 and 3.

5·33·51515

Again, the results are the same! 5·3=3·5. The order in which we multiply does not matter.

These examples illustrate the commutative properties of addition and multiplication.

Commutative Properties

Commutative Property of Addition: if a and b are real numbers, then

a+b=b+a

Commutative Property of Multiplication: if a and b are real numbers, then

a·b=b·a

The commutative properties have to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.

Use the commutative properties to rewrite the following expressions:

  1. ⓐ −1+3=_____

  2. ⓑ 4·9=_____

Solution

Solution

This table demonstrates the commutative property of addition using a numerical example.
ⓐ
−1+3=_____
Use the commutative property of addition to change the order. −1+3=3+(−1)
This table demonstrates the commutative property of multiplication with an example, showing how the order of factors can be changed.
ⓑ
4·9=_____
Use the commutative property of multiplication to change the order. 4·9=9·4
Use the commutative properties to rewrite the following:
  1. ⓐ −4+7=_____
  2. ⓑ 6·12=_____
Solution
  1. ⓐ −4 + 7 = 7 + (−4)
  2. ⓑ 6 · 12 = 12 · 6
Use the commutative properties to rewrite the following:
  1. ⓐ 14+(−2)=_____
  2. ⓑ 3(−5)=_____
Solution
  1. ⓐ 14 + (−2) = −2 + 14
  2. ⓑ 3(−5) = (−5)3

What about subtraction? Does order matter when we subtract numbers? Does 7−3 give the same result as 3−7?

7−33−74−44≠−4
The results are not the same.7−3≠3−7

Since changing the order of the subtraction did not give the same result, we can say that subtraction is not commutative.

Let’s see what happens when we divide two numbers. Is division commutative?

12÷44÷121244123133≠13
The results are not the same. So12÷4≠4÷12

Since changing the order of the division did not give the same result, division is not commutative.

Addition and multiplication are commutative. Subtraction and division are not commutative.

Suppose you were asked to simplify this expression.

7+8+2

How would you do it and what would your answer be?

Some people would think 7+8is15 and then 15+2is17. Others might start with 8+2makes10 and then 7+10makes17.

Both ways give the same result, as shown in Figure 1. (Remember that parentheses are grouping symbols that indicate which operations should be done first.)

The image shows an equation. The left side of the equation shows the quantity 7 plus 8 in parentheses plus 2. The right side of the equation show 7 plus the quantity 8 plus 2. Each side of the equation is boxed separately in red. Each box has an arrow pointing from the box to the number 17 below.

When adding three numbers, changing the grouping of the numbers does not change the result. This is known as the Associative Property of Addition.

The same principle holds true for multiplication as well. Suppose we want to find the value of the following expression:

5·13·3

Changing the grouping of the numbers gives the same result, as shown in Figure 2.

The image shows an equation. The left side of the equation shows the quantity 5 times 1 third in parentheses times 3. The right side of the equation show 5 times the quantity 1 third times 3. Each side of the equation is boxed separately in red. Each box has an arrow pointing from the box to the number 5 below.

When multiplying three numbers, changing the grouping of the numbers does not change the result. This is known as the Associative Property of Multiplication.

If we multiply three numbers, changing the grouping does not affect the product.

You probably know this, but the terminology may be new to you. These examples illustrate the Associative Properties.

Associative Properties

Associative Property of Addition: if a,b, and c are real numbers, then

(a+b)+c=a+(b+c)

Associative Property of Multiplication: if a,b, and c are real numbers, then

(a·b)·c=a·(b·c)

Use the associative properties to rewrite the following:

  1. ⓐ (3+0.6)+0.4=__________

  2. ⓑ (−4·25)·15=__________

Solution

Solution

Illustrates the associative property of addition with an example, showing how changing number grouping doesn't alter the sum.
ⓐ
(3+0.6)+0.4=__________
Change the grouping. (3+0.6)+0.4=3+(0.6+0.4)

Notice that 0.6+0.4 is 1, so the addition will be easier if we group as shown on the right.

Illustrates applying an instruction to change the grouping in a mathematical expression, demonstrating problem-solving steps.
ⓑ
(−4·25)·15=__________
Change the grouping. (−4·25)·15=−4·(25·15)

Notice that 25·15 is 6. The multiplication will be easier if we group as shown on the right.

Use the associative properties to rewrite the following:
ⓐ (1+0.7)+0.3=__________ ⓑ (−9·8)·34=__________

Solution
  1. ⓐ (1+0.7)+0.3=1+(0.7+0.3)
  2. ⓑ (−9·8)·34=−9(8·34)

Use the associative properties to rewrite the following:
ⓐ (4+0.6)+0.4=__________ ⓑ (−2·12)·56=__________

Solution
  1. ⓐ (4+0.6)+0.4=4+(0.6+0.4)
  2. ⓑ (−2·12)·56=−2(12·56)

Besides using the associative properties to make calculations easier, we will often use it to simplify expressions with variables.

Use the Associative Property of Multiplication to simplify: 6(3x).

Solution

Solution

Steps demonstrating the simplification of the expression 6(3x) to 18x using the associative property of multiplication.
6(3x)
Change the grouping. (6·3)x
Multiply in the parentheses. 18x

Notice that we can multiply 6·3, but we could not multiply 3·x without having a value for x.

Use the Associative Property of Multiplication to simplify the given expression: 8(4x).

Solution

8(4x) = (8 · 4)x = 32x

Use the Associative Property of Multiplication to simplify the given expression: −9(7y).

Solution

−9(7y) = (−9 · 7)y = −63y

Evaluate Expressions using the Commutative and Associative Properties

The commutative and associative properties can make it easier to evaluate some algebraic expressions. Since order does not matter when adding or multiplying three or more terms, we can rearrange and re-group terms to make our work easier, as the next several examples illustrate.

Evaluate each expression when x=78.

  1. ⓐ x+0.37+(−x)
  2. ⓑ x+(−x)+0.37
Solution

Solution

ⓐ
The expression x + 0.37 + (-x) illustrates the concept of additive inverses, where x and -x cancel out, leaving 0.37.
Substitute 78 for x. The equation 7/8 + 0.37 + (-7/8) is displayed, where 7/8 and -7/8 are additive inverses, simplifying the sum to 0.37.
Convert fractions to decimals. The image shows the mathematical expression: 0.875 + 0.37 + (-0.875).
Add left to right. The image displays a subtraction problem with decimal numbers: 1.245 - 0.875, presented in a clean, straightforward manner against a white background.
Subtract. The number 0.37 is displayed in black text against a plain white background.
ⓑ
A mathematical expression displays x plus the quantity negative x, plus 0.37, illustrating the additive inverse property where x and -x sum to zero, leaving only 0.37.
Substitute 78 for x. A mathematical expression showing the sum of 7/8, its additive inverse -7/8, and the decimal 0.37. The fractions are displayed in red text, while the operations and the decimal are in black.
Add opposites first. The number 0.37 is displayed in black text on a white background.

What was the difference between part ⓐ and part ⓑ ? Only the order changed. By the Commutative Property of Addition, x+0.37+(−x)=x+(−x)+0.37. But wasn’t part ⓑ much easier?

Evaluate each expression when y=38:ⓐ y+0.84+(−y) ⓑ y+(−y)+0.84.

Solution
  1. ⓐ 0.84
  2. ⓑ 0.84

Evaluate each expression when f=1720:ⓐ f+0.975+(−f) ⓑ f+(−f)+0.975.

Solution
  1. ⓐ 0.975
  2. ⓑ 0.975

Let’s do one more, this time with multiplication.

Evaluate each expression when n=17.

  1. ⓐ 43(34n)

  2. ⓑ (43·34)n

Solution

Solution

ⓐ
A mathematical expression shows the fraction 4/3 multiplied by the quantity (3/4 multiplied by n), enclosed in parentheses.
Substitute 17 for n. A mathematical expression showing the fraction 4/3 multiplied by a parenthesized term (3/4 multiplied by the number 17, which is highlighted in red).
Multiply in the parentheses first. A mathematical expression showing the fraction 4/3 multiplied by the fraction 51/4, enclosed in parentheses, against a white background.
Multiply again. The number 17 is displayed in black sans-serif font against a plain white background.
ⓑ
The image shows a mathematical expression: an open parenthesis, the fraction 4/3, a multiplication dot, the fraction 3/4, a close parenthesis, and the variable 'n'. The expression is (4/3 * 3/4)n.
Substitute 17 for n. A mathematical expression showing the product of two fractions, (4/3) and (3/4), inside parentheses, which is then multiplied by 17, with 17 highlighted in red.
Multiply. The product of reciprocals is 1. (1) * 17 is displayed on a white background.
Multiply again. The number '17' is displayed.

What was the difference between part ⓐ and part ⓑ here? Only the grouping changed. By the Associative Property of Multiplication, 43(34n)=(43·34)n. By carefully choosing how to group the factors, we can make the work easier.

Evaluate each expression when p=24:ⓐ 59(95p) ⓑ (59·95)p.

Solution
  1. ⓐ 24
  2. ⓑ 24

Evaluate each expression when q=15:ⓐ 711(117q) ⓑ (711·117)q

Solution
  1. ⓐ 15
  2. ⓑ 15

Simplify Expressions Using the Commutative and Associative Properties

When we have to simplify algebraic expressions, we can often make the work easier by applying the Commutative or Associative Property first instead of automatically following the order of operations. Notice that in Example 4 part ⓑ was easier to simplify than part ⓐ because the opposites were next to each other and their sum is 0. Likewise, part ⓑ in Example 5 was easier, with the reciprocals grouped together, because their product is 1. In the next few examples, we’ll use our number sense to look for ways to apply these properties to make our work easier.

Simplify: −84n+(−73n)+84n.

Solution

Solution

Notice the first and third terms are opposites, so we can use the commutative property of addition to reorder the terms.

Steps to simplify the expression -84n + (-73n) + 84n, demonstrating re-ordering and addition of terms.
−84n+(−73n)+84n
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

Solution

Notice the first and third terms are reciprocals, so we can use the Commutative Property of Multiplication to reorder the factors.

Step-by-step simplification of a fractional multiplication problem, demonstrating how to reorder terms to simplify the expression.
715·823·157
Re-order the terms. 715·157·823
Multiply left to right. 1·823
Multiply. 823

Simplify: 916·549·169.

Solution

549

Simplify: 617·1125·176.

Solution

1125

In expressions where we need to add or subtract three or more fractions, combine those with a common denominator first.

Simplify: (513+34)+14.

Solution

Solution

Notice that the second and third terms have a common denominator, so this work will be easier if we change the grouping.

Step-by-step simplification of a fractional mathematical expression.
(513+34)+14
Group the terms with a common denominator. 513+(34+14)
Add in the parentheses first. 513+(44)
Simplify the fraction. 513+1
Add. 1513
Convert to an improper fraction. 1813

Simplify: (715+58)+38.

Solution

2215

Simplify: (29+712)+512.

Solution

119

When adding and subtracting three or more terms involving decimals, look for terms that combine to give whole numbers.

Simplify: (6.47q+9.99q)+1.01q.

Solution

Solution

Notice that the sum of the second and third coefficients is a whole number.

This table demonstrates the step-by-step simplification of an algebraic expression, illustrating the associative property of addition.
(6.47q+9.99q)+1.01q
Change the grouping. 6.47q+(9.99q+1.01q)
Add in the parentheses first. 6.47q+(11.00q)
Add. 17.47q

Many people have good number sense when they deal with money. Think about adding 99 cents and 1 cent. Do you see how this applies to adding 9.99+1.01?

Simplify: (5.58c+8.75c)+1.25c.

Solution

15.58c

Simplify: (8.79d+3.55d)+5.45d.

Solution

17.79d

No matter what you are doing, it is always a good idea to think ahead. When simplifying an expression, think about what your steps will be. The next example will show you how using the Associative Property of Multiplication can make your work easier if you plan ahead.

Simplify the expression: [ 1.67(8) ] (0.25).

Solution

Solution

Notice that multiplying (8)(0.25) is easier than multiplying 1.67(8) because it gives a whole number. (Think about having 8 quarters—that makes $2.)

Step-by-step simplification of a mathematical expression, demonstrating regrouping and multiplication.
[1.67(8)](0.25)
Regroup. 1.67[(8)(0.25)]
Multiply in the brackets first. 1.67[2]
Multiply. 3.34

Simplify: [1.17(4)](2.25).

Solution

10.53

Simplify: [3.52(8)](2.5).

Solution

70.4

When simplifying expressions that contain variables, we can use the commutative and associative properties to re-order or regroup terms, as shown in the next pair of examples.

Simplify: 6(9x).

Solution

Solution

Demonstrates simplifying 6(9x) to 54x using the associative property of multiplication, showing each step.
6(9x)
Use the associative property of multiplication to re-group. (6·9)x
Multiply in the parentheses. 54x

Simplify: 8(3y).

Solution

24y

Simplify: 12(5z).

Solution

60z

In The Language of Algebra, we learned to combine like terms by rearranging an expression so the like terms were together. We simplified the expression 3x+7+4x+5 by rewriting it as 3x+4x+7+5 and then simplified it to 7x+12. We were using the Commutative Property of Addition.

Simplify: 18p+6q+(−15p)+5q.

Solution

Solution

Use the Commutative Property of Addition to re-order so that like terms are together.

This table demonstrates the step-by-step simplification of an algebraic expression by re-ordering and combining like terms to reach a final simplified form.
18p+6q+(−15p)+5q
Re-order terms. 18p+(−15p)+6q+5q
Combine like terms. 3p+11q

Simplify: 23r+14s+9r+(−15s).

Solution

32r − s

Simplify: 37m+21n+4m+(−15n).

Solution

41m + 6n

The Links to Literacy activity, "Each Orange Had 8 Slices" will provide you with another view of the topics covered in this section.

Key Concepts

  • Commutative Properties
    • Commutative Property of Addition:
      • If a,b are real numbers, then a+b=b+a
    • Commutative Property of Multiplication:
      • If a,b are real numbers, then a⋅b=b⋅a
  • Associative Properties
    • Associative Property of Addition:
      • If a,b,c are real numbers then (a+b)+c=a+(b+c)
    • Associative Property of Multiplication:
      • If a,b,c are real numbers then (a⋅b)⋅c=a⋅(b⋅c)

Practice Makes Perfect

Use the Commutative and Associative Properties

In the following exercises, use the commutative properties to rewrite the given expression.

8+9=___

7+6=___

Solution

7 + 6 = 6 + 7

8(−12)=___

7(−13)=___

Solution

7(−13) = (−13)7

(−19)(−14)=___

(−12)(−18)=___

Solution

(−12)(−18) = (−18)(−12)

−11+8=___

−15+7=___

Solution

−15 + 7 = 7 + (−15)

x+4=___

y+1=___

Solution

y + 1 = 1 + y

−2a=___

−3m=___

Solution

−3m = m(−3)

In the following exercises, use the associative properties to rewrite the given expression.

(11+9)+14=___

(21+14)+9=___

Solution

(21 + 14) + 9 = 21 + (14 + 9)

(12·5)·7=___

(14·6)·9=___

Solution

(14 · 6) · 9 = 14(6 · 9)

(−7+9)+8=___

(−2+6)+7=___

Solution

(−2 + 6) + 7 = −2 + (6 + 7)

(16·45)·15=___

(13·23)·18=___

Solution

(13·23)·18=13(23·18)

3(4x)=___

4(7x)=___

Solution

4(7x) = (4 · 7)x

(12+x)+28=___

(17+y)+33=___

Solution

(17 + y) + 33 = 17 + (y + 33)

Evaluate Expressions using the Commutative and Associative Properties

In the following exercises, evaluate each expression for the given value.

If y=58, evaluate:
  1. ⓐ y+0.49+(−y)
  2. ⓑ y+(−y)+0.49
If z=78, evaluate:
  • ⓐ z+0.97+(−z)
  • ⓑ z+(−z)+0.97
Solution
  1. ⓐ 0.97
  2. ⓑ 0.97
If c=−114, evaluate:
  1. ⓐ c+3.125+(−c)
  2. ⓑ c+(−c)+3.125
If d=−94, evaluate:
  1. ⓐ d+2.375+(−d)
  2. ⓑ d+(−d)+2.375
Solution
  1. ⓐ 2.375
  2. ⓑ 2.375
If j=11, evaluate:
  1. ⓐ 56(65j)
  2. ⓑ (56·65)j
If k=21, evaluate:
  1. ⓐ 413(134k)
  2. ⓑ (413·134)k
Solution
  1. ⓐ 21
  2. ⓑ 21
If m=−25, evaluate:
  1. ⓐ −37(73m)
  2. ⓑ (−37·73)m
If n=−8, evaluate:
  1. ⓐ −521(215n)
  2. ⓑ (−521·215)n
Solution
  1. ⓐ 8
  2. ⓑ 8

Simplify Expressions Using the Commutative and Associative Properties

In the following exercises, simplify.

−45a+15+45a

9y+23+(−9y)

Solution

23

12+78+(−12)

25+512+(−25)

Solution

512

320·4911·203

1318·257·1813


Solution

257

712·917·247

310·1323·503

Solution

6523

−24·7·38

−36·11·49

Solution

−176

(56+815)+715

(112+49)+59

Solution

1312

513+34+14

815+57+27

Solution

2315

(4.33p+1.09p)+3.91p

(5.89d+2.75d)+1.25d

Solution

9.89d

17(0.25)(4)

36(0.2)(5)

Solution

36

[2.48(12)](0.5)

[9.731(4)](0.75)

Solution

29.193

7(4a)

9(8w)

Solution

72w

−15(5m)

−23(2n)

Solution

−46n

12(56p)

20(35q)

Solution

12q

14x+19y+25x+3y

15u+11v+27u+19v

Solution

42u + 30v

43m+(−12n)+​(−16m)+(−9n)

−22p+17q+(−35p)+(−27q)

Solution

−57p + (−10q)

38g+112h+78g+512h

56a+310b+16a+910b

Solution

a+65b

6.8p+9.14q+(−4.37p)+(−0.88q)

9.6m+7.22n+(−2.19m)+(−0.65n)

Solution

7.41m + 6.57n

Everyday Math

Stamps Allie and Loren need to buy stamps. Allie needs four $0.49 stamps and nine $0.02 stamps. Loren needs eight $0.49 stamps and three $0.02 stamps.

  1. ⓐ How much will Allie’s stamps cost?

  2. ⓑ How much will Loren’s stamps cost?

  3. ⓒ What is the total cost of the girls’ stamps?

  4. ⓓ How many $0.49 stamps do the girls need altogether? How much will they cost?

  5. ⓔ How many $0.02 stamps do the girls need altogether? How much will they cost?

Counting Cash Grant is totaling up the cash from a fundraising dinner. In one envelope, he has twenty-three $5 bills, eighteen $10 bills, and thirty-four $20 bills. In another envelope, he has fourteen $5 bills, nine $10 bills, and twenty-seven $20 bills.

  1. ⓐ How much money is in the first envelope?

  2. ⓑ How much money is in the second envelope?

  3. ⓒ What is the total value of all the cash?

  4. ⓓ What is the value of all the $5 bills?

  5. ⓔ What is the value of all $10 bills?

  6. ⓕ What is the value of all $20 bills?

Solution
  1. ⓐ $975
  2. ⓑ $700
  3. ⓒ $1675
  4. ⓓ $185
  5. ⓔ $270
  6. ⓕ $1220

Writing Exercises

In your own words, state the Commutative Property of Addition and explain why it is useful.

In your own words, state the Associative Property of Multiplication and explain why it is useful.

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 understanding of Commutative and Associative Properties, with options: 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?

Distributive Property

Learning Objectives

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

  • Simplify expressions using the distributive property
  • Evaluate expressions using the distributive property

Before you get started, take this readiness quiz.

Multiply: 3(0.25).
If you missed this problem, review Example 5 in Decimal Operations

Solution

0.75

Simplify: 10−(−2)(3).
If you missed this problem, review Example 5 in Multiply and Divide Integers

Solution

16

Combine like terms: 9y+17+3y−2.
If you missed this problem, review Example 10 in Evaluate, Simplify, and Translate Expressions.

Solution

12y+15

Simplify Expressions Using the Distributive Property

Suppose three friends are going to the movies. They each need $9.25; that is, 9 dollars and 1 quarter. How much money do they need all together? You can think about the dollars separately from the quarters.

The image shows the equation 3 times 9 equal to 27. Below the 3 is an image of three people. Below the 9 is an image of 9 one dollar bills. Below the 27 is an image of three groups of 9 one dollar bills for a total of 27 one dollar bills. The image shows the equation 3 times 25 cents equal to 75 cents. Below the 3 is an image of three people. Below the 25 cents is an image of a quarter. Below the 75 cents is an image of three 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

If a,b,c are real numbers, then

a(b+c)=ab+ac

Back to our friends at the movies, we could show the math steps we take to find the total amount of money they need like this:

3(9.25) 3(9+0.25) 3(9)+3(0.25) 27+0.75 27.75

In algebra, we use the Distributive Property to remove parentheses as we simplify expressions. For example, if we are asked to simplify the expression 3(x+4), the order of operations says to work in the parentheses first. But we cannot add x and 4, since they are not like terms. So we use the Distributive Property, as shown in Example 1.

Simplify: 3(x+4).

Solution

Solution

Demonstrates simplifying the 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 1 would look like this:

The image shows the expression x plus 4 in parentheses with the number 3 outside the parentheses on the left. There are two arrows pointing from the top of the three. One arrow points to the top of the x. The other arrow points to the top of the 4. The image shows and equation. On the left side of the equation is the expression x plus 4 in parentheses with the number 3 outside the parentheses on the left. There are two arrows pointing from the top of the three. One arrow points to the top of the x. The other arrow points to the top of the 4. This is set equal to 3 times x plus 3 times 4.

Simplify: 6(5y+1).

Solution

Solution

A mathematical expression 6(5y + 1) with two blue arrows showing the distribution of 6 to both 5y and 1, illustrating the distributive property.
Distribute. A mathematical expression showing 6 multiplied by 5y, plus 6 multiplied by 1.
Multiply. The image shows the mathematical expression '30y + 6' in black text on a white background. The numbers '30' and '6' are visible, as is the variable 'y' and the plus sign '+'.

Simplify: 9(3y+8).

Solution

27y + 72

Simplify: 5(5w+9).

Solution

25w + 45

The distributive property can be used to simplify expressions that look slightly different from a(b+c). Here are two other forms.

Distributive Property

If a,b,c are real numbers, then

a(b+c)=ab+ac

Other forms

a(b−c)=ab−ac
(b+c)a=ba+ca

Simplify: 2(x−3).

Solution

Solution

An image illustrating the distributive property in algebra, showing the expression 2(x - 3) with two blue arrows indicating that 2 should be multiplied by both x and -3.
Distribute. The image shows the mathematical expression 2 multiplied by x minus 2 multiplied by 3.
Multiply. The image shows the mathematical expression '2x - 6' in bold, black text on a white background. The terms '2x' and '6' are separated by a minus sign.

Simplify: 7(x−6).

Solution

7x − 42

Simplify: 8(x−5).

Solution

8x − 40

Do you remember how to multiply a fraction by a whole number? We’ll need to do that in the next two examples.

Simplify: 34(n+12).

Solution

Solution

Illustration of the distributive property for the expression (3/4)(n + 12), with light blue arrows showing 3/4 being multiplied by 'n' and by '12'.
Distribute. The image shows the mathematical expression three-fourths times n plus three-fourths times twelve, written as (3/4) * n + (3/4) * 12.
Simplify. A mathematical expression displays three-fourths multiplied by 'n', with the result then added to nine, set against a plain white background.

Simplify: 25(p+10).

Solution

25p+4

Simplify: 37(u+21).

Solution

37u+9

Simplify: 8(38x+14).

Solution

Solution

An image illustrating the distributive property, showing how to multiply 8 by each term inside the parentheses (3/8x and 1/4) in the expression 8(3/8x + 1/4).
Distribute. A mathematical expression shows 8 multiplied by the fraction 3/8 times x, plus 8 multiplied by the fraction 1/4.
Multiply. The image displays the algebraic expression '3x + 2' in black characters 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 later.

Simplify: 100(0.3+0.25q).

Solution

Solution

Illustration of the distributive property with the expression 100(0.3 + 0.25q). Arrows show 100 multiplying 0.3 and 0.25q.
Distribute. An algebraic expression is shown, which reads as one hundred multiplied by zero point three, plus one hundred multiplied by zero point two five q.
Multiply. The image displays the mathematical expression '30 + 25q' in a clear, black font against a white background.

Simplify: 100(0.7+0.15p).

Solution

70 + 15p

Simplify: 100(0.04+0.35d).

Solution

4 + 35d

In the next example we’ll multiply by a variable. We’ll need to do this in a later chapter.

Simplify: m(n−4).

Solution

Solution

The image shows the algebraic expression m(n-4) with two blue arrows demonstrating the distributive property by indicating that 'm' multiplies both 'n' and '-4' within the parentheses.
Distribute. The mathematical expression m multiplied by n, minus m multiplied by 4, is displayed in bold black font on a white background.
Multiply. The image shows the mathematical expression 'mn - 4m' in a bold, italicized font on a white background.

Notice that we wrote m·4as4m. We can do this because of the Commutative Property of Multiplication. When a term is the product of a number and a variable, we write the number first.

Simplify: r(s−2).

Solution

rs − 2r

Simplify: y(z−8).

Solution

yz − 8y

The next example will use the ‘backwards’ form of the Distributive Property, (b+c)a=ba+ca.

Simplify: (x+8)p.

Solution

Solution

An algebraic expression (x + 8)p with arrows illustrating the distributive property, showing 'p' multiplying both 'x' and '8' within the parentheses.
Distribute. The mathematical expression 'px + 8p' is displayed in bold black text on a white background.

Simplify: (x+2)p.

Solution

xp + 2p

Simplify: (y+4)q.

Solution

yq + 4q

When you distribute a negative number, you need to be extra careful to get the signs correct.

Simplify: −2(4y+1).

Solution

Solution

The image displays the algebraic expression -2(4y + 1) with blue arrows illustrating the distributive property. The arrows indicate that -2 should be multiplied by both 4y and 1 inside the parentheses.
Distribute. A mathematical expression shows the sum of two products: -2 multiplied by 4y, and -2 multiplied by 1. It is written as -2   4y + (-2)   1.
Simplify. The mathematical expression '-8y - 2' is displayed in a dark gray font against a white background.

Simplify: −3(6m+5).

Solution

−18m − 15

Simplify: −6(8n+11).

Solution

−48n − 66

Simplify: −11(4−3a).

Solution

Solution

A mathematical expression shows -11 multiplied by the quantity (4 - 3a).
Distribute. A mathematical expression showing the subtraction of two terms, where the first term is -11 multiplied by 4, and the second term is -11 multiplied by 3a. The expression is -11 * 4 - (-11) * 3a.
Multiply. The image displays the mathematical expression '-44 - (-33a)' on a white background, demonstrating subtraction involving a negative number and a variable term enclosed in parentheses.
Simplify. The mathematical expression -44 + 33a is shown on a white background.

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. Remember, −a=−1·a.

Simplify: −(y+5).

Solution

Solution

A mathematical expression shows the negative of the sum of 'y' and '5', written as -(y+5).
Multiplying by −1 results in the opposite. The mathematical expression -1(y + 5) is displayed in black text on a white background, representing a negative one multiplied by the sum of y and five.
Distribute. A mathematical expression showing the term -1 multiplied by 'y', added to the product of -1 and 5. The full expression is -1*y + (-1)*5.
Simplify. The image shows the mathematical expression '-y + (-5)', which represents the addition of negative y and negative 5.
Simplify. The mathematical expression -y - 5 is displayed in black characters on a white background.

Simplify: −(z−11).

Solution

−z + 11

Simplify: −(x−4).

Solution

−x + 4

Sometimes we 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

Solution

The image shows the mathematical expression 8 - 2(x + 3).
Distribute. A mathematical expression reads 8 minus 2 multiplied by x, minus 2 multiplied by 3.
Multiply. The image shows the mathematical expression 8 - 2x - 6 in a clear, dark font against a white background.
Combine like terms. The mathematical expression -2x + 2 is shown on a white background. The text is rendered in a clear, bold black font, typical of mathematical notation.

Simplify: 9−3(x+2).

Solution

−3x + 3

Simplify: 7x−5(x+4).

Solution

2x − 20

Simplify: 4(x−8)−(x+3).

Solution

Solution

A mathematical expression featuring the terms 4(x - 8) - (x + 3) is displayed on a white background.
Distribute. The image shows a mathematical expression: 4x - 32 - x - 3. It displays a series of numbers and variables connected by subtraction signs, all in a black font against a white background.
Combine like terms. The image shows the mathematical expression '3x - 35' in black text on a white background.

Simplify: 6(x−9)−(x+12).

Solution

5x − 66

Simplify: 8(x−1)−(x+5).

Solution

7x − 13

Evaluate Expressions Using the Distributive Property

Some students need to be convinced that the Distributive Property always works.

In the examples below, we will practice evaluating some of the expressions from previous examples; in part ⓐ , we will evaluate the form with parentheses, and in part ⓑ we will evaluate the form we got after distributing. If we evaluate both expressions correctly, this will show that they are indeed equal.

When y=10 evaluate: ⓐ 6(5y+1) ⓑ 6·5y+6·1.

Solution

Solution

ⓐ
6(5y+1)
The text reads 'Substitute 10 for y.', with '10' highlighted in red. A mathematical expression 6(5 • 10 + 1) is displayed in black font, with the number 10 highlighted in red, on a white background. It represents a calculation following the order of operations.
Simplify in the parentheses. 6(51)
Multiply. 306
ⓑ
A mathematical expression showing the distributive property: 6 multiplied by 5y, plus 6 multiplied by 1. The expression is 6 * 5y + 6 * 1, rendered in a black font on a white background.
The image displays the instruction 'Substitute 10 for y.' written in a blue-green font, with the number '10' highlighted in red, all against a plain white background. A mathematical expression reads '6 times 5 times 10 plus 6 times 1', with the number 10 highlighted in red.
Simplify. The image shows the mathematical expression '300 + 6' in black text against a white background.
Add. The number '306' is displayed in black text on a plain white background.

Notice, the answers are the same. When y=10,

6(5y+1)=6·5y+6·1.

Try it yourself for a different value of y.

Evaluate when w=3:ⓐ 5(5w+9)ⓑ 5·5w+5·9.

Solution
  1. ⓐ 120
  2. ⓑ 120

Evaluate when y=2:ⓐ 9(3y+8)ⓑ 9·3y+9·8.

Solution
  1. ⓐ 126
  2. ⓑ 126

When y=3, evaluate ⓐ −2(4y+1)ⓑ −2·4y+(−2)·1.

Solution

Solution

ⓐ
−2(4y+1)
The image displays the instruction 'Substitute 3 for y.' in a sans-serif font, with the number 3 highlighted in red, on a plain white background. A mathematical expression is displayed, showing a negative two multiplied by the sum of four times three and one, enclosed in parentheses: -2(4 * 3 + 1). The number '3' is highlighted in red.
Simplify in the parentheses. −2(13)
Multiply. −26
ⓑ
−2·4y+(−2)·1
The text reads 'Substitute 3 for y.' in a sans-serif font, with the number '3' highlighted in red and the rest of the text in a dark teal color. The text is centered on a white background. A mathematical expression shows the calculation '-2 multiplied by 4 multiplied by 3, plus -2 multiplied by 1', with the number 3 highlighted in red.
Multiply. −24−2
Subtract. −26
The answers are the same. When y=3, −2(4y+1)=−8y−2

Evaluate when n=−2:ⓐ −6(8n+11)ⓑ −6·8n+(−6)·11.

Solution
  1. ⓐ 30
  2. ⓑ 30

Evaluate when m=−1:ⓐ −3(6m+5)ⓑ −3·6m+(−3)·5.

Solution
  1. ⓐ 3
  2. ⓑ 3

When y=35 evaluate ⓐ −(y+5) and ⓑ −y−5 to show that −(y+5)=−y−5.

Solution

Solution

ⓐ
−(y+5)
The image displays the text The image displays the mathematical expression -(35 + 5), where 35 is highlighted in red, signifying that 35 and 5 are enclosed within parentheses and preceded by a negative sign.
Add in the parentheses. −(40)
Simplify. −40
ⓑ
−y−5
The text reads 'Substitute 35 for y.' with the number 35 highlighted in red. A mathematical expression showing '-35 - -5' with '-35' in red and '- -5' in black, illustrating subtraction of a negative number.
Simplify. −40
The answers are the same when y=35, demonstrating that −(y+5)=−y−5

Evaluate when x=36:ⓐ −(x−4)ⓑ −x+4 to show that −(x−4)=−x+4.

Solution
  1. ⓐ −32
  2. ⓑ −32

Evaluate when z=55:ⓐ −(z−10)ⓑ −z+10 to show that −(z−10)=−z+10.

Solution
  1. ⓐ −45
  2. ⓑ −45

ACCESS ADDITIONAL ONLINE RESOURCES

  • Model Distribution
  • The Distributive Property

Key Concepts

  • Distributive Property:
    • If a,b,c are real numbers then
      • a(b+c)=ab+ac
      • (b+c)a=ba+ca
      • a(b-c)=ab-ac

Practice Makes Perfect

Simplify Expressions Using the Distributive Property

In the following exercises, simplify using the distributive property.

4(x+8)

3(a+9)

Solution

3a + 27

8(4y+9)

9(3w+7)

Solution

27w + 63

6(c−13)

7(y−13)

Solution

7y − 91

7(3p−8)

5(7u−4)

Solution

35u − 20

12(n+8)

13(u+9)

Solution

13u+3

14(3q+12)

15(4m+20)

Solution

45m+4

9(59y−13)

10(310x−25)

Solution

3x − 4

12(14+23r)

12(16+34s)

Solution

2 + 9s

r(s−18)

u(v−10)

Solution

uv − 10u

(y+4)p

(a+7)x

Solution

ax + 7x

−2(y+13)

−3(a+11)

Solution

−3a − 33

−7(4p+1)

−9(9a+4)

Solution

−81a − 36

−3(x−6)

−4(q−7)

Solution

−4q + 28

−9(3a−7)

−6(7x−8)

Solution

−42x + 48

−(r+7)

−(q+11)

Solution

−q − 11

−(3x−7)

−(5p−4)

Solution

−5p + 4

5+9(n−6)

12+8(u−1)

Solution

8u + 4

16−3(y+8)

18−4(x+2)

Solution

−4x + 10

4−11(3c−2)

9−6(7n−5)

Solution

−42n + 39

22−(a+3)

8−(r−7)

Solution

−r + 15

−12−(u+10)

−4−(c−10)

Solution

−c + 6

(5m−3)−(m+7)

(4y−1)−(y−2)

Solution

3y + 1

5(2n+9)+12(n−3)

9(5u+8)+2(u−6)

Solution

47u + 60

9(8x−3)−(−2)

4(6x−1)−(−8)

Solution

24x + 4

14(c−1)−8(c−6)

11(n−7)−5(n−1)

Solution

6n − 72

6(7y+8)−(30y−15)

7(3n+9)−(4n−13)

Solution

17n + 76

Evaluate Expressions Using the Distributive Property

In the following exercises, evaluate both expressions for the given value.

If v=−2, evaluate
  1. ⓐ 6(4v+7)
  2. ⓑ 6·4v+6·7
If u=−1, evaluate
  1. ⓐ 8(5u+12)
  2. ⓑ 8·5u+8·12
Solution
  1. ⓐ 56
  2. ⓑ 56
If n=23, evaluate
  1. ⓐ 3(n+56)
  2. ⓑ 3·n+3·56
If y=34, evaluate
  1. ⓐ 4(y+38)
  2. ⓑ 4·y+4·38
Solution
  1. ⓐ 92
  2. ⓑ 92
If y=712, evaluate
  1. ⓐ −3(4y+15)
  2. ⓑ −3·4y+(−3)·15
If p=2330, evaluate
  1. ⓐ −6(5p+11)
  2. ⓑ −6·5p+(−6)·11
Solution
  1. ⓐ −89
  2. ⓑ −89
If m=0.4, evaluate
  1. ⓐ −10(3m−0.9)
  2. ⓑ −10·3m−(−10)(0.9)
If n=0.75, evaluate
  1. ⓐ −100(5n+1.5)
  2. ⓑ −100·5n+(−100)(1.5)
Solution
  1. ⓐ −525
  2. ⓑ −525
If y=−25, evaluate
  1. ⓐ −(y−25)
  2. ⓑ −y+25
If w=−80, evaluate
  1. ⓐ −(w−80)
  2. ⓑ −w+80
Solution
  1. ⓐ 160
  2. ⓑ 160
If p=0.19, evaluate
  1. ⓐ −(p+0.72)
  2. ⓑ −p−0.72
If q=0.55, evaluate
  1. ⓐ −(q+0.48)
  2. ⓑ −q−0.48
Solution
  1. ⓐ −1.03
  2. ⓑ −1.03

Everyday Math

Buying by the case Joe can buy his favorite ice tea at a convenience store for $1.99 per bottle. At the grocery store, he can buy a case of 12 bottles for $23.88.

  1. ⓐ Use the distributive property to find the cost of 12 bottles bought individually at the convenience store. (Hint: notice that $1.99 is $2−$0.01.)

  2. ⓑ Is it a bargain to buy the iced tea at the grocery store by the case?

Multi-pack purchase Adele’s shampoo sells for $3.97 per bottle at the drug store. At the warehouse store, the same shampoo is sold as a 3-pack for $10.49.

  1. ⓐ Show how you can use the distributive property to find the cost of 3 bottles bought individually at the drug store.

  2. ⓑ How much would Adele save by buying the 3-pack at the warehouse store?

Solution
  1. ⓐ 3(4 − 0.03) = 11.91
  2. ⓑ $1.42

Writing Exercises

Simplify 8(x−14) using the distributive property and explain each step.

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

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 table for students to rate their understanding of simplifying and evaluating expressions using the Distributive Property.

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

Properties of Identity, Inverses, and Zero

Learning Objectives

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

  • Recognize the identity properties of addition and multiplication
  • Use the inverse properties of addition and multiplication
  • Use the properties of zero
  • Simplify expressions using the properties of identities, inverses, and zero

Before you get started, take this readiness quiz.

Find the opposite of −4.
If you missed this problem, review Example 3 in Introduction to Integers.

Solution

4

Find the reciprocal of 52.
If you missed this problem, review Example 11 in Multiply and Divide Fractions.

Solution

25

Multiply: 3a5·92a.
If you missed this problem, review Example 9 in Multiply and Divide Fractions.

Solution

2710

Recognize the Identity Properties of Addition and Multiplication

What happens when we add zero to any number? Adding zero doesn’t change the value. For this reason, we call 0 the additive identity.

For example,

13+0−14+00+(−3x) 13−14−3x

What happens when you multiply any number by one? Multiplying by one doesn’t change the value. So we call 1 the multiplicative identity.

For example,

43·1−27·11·6y543−276y5

Identity Properties

The identity property of addition: for any real number a,

a+0=a0+a=a0 is called the additive identity

The identity property of multiplication: for any real number a

a·1=a1·a=a1 is called the multiplicative identity

Identify whether each equation demonstrates the identity property of addition or multiplication.

  1. ⓐ 7+0=7

  2. ⓑ −16(1)=−16

Solution

Solution

This table illustrates the identity property of addition, showing an example where adding zero to a number results in the same number.
ⓐ
7+0=7
We are adding 0. We are using the identity property of addition.
This table illustrates the identity property of multiplication, showing that multiplying any number by one results in the original number.
ⓑ
−16(1)=−16
We are multiplying by 1. We are using the identity property of multiplication.

Identify whether each equation demonstrates the identity property of addition or multiplication:

ⓐ 23+0=23ⓑ −37(1)=−37.

Solution
  1. ⓐ identity property of addition
  2. ⓑ identity property of multiplication

Identify whether each equation demonstrates the identity property of addition or multiplication:

ⓐ 1·29=29ⓑ 14+0=14.

Solution
  1. ⓐ identity property of multiplication
  2. ⓑ identity property of addition

Use the Inverse Properties of Addition and Multiplication

What number added to 5 gives the additive identity, 0?
5+_____=0 The text reads 'We know 5 + (-5) = 0' illustrating the concept of additive inverses where a number and its negative sum to zero.
What number added to −6 gives the additive identity, 0?
−6+_____=0 The image displays the equation 'We know -6 + 6 = 0', written in a dark teal color, with the second '6' in the equation highlighted in red to emphasize its role in the sum.

Notice that in each case, 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 0, which is the additive identity.

What number multiplied by 23 gives the multiplicative identity, 1? In other words, two-thirds times what results in 1?

23·___=1 The image displays the equation 'We know (2/3) * (3/2) = 1', demonstrating that a fraction multiplied by its reciprocal results in 1.

What number multiplied by 2 gives the multiplicative identity, 1? In other words two times what results in 1?

2·___=1 The equation illustrates that two multiplied by one half equals one.

Notice that in each case, the missing number was the reciprocal of the number.

We call 1a the multiplicative inverse of a(a≠0). The reciprocal of a number is its multiplicative inverse. A number and its reciprocal multiply to 1, which is the multiplicative identity.

We’ll formally state the Inverse Properties here:

Inverse Properties

Inverse Property of Addition for any real number a,

a+(−a)=0−ais the additive inverse ofa.

Inverse Property of Multiplication for any real number a≠0,

a·1a=11ais the multiplicative inverse ofa.

Find the additive inverse of each expression: ⓐ 13 ⓑ −58 ⓒ 0.6.

Solution

Solution

To find the additive inverse, we find the opposite.

  1. ⓐ The additive inverse of 13 is its opposite, −13.

  2. ⓑ The additive inverse of −58 is its opposite, 58.

  3. ⓒ The additive inverse of 0.6 is its opposite, −0.6.

Find the additive inverse: ⓐ 18 ⓑ 79 ⓒ 1.2.

Solution
  1. ⓐ −18
  2. ⓑ −79
  3. ⓒ −1.2

Find the additive inverse: ⓐ 47 ⓑ 713 ⓒ 8.4.

Solution
  1. ⓐ −47
  2. ⓑ −713
  3. ⓒ −8.4

Find the multiplicative inverse: ⓐ 9 ⓑ −19 ⓒ 0.9.

Solution

Solution

To find the multiplicative inverse, we find the reciprocal.

  1. ⓐ The multiplicative inverse of 9 is its reciprocal, 19.

  2. ⓑ The multiplicative inverse of −19 is its reciprocal, −9.

  3. ⓒ To find the multiplicative inverse of 0.9, we first convert 0.9 to a fraction, 910. Then we find the reciprocal, 109.

Find the multiplicative inverse: ⓐ 5 ⓑ −17 ⓒ 0.3.

Solution
  1. ⓐ 15
  2. ⓑ −7
  3. ⓒ 103

Find the multiplicative inverse: ⓐ 18 ⓑ −45 ⓒ 0.6.

Solution
  1. ⓐ 118
  2. ⓑ −54
  3. ⓒ 53

Use the Properties of Zero

We have already learned that zero is the additive identity, since it can be added to any number without changing the number’s identity. But zero also has some special properties when it comes to multiplication and division.

Multiplication by Zero

What happens when you multiply a number by 0? Multiplying by 0 makes the product equal zero. The product of any real number and 0 is 0.

Multiplication by Zero

For any real number a,

a·0=00·a=0

Simplify: ⓐ −8·0 ⓑ 512·0 ⓒ 0(2.94).

Solution
Solution
This table illustrates the mathematical property that the product of any real number and zero is zero, accompanied by an example.
ⓐ
−8⋅0
The product of any real number and 0 is 0. 0
An example illustrating the multiplication property of zero: any number multiplied by zero equals zero.
ⓑ
512·0
The product of any real number and 0 is 0. 0
Table demonstrating the Zero Property of Multiplication with its statement and an example.
ⓒ
0(2.94)
The product of any real number and 0 is 0. 0

Simplify: ⓐ −14·0 ⓑ 0·23 ⓒ (16.5)·0.

Solution
  1. ⓐ 0
  2. ⓑ 0
  3. ⓒ 0

Simplify: ⓐ (1.95)·0 ⓑ 0(−17) ⓒ 0·54.

Solution
  1. ⓐ 0
  2. ⓑ 0
  3. ⓒ 0

Dividing with Zero

What about dividing with 0? Think about a real example: if there are no cookies in the cookie jar and three people want to share them, how many cookies would each person get? There are 0 cookies to share, so each person gets 0 cookies.

0÷3=0

Remember that we can always check division with the related multiplication fact. So, we know that

0÷3=0because0·3=0.

Division of Zero

For any real number a, except 0,0a=0 and 0÷a=0.

Zero divided by any real number except zero is zero.

Simplify: ⓐ 0÷5 ⓑ 0−2 ⓒ 0÷78.

Solution
Solution
Illustrates the rule that zero divided by any non-zero real number results in zero, with a mathematical example.
ⓐ
0÷5
Zero divided by any real number, except 0, is zero. 0
Mathematical rule: Zero divided by any non-zero real number equals zero, illustrated with an example.
ⓑ
0−2
Zero divided by any real number, except 0, is zero. 0
Demonstrates the rule: zero divided by any non-zero real number equals zero, with an example.
ⓒ
0÷78
Zero divided by any real number, except 0, is zero. 0

Simplify: ⓐ 0÷11 ⓑ 0−6 ⓒ 0÷310.

Solution
  1. ⓐ 0
  2. ⓑ 0
  3. ⓒ 0

Simplify: ⓐ 0÷83 ⓑ 0÷(−10) ⓒ 0÷12.75.

Solution
  1. ⓐ 0
  2. ⓑ 0
  3. ⓒ 0

Now let’s think about dividing a number by zero. What is the result of dividing 4 by 0? Think about the related multiplication fact. Is there a number that multiplied by 0 gives 4?

4÷0=___means___·0=4

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

Division by Zero

For any real number a,a0, and a÷0 are undefined.

Division by zero is undefined.

Simplify: ⓐ 7.5÷0 ⓑ −320 ⓒ 49÷0.

Solution
Solution
This table demonstrates the undefined outcome when performing division by zero with a mathematical expression.
ⓐ
7.5÷0
Division by zero is undefined. undefined
This table illustrates that division by zero is undefined, presenting an example expression and its corresponding status.
ⓑ
−320
Division by zero is undefined. undefined
Demonstrates that division by zero is undefined, with a mathematical example (4/9 0) and the general rule.
ⓒ
49÷0
Division by zero is undefined. undefined

Simplify: ⓐ 16.4÷0 ⓑ −20 ⓒ 15÷0.

Solution
  1. ⓐ undefined
  2. ⓑ undefined
  3. ⓒ undefined

Simplify: ⓐ −50 ⓑ 96.9÷0 ⓒ 415÷0

Solution
  1. ⓐ undefined
  2. ⓑ undefined
  3. ⓒ undefined

We summarize the properties of zero.

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=0 Zero divided by any real number, except itself, is zero.

a0 is undefined. Division by zero is undefined.

Simplify Expressions using the Properties of Identities, Inverses, and Zero

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

Simplify: 3x+15−3x.

Solution

Solution

Simplification of the algebraic expression 3x + 15 - 3x, demonstrating the use of additive inverses.
3x+15−3x
Notice the additive inverses, 3x and −3x. 0+15
Add. 15

Simplify: −12z+9+12z.

Solution

9

Simplify: −25u−18+25u.

Solution

−18

Simplify: 4(0.25q).

Solution

Solution

This table illustrates the step-by-step simplification of the expression 4(0.25q) to q, applying mathematical properties.
4(0.25q)
Regroup, using the associative property. [4(0.25)]q
Multiply. 1.00q
Simplify; 1 is the multiplicative identity. q

Simplify: 2(0.5p).

Solution

p

Simplify: 25(0.04r).

Solution

r

Simplify: 0n+5, where n≠−5.

Solution

Solution

This table illustrates the mathematical property: zero divided by any non-zero real number is zero.
0n+5
Zero divided by any real number except itself is zero. 0

Simplify: 0m+7, where m≠−7.

Solution

0

Simplify: 0d−4, where d≠4.

Solution

0

Simplify: 10−3p0.

Solution

Solution

This table explains that division by zero is undefined, showing a mathematical example and its corresponding status.
10−3p0
Division by zero is undefined. undefined

Simplify: 18−6c0.

Solution

undefined

Simplify: 15−4q0.

Solution

undefined

Simplify: 34·43(6x+12).

Solution

Solution

We cannot combine the terms in parentheses, so we multiply the two fractions first.

Illustrates simplifying an algebraic expression step-by-step using reciprocals and the multiplicative identity.
34·43(6x+12)
Multiply; the product of reciprocals is 1. 1(6x+12)
Simplify by recognizing the multiplicative identity. 6x+12

Simplify: 25·52(20y+50).

Solution

20y + 50

Simplify: 38·83(12z+16).

Solution

12z + 16

All the properties of real numbers we have used in this chapter are summarized in Table 20.

Properties of Real Numbers
Property Of Addition Of Multiplication
Commutative Property
If a and b are real numbers then… a+b=b+a a·b=b·a
Associative Property
If a, b, and c are real numbers then… (a+b)+c=a+(b+c) (a·b)·c=a·(b·c)
Identity Property 0 is the additive identity 1 is the multiplicative identity
For any real number a, a+0=a0+a=a a·1=a1·a=a
Inverse Property −ais the additive inverse of a a,a≠0
1/a is the multiplicative inverse of a
For any real number a, a+(−a)=0 a·1a=1
Distributive Property
If a,b,c are real numbers, then a(b+c)=ab+ac
Properties of Zero
For any real number a,
a⋅0=00⋅a=0
For any real number a,a≠0 0a=0
a0 is undefined

ACCESS ADDITIONAL ONLINE RESOURCES

  • Multiplying and Dividing Involving Zero

Key Concepts

  • Identity Properties
    • Identity Property of Addition: For any real number a: a+0=a0+a=a 0 is the additive identity
    • Identity Property of Multiplication: For any real number a: a⋅1=a1⋅a=a 1 is the multiplicative identity
  • Inverse Properties
    • Inverse Property of Addition: For any real number a: a+(-a)=0-a is the additive inverse of a
    • Inverse Property of Multiplication: For any real number a: (a≠0)a⋅1a=11a is the multiplicative inverse of a
  • 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 of Zero: For any real number a, 0a=0Zero divided by any real number, except itself, is zero.
    • Division by Zero: For any real number a, a0 is undefined and a÷0 is undefined. Division by zero is undefined.

Practice Makes Perfect

Recognize the Identity Properties of Addition and Multiplication

In the following exercises, identify whether each example is using the identity property of addition or multiplication.

101+0=101

35(1)=35

Solution

identity property of multiplication

−9·1=−9

0+64=64

Solution

identity property of addition

Use the Inverse Properties of Addition and Multiplication

In the following exercises, find the multiplicative inverse.

8

14

Solution

114

−17

−19

Solution

−119

712

813

Solution

138

−310

−512

Solution

−125

0.8

0.4

Solution

52

−0.2

−0.5

Solution

−2

Use the Properties of Zero

In the following exercises, simplify using the properties of zero.

48·0

06

Solution

0

30

22·0

Solution

0

0÷1112

60

Solution

undefined

03

0÷715

Solution

0

0·815

(−3.14)(0)

Solution

0

5.72÷0

1100

Solution

undefined

Simplify Expressions using the Properties of Identities, Inverses, and Zero

In the following exercises, simplify using the properties of identities, inverses, and zero.

19a+44−19a

27c+16−27c

Solution

16

38+11r−38

92+31s−92

Solution

31s

10(0.1d)

100(0.01p)

Solution

p

5(0.6q)

40(0.05n)

Solution

2n

0r+20, where r≠−20

0s+13, where s≠−13

Solution

0

0u−4.99, where u≠4.99

0v−65.1, where v≠65.1

Solution

0

0÷(x−12), where x≠12

0÷(y−16), where y≠16

Solution

0

32−5a0, where 32−5a≠0

28−9b0, where 28−9b≠0

Solution

undefined

2.1+0.4c0, where 2.1+0.4c≠0

1.75+9f0, where 1.75+9f≠0

Solution

undefined

(34+910m)÷0, where 34+910m≠0

(516n−37)÷0, where 516n−37≠0

Solution

undefined

910·109(18p−21)

57·75(20q−35)

Solution

20q − 35

15·35(4d+10)

18·56(15h+24)

Solution

225h + 360

Everyday Math

Insurance copayment Carrie had to have 5 fillings done. Each filling cost $80. Her dental insurance required her to pay 20% of the cost. Calculate Carrie’s cost

  1. ⓐ by finding her copay for each filling, then finding her total cost for 5 fillings, and

  2. ⓑ by multiplying 5(0.20)(80).

  3. ⓒ Which of the Properties of Real Numbers did you use for part (b)?

Cooking time Helen bought a 24-pound turkey for her family’s Thanksgiving dinner and wants to know what time to put the turkey in the oven. She wants to allow 20 minutes per pound cooking time.

  1. ⓐ Calculate the length of time needed to roast the turkey by multiplying 24·20 to find the number of minutes and then multiplying the product by 160 to convert minutes into hours.

  2. ⓑ Multiply 24(20·160).

  3. ⓒ Which of the Properties of Real Numbers allows you to multiply 24(20·160) instead of (24·20)160?

Solution
  1. ⓐ 8 hours
  2. ⓑ 8
  3. ⓒ associative property of multiplication

Writing Exercises

In your own words, describe the difference between the additive inverse and the multiplicative inverse of a number.

How can the use of the properties of real numbers make it easier to simplify expressions?

Solution

Answers will vary.

Self Check

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

Math self-assessment grid for properties of addition and multiplication (identity, inverse, zero). Learners can indicate if they understand '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?

Additive Identity
The additive identity is 0. When zero is added to any number, it does not change the value.
Additive Inverse
The opposite of a number is its additive inverse. The additive inverse of a is −a.
Multiplicative Identity
The multiplicative identity is 1. When one multiplies any number, it does not change the value.
Multiplicative Inverse
The reciprocal of a number is its multiplicative inverse. The multiplicative inverse of a is 1a.

Systems of Measurement

Learning Objectives

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

  • Make unit conversions in the U.S. system
  • Use mixed units of measurement in the U.S. system
  • Make unit conversions in the metric system
  • Use mixed units of measurement in the metric system
  • Convert between the U.S. and the metric systems of measurement
  • Convert between Fahrenheit and Celsius temperatures

Before you get started, take this readiness quiz.

Multiply: 4.29(1000).
If you missed this problem, review Example 8 in Decimal Operations.

Solution

4,290

Simplify: 3054.
If you missed this problem, review Example 2 in Multiply and Divide Fractions.

Solution

59

Multiply: 715·2528.
If you missed this problem, review Example 9 in Multiply and Divide Fractions.

Solution

512

In this section we will see how to convert among different types of units, such as feet to miles or kilograms to pounds. The basic idea in all of the unit conversions will be to use a form of 1, the multiplicative identity, to change the units but not the value of a quantity.

Make Unit Conversions in the U.S. System

There are two systems of measurement commonly used around the world. Most countries use the metric system. The United States uses a different system of measurement, usually called the U.S. system. We will look at the U.S. system first.

The U.S. system of measurement uses units of inch, foot, yard, and mile to measure length and pound and ton to measure weight. For capacity, the units used are cup, pint, quart and gallons. Both the U.S. system and the metric system measure time in seconds, minutes, or hours.

The equivalencies among the basic units of the U.S. system of measurement are listed in Table 1. The table also shows, in parentheses, the common abbreviations for each measurement.

U.S. System Units
Length Volume
1 foot (ft) = 12 inches (in)
1 yard (yd) = 3 feet (ft)
1 mile (mi) = 5280 feet (ft)
3 teaspoons (t) = 1 tablespoon (T)
16 Tablespoons (T) = 1 cup (C)
1 cup (C) = 8 fluid ounces (fl oz)
1 pint (pt) = 2 cups (C)
1 quart (qt) = 2 pints (pt)
1 gallon (gal) = 4 quarts (qt)
Weight Time
1 pound (lb) = 16 ounces (oz)
1 ton = 2000 pounds (lb)
1 minute (min) = 60 seconds (s)
1 hour (h) = 60 minutes (min)
1 day = 24 hours (h)
1 week (wk) = 7 days
1 year (yr) = 365 days

In many real-life applications, we need to convert between units of measurement. We will use the identity property of multiplication to do these conversions. We’ll restate the Identity Property of Multiplication here for easy reference.

For any real numbera,a·1=a1·a=a

To use the identity property of multiplication, we write 1 in a form that will help us convert the units. For example, suppose we want to convert inches to feet. We know that 1 foot is equal to 12 inches, so we can write 1 as the fraction 1 ft12 in. When we multiply by this fraction, we do not change the value but just change the units.

But 12 in1 ft also equals 1. How do we decide whether to multiply by 1 ft12 in or 12 in1 ft? We choose the fraction that will make the units we want to convert from divide out. For example, suppose we wanted to convert 60 inches to feet. If we choose the fraction that has inches in the denominator, we can eliminate the inches.

60in·1 ft12in=5 ft

On the other hand, if we wanted to convert 5 feet to inches, we would choose the fraction that has feet in the denominator.

5 ft·12 in1ft=60 in

We treat the unit words like factors and ‘divide out’ common units like we do common factors.

Make unit conversions.

  1. Multiply the measurement to be converted by 1; write 1 as a fraction relating the units given and the units needed.
  2. Multiply.
  3. Simplify the fraction, performing the indicated operations and removing the common units.

Mary Anne is 66 inches tall. What is her height in feet?

Solution

Solution

This table demonstrates the step-by-step process and calculations for converting 66 inches into feet.
Convert 66 inches into feet.
Multiply the measurement to be converted by 1. 66 inches ·1
Write 1 as a fraction relating the units given and the units needed. 66 inches·1 foot12 inches
Multiply. 66 inches·1 foot12 inches
Simplify the fraction. 66inches·1 foot12inches
66 feet12
5.5 feet

Notice that the when we simplified the fraction, we first divided out the inches.

Mary Anne is 5.5 feet tall.

Lexie is 30 inches tall. Convert her height to feet.

Solution

2.5 feet

Rene bought a hose that is 18 yards long. Convert the length to feet.

Solution

54 feet

When we use the Identity Property of Multiplication to convert units, we need to make sure the units we want to change from will divide out. Usually this means we want the conversion fraction to have those units in the denominator.

Ndula, an elephant at the San Diego Safari Park, weighs almost 3.2 tons. Convert her weight to pounds.

A photograph of an adult elephant.
(credit: Guldo Da Rozze, Flickr)
Solution

Solution

We will convert 3.2 tons into pounds, using the equivalencies in Table 1. We will use the Identity Property of Multiplication, writing 1 as the fraction 2000 pounds1 ton.

This table demonstrates the step-by-step conversion of 3.2 tons to pounds, culminating in a real-world example.
3.2 tons
Multiply the measurement to be converted by 1. 3.2 tons·1
Write 1 as a fraction relating tons and pounds. 3.2 tons·2000 lbs1 ton
Simplify. 3.2tons·2000 lbs1ton
Multiply. 6400 lbs
Ndula weighs almost 6,400 pounds.

Arnold’s SUV weighs about 4.3 tons. Convert the weight to pounds.

Solution

8600 pounds

A cruise ship weighs 51,000 tons. Convert the weight to pounds.

Solution

102,000,000 pounds

Sometimes to convert from one unit to another, we may need to use several other units in between, so we will need to multiply several fractions.

Juliet is going with her family to their summer home. She will be away for 9 weeks. Convert the time to minutes.

Solution

Solution

To convert weeks into minutes, we will convert weeks to days, days to hours, and then hours to minutes. To do this, we will multiply by conversion factors of 1.
9 weeks
Write 1 as 7days1week,24hours1day,60minutes1hour. A dimensional analysis problem converting 9 weeks into minutes using conversion factors for days in a week, hours in a day, and minutes in an hour.
Cancel common units. A clear demonstration of unit conversion, illustrating how 9 weeks are converted to minutes by sequentially multiplying conversion factors for days, hours, and minutes, with units canceled at each step.
Multiply. 9·7·24·60min1·1·1·1=90,720min
Juliet will be away for 90,720 minutes.

The distance between Earth and the moon is about 250,000 miles. Convert this length to yards.

Solution

440,000,000 yards

A team of astronauts spends 15 weeks in space. Convert the time to minutes.

Solution

151,200 minutes

How many fluid ounces are in 1 gallon of milk?

A photograph of a milk display in a grocery store.
(credit: www.bluewaikiki.com, Flickr)
Solution

Solution

Use conversion factors to get the right units: convert gallons to quarts, quarts to pints, pints to cups, and cups to fluid ounces.

1 gallon
Multiply the measurement to be converted by 1. 1 gal1·4 qt1 gal·2 pt1 qt·2 C1 pt·8 fl oz1 C
Simplify. 1gal1·4qt1gal·2pt1qt·2C1pt·8 fl oz1C
Multiply. 1·4·2·2·8 fl oz1·1·1·1·1
Simplify. 128 fluid ounces
There are 128 fluid ounces in a gallon.

How many cups are in 1 gallon?

Solution

16 cups

How many teaspoons are in 1 cup?

Solution

48 teaspoons

Use Mixed Units of Measurement in the U.S. System

Performing arithmetic operations on measurements with mixed units of measures requires care. Be sure to add or subtract like units.

Charlie bought three steaks for a barbecue. Their weights were 14 ounces, 1 pound 2 ounces, and 1 pound 6 ounces. How many total pounds of steak did he buy?

A photograph of meat being cooked on a charcoal grill.
(credit: Helen Penjam, Flickr)
Solution

Solution

We will add the weights of the steaks to find the total weight of the steaks.
Add the ounces. Then add the pounds. An arithmetic problem showing the addition of weights: 1 pound 2 ounces plus 1 pound 6 ounces equals 2 pounds 22 ounces, with 14 ounces noted above.
Convert 22 ounces to pounds and ounces.
Add the pounds. 2 pounds + 1 pound, 6 ounces
3 pounds, 6 ounces
Charlie bought 3 pounds 6 ounces of steak.

Laura gave birth to triplets weighing 3 pounds 12 ounces, 3 pounds 3 ounces, and 2 pounds 9 ounces. What was the total birth weight of the three babies?

Solution

9 lbs. 8 oz

Seymour cut two pieces of crown molding for his family room that were 8 feet 7 inches and 12 feet 11 inches. What was the total length of the molding?

Solution

21 ft. 6 in.

Anthony bought four planks of wood that were each 6 feet 4 inches long. If the four planks are placed end-to-end, what is the total length of the wood?

The image shows 4 planks of wood placed end-to-end horizontally. Each plank is labeled 6 feet 4 inches. A line starts at the left of the first plank and runs horizontally to the right of the fourth plank. The line is labeled with the letter l to represent length.
Solution

Solution

We will multiply the length of one plank by 4 to find the total length.
Multiply the inches and then the feet. A multiplication problem is displayed, showing '6 feet 4 inches' multiplied by 4, resulting in '24 feet 16 inches' as the product.
Convert 16 inches to feet. 24 feet + 1 foot 4 inches
Add the feet. 25 feet 4 inches
Anthony bought 25 feet 4 inches of wood.

Henri wants to triple his spaghetti sauce recipe, which calls for 1 pound 8 ounces of ground turkey. How many pounds of ground turkey will he need?

Solution

4 lbs. 8 oz.

Joellen wants to double a solution of 5 gallons 3 quarts. How many gallons of solution will she have in all?

Solution

11 gal. 2 qts.

Make Unit Conversions in the Metric System

In the metric system, units are related by powers of 10. The root words of their names reflect this relation. For example, the basic unit for measuring length is a meter. One kilometer is 1000 meters; the prefix kilo- means thousand. One centimeter is 1100 of a meter, because the prefix centi- means one one-hundredth (just like one cent is 1100 of one dollar).

The equivalencies of measurements in the metric system are shown in Table 8. The common abbreviations for each measurement are given in parentheses.

Metric Measurements
Length Mass Volume/Capacity
1 kilometer (km) = 1000 m
1 hectometer (hm) = 100 m
1 dekameter (dam) = 10 m
1 meter (m) = 1 m
1 decimeter (dm) = 0.1 m
1 centimeter (cm) = 0.01 m
1 millimeter (mm) = 0.001 m
1 kilogram (kg) = 1000 g
1 hectogram (hg) = 100 g
1 dekagram (dag) = 10 g
1 gram (g) = 1 g
1 decigram (dg) = 0.1 g
1 centigram (cg) = 0.01 g
1 milligram (mg) = 0.001 g
1 kiloliter (kL) = 1000 L
1 hectoliter (hL) = 100 L
1 dekaliter (daL) = 10 L
1 liter (L) = 1 L
1 deciliter (dL) = 0.1 L
1 centiliter (cL) = 0.01 L
1 milliliter (mL) = 0.001 L
1 meter = 100 centimeters
1 meter = 1000 millimeters
1 gram = 100 centigrams
1 gram = 1000 milligrams
1 liter = 100 centiliters
1 liter = 1000 milliliters

To make conversions in the metric system, we will use the same technique we did in the U.S. system. Using the identity property of multiplication, we will multiply by a conversion factor of one to get to the correct units.

Have you ever run a 5 k or 10 k race? The lengths of those races are measured in kilometers. The metric system is commonly used in the United States when talking about the length of a race.

Nick ran a 10-kilometer race. How many meters did he run?

A photograph of 8 male runners in a track race.
(credit: William Warby, Flickr)
Solution

Solution

We will convert kilometers to meters using the Identity Property of Multiplication and the equivalencies in Table 8.
10 kilometers
Multiply the measurement to be converted by 1. 10 km times 1. The unit, km, is colored red.
Write 1 as a fraction relating kilometers and meters. A mathematical expression shows 10 kilometers multiplied by the conversion factor of 1000 meters per 1 kilometer, illustrating how to convert kilometers to meters.
Simplify. An image illustrating unit conversion, showing 10 kilometers multiplied by 1000 meters divided by 1 kilometer, with the 'km' units crossed out to demonstrate cancellation, resulting in meters.
Multiply. 10,000 m
Nick ran 10,000 meters.

Sandy completed her first 5-km race. How many meters did she run?

Solution

5000 m

Herman bought a rug 2.5 meters in length. How many centimeters is the length?

Solution

250 cm

Eleanor’s newborn baby weighed 3200 grams. How many kilograms did the baby weigh?

Solution

Solution

We will convert grams to kilograms.
The image displays the text '3200 grams' in a clean, legible font, with the numbers '3200' in a bluish-gray hue and 'grams' in a distinct red color, all set against a white background.
Multiply the measurement to be converted by 1. A mathematical expression shows '3200' in dark teal, followed by a lowercase 'g' in red, then a dark teal multiplication dot, and finally the number '1' in dark teal, all on a white background.
Write 1 as a fraction relating kilograms and grams. A mathematical expression for converting 3200 grams to kilograms, showing 3200g multiplied by 1kg/1000g. The 'g' units are highlighted in red for clarity during cancellation.
Simplify. A unit conversion showing 3200 grams being converted to kilograms by multiplying by (1 kg / 1000 g), with the 'g' units crossed out for cancellation.
Multiply. A fraction shows 3200 kilograms divided by 1000, illustrating a mathematical expression or unit conversion.
Divide. 3.2 kilograms
The baby weighed 3.2 kilograms.

Kari’s newborn baby weighed 2800 grams. How many kilograms did the baby weigh?

Solution

2.8 kilograms

Anderson received a package that was marked 4500 grams. How many kilograms did this package weigh?

Solution

4.5 kilograms

Since the metric system is based on multiples of ten, conversions involve multiplying by multiples of ten. In Decimal Operations, we learned how to simplify these calculations by just moving the decimal.

To multiply by 10,100,or1000, we move the decimal to the right 1,2,or3 places, respectively. To multiply by 0.1,0.01,or0.001 we move the decimal to the left 1,2,or3 places respectively.

We can apply this pattern when we make measurement conversions in the metric system.

In Example 8, we changed 3200 grams to kilograms by multiplying by 11000(or0.001). This is the same as moving the decimal 3 places to the left.

Multiplying 3200 by 1 over 1000 gives 3.2. Notice that the answer, 3.2, is similar to the original value, 3200, just with the decimal moved three places to the left.

Convert:

  1. ⓐ 350 liters to kiloliters
  2. ⓑ 4.1 liters to milliliters.
Solution

Solution

ⓐ We will convert liters to kiloliters. In Table 8, we see that 1 kiloliter=1000 liters.
350 L
Multiply by 1, writing 1 as a fraction relating liters to kiloliters. A mathematical expression showing the conversion of 350 liters to kiloliters: 350 L multiplied by the fraction 1 kL over 1000 L.
Simplify. A calculation converting 350 liters (L) to kiloliters (kL), showing the multiplication by a conversion factor of 1 kL over 1000 L, with the L units crossed out to indicate cancellation.
Move the decimal 3 units to the left. A mathematical expression showing the conversion of 350 liters to kiloliters, multiplying 350 L by the ratio 1 kL/1000 L, with the liter units crossed out to indicate cancellation.
0.35 kL
ⓑ We will convert liters to milliliters. In Table 8, we see that 1 liter=1000milliliters.
4.1 L
Multiply by 1, writing 1 as a fraction relating milliliters to liters. A mathematical expression showing the conversion of 4.1 liters to milliliters: 4.1 L * (1000 mL / 1 L).
Simplify. An image displays the conversion of 4.1 liters to milliliters. The expression shows 4.1 L multiplied by a fraction (1000 mL / 1 L), with the 'L' unit diagonally crossed out in both the numerator and denominator.
Move the decimal 3 units to the left. A graphic displays '4.100 mL' in dark teal text, with a lighter teal wavy arrow pointing upwards from beneath the text.
4100 mL

Convert: ⓐ 7.25 L to kL ⓑ 6.3 L to mL.

Solution
  1. ⓐ 0.00725 kL
  2. ⓑ 6300 mL

Convert: ⓐ 350 hL to L ⓑ 4.1 L to cL.

Solution
  1. ⓐ 35,000 L
  2. ⓑ 410 cL

Use Mixed Units of Measurement in the Metric System

Performing arithmetic operations on measurements with mixed units of measures in the metric system requires the same care we used in the U.S. system. But it may be easier because of the relation of the units to the powers of 10. We still must make sure to add or subtract like units.

Ryland is 1.6 meters tall. His younger brother is 85 centimeters tall. How much taller is Ryland than his younger brother?

Solution

Solution

We will subtract the lengths in meters. Convert 85 centimeters to meters by moving the decimal 2 places to the left; 85 cm is the same as 0.85 m.

Now that both measurements are in meters, subtract to find out how much taller Ryland is than his brother.

1.60 m−0.85 m_______0.75 m

Ryland is 0.75 meters taller than his brother.

Mariella is 1.58 meters tall. Her daughter is 75 centimeters tall. How much taller is Mariella than her daughter? Write the answer in centimeters.

Solution

83 cm

The fence around Hank’s yard is 2 meters high. Hank is 96 centimeters tall. How much shorter than the fence is Hank? Write the answer in meters.

Solution

1.04 m

Dena’s recipe for lentil soup calls for 150 milliliters of olive oil. Dena wants to triple the recipe. How many liters of olive oil will she need?

Solution

Solution

We will find the amount of olive oil in milliliters then convert to liters.

This table demonstrates the step-by-step calculation of triple 150 mL, including unit conversion to liters, and presents the final solution.
Triple 150 mL
Translate to algebra. 3·150mL
Multiply. 450mL
Convert to liters. 450mL·0.001L1mL
Simplify. 0.45L
Dena needs 0.45 liter of olive oil.

A recipe for Alfredo sauce calls for 250 milliliters of milk. Renata is making pasta with Alfredo sauce for a big party and needs to multiply the recipe amounts by 8. How many liters of milk will she need?

Solution

2 L

To make one pan of baklava, Dorothea needs 400 grams of filo pastry. If Dorothea plans to make 6 pans of baklava, how many kilograms of filo pastry will she need?

Solution

2.4 kg

Convert Between U.S. and Metric Systems of Measurement

Many measurements in the United States are made in metric units. A drink may come in 2-liter bottles, calcium may come in 500-mg capsules, and we may run a 5-K race. To work easily in both systems, we need to be able to convert between the two systems.

Table 14 shows some of the most common conversions.

Conversion Factors Between U.S. and Metric Systems
Length Weight Volume
1 in = 2.54 cm
1 ft = 0.305 m
1 yd = 0.914 m
1 mi = 1.61 km


1 m = 3.28 ft
1 lb = 0.45 kg
1 oz = 28 g




1 kg = 2.2 lb
1 qt = 0.95 L
1 fl oz = 30 mL




1 L = 1.06 qt

We make conversions between the systems just as we do within the systems—by multiplying by unit conversion factors.

Lee’s water bottle holds 500 mL of water. How many fluid ounces are in the bottle? Round to the nearest tenth of an ounce.

Solution

Solution

This table demonstrates the step-by-step conversion of 500 mL to fluid ounces, resulting in 16.7 fl. oz.
500 mL
Multiply by a unit conversion factor relating mL and ounces. 500mL·1fl oz30mL
Simplify. 500fl oz30
Divide. 16.7fl. oz.
The water bottle holds 16.7 fluid ounces.

How many quarts of soda are in a 2-liter bottle?

Solution

2.12 quarts

How many liters are in 4 quarts of milk?

Solution

3.8 liters

The conversion factors in Table 14 are not exact, but the approximations they give are close enough for everyday purposes. In Example 12, we rounded the number of fluid ounces to the nearest tenth.

Soleil lives in Minnesota but often travels in Canada for work. While driving on a Canadian highway, she passes a sign that says the next rest stop is in 100 kilometers. How many miles until the next rest stop? Round your answer to the nearest mile.

Solution

Solution

This table outlines the step-by-step conversion of 100 kilometers to approximately 62 miles, detailing the calculation process.
100 kilometers
Multiply by a unit conversion factor relating kilometers and miles. 100kilometers·1mile1.61kilometers
100·1mi1.61km
Simplify. 100mi1.61
Divide. 62 mi
It is about 62 miles to the next rest stop.

The height of Mount Kilimanjaro is 5,895 meters. Convert the height to feet. Round to the nearest foot.

Solution

19,336 ft

The flight distance from New York City to London is 5,586 kilometers. Convert the distance to miles. Round to the nearest mile.

Solution

3,470 mi

Convert Between Fahrenheit and Celsius Temperatures

Have you ever been in a foreign country and heard the weather forecast? If the forecast is for 22°C. What does that mean?

The U.S. and metric systems use different scales to measure temperature. The U.S. system uses degrees Fahrenheit, written °F. The metric system uses degrees Celsius, written °C. Figure 5 shows the relationship between the two systems.

On the left side of the figure is a thermometer marked in degrees Celsius. The bottom of the thermometer begins with negative 20 degrees Celsius and ranges up to 100 degrees Celsius. There are tick marks on the thermometer every 5 degrees with every 10 degrees labeled. On the right side is a thermometer marked in degrees Fahrenheit. The bottom of the thermometer begins with negative 10 degrees Fahrenheit and ranges up to 212 degrees Fahrenheit. There are tick marks on the thermometer every 2 degrees with every 10 degrees labeled. Between the thermometers there is an arrow pointing on the left to 0 degrees Celsius and on the right to 32 degrees Fahrenheit. This is the temperature at which water freezes. Another arrow points on the left to 37 degrees Celsius and on the right to 98.6 degrees Fahrenheit. This is normal body temperature. A third arrow points on the left to 100 degrees Celsius and on the right to 212 degrees Fahrenheit. This is the temperature at which water boils.
A temperature of 37°C is equivalent to 98.6°F.

If we know the temperature in one system, we can use a formula to convert it to the other system.

Temperature Conversion

To convert from Fahrenheit temperature, F, to Celsius temperature, C, use the formula

C=59(F−32)

To convert from Celsius temperature, C, to Fahrenheit temperature, F, use the formula

F=95C+32

Convert 50°F into degrees Celsius.

Solution

Solution

We will substitute 50°F into the formula to find C.
Use the formula for converting °F to °C C=59(F−32)
The image displays the instruction 'Substitute 50 for F.' in a dark blue font, with the number 50 highlighted in red. The image shows the formula for converting temperature from Fahrenheit to Celsius, C = 5/9(F - 32), with the Fahrenheit temperature F given as 50 (highlighted in red) to calculate C = 5/9(50 - 32).
Simplify in parentheses. C=59(18)
Multiply. C=10
A temperature of 50°F is equivalent to 10°C.

Convert the Fahrenheit temperatures to degrees Celsius: 59°F.

Solution

15°C

Convert the Fahrenheit temperatures to degrees Celsius: 41°F.

Solution

5°C

The weather forecast for Paris predicts a high of 20°C. Convert the temperature into degrees Fahrenheit.

Solution

Solution

We will substitute 20°C into the formula to find F.
Use the formula for converting °F to °C F=95C+32
The text 'Substitute 20 for C.' is displayed in a dark teal color, with the number '20' highlighted in red. A mathematical equation for converting Celsius to Fahrenheit is displayed: F = (9/5)(20) + 32. The number 20, representing the Celsius temperature, is highlighted in red.
Multiply. F=36+32
Add. F=68
So 20°C is equivalent to 68°F.

Convert the Celsius temperatures to degrees Fahrenheit:

The temperature in Helsinki, Finland was 15°C.

Solution

59°F

Convert the Celsius temperatures to degrees Fahrenheit:

The temperature in Sydney, Australia was 10°C.

Solution

50°F

ACCESS ADDITIONAL ONLINE RESOURCES

  • American Unit Conversion
  • Time Conversions
  • Metric Unit Conversions
  • American and Metric Conversions
  • Convert from Celsius to Fahrenheit
  • Convert from Fahrenheit to Celsius

Section Exercises

Practice Makes Perfect

Make Unit Conversions in the U.S. System

In the following exercises, convert the units.

A park bench is 6 feet long. Convert the length to inches.

A floor tile is 2 feet wide. Convert the width to inches.

Solution

24 inches

A ribbon is 18 inches long. Convert the length to feet.

Carson is 45 inches tall. Convert his height to feet.

Solution

3.75 feet

Jon is 6 feet 4 inches tall. Convert his height to inches.

Faye is 4 feet 10 inches tall. Convert her height to inches.

Solution

58 inches

A football field is 160 feet wide. Convert the width to yards.

On a baseball diamond, the distance from home plate to first base is 30 yards. Convert the distance to feet.

Solution

90 feet

Ulises lives 1.5 miles from school. Convert the distance to feet.

Denver, Colorado, is 5,183 feet above sea level. Convert the height to miles.

Solution

0.98 miles

A killer whale weighs 4.6 tons. Convert the weight to pounds.

Blue whales can weigh as much as 150 tons. Convert the weight to pounds.

Solution

300,000 pounds

An empty bus weighs 35,000 pounds. Convert the weight to tons.

At take-off, an airplane weighs 220,000 pounds. Convert the weight to tons.

Solution

110 tons

The voyage of the Mayflower took 2 months and 5 days. Convert the time to days (30 days = 1 month).

Lynn’s cruise lasted 6 days and 18 hours. Convert the time to hours.

Solution

162 hours

Rocco waited 112 hours for his appointment. Convert the time to seconds.

Misty’s surgery lasted 214 hours. Convert the time to seconds.

Solution

8100 seconds

How many teaspoons are in a pint?

How many tablespoons are in a gallon?

Solution

256 tablespoons

JJ’s cat, Posy, weighs 14 pounds. Convert her weight to ounces.

April’s dog, Beans, weighs 8 pounds. Convert his weight to ounces.

Solution

128 ounces

Baby Preston weighed 7 pounds 3 ounces at birth. Convert his weight to ounces.

Baby Audrey weighed 6 pounds 15 ounces at birth. Convert her weight to ounces.

Solution

111 ounces

Crista will serve 20 cups of juice at her son’s party. Convert the volume to gallons.

Lance needs 500 cups of water for the runners in a race. Convert the volume to gallons.

Solution

31.25 gallons

Use Mixed Units of Measurement in the U.S. System

In the following exercises, solve and write your answer in mixed units.

Eli caught three fish. The weights of the fish were 2 pounds 4 ounces, 1 pound 11 ounces, and 4 pounds 14 ounces. What was the total weight of the three fish?

Judy bought 1 pound 6 ounces of almonds, 2 pounds 3 ounces of walnuts, and 8 ounces of cashews. What was the total weight of the nuts?

Solution

4 lbs. 1 oz.

One day Anya kept track of the number of minutes she spent driving. She recorded trips of 45,10,8,65,20,and 35 minutes. How much time (in hours and minutes) did Anya spend driving?

Last year Eric went on 6 business trips. The number of days of each was 5,2,8,12,6,and 3. How much time (in weeks and days) did Eric spend on business trips last year?

Solution

5 weeks and 1 day

Renee attached a 6-foot-6-inch extension cord to her computer’s 3-foot-8-inch power cord. What was the total length of the cords?

Fawzi’s SUV is 6 feet 4 inches tall. If he puts a 2-foot-10-inch box on top of his SUV, what is the total height of the SUV and the box?

Solution

9 ft 2 in

Leilani wants to make 8 placemats. For each placemat she needs 18 inches of fabric. How many yards of fabric will she need for the 8 placemats?

Mireille needs to cut 24 inches of ribbon for each of the 12 girls in her dance class. How many yards of ribbon will she need altogether?

Solution

8 yards

Make Unit Conversions in the Metric System

In the following exercises, convert the units.

Ghalib ran 5 kilometers. Convert the length to meters.

Kitaka hiked 8 kilometers. Convert the length to meters.

Solution

8000 meters

Estrella is 1.55 meters tall. Convert her height to centimeters.

The width of the wading pool is 2.45 meters. Convert the width to centimeters.

Solution

245 centimeters

Mount Whitney is 3,072 meters tall. Convert the height to kilometers.

The depth of the Mariana Trench is 10,911 meters. Convert the depth to kilometers.

Solution

10.911 kilometers

June’s multivitamin contains 1,500 milligrams of calcium. Convert this to grams.

A typical ruby-throated hummingbird weights 3 grams. Convert this to milligrams.

Solution

3000 milligrams

One stick of butter contains 91.6 grams of fat. Convert this to milligrams.

One serving of gourmet ice cream has 25 grams of fat. Convert this to milligrams.

Solution

25,000 milligrams

The maximum mass of an airmail letter is 2 kilograms. Convert this to grams.

Dimitri’s daughter weighed 3.8 kilograms at birth. Convert this to grams.

Solution

3800 grams

A bottle of wine contained 750 milliliters. Convert this to liters.

A bottle of medicine contained 300 milliliters. Convert this to liters.

Solution

0.3 liters

Use Mixed Units of Measurement in the Metric System

In the following exercises, solve and write your answer in mixed units.

Matthias is 1.8 meters tall. His son is 89 centimeters tall. How much taller, in centimeters, is Matthias than his son?

Stavros is 1.6 meters tall. His sister is 95 centimeters tall. How much taller, in centimeters, is Stavros than his sister?

Solution

65 centimeters

A typical dove weighs 345 grams. A typical duck weighs 1.2 kilograms. What is the difference, in grams, of the weights of a duck and a dove?

Concetta had a 2-kilogram bag of flour. She used 180 grams of flour to make biscotti. How many kilograms of flour are left in the bag?

Solution

1.82 kilograms

Harry mailed 5 packages that weighed 420 grams each. What was the total weight of the packages in kilograms?

One glass of orange juice provides 560 milligrams of potassium. Linda drinks one glass of orange juice every morning. How many grams of potassium does Linda get from her orange juice in 30 days?

Solution

16.8 grams

Jonas drinks 200 milliliters of water 8 times a day. How many liters of water does Jonas drink in a day?

One serving of whole grain sandwich bread provides 6 grams of protein. How many milligrams of protein are provided by 7 servings of whole grain sandwich bread?

Solution

42,000 milligrams

Convert Between U.S. and Metric Systems

In the following exercises, make the unit conversions. Round to the nearest tenth.

Bill is 75 inches tall. Convert his height to centimeters.

Frankie is 42 inches tall. Convert his height to centimeters.

Solution

106.7 centimeters

Marcus passed a football 24 yards. Convert the pass length to meters.

Connie bought 9 yards of fabric to make drapes. Convert the fabric length to meters.

Solution

8.2 meters

Each American throws out an average of 1,650 pounds of garbage per year. Convert this weight to kilograms (2.20 pounds = 1 kilogram).

An average American will throw away 90,000 pounds of trash over his or her lifetime. Convert this weight to kilograms (2.20 pounds = 1 kilogram).

Solution

40,900 kilograms

A 5K run is 5 kilometers long. Convert this length to miles.

Kathryn is 1.6 meters tall. Convert her height to feet.

Solution

5.2 feet

Dawn’s suitcase weighed 20 kilograms. Convert the weight to pounds.

Jackson’s backpack weighs 15 kilograms. Convert the weight to pounds.

Solution

33 pounds

Ozzie put 14 gallons of gas in his truck. Convert the volume to liters.

Bernard bought 8 gallons of paint. Convert the volume to liters.

Solution

30.2 liters

Convert between Fahrenheit and Celsius

In the following exercises, convert the Fahrenheit temperature to degrees Celsius. Round to the nearest tenth.

86°F

77°F

Solution

25°C

104°F

14°F

Solution

−10°C

72°F

4°F

Solution

−15.6°C

0°F

120°F

Solution

48.9°C

In the following exercises, convert the Celsius temperatures to degrees Fahrenheit. Round to the nearest tenth.

5°C

25°C

Solution

77°F

−10°C

−15°C

Solution

5°F

22°C

8°C

Solution

46.4°F

43°C

16°C

Solution

60.8°F

Everyday Math

Nutrition Julian drinks one can of soda every day. Each can of soda contains 40 grams of sugar. How many kilograms of sugar does Julian get from soda in 1 year?

Reflectors The reflectors in each lane-marking stripe on a highway are spaced 16 yards apart. How many reflectors are needed for a one-mile-long stretch of highway?

Solution

110 reflectors

Writing Exercises

Some people think that 65° to 75° Fahrenheit is the ideal temperature range.

  1. ⓐ What is your ideal temperature range? Why do you think so?

  2. ⓑ Convert your ideal temperatures from Fahrenheit to Celsius.

ⓐ Did you grow up using the U.S. customary or the metric system of measurement? ⓑ Describe two examples in your life when you had to convert between systems of measurement. ⓒ Which system do you think is easier to use? Explain.

Self Check

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

A self-assessment checklist for students to evaluate their proficiency in unit conversions and measurements across U.S. and metric systems, including temperature conversions.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next chapter? Why or why not?

Chapter Review Exercises

Rational and Irrational Numbers

In the following exercises, write as the ratio of two integers.

6

−5

Solution

−51

2.9

1.8

Solution

1810

In the following exercises, determine which of the numbers is rational.

0.42,0.3–,2.56813…

0.75319…,0.16—,1.95

Solution

0.16—,1.95

In the following exercises, identify whether each given number is rational or irrational.

ⓐ 49 ⓑ 55

ⓐ 72 ⓑ 64

Solution
  1. ⓐ irrational
  2. ⓑ rational

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

−9,0,0.361....,89,16,9

−5,−214,−4,0.25—,135,4

Solution
  1. ⓐ 4
  2. ⓑ −5,−4,4
  3. ⓒ −5,−214,−4,0.25—,135,4
  4. ⓓ none
  5. ⓔ −5,−214,−4,0.25—,135,4

Commutative and Associative Properties

In the following exercises, use the commutative property to rewrite the given expression.

6+4=____

−14·5=____

Solution

−14·5 = 5(−14)

3n=____

a+8=____

Solution

a + 8 = 8 + a

In the following exercises, use the associative property to rewrite the given expression.

(13·5)·2=_____

(22+7)+3=_____

Solution

(22 + 7) + 3 = 22 + (7 + 3)

(4+9x)+x=_____

12(22y)=_____

Solution

12(22y)=(12·22)y

In the following exercises, evaluate each expression for the given value.

If y=1112, evaluate:
ⓐ y+0.7+(−y)
ⓑ y+(−y)+0.7

If z=−53, evaluate:
ⓐ z+5.39+(−z)
ⓑ z+(−z)+5.39

Solution
  1. ⓐ 5.39
  2. ⓑ 5.39

If k=65, evaluate:
ⓐ 49(94k)
ⓑ (49·94)k

If m=−13, evaluate:
ⓐ −25(52m)
ⓑ (−25·52)m

Solution
  1. ⓐ 13
  2. ⓑ 13

In the following exercises, simplify using the commutative and associative properties.

6y+37+(−6y)

14+1115+(−14)

Solution

1115

1411·359·1114

−18·15·29

Solution

−60

(712+45)+15

(3.98d+0.75d)+1.25d

Solution

5.98 d

−12(4m)

30(56q)

Solution

25 q

11x+8y+16x+15y

52m+(−20n)+(−18m)+(−5n)

Solution

34 m + (−25 n)

Distributive Property

In the following exercises, simplify using the distributive property.

7(x+9)

9(u−4)

Solution

9y − 36

−3(6m−1)

−8(−7a−12)

Solution

56a + 96

13(15n−6)

(y+10)·p

Solution

yp + 10p

(a−4)−(6a+9)

4(x+3)−8(x−7)

Solution

−4x + 68

In the following exercises, evaluate using the distributive property.

If u=2, evaluate
ⓐ 3(8u+9)and
ⓑ 3·8u+3·9 to show that 3(8u+9)=3·8u+3·9

If n=78, evaluate
ⓐ 8(n+14) and
ⓑ 8·n+8·14 to show that 8(n+14)=8·n+8·14

Solution
  1. ⓐ 9
  2. ⓑ 9

If d=14, evaluate
ⓐ −100(0.1d+0.35) and
ⓑ −100·(0.1d)+(−100)(0.35) to show that −100(0.1d+0.35)=−100·(0.1d)+(−100)(0.35)

If y=−18, evaluate
ⓐ −(y−18) and
ⓑ −y+18 to show that −(y−18)=−y+18

Solution
  1. ⓐ 36
  2. ⓑ 36

Properties of Identities, Inverses, and Zero

In the following exercises, identify whether each example is using the identity property of addition or multiplication.

−35(1)=−35

29+0=29

Solution

identity property of addition

(6x+0)+4x=6x+4x

9·1+(−3)=9+(−3)

Solution

identity property of multiplication

In the following exercises, find the additive inverse.

−32

19.4

Solution

−19.4

35

−715

Solution

715

In the following exercises, find the multiplicative inverse.

92

−5

Solution

−15

110

−49

Solution

−94

In the following exercises, simplify.

83·0

09

Solution

0

50

0÷23

Solution

0

43+39+(−43)

(n+6.75)+0.25

Solution

n + 7

513·57·135

16·17·12

Solution

34

23·28·37

9(6x−11)+15

Solution

54x − 84

Systems of Measurement

In the following exercises, convert between U.S. units. Round to the nearest tenth.

A floral arbor is 7 feet tall. Convert the height to inches.

A picture frame is 42 inches wide. Convert the width to feet.

Solution

3.5 feet

Kelly is 5 feet 4 inches tall. Convert her height to inches.

A playground is 45 feet wide. Convert the width to yards.

Solution

15 yards

The height of Mount Shasta is 14,179 feet. Convert the height to miles.

Shamu weighs 4.5 tons. Convert the weight to pounds.

Solution

9000 pounds

The play lasted 134 hours. Convert the time to minutes.

How many tablespoons are in a quart?

Solution

64 tablespoons

Naomi’s baby weighed 5 pounds 14 ounces at birth. Convert the weight to ounces.

Trinh needs 30 cups of paint for her class art project. Convert the volume to gallons.

Solution

1.9 gallons

In the following exercises, solve, and state your answer in mixed units.

John caught 4 lobsters. The weights of the lobsters were 1 pound 9 ounces, 1 pound 12 ounces, 4 pounds 2 ounces, and 2 pounds 15 ounces. What was the total weight of the lobsters?

Every day last week, Pedro recorded the amount of time he spent reading. He read for 50,25,83,45,32,60,and135 minutes. How much time, in hours and minutes, did Pedro spend reading?

Solution

7 hours 10 minutes

Fouad is 6 feet 2 inches tall. If he stands on a rung of a ladder 8 feet 10 inches high, how high off the ground is the top of Fouad’s head?

Dalila wants to make pillow covers. Each cover takes 30 inches of fabric. How many yards and inches of fabric does she need for 4 pillow covers?

Solution

3 yards, 12 inches

In the following exercises, convert between metric units.

Donna is 1.7 meters tall. Convert her height to centimeters.

Mount Everest is 8,850 meters tall. Convert the height to kilometers.

Solution

8.85 kilometers

One cup of yogurt contains 488 milligrams of calcium. Convert this to grams.

One cup of yogurt contains 13 grams of protein. Convert this to milligrams.

Solution

13,000 milligrams

Sergio weighed 2.9 kilograms at birth. Convert this to grams.

A bottle of water contained 650 milliliters. Convert this to liters.

Solution

0.65 liters

In the following exercises, solve.

Minh is 2 meters tall. His daughter is 88 centimeters tall. How much taller, in meters, is Minh than his daughter?

Selma had a 1-liter bottle of water. If she drank 145 milliliters, how much water, in milliliters, was left in the bottle?

Solution

855 milliliters

One serving of cranberry juice contains 30 grams of sugar. How many kilograms of sugar are in 30 servings of cranberry juice?

One ounce of tofu provides 2 grams of protein. How many milligrams of protein are provided by 5 ounces of tofu?

Solution

10,000 milligrams

In the following exercises, convert between U.S. and metric units. Round to the nearest tenth.

Majid is 69 inches tall. Convert his height to centimeters.

A college basketball court is 84 feet long. Convert this length to meters.

Solution

25.6 meters

Caroline walked 2.5 kilometers. Convert this length to miles.

Lucas weighs 78 kilograms. Convert his weight to pounds.

Solution

171.6 pounds

Steve’s car holds 55 liters of gas. Convert this to gallons.

A box of books weighs 25 pounds. Convert this weight to kilograms.

Solution

11.4 kilograms

In the following exercises, convert the Fahrenheit temperatures to degrees Celsius. Round to the nearest tenth.

95°F

23°F

Solution

−5°C

20°F

64°F

Solution

17.8°C

In the following exercises, convert the Celsius temperatures to degrees Fahrenheit. Round to the nearest tenth.

30°C

−5°C

Solution

23°F

−12°C

24°C

Solution

75.2°F

Chapter Practice Test

For the numbers 0.18349…,0.2–,1.67, list the ⓐ rational numbers and ⓑ irrational numbers.

Is 144 rational or irrational?

Solution

144=12therefore rational.

From the numbers −4,−112,0,58,2,7, which are ⓐ integers ⓑ rational ⓒ irrational ⓓ real numbers?

Rewrite using the commutative property: x·14=_________

Solution

x·14 = 14·x

Rewrite the expression using the associative property: (y+6)+3=_______________

Rewrite the expression using the associative property: (8·2)·5=___________

Solution

(8·2)·5 = 8·(2·5)

Evaluate 316(163n) when n=42.

For the number 25 find the ⓐ additive inverse ⓑ multiplicative inverse.

Solution
  1. ⓐ −25
  2. ⓑ 52

In the following exercises, simplify the given expression.

34(−29)(43)

−3+15y+3

Solution

15y

(1.27q+0.25q)+0.75q

(815+29)+79

Solution

2315

−18(32n)

14y+(−6z)+16y+2z

Solution

30y − 4z

9(q+9)

6(5x−4)

Solution

30x − 24

−10(0.4n+0.7)

14(8a+12)

Solution

2a + 3

m(n+2)

8(6p−1)+2(9p+3)

Solution

66p − 2

(12a+4)−(9a+6)

08

Solution

0

4.50

0÷(23)

Solution

0

In the following exercises, solve using the appropriate unit conversions.

Azize walked 412 miles. Convert this distance to feet. (1 mile=5,280 feet).

One cup of milk contains 276 milligrams of calcium. Convert this to grams. (1 milligram=0.001 gram)

Solution

.276 grams

Larry had 5 phone customer phone calls yesterday. The calls lasted 28,44,9,75,and55 minutes. How much time, in hours and minutes, did Larry spend on the phone? (1 hour=60 minutes)

Janice ran 15 kilometers. Convert this distance to miles. Round to the nearest hundredth of a mile. (1 mile=1.61 kilometers)

Solution

9.317 miles

Yolie is 63 inches tall. Convert her height to centimeters. Round to the nearest centimeter. (1 inch=2.54 centimeters)

Use the formula F=95C+32 to convert 35°C to degrees F

Solution

95°F

Introduction to Solving Linear Equations

An image of a calder mobile is shown. It has several black and red geometric shapes hanging down.
A Calder mobile is balanced and has several elements on each side. (credit: paurian, Flickr)

Teetering high above the floor, this amazing mobile remains aloft thanks to its carefully balanced mass. Any shift in either direction could cause the mobile to become lopsided, or even crash downward. In this chapter, we will solve equations by keeping quantities on both sides of an equal sign in perfect balance.

Solve Equations Using the Subtraction and Addition Properties of Equality

Learning Objectives

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

  • Solve equations using the Subtraction and Addition Properties of Equality
  • Solve equations that need to be simplified
  • Translate an equation and solve
  • Translate and solve applications

Before you get started, take this readiness quiz.

Solve: n−12=16.
If you missed this problem, review Example 6 in Solving Equations Using the Subtraction and Addition Properties of Equality.

Solution

28

Translate into algebra ‘five less than x.’
If you missed this problem, review Example 12 in Evaluate, Simplify, and Translate Expressions.

Solution

x−5

Is x=2 a solution to 5x−3=7?
If you missed this problem, review Example 1 in Solving Equations Using the Subtraction and Addition Properties of Equality.

Solution

yes

We are now ready to “get to the good stuff.” You have the basics down and are ready to begin one of the most important topics in algebra: solving equations. The applications are limitless and extend to all careers and fields. Also, the skills and techniques you learn here will help improve your critical thinking and problem-solving skills. This is a great benefit of studying mathematics and will be useful in your life in ways you may not see right now.

Solve Equations Using the Subtraction and Addition Properties of Equality

We began our work solving equations in previous chapters. It has been a while since we have seen an equation, so we will review some of the key concepts before we go any further.

We said that 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 make each side of the equation the same. 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.

In the earlier sections, we listed the steps to determine if a value is a solution. We restate them here.

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 y=34 is a solution for 4y+3=8y.

Solution

Solution

An image displaying the algebraic equation '4y + 3 = 8y' on a white background. The equation is rendered in a black, clear, serif-like font.
The text 'Substitute 3/4 for y.' is displayed on a white background. The word 'Substitute' is in dark blue, the fraction '3/4' is in red, and 'for y.' is in dark blue. A mathematical equation is displayed: 4(3/4) + 3 ?= 8(3/4). The question mark above the equals sign asks to verify if the statement is true. The numbers 3/4 are highlighted in red.
Multiply. A mathematical equation '3 + 3 = 6' is displayed with a question mark above the equals sign, implying a query about its truth or an unknown.
Add. The image shows the mathematical statement '6 = 6' followed by a checkmark, indicating that the equality is true. The text is rendered in a black sans-serif font against a plain white background.

Since y=34 results in a true equation, 34 is a solution to the equation 4y+3=8y.

Is y=23 a solution for 9y+2=6y?

Solution

no

Is y=25 a solution for 5y−3=10y?

Solution

no

We introduced the Subtraction and Addition Properties of Equality in Solving Equations Using the Subtraction and Addition Properties of Equality. In that section, we modeled how these properties work and then applied them to solving equations with whole numbers. We used these properties again each time we introduced a new system of numbers. Let’s review those properties here.

Subtraction and Addition Properties of Equality

Subtraction Property of Equality

For all real numbers a,b, and c, if a=b, then a−c=b−c.

Addition Property of Equality

For all real numbers a,b, and c, if a=b, then a+c=b+c.

When you add or subtract the same quantity from both sides of an equation, you still have equality.

We introduced the Subtraction Property of Equality earlier by modeling equations with envelopes and counters. Figure 1 models the equation x+3=8.

An envelope and three yellow counters are shown on the left side. On the right side are eight yellow counters.

The goal is to isolate the variable on one side of the equation. So we ‘took away’ 3 from both sides of the equation and found the solution x=5.

Some people picture a balance scale, as in Figure 2, when they solve equations.

Three balance scales are shown. The top scale has one red weight on each side and is balanced. Beside it is “1 mass on each side equals balanced.” The next scale has two weights on each side and is balanced. Beside it is “2 masses on each side equals balanced.” The bottom scale has one weight on the left and two on the right. The right side is lower than the left. Beside the image is “1 mass on one side and 2 masses on the other equals unbalanced.”

The quantities on both sides of the equal sign in an equation are equal, or balanced. Just as with the balance scale, whatever you do to one side of the equation you must also do to the other to keep it balanced.

Let’s review how to use Subtraction and Addition Properties of Equality to solve equations. We need to isolate the variable on one side of the equation. And we check our solutions by substituting the value into the equation to make sure we have a true statement.

Solve: x-11=−3.

Solution

Solution

To isolate x, we undo the addition of 11 by using the Subtraction Property of Equality.

A mathematical equation is displayed, showing 'x - 11 = -3' in black characters against a white background.
We "undo" the subtraction of 11 by adding 11 to each side. The equation x - 11 + 11 = -3 + 11, illustrating the step of adding 11 to both sides to solve for x. The added '11's are highlighted in red.
Simplify. The equation x = 8 is displayed in black text on a white background.
Check: A mathematical equation is displayed, showing 'x - 11 = -3' in black text against a white background. This is a linear equation with one variable, 'x'.
Substitute x=8. A mathematical equation reads 8 - 11 = -3, with a question mark positioned above the equals sign, suggesting a verification of its correctness.
A mathematical equation shows '-3 = -3' with a black checkmark next to it on a white background, confirming the equality is correct.

Since x=8 makes x-11=−3 a true statement, we know that it is a solution to the equation.

Solve: x+9=−7.

Solution

x = −16

Solve: x+16=−4.

Solution

x = −20

In the original equation in the previous example, 11 was added to the x, so we subtracted 11 to ‘undo’ the addition. In the next example, we will need to ‘undo’ subtraction by using the Addition Property of Equality.

Solve: m+4=−5.

Solution

Solution

The equation m + 4 = -5 is displayed in black text on a white background.
Subtract 4 from each side to "undo" the addition. A math problem demonstrating a step in solving for 'm', where 4 is subtracted from both sides of the equation m + 4 = -5. The subtractions are highlighted in red.
Simplify. The image displays the algebraic equation m = -9, set against a plain white background.
Check: The image displays the algebraic equation m + 4 = -5 in black text against a white background.
Substitute m=−9. A mathematical equation displays '-9 + 4' followed by an equals sign with a question mark above it, and then '-5', querying whether -9 plus 4 is equal to -5.
A close-up shot of a mathematical equation '-5 = -5' followed by a checkmark, indicating correctness or verification.
The solution to m−4=−5 is m=−1.

Solve: n−6=−7.

Solution

n = −1

Solve: x−5=−9.

Solution

x = −4

Now let’s review solving equations with fractions.

Solve: n−38=12.

Solution

Solution

A mathematical equation is displayed on a white background, which reads 'n - 3/8 = 1/2'. The variable 'n' is followed by a minus sign, then the fraction three-eighths, an equals sign, and finally the fraction one-half.
Use the Addition Property of Equality. A mathematical equation shows 'n minus three-eighths plus three-eighths equals one-half plus three-eighths.' The plus signs and the last 'three-eighths' are in red, while the rest are black.
Find the LCD to add the fractions on the right. A mathematical equation showing 'n minus three-eighths plus three-eighths equals four-eighths plus three-eighths'.
Simplify The mathematical equation n = 7/8 is displayed in the center of a white background.
Check: A mathematical equation shows 'n minus three eighths equals one half'.
The text reads 'Substitute n = 7/8.' The word 'Substitute' and the variable 'n' are in a dark teal color, while the fraction '7/8' is in red. The equality sign '=' is also in teal. A math equation reads '7/8 - 3/8 = ? 1/2'. The problem involves subtracting two fractions with the same denominator and comparing the result to one-half.
Subtract. A mathematical equation showing the fractions 4/8 and 1/2 separated by a question mark, indicating a missing operator. The question asks what symbol should replace the question mark.
Simplify. The image displays the equation 1/2 = 1/2, followed by a checkmark, indicating that the equality is correct or verified.
The solution checks.

Solve: p−13=56.

Solution

p=76

Solve: q−12=16.

Solution

q=23

In Solve Equations with Decimals, we solved equations that contained decimals. We’ll review this next.

Solve a−3.7=4.3.

Solution

Solution

A mathematical equation is displayed, reading 'a - 3.7 = 4.3'. The variable 'a' is followed by a minus sign, then the decimal number '3.7', an equals sign, and finally the decimal number '4.3'.
Use the Addition Property of Equality. The equation a - 3.7 + 3.7 = 4.3 + 3.7 illustrates adding 3.7 (highlighted in red) to both sides to isolate the variable 'a', applying the addition property of equality.
Add. The mathematical equation 'a = 8' is displayed on a white background, representing that the variable 'a' is equal to the number 8.
Check: A mathematical equation shows 'a minus 3.7 equals 4.3' in black text on a white background.
Substitute a=8. A math problem is displayed: 8 - 3.7 =? 4.3. The number '8' is colored red, and a question mark is placed directly above the equals sign, suggesting an inquiry into whether the subtraction equals 4.3.
Simplify. The mathematical equation '4.3 = 4.3' is displayed in black text on a white background, accompanied by a black checkmark indicating correctness.
The solution checks.

Solve: b−2.8=3.6.

Solution

b = 6.4

Solve: c−6.9=7.1.

Solution

c = 14

Solve Equations That Need to Be Simplified

In the examples up to this point, we have been able to isolate the variable with just one operation. Many of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before trying to isolate the variable.

Solve: 3x−7−2x−4=1.

Solution

Solution

The left side of the equation has an expression that we should simplify before trying to isolate the variable.

A mathematical equation is displayed against a white background, reading '3x - 7 - 2x - 4 = 1'.
Rearrange the terms, using the Commutative Property of Addition. A mathematical equation on a white background: 3x - 2x - 7 - 4 = 1.
Combine like terms. A mathematical equation is displayed with a white background. The equation reads 'x - 11 = 1' with the characters rendered in black.
Add 11 to both sides to isolate x. A mathematical equation x - 11 + 11 = 1 + 11 is presented. The terms + 11 on both sides of the equation are highlighted in red, while the rest of the numbers, variables, and operators are in black.
Simplify. The equation x = 12 is displayed in black text against a white background.
Check.
Substitute x=12 into the original equation.
The top line shows 3x minus 7 minus 2x minus 4 equals 1. Below this is 3 times a red 12 minus 7 minus 2 times a red 12 minus 4 equals 1. Next is 36 minus 7 minus 24 minus 4 equals 1. Below is 29 minus 24 minus 4 equals 1. Next is 5 minus 4 equals 1. Last is 1 equals 1.

The solution checks.

Solve: 8y−4−7y−7=4.

Solution

y = 15

Solve: 6z+5−5z−4=3.

Solution

z = 2

Solve: 3(n−4)−2n=−3.

Solution

Solution

The left side of the equation has an expression that we should simplify.

A mathematical equation is displayed: 3(n - 4) - 2n = -3. It's an algebraic problem requiring solving for the variable 'n', potentially involving distribution and combining like terms.
Distribute on the left. A mathematical equation is displayed with the expression 3n - 12 - 2n = -3, showing variables, constants, and operators in a standard algebraic format.
Use the Commutative Property to rearrange terms. A mathematical equation is displayed on a white background, which reads '3n - 2n - 12 = -3'.
Combine like terms. A simple algebraic equation is displayed, showing 'n - 12 = -3' with the variable 'n' to be solved for.
Isolate n using the Addition Property of Equality. The mathematical equation n - 12 + 12 = -3 + 12, showing 12 added to both sides in red to solve for n.
Simplify. The mathematical equation 'n=9' is displayed in bold black text on a white background. The letter 'n' is lowercase and italicized, followed by an equals sign and the numeral '9'.
Check.
Substitute n=9 into the original equation.
The top line says 3 times parentheses n minus 4 minus 2n equals negative 3. The next line says 3 times parentheses red 9 minus 3 minus 2 times red 9 equals negative 3. The next line says 3 times 5 minus 18 equals negative 3. Below this is 15 minus 18 equals negative 3. Last is negative 3 equals negative 3.
The solution checks.

Solve: 5(p−3)−4p=−10.

Solution

p = 5

Solve: 4(q+2)−3q=−8.

Solution

q = −16

Solve: 2(3k−1)−5k=−2−7.

Solution

Solution

Both sides of the equation have expressions that we should simplify before we isolate the variable.

A mathematical equation is displayed: 2(3k - 1) - 5k = -2 - 7.
Distribute on the left, subtract on the right. A mathematical equation is displayed, showing '6k - 2 - 5k = -9'. This is an algebraic expression involving the variable 'k' that needs to be solved.
Use the Commutative Property of Addition. The image shows the algebraic equation 6k - 5k - 2 = -9, which simplifies to k - 2 = -9.
Combine like terms. An algebraic equation showing 'k - 2 = -9' against a white background.
Undo subtraction by using the Addition Property of Equality. An algebraic equation k - 2 + 2 = -9 + 2, illustrating the step of adding 2 to both sides to solve for k.
Simplify. The equation k = -7 is displayed in black text on a white background.
Check.
Let k=−7.
The top line says 2 times parentheses 3k minus 1 minus 5k equals negative 2 minus 7. Below this is 2 times parentheses red negative 7 minus 1 minus 5 times red negative 7 equals negative 2 minus 7. The next line says 2 times parentheses negative 21 minus 1 minus 5 times negative 7 equals negative 9. Below that is 2 times negative 22 plus 35 equals negative 9. Next is negative 44 plus 35 equals negative 9. The last line says negative 9 equals negative 9.
The solution checks.

Solve: 4(2h−3)−7h=−6−7.

Solution

h = −1

Solve: 2(5x+2)−9x=−2+7.

Solution

x = 1

Translate an Equation and Solve

In previous chapters, we translated word sentences into equations. The first step is to look for the word (or words) that translate(s) to the equal sign. Table 9 reminds us of some of the words that translate to the equal sign.

Equals (=)
is is equal to is the same as the result is gives was will be

Let’s review the steps we used to translate a sentence into an equation.

Translate a word sentence to an algebraic equation.

  1. Locate the "equals" word(s). Translate to an equal sign.
  2. Translate the words to the left of the "equals" word(s) into an algebraic expression.
  3. Translate the words to the right of the "equals" word(s) into an algebraic expression.

Now we are ready to try an example.

Translate and solve: five more than x is equal to 26.

Solution

Solution

Translate. The image translates the phrase 'Five more than x is equal to 26' into the algebraic equation 'x + 5 = 26', using brackets to show the corresponding parts.
Subtract 5 from both sides. A mathematical equation shows 'x + 5 - 5 = 26 - 5' with the subtracted '5' highlighted in red on both sides of the equality sign, illustrating a step in solving for x.
Simplify. A simple mathematical equation 'x = 21' is displayed in black text against a plain white background, centered within the frame.
Check:
Is 26 five more than 21?
A mathematical equation is displayed: '21 + 5 =? 26'. The question mark above the equals sign indicates that the equation is a query, asking if 21 plus 5 is equal to 26.
The image displays the equation '26 = 26' followed by a checkmark, indicating that the statement is correct.
The solution checks.

Translate and solve: Eleven more than x is equal to 41.

Solution

x + 11 = 41; x = 30

Translate and solve: Twelve less than y is equal to 51.

Solution

y − 12 = 51; y = 63

Translate and solve: The difference of 5p and 4p is 23.

Solution

Solution

Translate. Translating the word problem 'The difference of 5p and 4p is 23' into the algebraic equation 5p - 4p = 23.
Simplify. The image shows the mathematical equation 'p = 23' written in black font on a plain white background, isolated in the center-right of the frame.
Check:
The image shows a mathematical equation: 5p - 4p = 23. This equation simplifies to p = 23.
A math problem showing the expression 5(23) - 4(23) =? 23, which simplifies to 23 = 23, demonstrating a basic algebraic property with numerical values.
A mathematical expression is displayed: 115 minus 92, followed by an equals sign with a question mark above it, and then the number 23. It prompts to verify if 115 - 92 is indeed equal to 23.
A mathematical equation '23 = 23' is prominently displayed on a white background, followed by a clear checkmark, indicating its correctness.
The solution checks.

Translate and solve: The difference of 4x and 3x is 14.

Solution

4x − 3x = 14; x = 14

Translate and solve: The difference of 7a and 6a is −8.

Solution

7a − 6a = −8; a = −8

Translate and Solve Applications

In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for.

The Robles family has two dogs, Buster and Chandler. Together, they weigh 71 pounds.

Chandler weighs 28 pounds. How much does Buster weigh?

Solution

Solution

A step-by-step guide demonstrating how to solve word problems using algebraic equations, exemplified by calculating a dog's weight.
Read the problem carefully.
Identify what you are asked to find, and choose a variable to represent it. How much does Buster weigh?
Let b= Buster's weight
Write a sentence that gives the information to find it. Buster's weight plus Chandler's weight equals 71 pounds.
We will restate the problem, and then include the given information. Buster's weight plus 28 equals 71.
Translate the sentence into an equation, using the variable b. The image displays the algebraic equation b + 28 = 71, presented in black text on a white background.
Solve the equation using good algebraic techniques. A mathematical equation is displayed on a white background, which reads 'b + 28 - 28 = 71 - 28'.
The image displays the equation 'b = 43' in black text against a white background.
Check the answer in the problem and make sure it makes sense.
Is 43 pounds a reasonable weight for a dog? Yes. Does Buster's weight plus Chandler's weight equal 71 pounds?
43+28=?71
71=71✓
Write a complete sentence that answers the question, "How much does Buster weigh?" Buster weighs 43 pounds

Translate into an algebraic equation and solve: The Pappas family has two cats, Zeus and Athena. Together, they weigh 13 pounds. Zeus weighs 6 pounds. How much does Athena weigh?

Solution

a + 6 = 13; Athena weighs 7 pounds.

Translate into an algebraic equation and solve: Sam and Henry are roommates. Together, they have 68 books. Sam has 26 books. How many books does Henry have?

Solution

26 + h = 68; Henry has 42 books.

Devise a problem-solving strategy.

  1. Read the problem. Make sure you understand all the words and ideas.
  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 good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Shayla paid $24,575 for her new car. This was $875 less than the sticker price. What was the sticker price of the car?

Solution

Solution

Step-by-step solution for a word problem involving finding a car's sticker price, demonstrating variable assignment, equation translation, and verification.
What are you asked to find? "What was the sticker price of the car?"
Assign a variable. Let s= the sticker price of the car.
Write a sentence that gives the information to find it. $24,575 is $875 less than the sticker price
$24,575 is $875 less than s
Translate into an equation. A mathematical equation is displayed with the numbers 24,575 and 875, separated by an equals sign and a subtraction operation involving the variable 's'. The equation reads as 24,575 = s - 875.
Solve. An algebraic equation: 24,575 + 875 = s - 875 + 875. The right side simplifies to 's', so the equation becomes 24,575 + 875 = s.
The number 25,450 is shown followed by an equals sign and the letter 's', forming the equation 25,450 = s.
Check:
Is $875 less than $25,450 equal to $24,575?
25,450−875=?24,575
24,575=24,575✓
Write a sentence that answers the question. The sticker price was $25,450.

Translate into an algebraic equation and solve: Eddie paid $19,875 for his new car. This was $1,025 less than the sticker price. What was the sticker price of the car?

Solution

19,875 = s − 1025; the sticker price is $20,900.

Translate into an algebraic equation and solve: The admission price for the movies during the day is $7.75. This is $3.25 less than the price at night. How much does the movie cost at night?

Solution

7.75 = n − 3.25; the price at night is $11.00.

The Links to Literacy activity, "The 100-pound Problem", will provide you with another view of the topics covered in this section.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Solving One Step Equations By Addition and Subtraction
  • Solve One Step Equations By Add and Subtract Whole Numbers (Variable on Left)
  • Solve One Step Equations By Add and Subtract Whole Numbers (Variable on Right)

Key Concepts

  • 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.
  • Subtraction and Addition Properties of Equality
    • Subtraction Property of Equality
      For all real numbers a, b, and c,
      if a = b then a-c=b-c.
    • Addition Property of Equality
      For all real numbers a, b, and c,
      if a = b then a+c=b+c.
  • Translate a word sentence to an algebraic equation.
    1. Locate the “equals” word(s). Translate to an equal sign.
    2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
    3. Translate the words to the right of the “equals” word(s) into an algebraic expression.
  • Problem-solving strategy
    1. Read the problem. Make sure you understand all the words and ideas.
    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 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 Equations Using the Subtraction and Addition Properties of Equality

In the following exercises, determine whether the given value is a solution to the equation.

Is y=13 a solution of 4y+2=10y?

Solution

yes

Is x=34 a solution of 5x+3=9x?

Is u=−12 a solution of 8u−1=6u?

Solution

no

Is v=−13 a solution of 9v−2=3v?

In the following exercises, solve each equation.

x+7=12

Solution

x = 5

y+5=−6

b+14=34

Solution

b=12

a+25=45

p+2.4=−9.3

Solution

p = −11.7

m+7.9=11.6

a−3=7

Solution

a = 10

m−8=−20

x−13=2

Solution

x=73

x−15=4

y−3.8=10

Solution

y = 13.8

y−7.2=5

x−15=−42

Solution

x = −27

z+5.2=−8.5

q+34=12

Solution

q=−14

p−25=23

y−34=35

Solution

y=2720

Solve Equations that Need to be Simplified

In the following exercises, solve each equation.

c+3−10=18

m+6−8=15

Solution

m = 17

9x+5−8x+14=20

6x+8−5x+16=32

Solution

x = 8

−6x−11+7x−5=−16

−8n−17+9n−4=−41

Solution

n = −20

3(y−5)−2y=−7

4(y−2)−3y=−6

Solution

y = 2

8(u+1.5)−7u=4.9

5(w+2.2)−4w=9.3

Solution

w = −1.7

−5(y−2)+6y=−7+4

−8(x−1)+9x=−3+9

Solution

x = −2

3(5n−1)−14n+9=1−2

2(8m+3)−15m−4=3−5

Solution

m = −4

−(j+2)+2j−1=5

−(k+7)+2k+8=7

Solution

k = 6

6a−5(a−2)+9=−11

8c−7(c−3)+4=−16

Solution

c = −41

8(4x+5)−5(6x)−x=53

6(9y−1)−10(5y)−3y=22

Solution

y = 28

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve.

Five more than x is equal to 21.

The sum of x and −5 is 33.

Solution

x + (−5) = 33; x = 38

Ten less than m is −14.

Three less than y is −19.

Solution

y − 3 = −19; y = −16

The sum of y and −3 is 40.

Eight more than p is equal to 52.

Solution

p + 8 = 52; p = 44

The difference of 9x and 8x is 17.

The difference of 5c and 4c is 60.

Solution

5c − 4c = 60; c = 60

The difference of n and 16 is 12.

The difference of f and 13 is 112.

Solution

f−13=112;f=512

The sum of −4n and 5n is −32.

The sum of −9m and 10m is −25.

Solution

−9m + 10m = −25; m = −25

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Pilar drove from home to school and then to her aunt’s house, a total of 18 miles. The distance from Pilar’s house to school is 7 miles. What is the distance from school to her aunt’s house?

Jeff read a total of 54 pages in his English and Psychology textbooks. He read 41 pages in his English textbook. How many pages did he read in his Psychology textbook?

Solution

Let p equal the number of pages read in the Psychology book. 41 + p = 54. Jeff read 13 pages in his Psychology book.

Pablo’s father is 3 years older than his mother. Pablo’s mother is 42 years old. How old is his father?

Eva’s daughter is 5 years younger than her son. Eva’s son is 12 years old. How old is her daughter?

Solution

Let d equal the daughter’s age. d = 12 − 5. Eva’s daughter’s age is 7 years old.

Allie weighs 8 pounds less than her twin sister Lorrie. Allie weighs 124 pounds. How much does Lorrie weigh?

For a family birthday dinner, Celeste bought a turkey that weighed 5 pounds less than the one she bought for Thanksgiving. The birthday dinner turkey weighed 16 pounds. How much did the Thanksgiving turkey weigh?

Solution

21 pounds

The nurse reported that Tricia’s daughter had gained 4.2 pounds since her last checkup and now weighs 31.6 pounds. How much did Tricia’s daughter weigh at her last checkup?

Connor’s temperature was 0.7 degrees higher this morning than it had been last night. His temperature this morning was 101.2 degrees. What was his temperature last night?

Solution

100.5 degrees

Melissa’s math book cost $22.85 less than her art book cost. Her math book cost $93.75. How much did her art book cost?

Ron’s paycheck this week was $17.43 less than his paycheck last week. His paycheck this week was $103.76. How much was Ron’s paycheck last week?

Solution

$121.19

Everyday Math

Baking Kelsey needs 23 cup of sugar for the cookie recipe she wants to make. She only has 14 cup of sugar and will borrow the rest from her neighbor. Let s equal the amount of sugar she will borrow. Solve the equation 14+s=23 to find the amount of sugar she should ask to borrow.

Construction Miguel wants to drill a hole for a 58-inch screw. The screw should be 112 inch larger than the hole. Let d equal the size of the hole he should drill. Solve the equation d+112=58 to see what size the hole should be.

Solution

d=1324

Writing Exercises

Is −18 a solution to the equation 3x=16−5x? How do you know?

Write a word sentence that translates the equation y−18=41 and then make up an application that uses this equation in its solution.

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 with 'I can...' statements for solving equations, including using properties of equality, simplifying, translating, and solving applications. Options are 'Confidently,' 'With some help,' and '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.

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.

Solve Equations Using the Division and Multiplication Properties of Equality

Learning Objectives

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

  • Solve equations using the Division and Multiplication Properties of Equality
  • Solve equations that need to be simplified

Before you get started, take this readiness quiz.

Simplify: −7(1−7).
If you missed this problem, review Example 10 in Multiply and Divide Fractions.

Solution

1

What is the reciprocal of −38?
If you missed this problem, review Example 11 in Multiply and Divide Fractions.

Solution

−83

Evaluate 9x+2 when x=−3.
If you missed this problem, review Example 10 in Multiply and Divide Integers.

Solution

−25

Solve Equations Using the Division and Multiplication Properties of Equality

We introduced the Multiplication and Division Properties of Equality in Solve Equations Using Integers; The Division Property of Equality and Solve Equations with Fractions. We modeled how these properties worked using envelopes and counters and then applied them to solving equations (See Solve Equations Using Integers; The Division Property of Equality). We restate them again here as we prepare to use these properties again.

Division and Multiplication Properties of Equality

Division Property of Equality: For all real numbers a,b,c, and c≠0, if a=b, then ac=bc.

Multiplication Property of Equality: For all real numbers a,b,c, if a=b, then ac=bc.

When you divide or multiply both sides of an equation by the same quantity, you still have equality.

Let’s review how these properties of equality can be applied in order to solve equations. Remember, the goal is to ‘undo’ the operation on the variable. In the example below the variable is multiplied by 4, so we will divide both sides by 4 to ‘undo’ the multiplication.

Solve: 4x=−28.

Solution

Solution

We use the Division Property of Equality to divide both sides by 4.

A mathematical equation displays '4x = -28' in black font against a white background.
Divide both sides by 4 to undo the multiplication. A mathematical equation showing 4x divided by 4 equals -28 divided by 4. The number 4 in the denominators is highlighted in red, indicating division on both sides of the equation.
Simplify. The equation x = -7 is displayed in black text on a plain white background, presenting a simple algebraic expression.
Check your answer. Let x=−7.
The image shows the mathematical equation 4x = -28, rendered in a bold, metallic-looking font against a white background.
A mathematical equation, 4 multiplied by -7, is shown with a question mark above the equals sign, asking to verify if the product is -28. The statement 4(-7) = -28 is true.
A mathematical equation displays '-28 = -28' followed by a checkmark, indicating that the statement is true and verified.

Since this is a true statement, x=−7 is a solution to 4x=−28.

Solve: 3y=−48.

Solution

y = −16

Solve: 4z=−52.

Solution

z = −13

In the previous example, to ‘undo’ multiplication, we divided. How do you think we ‘undo’ division?

Solve: a−7=−42.

Solution

Solution

Here a is divided by −7. We can multiply both sides by −7 to isolate a.

A mathematical equation showing 'a' divided by '-7' equals '-42'. The equation is written as a fraction: a over -7 = -42.
Multiply both sides by −7. A mathematical equation shows '-7 multiplied by (a divided by -7) equals -7 multiplied by -42'. The -7 on both sides of the equation is highlighted in red.
A mathematical equation is displayed, showing -7a divided by -7 equals 294.
Simplify. The variable 'a' is assigned the value 294.
Check your answer. Let a=294.
A mathematical equation is displayed on a white background. The equation reads as 'a divided by negative seven equals negative forty-two' (a/-7 = -42).
A mathematical equation shows a fraction 294 divided by -7, with a question mark over the equals sign, comparing the result to -42. This asks whether 294 / -7 is indeed equal to -42.
A mathematical equation displays '-42 = -42' with a checkmark next to it, indicating that the statement is correct and verified.

Solve: b−6=−24.

Solution

b = 144

Solve: c−8=−16.

Solution

c = 128

Solve: −r=2.

Solution

Solution

Remember −r is equivalent to −1r.

A mathematical equation is displayed with a black text on a white background. The equation reads '-r = 2'.
Rewrite −r as −1r. A mathematical equation is displayed on a white background, reading '-1r = 2' in bold, black characters, appearing as if a variable 'r' is being solved for.
Divide both sides by −1. The equation shows dividing -1r by -1 on the left side and 2 by -1 on the right side, resulting in r = -2.
The image shows the mathematical equation 'r = -2' displayed in a bold, black font against a white background.
Check. The image displays a mathematical equation written in black text on a white background: -r=2.
Substitute r=−2 Mathematical expression -(-2) =? 2, questioning if a double negative results in a positive. It does, so the statement is true.
Simplify. A simple equation '2=2' with a checkmark, denoting a verified truth.

In Solve Equations with Fractions, we saw that there are two other ways to solve −r=2.

We could multiply both sides by −1.

We could take the opposite of both sides.

Solve: −k=8.

Solution

k = −8

Solve: −g=3.

Solution

g = −3

Solve: 23x=18.

Solution

Solution

Since the product of a number and its reciprocal is 1, our strategy will be to isolate x by multiplying by the reciprocal of 23.

A mathematical equation is displayed on a white background, reading '(2/3)x = 18'.
Multiply by the reciprocal of 23. An algebraic equation showing 3/2 multiplied by 2/3x equals 3/2 multiplied by 18, demonstrating the isolation of 'x' by multiplying both sides by the reciprocal 3/2, highlighted in red.
Reciprocals multiply to one. A mathematical equation shows '1x = 3/2 multiplied by 18/1' on a white background.
Multiply. The equation 'x = 27' is displayed in a black, sans-serif font on a plain white background, centrally positioned within the frame.
Check your answer. Let x=27
A mathematical equation shows (2/3)x = 18.
A mathematical expression shows 2/3 multiplied by 27 (in red), followed by an equals sign with a question mark above it, and then the number 18, prompting a check of the equality.
The image displays the equation '18 = 18' followed by a checkmark, confirming its accuracy.

Notice that we could have divided both sides of the equation 23x=18 by 23 to isolate x. While this would work, multiplying by the reciprocal requires fewer steps.

Solve: 25n=14.

Solution

n = 35

Solve: 56y=15.

Solution

y = 18

Solve Equations That Need to be Simplified

Many equations start out more complicated than the ones we’ve just solved. First, we need to simplify both sides of the equation as much as possible

Solve: 8x+9x−5x=−3+15.

Solution

Solution

Start by combining like terms to simplify each side.

A mathematical equation is displayed, reading '8x + 9x - 5x = -3 + 15'.
Combine like terms. The mathematical equation 12x = 12 is displayed in black text on a white background.
Divide both sides by 12 to isolate x. An algebraic equation showing a step in solving for x: 12x/12 = 12/12, where both sides are divided by 12, with the denominator 12 highlighted in red.
Simplify. The equation 'x=1' is displayed in a serif font, rendered in black text against a plain white background.
Check your answer. Let x=1
The image displays the algebraic equation 8x + 9x - 5x = -3 + 15.
A mathematical equation, '8 ×× ×× 1 + 9 ×× ×× 1 - 5 ×× ×× 1 = -3 + 15', with a question mark above the equals sign, indicating a verification task.
A math problem displays '8 + 9 - 5 = -3 + 15' with a question mark above the equals sign, asking if the statement is true. Calculating both sides reveals that 12 = 12, making the equation true.
A mathematical equation '12 = 12' is displayed, followed by a checkmark, indicating correctness or validation.

Solve: 7x+6x−4x=−8+26.

Solution

x = 2

Solve: 11n−3n−6n=7−17.

Solution

n = −5

Solve: 11−20=17y−8y−6y.

Solution

Solution

Simplify each side by combining like terms.

A mathematical equation is displayed, showing '11 - 20 = 17y - 8y - 6y' in black text on a white background, demonstrating an algebraic problem to be solved for the variable 'y'.
Simplify each side. A mathematical equation is displayed against a white background, which reads '-9 = 3y' in black text.
Divide both sides by 3 to isolate y. The algebraic equation -9/3 = 3y/3, demonstrating division by 3 on both sides. The number 3 in the denominator is highlighted in red.
Simplify. -3 = y is an equation showing that the variable 'y' is equal to the constant value -3. It represents a horizontal line in a Cartesian coordinate system, where all points on the line have a y-coordinate of -3.
Check your answer. Let y=−3
A mathematical equation is displayed, reading '11 - 20 = 17y - 8y - 6y' on a white background.
A mathematical equation asks whether 11 - 20 is equal to 17(-3) - 8(-3) - 6(-3), with a question mark positioned above the equality sign.
A mathematical equation with a question mark over the equals sign asks to verify if 11 - 20 is equal to -51 + 24 + 18.
The equation -9 = -9 is shown with a checkmark, indicating its correctness.

Notice that the variable ended up on the right side of the equal sign when we solved the equation. You may prefer to take one more step to write the solution with the variable on the left side of the equal sign.

Solve: 18−27=15c−9c−3c.

Solution

c = −3

Solve: 18−22=12x−x−4x.

Solution

x=−47

Solve: −3(n−2)−6=21.

Solution

Solution

Remember—always simplify each side first.

A mathematical equation is displayed: -3(n - 2) - 6 = 21. This algebraic expression involves multiplication, subtraction, and a variable 'n', set equal to 21.
Distribute. A mathematical equation displays '-3n + 6 - 6 = 21' in black text on a white background.
Simplify. A mathematical equation shows '-3n = 21' in black text against a white background.
Divide both sides by -3 to isolate n. A mathematical equation shows '-3n divided by -3 equals 21 divided by -3'. The denominator, -3, is highlighted in red on both sides of the equation, indicating division as a step in solving for 'n'.
The mathematical equation 'n = -7' is displayed in bold black text on a plain white background.
Check your answer. Let n=−7.
A mathematical equation is displayed, showing -3(n-2) - 6 = 21, centered against a white background.
A mathematical equation showing -3 multiplied by the quantity (-7 minus 2), then minus 6, with a question mark over the equals sign before 21, asking if the expression equals 21.
A mathematical equation is displayed: -3(-9) - 6 =? 21. It asks to verify if the expression on the left equals 21.
A mathematical equation '27 - 6 =? 21' is displayed, with a question mark positioned above the equals sign, implying an inquiry into the truth of the statement.
A mathematical equation '21 = 21' is displayed on a white background, followed by a black checkmark, confirming the equality's correctness.

Solve: −4(n−2)−8=24.

Solution

n = −6

Solve: −6(n−2)−12=30.

Solution

n = −5

The Links to Literacy activity, "Everybody Wins" will provide you with another view of the topics covered in this section.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Solving One Step Equation by Mult/Div. Integers (Var on Left)
  • Solving One Step Equation by Mult/Div. Integers (Var on Right)
  • Solving One Step Equation in the Form: −x = −a

Key Concepts

  • Division and Multiplication Properties of Equality
    • Division Property of Equality: For all real numbers a, b, c, and c≠0, if a=b, then ac=bc.
    • Multiplication Property of Equality: For all real numbers a, b, c, if a=b, then ac=bc.

Practice Makes Perfect

Solve Equations Using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation for the variable using the Division Property of Equality and check the solution.

8x=32

7p=63

Solution

p = 9

−5c=55

−9x=−27

Solution

x = 3

−90=6y

−72=12y

Solution

y = −6

−16p=−64

−8m=−56

Solution

m = 7

0.25z=3.25

0.75a=11.25

Solution

a = 15

−3x=0

4x=0

Solution

x = 0

In the following exercises, solve each equation for the variable using the Multiplication Property of Equality and check the solution.

x4=15

z2=14

Solution

z = 28

−20=q−5

c−3=−12

Solution

c = 36

y9=−6

q6=−8

Solution

q = −48

m−12=5

−4=p−20

Solution

p = 80

23y=18

35r=15

Solution

r = 25

−58w=40

24=−34x

Solution

x = −32

−25=110a

−13q=−56

Solution

q=52

Solve Equations That Need to be Simplified

In the following exercises, solve the equation.

8a+3a−6a=−17+27

6y−3y+12y=−43+28

Solution

y = −1

−9x−9x+2x=50−2

−5m+7m−8m=−6+36

Solution

m = −5

100−16=4p−10p−p

−18−7=5t−9t−6t

Solution

t=52

78n−34n=9+2

512q+12q=25−3

Solution

q = 24

0.25d+0.10d=6−0.75

0.05p−0.01p=2+0.24

Solution

p = 56

Everyday Math

Balloons Ramona bought 18 balloons for a party. She wants to make 3 equal bunches. Find the number of balloons in each bunch, b, by solving the equation 3b=18.

Teaching Connie’s kindergarten class has 24 children. She wants them to get into 4 equal groups. Find the number of children in each group, g, by solving the equation 4g=24.

Solution

6 children

Ticket price Daria paid $36.25 for 5 children’s tickets at the ice skating rink. Find the price of each ticket, p, by solving the equation 5p=36.25.

Unit price Nishant paid $12.96 for a pack of 12 juice bottles. Find the price of each bottle, b, by solving the equation 12b=12.96.

Solution

$1.08

Fuel economy Tania’s SUV gets half as many miles per gallon (mpg) as her husband’s hybrid car. The SUV gets 18 mpg. Find the miles per gallons, m, of the hybrid car, by solving the equation 12m=18.

Fabric The drill team used 14 yards of fabric to make flags for one-third of the members. Find how much fabric, f, they would need to make flags for the whole team by solving the equation 13f=14.

Solution

42 yards

Writing Exercises

Frida started to solve the equation −3x=36 by adding 3 to both sides. Explain why Frida’s method will result in the correct solution.

Emiliano thinks x=40 is the solution to the equation 12x=80. Explain why he is wrong.

Solution

Answer 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 math skills, asking students to rate their ability to solve equations using division/multiplication properties and simplifying them, across three confidence levels.

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

Solve Equations with Variables and Constants on Both Sides

Learning Objectives

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

  • Solve an equation with constants on both sides
  • Solve an equation with variables on both sides
  • Solve an equation with variables and constants on both sides
  • Solve equations using a general strategy

Before you get started, take this readiness quiz.

Simplify: 4y−9+9.
If you missed this problem, review Example 10 in Evaluate, Simplify, and Translate Expressions.

Solution

4y

Solve: y+12=16.
If you missed this problem, review Example 4 in Solving Equations Using the Subtraction and Addition Properties of Equality.

Solution

4

Solve: −3y=63.
If you missed this problem, review Example 6 in Solve Equations Using Integers; The Division Property of Equality.

Solution

−21

Solve an Equation with Constants on Both Sides

You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the equation.

Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. Then, we will use the Subtraction and Addition Properties of Equality, step by step, to get all the variable terms together on one side of the equation and the constant terms together on the other side.

By doing this, we will transform the equation that started with variables and constants on both sides into the form ax=b. We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Solve: 4x+6=−14.

Solution

Solution

In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side. We’ll write the labels above the equation to help us remember what goes where.

The equation 4x + 6 = -14 is displayed, with 'variable' labeling 4x and 'constant' labeling 6 and -14, illustrating basic algebraic components.
Since the left side is the variable side, the 6 is out of place.
We must "undo" adding 6 by subtracting 6,
and to keep the equality we must subtract 6 from both sides.
Use the Subtraction Property of Equality.
The equation 4x + 6 - 6 = -14 - 6, illustrating a step in solving for x by subtracting 6 from both sides, with the subtracted '6's highlighted in red.
Simplify. A mathematical equation is displayed on a white background: 4x = -20.
Now all the xs are on the left and the constant on the right.
Use the Division Property of Equality. An algebraic equation showing 4x/4 = -20/4, with the denominator '4' in red, indicating division by 4 on both sides to solve for x.
Simplify. The image displays the mathematical equation 'x = -5' in black font against a plain white background, centered in the frame. The 'x' is a lowercase variable, followed by an equals sign, and then a negative sign preceding the numeral '5'.
Check: A mathematical equation is displayed: 4x + 6 = -14. The equation involves a variable 'x', addition, and negative numbers, representing a basic algebraic problem.
Let x=−5. The image displays the mathematical equation 4(-5) + 6 = -14, showing the multiplication of 4 by negative 5, followed by the addition of 6, resulting in negative 14. The negative 5 is highlighted in red.
A mathematical equation is displayed on a white background, which reads '-20 + 6 = -14'. The numbers and symbols are in a dark gray font, creating a clear contrast.
The equation -14 = -14 is correctly displayed with a checkmark.

Solve: 3x+4=−8.

Solution

x = −4

Solve: 5a+3=−37.

Solution

a = −8

Solve: 2y−7=15.

Solution

Solution

Notice that the variable is only on the left side of the equation, so this will be the variable side and the right side will be the constant side. Since the left side is the variable side, the 7 is out of place. It is subtracted from the 2y, so to ‘undo’ subtraction, add 7 to both sides.

An algebraic equation 2y - 7 = 15 with terms labeled as 'variable' (2y) and 'constant' (-7 and 15) to illustrate basic algebraic components.
Add 7 to both sides. The equation 2y - 7 + 7 = 15 + 7 illustrates the step of adding 7 to both sides to simplify the expression and solve for y, with the added 7 highlighted in red.
Simplify. The mathematical equation '2y = 22' is displayed in black text on a white background.
The variables are now on one side and the constants on the other.
Divide both sides by 2. A mathematical equation showing 2y/2 = 22/2, with the denominator '2' highlighted in red on both sides.
Simplify. The image displays a simple mathematical equation, 'y = 11', written in black text on a plain white background. The equation indicates that the variable 'y' is equal to the numerical value of 11.
Check: A mathematical equation, 2y - 7 = 15, is displayed in black text against a white background.
Substitute: y=11. A math problem asking if 2 * 11 - 7 equals 15. The number 11 is highlighted in red. The equation is true: 22 - 7 = 15.
The mathematical expression '22 - 7 ?= 15' is shown, verifying if 22 minus 7 is equal to 15.
The equation '15 = 15' is displayed, followed by a checkmark, confirming its correctness.

Solve: 5y−9=16.

Solution

y = 5

Solve: 3m−8=19.

Solution

m = 9

Solve an Equation with Variables on Both Sides

What if there are variables on both sides of the equation? We will start like we did above—choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what you do to the left side of the equation, you must do to the right side too.

Solve: 5x=4x+7.

Solution

Solution

Here the variable, x, is on both sides, but the constants appear only on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.

An image showing the equation 5x = 4x + 7, with 'variable' labeled above 5x and 'constant' labeled above 7.
We don't want any variables on the right, so subtract the 4x. An algebraic equation showing 5x minus 4x equals 4x minus 4x plus 7. The 4x terms are highlighted in red, indicating they are being subtracted or canceled out from both sides of the equation.
Simplify. The mathematical equation x=7 is displayed in black font against a white background.
We have all the variables on one side and the constants on the other. We have solved the equation.
Check: A basic algebraic equation is displayed, reading 5x = 4x + 7, shown in a bold, sans-serif font against a white background.
Substitute 7 for x. A mathematical equation is displayed: 5(7) with a question mark over an equals sign, then 4(7) + 7. The number 7 is highlighted in red in each instance within parentheses.
A mathematical equation reads '35 =? 28 + 7'. The question mark above the equals sign indicates that the equation is being posed as a question or problem to be solved, asking if 35 is indeed equal to 28 plus 7.
A mathematical equation shows '35 = 35' with a checkmark, signifying that the statement is correct.

Solve: 6n=5n+10.

Solution

n = 10

Solve: −6c=−7c+1.

Solution

c = 1

Solve: 5y−8=7y.

Solution

Solution

The only constant, −8, is on the left side of the equation and variable, y, is on both sides. Let’s leave the constant on the left and collect the variables to the right.

The equation 5y - 8 = 7y is shown, with the word 'constant' in red text above '5y - 8', and 'variable' in red text above '7y'.
Subtract 5y from both sides. A mathematical equation is displayed: 5y - 5y - 8 = 7y - 5y. The terms -5y on the left side and -5y on the right side are highlighted in red, indicating they are part of a simplification step.
Simplify. A mathematical equation is displayed with the expression -8 = 2y, representing a simple linear equation where the variable 'y' needs to be solved.
We have the variables on the right and the constants on the left. Divide both sides by 2. A mathematical equation shows '-8' divided by '2' (in red) equals '2y' divided by '2' (in red), indicating a step in solving for 'y' where both sides of the equation are divided by 2.
Simplify. The image displays the simple algebraic equation '-4 = y' against a white background.
Rewrite with the variable on the left. The equation y = -4 is displayed on a white background.
Check: Let y=−4.
A mathematical equation is displayed with black text on a white background, reading '5y - 8 = 7y'.
A math equation reads 5(-4) - 8 =? 7(-4). Negative numbers are highlighted in red, suggesting a comparison or calculation of both sides.
A mathematical equation reads -20 - 8 = -28, with a question mark above the equals sign, asking if the statement is true.
A mathematical equation shows '-28 = -28' with a checkmark, indicating the equality is correct.

Solve: 3p−14=5p.

Solution

p = −7

Solve: 8m+9=5m.

Solution

m = −3

Solve: 7x=−x+24.

Solution

Solution

The only constant, 24, is on the right, so let the left side be the variable side.

An algebraic equation 7x = -x + 24 is displayed. The left side (7x) is labeled 'variable side' and the right side (-x + 24) is labeled 'constant side' in red text.
Remove the −x from the right side by adding x to both sides. A mathematical equation shows 7x + x = -x + x + 24. The plus and x symbols are highlighted in red on both sides of the equation.
Simplify. The mathematical equation '8x = 24' is displayed in bold black text on a white background.
All the variables are on the left and the constants are on the right. Divide both sides by 8. The equation 8x/8 = 24/8, demonstrating dividing both sides by 8 (highlighted in red) to solve for x.
Simplify. The mathematical equation 'x = 3' is displayed in black font against a plain white background.
Check: Substitute x=3.
The image demonstrates the verification of the solution to the algebraic equation 7x = -x + 24. By substituting x=3, the equation simplifies to 21 = 21, confirming that 3 is indeed the correct solution.

Solve: 12j=−4j+32.

Solution

j = 2

Solve: 8h=−4h+12.

Solution

h = 1

Solve Equations with Variables and Constants on Both Sides

The next example will be the first to have variables and constants on both sides of the equation. As we did before, we’ll collect the variable terms to one side and the constants to the other side.

Solve: 7x+5=6x+2.

Solution

Solution

Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are 7x and 6x. Since 7 is greater than 6, make the left side the variable side and so the right side will be the constant side.

A mathematical equation is displayed, showing '7x + 5 = 6x + 2' in black text on a white background, representing a linear equation in one variable.
Collect the variable terms to the left side by subtracting 6x from both sides. A mathematical equation, 7x - 6x + 5 = 6x - 6x + 2, is displayed. The terms '-6x' on the left side and '6x - 6x' on the right side are highlighted in red.
Simplify. The image shows a mathematical equation: x + 5 = 2.
Now, collect the constants to the right side by subtracting 5 from both sides. The equation x + 5 - 5 = 2 - 5 is shown, illustrating a step in solving for x by subtracting 5 from both sides, with the subtracted 5s in red.
Simplify. The equation x = -3 is displayed in black font against a white background.
The solution is x=−3.
Check: Let x=−3.
Step-by-step solution and verification of the linear equation 7x + 5 = 6x + 2, demonstrating that x = -3 is the correct solution by substituting it back into the equation.

Solve: 12x+8=6x+2.

Solution

x = −1

Solve: 9y+4=7y+12.

Solution

y = 4

We’ll summarize the steps we took so you can easily refer to them.

Solve an equation with variables and constants on both sides.

  1. Choose one side to be the variable side and then the other will be the constant side.
  2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
  3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable 1, using the Multiplication or Division Property of Equality.
  5. Check the solution by substituting it into the original equation.

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

Solve: 6n−2=−3n+7.

Solution

Solution

We have 6n on the left and −3n on the right. Since 6>−3, make the left side the “variable” side.

A mathematical equation is displayed, showing 6n - 2 = -3n + 7. It is a linear equation with one variable, 'n', on both sides of the equals sign.
We don't want variables on the right side—add 3n to both sides to leave only constants on the right. An algebraic equation 6n + 3n - 2 = -3n + 3n + 7, with variables and constants. Some terms are highlighted in red.
Combine like terms. The image displays the algebraic equation 9n - 2 = 7, written in a clear, dark font against a white background.
We don't want any constants on the left side, so add 2 to both sides. A mathematical equation shows '9n - 2 + 2 = 7 + 2', where the '+ 2' on both sides of the equals sign is highlighted in red, demonstrating the addition property of equality.
Simplify. A mathematical equation shows '9n = 9' in black text against a white background.
The variable term is on the left and the constant term is on the right.
To get the coefficient of n to be one, divide both sides by 9.
A mathematical equation showing 9n/9 = 9/9, with the number 9 in the denominators highlighted in red.
Simplify. The mathematical equation 'n = 1' is shown in a black serif font on a plain white background, symbolizing a foundational value or a starting point in a sequence.
Check: Substitute 1 for n.
The solution to the equation 6n - 2 = -3n + 7 is verified by substituting n=1, resulting in 4 = 4, confirming its correctness.

Solve: 8q−5=−4q+7.

Solution

q = 1

Solve: 7n−3=n+3.

Solution

n = 1

Solve: 2a−7=5a+8.

Solution

Solution

This equation has 2a on the left and 5a on the right. Since 5>2, make the right side the variable side and the left side the constant side.

A mathematical equation displays '2a - 7 = 5a + 8' in a bold, black serif font on a plain white background.
Subtract 2a from both sides to remove the variable term from the left. A math problem demonstrating simplification: 2a - 2a - 7 = 5a - 2a + 8. The red '2a' terms on each side are being subtracted, indicating their removal.
Combine like terms. A mathematical equation is displayed, reading '-7 = 3a + 8' in black text against a white background.
Subtract 8 from both sides to remove the constant from the right. An algebraic equation: -7 - 8 = 3a + 8 - 8. The '8's subtracted from each side are highlighted in red, showing a step to simplify the equation and solve for the variable 'a'.
Simplify. A mathematical equation shows '-15 = 3a' on a white background.
Divide both sides by 3 to make 1 the coefficient of a. A mathematical equation is shown with fractions. On the left, -15 is divided by 3, with the 3 in red. On the right, 3a is divided by 3, with the 3 in red. An equals sign separates the two fractions.
Simplify. A mathematical equation is displayed on a white background, reading '-5 = a' in black text, indicating that the variable 'a' is equal to negative five.
Check: Let a=−5.
Step-by-step verification that a = -5 is the correct solution for the equation 2a - 7 = 5a + 8, resulting in -17 = -17.

Note that we could have made the left side the variable side instead of the right side, but it would have led to a negative coefficient on the variable term. While we could work with the negative, there is less chance of error when working with positives. The strategy outlined above helps avoid the negatives!

Solve: 2a−2=6a+18.

Solution

a = −5

Solve: 4k−1=7k+17.

Solution

k = −6

To solve an equation with fractions, we still follow the same steps to get the solution.

Solve: 32x+5=12x−3.

Solution

Solution

Since 32>12, make the left side the variable side and the right side the constant side.

A mathematical equation is displayed: three-halves x plus five equals one-half x minus three.
Subtract 12x from both sides. A mathematical equation displayed as 3/2x - 1/2x + 5 = 1/2x - 1/2x - 3, featuring fractional coefficients, the variable 'x', and integer constants, with some terms highlighted in red.
Combine like terms. A mathematical equation is displayed on a white background, reading 'x + 5 = -3'.
Subtract 5 from both sides. A mathematical equation is displayed on a white background: x + 5 - 5 = -3 - 5. The two '5's on the left side are in black, while the '-5' after the 'x+5' and the '-5' after the '-3' are in red.
Simplify. A mathematical expression on a white background, displaying the equation 'x = -8' in black serif font.
Check: Let x=−8.
Verification of a linear equation where x = -8 is substituted into (3/2)x + 5 = (1/2)x - 3, demonstrating that both sides simplify to -7, thereby confirming the solution.

Solve: 78x−12=−18x−2.

Solution

x = 10

Solve: 76y+11=16y+8.

Solution

y = −3

We follow the same steps when the equation has decimals, too.

Solve: 3.4x+4=1.6x−5.

Solution

Solution

Since 3.4>1.6, make the left side the variable side and the right side the constant side.

An algebraic equation is shown on a white background: 3.4x + 4 = 1.6x - 5. The equation involves decimal coefficients, a variable 'x', addition, and subtraction, presented in a standard mathematical format.
Subtract 1.6x from both sides. A mathematical equation is displayed: 3.4x - 1.6x + 4 = 1.6x - 1.6x - 5. The terms -1.6x (left side) and 1.6x - 1.6x (right side) are highlighted in red.
Combine like terms. A mathematical equation, 1.8x + 4 = -5, is displayed in black text on a white background.
Subtract 4 from both sides. A mathematical equation '1.8x + 4 - 4 = -5 - 4' is displayed, demonstrating a step in solving for 'x' where the subtraction of '4' from both sides is highlighted in red.
Simplify. A mathematical equation, 1.8x = -9, is displayed in a black font against a white background.
Use the Division Property of Equality. A mathematical equation shows both sides being divided by 1.8: '1.8x / 1.8 = -9 / 1.8', with '1.8' in the denominators highlighted in red, indicating a step to solve for x.
Simplify. The image displays a mathematical equation in black text on a white background, which reads 'x = -5'.
Check: Let x=−5.
The image shows the verification of the solution for the equation 3.4x + 4 = 1.6x - 5. By substituting x with -5, both sides of the equation simplify to -13, confirming -5 as the correct solution.

Solve: 2.8x+12=−1.4x−9.

Solution

x = −5

Solve: 3.6y+8=1.2y−4.

Solution

y = −5

Solve Equations Using a General Strategy

Each of the first few sections of this chapter has dealt with solving one specific form of a linear equation. It’s time now to lay out an overall strategy that can be used to solve any linear equation. We call this the general strategy. Some equations won’t require all the steps to solve, but many will. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Use a general strategy for solving linear equations.

  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 to one side of the equation. Use the Addition or Subtraction Property of Equality.
  3. Collect all the constant terms to the other side of the equation. Use the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable term to 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: 3(x+2)=18.

Solution

Solution

The algebraic equation 3(x + 2) = 18 is displayed on a white background.
Simplify each side of the equation as much as possible.
Use the Distributive Property.
A mathematical equation reads '3x + 6 = 18' on a white background. It represents a simple linear equation where the variable 'x' needs to be solved.
Collect all variable terms on one side of the equation—all xs are already on the left side.
Collect constant terms on the other side of the equation.
Subtract 6 from each side
An algebraic equation showing the subtraction of 6 from both sides to solve for x: 3x + 6 - 6 = 18 - 6.
Simplify. A mathematical equation, 3x = 12, is displayed in bold black text on a white background.
Make the coefficient of the variable term equal to 1. Divide each side by 3. A mathematical equation showing 3x divided by 3 equals 12 divided by 3, with the denominator '3' highlighted in red, representing a step in solving for x.
Simplify. The equation x = 4 is displayed in black text on a white background.
Check: Let x=4.
A step-by-step verification of the equation 3(x+2)=18. Substituting x=4 into the equation results in 3(4+2)=18, which simplifies to 3(6)=18, and finally 18=18, confirming x=4 as the solution.

Solve: 5(x+3)=35.

Solution

x = 4

Solve: 6(y−4)=−18.

Solution

y = 1

Solve: −(x+5)=7.

Solution

Solution

A mathematical equation is displayed on a white background: -(x + 5) = 7. The equation shows a negative sign outside parentheses, which enclose the sum of x and 5, equated to the number 7.
Simplify each side of the equation as much as possible by distributing.
The only x term is on the left side, so all variable terms are on the left side of the equation.
A mathematical equation is displayed on a white background. The equation reads: -x - 5 = 7.
Add 5 to both sides to get all constant terms on the right side of the equation. A mathematical equation shows '-x - 5 + 5 = 7 + 5' with the addition of 5 on both sides highlighted in red, demonstrating the first step in solving for x by isolating the variable.
Simplify. The image displays a mathematical equation: -x = 12. It's a simple algebraic expression where a negative variable 'x' is set equal to the number 12, prompting for the solution of 'x'.
Make the coefficient of the variable term equal to 1 by multiplying both sides by -1. A mathematical equation showing -1 multiplied by -x equals -1 multiplied by 12, or -1(-x) = -1(12), with the -1 highlighted in red on both sides.
Simplify. The image displays a mathematical equation written in a black font on a white background, stating 'x = -12'.
Check: Let x=−12.
A mathematical equation shown as -(x + 5) = 7.
A mathematical equation reads 'minus open parenthesis minus twelve plus five close parenthesis equals question mark seven' to check if the statement is true.
A mathematical expression showing a question: -(-7) =? 7. This checks understanding of negative numbers and double negatives, where -(-7) simplifies to 7, making the statement true.
The equation '7=7' is displayed, followed by a checkmark, signifying that the statement is correct and validated.

Solve: −(y+8)=−2.

Solution

y = −6

Solve: −(z+4)=−12.

Solution

z = 8

Solve: 4(x−2)+5=−3.

Solution

Solution

A mathematical equation is displayed on a white background: 4(x - 2) + 5 = -3.
Simplify each side of the equation as much as possible.
Distribute.
The image displays a mathematical equation: 4x - 8 + 5 = -3. It is an algebraic equation where 'x' is the unknown variable, and the equation needs to be solved to find the value of 'x'.
Combine like terms A mathematical equation is displayed on a white background: 4x - 3 = -3.
The only x is on the left side, so all variable terms are on one side of the equation.
Add 3 to both sides to get all constant terms on the other side of the equation. A mathematical equation shows '4x - 3 + 3 = -3 + 3'. The red '+ 3' on both sides indicates the addition of 3 to balance the equation and isolate '4x'.
Simplify. The equation 4x = 0 is displayed on a white background, representing a simple algebraic problem.
Make the coefficient of the variable term equal to 1 by dividing both sides by 4. The equation 4x/4 = 0/4 is displayed, with the number 4 in the denominators highlighted in red.
Simplify. The mathematical equation x=0 is displayed in black text on a white background.
Check: Let x=0.
Step-by-step solution verification for 4(x-2)+5=-3, showing that x=0 is the correct answer. Substituting x=0 simplifies the equation to -3=-3, confirmed by a checkmark.

Solve: 2(a−4)+3=−1.

Solution

a = 2

Solve: 7(n−3)−8=−15.

Solution

n = 2

Solve: 8−2(3y+5)=0.

Solution

Solution

Be careful when distributing the negative.

A mathematical equation is displayed: 8 - 2(3y + 5) = 0. The equation is rendered in black characters against a plain white background.
Simplify—use the Distributive Property. A mathematical equation is displayed, reading '8 - 6y - 10 = 0'.
Combine like terms. The image shows a mathematical equation on a white background: -6y - 2 = 0. The equation is displayed in black text.
Add 2 to both sides to collect constants on the right. A mathematical equation shows '-6y - 2 + 2 = 0 + 2'. The numbers 2 on both sides of the equation are highlighted in red, indicating an operation or a step in solving the equation.
Simplify. A mathematical equation shows '-6y = 2' written in black text on a white background.
Divide both sides by −6. The equation -6y/-6 = 2/-6 is presented, illustrating division by a negative number in red.
Simplify. The image displays the mathematical equation y = -1/3 in black text on a white background.
Check: Let y=−13.
This image demonstrates the verification process for the solution y = -1/3 in the equation 8 - 2(3y + 5) = 0, concluding with 0 = 0, confirming its accuracy.

Solve: 12−3(4j+3)=−17.

Solution

j=53

Solve: −6−8(k−2)=−10.

Solution

k=52

Solve: 3(x−2)−5=4(2x+1)+5.

Solution

Solution

A mathematical equation is displayed, showing 3 multiplied by the quantity x minus 2, then minus 5, which equals 4 multiplied by the quantity 2x plus 1, plus 5.
Distribute. A mathematical equation is displayed on a white background: 3x - 6 - 5 = 8x + 4 + 5.
Combine like terms. A mathematical equation is displayed, reading 3x minus 11 equals 8x plus 9, set against a plain white background.
Subtract 3x to get all the variables on the right since 8>3. An algebraic equation is shown: 3x - 3x - 11 = 8x - 3x + 9. The '3x' terms are highlighted in red on both sides of the equation, indicating terms that can be combined or cancelled.
Simplify. A mathematical equation is displayed on a white background, reading '-11 = 5x + 9'.
Subtract 9 to get the constants on the left. A mathematical equation showing the step to isolate a variable, where '-11 - 9 = 5x + 9 - 9' is displayed with the number 9 highlighted in red on both sides of the equation.
Simplify. A mathematical equation is displayed, stating '-20 = 5x' in a clear, dark font against a white background.
Divide by 5. A mathematical equation shows 'negative 20 over 5' equals '5x over 5'. The number 5 in the denominators is highlighted in red.
Simplify. The mathematical equation '-4 = x' is displayed in black font against a white background.
Check: Substitute: −4=x.
Step-by-step verification of an algebraic equation by substituting x = -4 and simplifying both sides to confirm the equality.

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: 12(6x−2)=5−x.

Solution

Solution

A mathematical equation is displayed, showing one-half multiplied by the quantity six-x minus two, which is set equal to five minus x. The equation is written as (1/2)(6x - 2) = 5 - x.
Distribute. A mathematical equation is displayed, reading '3x - 1 = 5 - x' in black text against a white background.
Add x to get all the variables on the left. A mathematical equation is displayed: 3x - 1 + x = 5 - x + x. On both sides of the equation, the red 'x' terms are added to simplify the expression, demonstrating a step in solving for 'x'.
Simplify. A mathematical equation, '4x - 1 = 5', is displayed on a white background. The equation is centered in the image.
Add 1 to get constants on the right. A mathematical equation shows 4x - 1 + 1 = 5 + 1, with the '+ 1' on both sides highlighted in red, illustrating the step of adding the same value to balance the equation and isolate the variable term.
Simplify. A mathematical equation is displayed on a white background, reading '4x = 6'.
Divide by 4. A mathematical equation is displayed on a white background. The equation reads '4x/4 = 6/4', where the '4' in the denominator of both fractions is colored red. The '4x' and '6' are in black.
Simplify. The image displays a mathematical equation, x = 3/2, set against a plain white background. The variable 'x' is shown equal to the fraction three halves, with a horizontal line separating the numerator '3' from the denominator '2'.
Check: Let x=32.
Step-by-step verification of x=3/2 as the solution for the equation 1/2(6x-2)=5-x, proving equality on both sides.

Solve: 13(6u+3)=7−u.

Solution

u = 2

Solve: 23(9x−12)=8+2x.

Solution

x = 4

In many applications, we will have to solve equations with decimals. The same general strategy will work for these equations.

Solve: 0.24(100x+5)=0.4(30x+15).

Solution

Solution

A mathematical equation is displayed: 0.24(100x + 5) = 0.4(30x + 15).
Distribute. A mathematical equation is displayed, reading '24x + 1.2 = 12x + 6' in a black serif font against a plain white background. It's a linear equation with variables on both sides.
Subtract 12x to get all the xs to the left. An algebraic equation: 24x + 1.2 - 12x = 12x + 6 - 12x. The term '12x' is highlighted in red, indicating it is being subtracted from both sides to simplify the equation.
Simplify. The image shows the linear equation 12x + 1.2 = 6.
Subtract 1.2 to get the constants to the right. A mathematical equation showing the subtraction of 1.2 from both sides of the equation. The equation reads: 12x + 1.2 - 1.2 = 6 - 1.2. The numbers being subtracted (1.2) are highlighted in red.
Simplify. A mathematical equation is displayed, showing '12x = 4.8' on a white background.
Divide. An image showing the algebraic equation 12x/12 = 4.8/12, which demonstrates how to isolate the variable x by dividing both sides of the equation by 12. The number 12 is colored red below the fraction bar.
Simplify. The mathematical equation x = 0.4 is displayed in black text against a white background.
Check: Let x=0.4.
The image displays the verification of the solution x=0.4 for the equation 0.24(100x+5) = 0.4(30x+15), showing a final equality of 10.8 = 10.8.

Solve: 0.55(100n+8)=0.6(85n+14).

Solution

n = 1

Solve: 0.15(40m−120)=0.5(60m+12).

Solution

m = −1

ACCESS ADDITIONAL ONLINE RESOURCES

  • Solving Multi-Step Equations
  • Solve an Equation with Variable Terms on Both Sides
  • Solving Multi-Step Equations (L5.4)
  • Solve an Equation with Variables and Parentheses on Both Sides

Key Concepts

  • Solve an equation with variables and constants on both sides
    1. Choose one side to be the variable side and then the other will be the constant side.
    2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
    3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
    4. Make the coefficient of the variable 1, using the Multiplication or Division Property of Equality.
    5. Check the solution by substituting into the original equation.
  • General strategy for solving linear equations
    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 to one side of the equation. Use the Addition or Subtraction Property of Equality.
    3. Collect all the constant terms to the other side of the equation. Use the Addition or Subtraction Property of Equality.
    4. Make the coefficient of the variable term to 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.

Practice Makes Perfect

Solve an Equation with Constants on Both Sides

In the following exercises, solve the equation for the variable.

6x−2=40

7x−8=34

Solution

x = 6

11w+6=93

14y+7=91

Solution

y = 6

3a+8=−46

4m+9=−23

Solution

m = −8

−50=7n−1

−47=6b+1

Solution

b = −8

25=−9y+7

29=−8x−3

Solution

x = −4

−12p−3=15

−14q−15=13

Solution

q = −2

Solve an Equation with Variables on Both Sides

In the following exercises, solve the equation for the variable.

8z=7z−7

9k=8k−11

Solution

k = −11

4x+36=10x

6x+27=9x

Solution

x = 9

c=−3c−20

b=−4b−15

Solution

b = −3

5q=44−6q

7z=39−6z

Solution

z = 3

3y+12=2y

8x+34=7x

Solution

x=−34

−12a−8=−16a

−15r−8=−11r

Solution

r = −2

Solve an Equation with Variables and Constants on Both Sides

In the following exercises, solve the equations for the variable.

6x−15=5x+3

4x−17=3x+2

Solution

x = 19

26+8d=9d+11

21+6f=7f+14

Solution

f = 7

3p−1=5p−33

8q−5=5q−20

Solution

q = −5

4a+5=−a−40

9c+7=−2c−37

Solution

c = −4

8y−30=−2y+30

12x−17=−3x+13

Solution

x = 2

2z−4=23−z

3y−4=12−y

Solution

y = 4

54c−3=14c−16

43m−7=13m−13

Solution

m = −6

8−25q=35q+6

11−14a=34a+4

Solution

a = 7

43n+9=13n−9

54a+15=34a−5

Solution

a = −40

14y+7=34y−3

35p+2=45p−1

Solution

p = 15

14n+8.25=9n+19.60

13z+6.45=8z+23.75

Solution

z = 3.46

2.4w−100=0.8w+28

2.7w−80=1.2w+10

Solution

w = 60

5.6r+13.1=3.5r+57.2

6.6x−18.9=3.4x+54.7

Solution

x = 23

Solve an Equation Using the General Strategy

In the following exercises, solve the linear equation using the general strategy.

5(x+3)=75

4(y+7)=64

Solution

y = 9

8=4(x−3)

9=3(x−3)

Solution

x = 6

20(y−8)=−60

14(y−6)=−42

Solution

y = 3

−4(2n+1)=16

−7(3n+4)=14

Solution

n = −2

3(10+5r)=0

8(3+3p)=0

Solution

p = −1

23(9c−3)=22

35(10x−5)=27

Solution

x = 5

5(1.2u−4.8)=−12

4(2.5v−0.6)=7.6

Solution

v = 1

0.2(30n+50)=28

0.5(16m+34)=−15

Solution

m = 0.25

−(w−6)=24

−(t−8)=17

Solution

t = −9

9(3a+5)+9=54

8(6b−7)+23=63

Solution

b = 2

10+3(z+4)=19

13+2(m−4)=17

Solution

m = 6

7+5(4−q)=12

−9+6(5−k)=12

Solution

k=32

15−(3r+8)=28

18−(9r+7)=−16

Solution

r = 3

11−4(y−8)=43

18−2(y−3)=32

Solution

y = −4

9(p−1)=6(2p−1)

3(4n−1)−2=8n+3

Solution

n = 2

9(2m−3)−8=4m+7

5(x−4)−4x=14

Solution

x = 34

8(x−4)−7x=14

5+6(3s−5)=−3+2(8s−1)

Solution

s = 10

−12+8(x−5)=−4+3(5x−2)

4(x−1)−8=6(3x−2)−7

Solution

x=12

7(2x−5)=8(4x−1)−9

Everyday Math

Making a fence Jovani has a fence around the rectangular garden in his backyard. The perimeter of the fence is 150 feet. The length is 15 feet more than the width. Find the width, w, by solving the equation 150=2(w+15)+2w.

Solution

30 feet

Concert tickets At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $6 and adult tickets sold for $9. The number of adult tickets sold was 5 less than 3 times the number of student tickets. Find the number of student tickets sold, s, by solving the equation 6s+9(3s−5)=1506.

Coins Rhonda has $1.90 in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, n, by solving the equation 0.05n+0.10(2n−1)=1.90.

Solution

8 nickels

Fencing Micah has 74 feet of fencing to make a rectangular dog pen in his yard. He wants the length to be 25 feet more than the width. Find the length, L, by solving the equation 2L+2(L−25)=74.

Writing Exercises

When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient as the variable side?

Solution

Answers will vary.

Solve the equation 10x+14=−2x+38, explaining all the steps of your solution.

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

Solution

Answers will vary.

Solve the equation 14(8x+20)=3x−4 explaining all the steps of your solution as in the examples in this section.

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.

Self Check

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

Self-evaluation grid for algebraic equation solving skills, covering equations with constants, variables, or both on both sides, rated by confidence level.

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

Solve Equations with Fraction or Decimal Coefficients

Learning Objectives

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

  • Solve equations with fraction coefficients
  • Solve equations with decimal coefficients

Before you get started, take this readiness quiz.

Multiply: 8·38.
If you missed this problem, review Example 10 in Multiply and Divide Fractions

Solution

3

Find the LCD of 56and14.
If you missed this problem, review Example 1 in Add and Subtract Fractions with Different Denominators

Solution

12

Multiply: 4.78 by 100.
If you missed this problem, review Example 8 in Decimal Operations

Solution

478

Solve Equations with Fraction Coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation 18x+12=14.

A mathematical equation is displayed: 1/8x + 1/2 = 1/4. This is a linear equation with one variable, x, involving fractions.
To isolate the x term, subtract 12 from both sides. An algebraic equation: 1/8x + 1/2 - 1/2 = 1/4 - 1/2, with the -1/2 terms highlighted in red, likely for simplification.
Simplify the left side. A mathematical equation is displayed, showing '1/8x = 1/4 - 1/2' in black text against a white background.
Change the constants to equivalent fractions with the LCD. A mathematical equation is displayed: 1/8x = 1/4 - 2/4. The equation involves fractions and a variable 'x'.
Subtract. The image displays the algebraic equation '1/8x = -1/4' centered on a white background.
Multiply both sides by the reciprocal of 18. The equation (8/1)*(1/8)x = (8/1)(-1/4) illustrates multiplying both sides by 8/1 (highlighted in red) to solve for the variable x.
Simplify. The equation x = -2 is displayed, representing a vertical line on a coordinate plane or a simple algebraic solution for x.

This method worked fine, but many students don’t 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 of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions. Let’s solve the same equation again, but this time use the method that clears the fractions.

Solve: 18x+12=14.

Solution

Solution

Find the least common denominator of all the fractions in the equation. A mathematical equation is shown: 1/8x + 1/2 = 1/4, with 'LCD = 8' indicating the Least Common Denominator.
Multiply both sides of the equation by that LCD, 8. This clears the fractions. A mathematical equation shows 8 multiplied by the sum of 1/8x and 1/2, which equals 8 multiplied by 1/4. This is a step in solving a linear equation, likely to clear denominators.
Use the Distributive Property. An algebraic equation is shown, displaying 8 multiplied by 1/8x, plus 8 multiplied by 1/2, equals 8 multiplied by 1/4.
Simplify — and notice, no more fractions! A mathematical equation is displayed, showing 'x + 4 = 2' in black text against a white background.
Solve using the General Strategy for Solving Linear Equations. An algebraic equation is shown where 4 is subtracted from both sides, with the subtracted 4s highlighted in red. The equation is x + 4 - 4 = 2 - 4.
Simplify. The image displays a simple mathematical equation, 'x = -2', written in black text on a white background. The equation indicates that the variable 'x' is equal to negative two.
Check: Let x=−2
Algebraic steps verifying that x=-2 satisfies the equation (1/8)x + 1/2 = 1/4, leading to the equality 1/4 = 1/4.

Solve: 14x+12=58.

Solution

x=12

Solve: 16y−13=16.

Solution

y = 3

Notice in Example 1 that 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 coefficients by clearing the fractions.

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

Solve: 7=12x+34x−23x.

Solution

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 least common denominator of all the fractions in the equation. A mathematical equation shows '7 = (1/2)x + (3/4)x - (2/3)x' followed by 'LCD = 12', indicating the least common denominator for the fractions in the expression.
Multiply both sides of the equation by 12. A mathematical equation is displayed, reading '12(7) = 12 * (1/2x + 3/4x - 2/3x)'. The numbers '12' on both sides of the equals sign are highlighted in red.
Distribute. A mathematical equation is displayed, showing 12 multiplied by 7 on the left side, equal to the sum and difference of three terms on the right side: 12 times 1/2x, plus 12 times 3/4x, minus 12 times 2/3x.
Simplify — and notice, no more fractions! A mathematical equation is displayed, reading 84 = 6x + 9x - 8x. It's a linear equation with one variable 'x' on the right side, ready to be solved.
Combine like terms. A mathematical equation is displayed with the number 84 on the left side of an equals sign, and '7x' on the right side. The equation reads '84 = 7x'.
Divide by 7. An algebraic equation showing 84/7 = 7x/7, with the common divisor 7 highlighted in red.
Simplify. A simple mathematical equation '12 = x' is displayed in black text on a plain white background, indicating that the variable 'x' is equal to the number 12.
Check: Let x=12.
Mathematical solution showing the steps to solve the equation 7 = 1/2x + 3/4x - 2/3x, concluding with x=12 and a verification that 7 = 7. A clear example of algebraic problem-solving.

Solve: 6=12v+25v−34v.

Solution

v = 40

Solve: −1=12u+14u−23u.

Solution

u = −12

In the next example, we’ll have variables and fractions on both sides of the equation.

Solve: x+13=16x−12.

Solution

Solution

Find the LCD of all the fractions in the equation. A mathematical equation showing x + 1/3 = 1/6x - 1/2, with LCD = 6. This represents a linear equation with fractions where the least common denominator is 6.
Multiply both sides by the LCD. A mathematical equation showing 6 multiplied by the sum of x and 1/3, which is equal to 6 multiplied by the difference of 1/6x and 1/2, with the number 6 highlighted in red.
Distribute. A mathematical equation showing the multiplication of fractions and a variable 'x': 6 * x + 6 * (1/3) = 6 * (1/6) * x - 6 * (1/2).
Simplify — no more fractions! A mathematical equation is displayed, reading '6x + 2 = x - 3' in black text against a white background.
Subtract x from both sides. A mathematical equation is displayed on a white background: 6x - x + 2 = x - x - 3. Some instances of the variable 'x' are highlighted in red, specifically the second 'x' on the left and the second 'x' on the right side of the equation.
Simplify. A mathematical equation is displayed on a white background. The equation reads '5x + 2 = -3' in black text, showing an algebraic expression set equal to a negative integer.
Subtract 2 from both sides. A mathematical equation shows '5x + 2 - 2 = -3 - 2', with the number '2' highlighted in red as it is subtracted from both sides, indicating a step in solving for 'x'.
Simplify. A mathematical equation is displayed on a white background, reading '5x = -5' in black font.
Divide by 5. A mathematical equation showing 5x divided by 5 equals -5 divided by 5. The number 5 in the denominator on both sides is highlighted in red, indicating a division step to solve for x.
Simplify. A mathematical equation is displayed on a white background, reading 'x = -1' in a dark, serif-like font. The equation is centrally aligned, representing a simple algebraic solution.
Check: Substitute x=−1.
Verifying a linear equation's solution: Substituting x = -1 into x + 1/3 = 1/6x - 1/2 and simplifying both sides to confirm the equality, resulting in -2/3 = -2/3.

Solve: a+34=38a−12.

Solution

a = −2

Solve: c+34=12c−14.

Solution

c = −2

In Example 4, we’ll start by using the Distributive Property. This step will clear the fractions right away!

Solve: 1=12(4x+2).

Solution

Solution

A mathematical equation is displayed, showing '1 = (1/2)(4x + 2)'. The equation is presented in black text against a white background.
Distribute. An algebraic equation is shown: 1 = (1/2) * 4x + (1/2) * 2.
Simplify. Now there are no fractions to clear! A mathematical equation is displayed, showing '1 = 2x + 1'.
Subtract 1 from both sides. An algebraic equation reads '1 - 1 = 2x + 1 - 1'. The '1's that are being subtracted on both sides of the equals sign are visually emphasized in red, suggesting cancellation for simplification.
Simplify. The image displays a simple algebraic equation, 0 = 2x, on a white background. This equation implies that the value of x must be zero.
Divide by 2. The image displays a mathematical equation: 0/2 = 2x/2. The denominator '2' is highlighted in red on both the left and right sides of the equality.
Simplify. A mathematical equation shows '0 = x' on a white background.
Check: Let x=0.
This image demonstrates the verification of the equation 1 = 1/2(4x + 2) by substituting x=0. The process shows that both sides of the equation simplify to 1, thus confirming the equality.

Solve: −11=12(6p+2).

Solution

p = −4

Solve: 8=13(9q+6).

Solution

q = 2

Many times, there will still be fractions, even after distributing.

Solve: 12(y−5)=14(y−1).

Solution

Solution

An algebraic equation showing one-half times the quantity y minus 5 equals one-fourth times the quantity y minus 1. The equation is: 1/2(y-5) = 1/4(y-1).
Distribute. A mathematical equation is displayed, reading: one-half times y minus one-half times five equals one-fourth times y minus one-fourth times one.
Simplify. A mathematical equation showing one-half y minus five-halves equals one-fourth y minus one-fourth.
Multiply by the LCD, 4. An algebraic equation showing 4 multiplied by the quantity one-half y minus five-halves, equals 4 multiplied by the quantity one-fourth y minus one-fourth.
Distribute. A mathematical equation shown in black text on a white background. The equation is '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' in black text against a plain white background.
Collect the y terms to the left. A mathematical equation is displayed, showing '2y - 10 - y = y - 1 - y'. The terms '-y' on both sides of the equation are highlighted in red, indicating they might be cancelled out.
Simplify. A mathematical equation is displayed, reading 'y - 10 = -1' in black text against a white background. It represents a simple algebraic problem.
Collect the constants to the right. An algebraic equation showing the step of adding 10 to both sides to solve for y: y - 10 + 10 = -1 + 10.
Simplify. The equation y=9 is displayed in black text on a white background.
Check: Substitute 9 for y.
Verification of the solution y=9 for the equation 1/2(y-5) = 1/4(y-1). The steps show substituting 9 for y, simplifying both sides to 2, thus confirming the equality.

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

Solution

n = 2

Solve: 12(m−3)=14(m−7).

Solution

m = −1

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, 0.3=310 and 0.17=17100. So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator.

Solve: 0.8x−5=7.

Solution

Solution

The only decimal in the equation is 0.8. Since 0.8=810, the LCD is 10. We can multiply both sides by 10 to clear the decimal.

A mathematical equation is displayed against a white background, reading '0.8x - 5 = 7'.
Multiply both sides by the LCD. The mathematical equation 10(0.8x - 5) = 10(7) is displayed, with the number 10 highlighted in red on both sides of the equality sign, indicating a common factor or a step in solving for x.
Distribute. A mathematical equation is shown: 10(0.8x) - 10(5) = 10(7). The equation involves multiplication and subtraction, with the number 10 being a common factor on both sides of the equality.
Multiply, and notice, no more decimals! A mathematical equation displays '8x - 50 = 70' in black text on a white background, representing a linear equation with one variable to be solved.
Add 50 to get all constants to the right. The equation 8x - 50 + 50 = 70 + 50 demonstrates adding 50 to both sides to isolate the term with x in a linear equation.
Simplify. A mathematical equation shows '8x = 120' in black text on a white background.
Divide both sides by 8. The equation 8x/8 = 120/8, showing division by 8 on both sides. The number 8 in the denominator is highlighted in red.
Simplify. The mathematical equation 'x = 15' is displayed in a clean, minimalist style against a white background.
Check: Let x=15.
A step-by-step mathematical solution demonstrates that 0.8(15) - 5 equals 7. The calculation shows 12 - 5 simplifies to 7, confirming the equality with a checkmark.

Solve: 0.6x−1=11.

Solution

x = 20

Solve: 1.2x−3=9.

Solution

x = 10

Solve: 0.06x+0.02=0.25x−1.5.

Solution

Solution

Look at the decimals and think of the equivalent fractions.

0.06=6100,0.02=2100,0.25=25100,1.5=1510

Notice, the LCD is 100.

By multiplying by the LCD we will clear the decimals.
A linear equation is displayed as '0.06x + 0.02 = 0.25x - 1.5'.
Multiply both sides by 100. An algebraic equation showing 100 multiplied by the sum of 0.06x and 0.02, equaling 100 multiplied by the difference of 0.25x and 1.5. The number 100 is highlighted in red.
Distribute. A mathematical equation is displayed: 100(0.06x) + 100(0.02) = 100(0.25x) - 100(1.5).
Multiply, and now no more decimals. A mathematical equation is displayed against a white background, reading 6x + 2 = 25x - 150.
Collect the variables to the right. An algebraic equation: 6x - 6x + 2 = 25x - 6x - 150, with specific '6x' terms highlighted in red for emphasis.
Simplify. A mathematical equation is displayed, showing '2 = 19x - 150'. The equation contains the numbers 2, 19, and 150, along with the variable 'x', an equals sign, and a minus sign.
Collect the constants to the left. A mathematical equation, 2 + 150 = 19x - 150 + 150, is displayed in black text with the numbers '+ 150' on both sides highlighted in red, indicating a step in solving for x.
Simplify. A simple algebraic equation is presented, showing '152 = 19x' against a white background.
Divide by 19. Algebraic equation: 152/19 = 19x/19, demonstrating division by 19 to isolate x.
Simplify. The image shows a mathematical equation on a white background, stating '8 = x', indicating that the value of the variable x is equal to 8.
Check: Let x=8.
A three-step solution to a mathematical equation, showing 0.06(8) + 0.02 = 0.25(8) - 1.5 simplifies to 0.50 = 0.50, verified with a checkmark.

Solve: 0.14h+0.12=0.35h−2.4.

Solution

h = 12

Solve: 0.65k−0.1=0.4k−0.35.

Solution

k = −1

The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation.

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

Solution

Solution

The image shows the algebraic equation 0.25x + 0.05(x + 3) = 2.85.
Distribute first. A mathematical equation is presented, reading '0.25x + 0.05x + 0.15 = 2.85' against a white background.
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 shows 100 multiplied by the sum of 0.30x and 0.15 on the left side, which is equal to 100 multiplied by 2.85 on the right side. The number 100 is highlighted in red.
Distribute. A mathematical equation is displayed on a white background, which reads '30x + 15 = 285'.
Subtract 15 from both sides. The equation 30x + 15 - 15 = 285 - 15, illustrating a step in solving for x where 15 is subtracted from both sides, with the subtracted 15 shown in red.
Simplify. A mathematical equation showing 30 multiplied by x equals 270.
Divide by 30. A mathematical equation shows '30x / 30 = 270 / 30' with the denominator '30' on both sides highlighted in red.
Simplify. A simple mathematical equation 'x = 9' is displayed on a white background.
Check: Let x=9.
Mathematical steps showing the verification of the solution x=9 for the equation 0.25x + 0.05(x + 3) = 2.85, concluding with a checked equality 2.85 = 2.85.

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

Solution

n = 9

Solve: 0.10d+0.05(d−5)=2.15.

Solution

d = 16

ACCESS ADDITIONAL ONLINE RESOURCES

  • Solve an Equation with Fractions with Variable Terms on Both Sides
  • Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides
  • Ex 2: Solve an Equation with Fractions with Variable Terms on Both Sides
  • Solving Multiple Step Equations Involving Decimals
  • Ex: Solve a Linear Equation With Decimals and Variables on Both Sides
  • Ex: Solve an Equation with Decimals and Parentheses

Key Concepts

  • Solve equations with fraction coefficients by clearing the fractions.
    1. Find the least common denominator of all the fractions in the equation.
    2. Multiply both sides of the equation by that LCD. This clears the fractions.
    3. Solve using the General Strategy for Solving Linear Equations.

Section Exercises

Practice Makes Perfect

Solve equations with fraction coefficients

In the following exercises, solve the equation by clearing the fractions.

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

14m−45m+12m=−1

Solution

m = 20

56n−14n−12n=−2

x+12=23x−12

Solution

x = −3

x+34=12x−54

13w+54=w−14

Solution

w=94

32z+13=z−23

12x−14=112x+16

Solution

x = 1

12a−14=16a+112

13b+15=25b−35

Solution

b = 12

13x+25=15x−25

1=16(12x−6)

Solution

x = 1

1=15(15x−10)

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

Solve Equations with Decimal Coefficients

In the following exercises, solve the equation by clearing the decimals.

0.6y+3=9

Solution

y = 10

0.4y−4=2

3.6j−2=5.2

Solution

j = 2

2.1k+3=7.2

0.4x+0.6=0.5x−1.2

Solution

x = 18

0.7x+0.4=0.6x+2.4

0.23x+1.47=0.37x−1.05

Solution

x = 18

0.48x+1.56=0.58x−0.64

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

0.05(q−5)+0.25q=3.05

Solution

q = 11

0.05(q−8)+0.25q=4.10

Everyday Math

Coins Taylor has $2.00 in dimes and pennies. The number of pennies is 2 more than the number of dimes. Solve the equation 0.10d+0.01(d+2)=2 for d, the number of dimes.

Solution

d = 18

Stamps Travis bought $9.45 worth of 49-cent stamps and 21-cent stamps. The number of 21-cent stamps was 5 less than the number of 49-cent stamps. Solve the equation 0.49s+0.21(s−5)=9.45 for s, to find the number of 49-cent stamps Travis bought.

Writing Exercises

Explain how to find the least common denominator of 38,16,and23.

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.

In the equation 0.35x+2.1=3.85, what is the LCD? How do you know?

Self Check

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

A self-assessment chart for math skills, asking students to rate their ability to solve equations using a general strategy, with fraction coefficients, and with decimal coefficients, across three confidence levels.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, determine whether the given number is a solution to the equation.

x+16=31,x=15

Solution

yes

w−8=5,w=3

−9n=45,n=54

Solution

no

4a=72,a=18

In the following exercises, solve the equation using the Subtraction Property of Equality.

x+7=19

Solution

12

y+2=−6

a+13=53

Solution

a=43

n+3.6=5.1

In the following exercises, solve the equation using the Addition Property of Equality.

u−7=10

Solution

u = 17

x−9=−4

c−311=911

Solution

c=1211

p−4.8=14

In the following exercises, solve the equation.

n−12=32

Solution

n = 44

y+16=−9

f+23=4

Solution

f=103

d−3.9=8.2

y+8−15=−3

Solution

y = 4

7x+10−6x+3=5

6(n−1)−5n=−14

Solution

n = −8

8(3p+5)−23(p−1)=35

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

The sum of −6 and m is 25.

Solution

−6 + m = 25; m = 31

Four less than n is 13.

In the following exercises, translate into an algebraic equation and solve.

Rochelle’s daughter is 11 years old. Her son is 3 years younger. How old is her son?

Solution

s = 11 − 3; 8 years old

Tan weighs 146 pounds. Minh weighs 15 pounds more than Tan. How much does Minh weigh?

Peter paid $9.75 to go to the movies, which was $46.25 less than he paid to go to a concert. How much did he pay for the concert?

Solution

c − 46.25 = 9.75; $56.00

Elissa earned $152.84 this week, which was $21.65 more than she earned last week. How much did she earn last week?

Solve Equations using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation using the Division Property of Equality.

8x=72

Solution

x = 9

13a=−65

0.25p=5.25

Solution

p = 21

−y=4

In the following exercises, solve each equation using the Multiplication Property of Equality.

n6=18

Solution

n = 108

y−10=30

36=34x

Solution

x = 48

58u=1516

In the following exercises, solve each equation.

−18m=−72

Solution

m = 4

c9=36

0.45x=6.75

Solution

x = 15

1112=23y

5r−3r+9r=35−2

Solution

r = 3

24x+8x−11x=−7−14

Solve Equations with Variables and Constants on Both Sides

In the following exercises, solve the equations with constants on both sides.

8p+7=47

Solution

p = 5

10w−5=65

3x+19=−47

Solution

x = −22

32=−4−9n

In the following exercises, solve the equations with variables on both sides.

7y=6y−13

Solution

y = −13

5a+21=2a

k=−6k−35

Solution

k = −5

4x−38=3x

In the following exercises, solve the equations with constants and variables on both sides.

12x−9=3x+45

Solution

x = 6

5n−20=−7n−80

4u+16=−19−u

Solution

u = −7

58c−4=38c+4

In the following exercises, solve each linear equation using the general strategy.

6(x+6)=24

Solution

x = −2

9(2p−5)=72

−(s+4)=18

Solution

s = −22

8+3(n−9)=17

23−3(y−7)=8

Solution

y = 12

13(6m+21)=m−7

8(r−2)=6(r+10)

Solution

r = 38

5+7(2−5x)=2(9x+1)−(13x−57)

4(3.5y+0.25)=365

Solution

y = 26

0.25(q−8)=0.1(q+7)

Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation by clearing the fractions.

25n−110=710

Solution

n = 2

13x+15x=8

34a−13=12a+56

Solution

a=143

12(k+3)=13(k+16)

In the following exercises, solve each equation by clearing the decimals.

0.8x−0.3=0.7x+0.2

Solution

x = 5

0.36u+2.55=0.41u+6.8

0.6p−1.9=0.78p+1.7

Solution

p = −20

0.10d+0.05(d−4)=2.05

Chapter Practice Test

Determine whether each number is a solution to the equation.
3x+5=23.
  1. ⓐ 6
  2. ⓑ 235
Solution
  1. ⓐ yes
  2. ⓑ no

In the following exercises, solve each equation.

n−18=31

9c=144

Solution

c = 16

4y−8=16

−8x−15+9x−1=−21

Solution

x = −5

−15a=120

23x=6

Solution

x = 9

x+3.8=8.2

10y=−5y+60

Solution

y = 4

8n+2=6n+12

9m−2−4m+m=42−8

Solution

m = 6

−5(2x+1)=45

−(d+9)=23

Solution

d = −32

13(6m+21)=m−7

2(6x+5)−8=−22

Solution

x = −2

8(3a+5)−7(4a−3)=20−3a

14p+13=12

Solution

p=23

0.1d+0.25(d+8)=4.1

Translate and solve: The difference of twice x and 4 is 16.

Solution

2x − 4 = 16; x = 10

Samuel paid $25.82 for gas this week, which was $3.47 less than he paid last week. How much did he pay last week?

Introduction

Part of a glass building is shown. The structure is made up of individual shapes.
Note the many individual shapes in this building. (credit: Bert Kaufmann, Flickr)

We are surrounded by all sorts of geometry. Architects use geometry to design buildings. Artists create vivid images out of colorful geometric shapes. Street signs, automobiles, and product packaging all take advantage of geometric properties. In this chapter, we will begin by considering a formal approach to solving problems and use it to solve a variety of common problems, including making decisions about money. Then we will explore geometry and relate it to everyday situations, using the problem-solving strategy we develop.

Use a Problem Solving Strategy

Learning Objectives

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

  • Approach word problems with a positive attitude
  • Use a problem solving strategy for word problems
  • Solve number problems

Before you get started, take this readiness quiz.

Translate “6 less than twice x” into an algebraic expression.
If you missed this problem, review Example 13 in Evaluate, Simplify, and Translate Expressions.

Solution

2x−6

Solve: 23x=24.
If you missed this problem, review Example 4 in Solve Equations Using the Division and Multiplication Properties of Equality.

Solution

36

Solve: 3x+8=14.
If you missed this problem, review Example 1 in Solve Equations with Variables and Constants on Both Sides.

Solution

2

Approach Word Problems with a Positive Attitude

The world is full of word problems. How much money do I need to fill the car with gas? How much should I tip the server at a restaurant? How many socks should I pack for vacation? How big a turkey do I need to buy for Thanksgiving dinner, and what time do I need to put it in the oven? If my sister and I buy our mother a present, how much will each of us pay?

Now that we can solve equations, we are ready to apply our new skills to word problems. Do you know anyone who has had negative experiences in the past with word problems? Have you ever had thoughts like the student in Figure 1?

A cartoon image of a girl with a sad expression writing on a piece of paper is shown. There are 5 thought bubbles. They read, “I don't know whether to add, subtract multiply, or divide!,” then “I don't understand word problems!,” then “My teachers never explained this!,” then “If I just skip all the word problems, I can probably still pass the class,” and lastly, “I just can't do this!”
Negative thoughts about word problems can be barriers to success.

When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. We need to calm our fears and change our negative feelings.

Start with a fresh slate and begin to think positive thoughts like the student in Figure 2. Read the positive thoughts and say them out loud.

A cartoon image of a girl with a confident expression holding some books is shown.  There are 4 thought bubbles. They read, “While word problems were hard in the past, I think I can try them now,” then “I am better prepared now. I think I will begin to understand word problems,” then “I think I can! I think I can!,” and lastly, “It may take time, but I can begin to solve word problems.”
When it comes to word problems, a positive attitude is a big step toward success.

If we take control and believe we can be successful, we will be able to master word problems.

Think of something that you can do now but couldn't do three years ago. Whether it's driving a car, snowboarding, cooking a gourmet meal, or speaking a new language, you have been able to learn and master a new skill. Word problems are no different. Even if you have struggled with word problems in the past, you have acquired many new math skills that will help you succeed now!

Use a Problem-solving Strategy for Word Problems

In earlier chapters, you translated word phrases into algebraic expressions, using some basic mathematical vocabulary and symbols. Since then you've increased your math vocabulary as you learned about more algebraic procedures, and you've had more practice translating from words into algebra.

You have also translated word sentences into algebraic equations and solved some word problems. The word problems applied math to everyday situations. You had to restate the situation in one sentence, assign a variable, and then write an equation to solve. This method works as long as the situation is familiar to you and the math is not too complicated.

Now we'll develop a strategy you can use to solve any word problem. This strategy will help you become successful with word problems. We'll demonstrate the strategy as we solve the following problem.

Pete bought a shirt on sale for $18, which is one-half the original price. What was the original price of the shirt?

Solution

Solution

Step 1. Read the problem. Make sure you understand all the words and ideas. You may need to read the problem two or more times. If there are words you don't understand, look them up in a dictionary or on the Internet.

  • In this problem, do you understand what is being discussed? Do you understand every word?

Step 2. Identify what you are looking for. It's hard to find something if you are not sure what it is! Read the problem again and look for words that tell you what you are looking for!

  • In this problem, the words “what was the original price of the shirt” tell you that what you are looking for: the original price of the shirt.

Step 3. Name what you are looking for. Choose a variable to represent that quantity. You can use any letter for the variable, but it may help to choose one that helps you remember what it represents.

  • Let p= the original price of the shirt

Step 4. Translate into an equation. It may help to first restate the problem in one sentence, with all the important information. Then translate the sentence into an equation.

The top line reads: “18 is one-half of the original price.” Below 18 is a brace and the number 18. Below “is” is a brace and an equal sign. Below “one-half” is a brace and the fraction 1 over 2. Below “of” is a brace and a multiplication dot. Below “the original price” is a brace and an italicized p.
Step 5. Solve the equation using good algebra techniques. Even if you know the answer right away, using algebra will better prepare you to solve problems that do not have obvious answers.
Write the equation. A mathematical equation is presented, showing '18 = (1/2)p' in black font against a white background.
Multiply both sides by 2. A mathematical equation shows '2 * 18 = 2 * (1/2)p'. The number 2 is highlighted in red on both sides of the equals sign, indicating it is a common factor used in solving for 'p'.
Simplify. The image displays a simple mathematical equation, '36 = p', where the number 36 is set equal to the variable 'p' on a plain white background.

Step 6. Check the answer in the problem and make sure it makes sense.

  • We found that p=36, which means the original price was $36. Does $36 make sense in the problem? Yes, because 18 is one-half of 36, and the shirt was on sale at half the original price.

Step 7. Answer the question with a complete sentence.

  • The problem asked “What was the original price of the shirt?” The answer to the question is: “The original price of the shirt was $36.”

If this were a homework exercise, our work might look like this:

The top reads, “Let p equal the original price. 18 is one-half the original price.” The next line shows the equation 18 equals one-half times p. The following line shows the same equation with each side being multiplied by 2. The next line shows 36 equals p. Below this, it reads, “Check: Is $36 a reasonable price for a shirt? Yes. Is 18 one-half of 36? Yes. The original price of the shirt as $36.

Joaquin bought a bookcase on sale for $120, which was two-thirds the original price. What was the original price of the bookcase?

Solution

$180

Two-fifths of the people in the senior center dining room are men. If there are 16 men, what is the total number of people in the dining room?

Solution

40

We list the steps we took to solve the previous example.

Problem-Solving Strategy

  1. Read the word problem. Make sure you understand all the words and ideas. You may need to read the problem two or more times. If there are words you don't understand, look them up in a dictionary or on the internet.
  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 first restate the problem in one sentence before translating.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem. Make sure it makes sense.
  7. Answer the question with a complete sentence.

Let's use this approach with another example.

Yash brought apples and bananas to a picnic. The number of apples was three more than twice the number of bananas. Yash brought 11 apples to the picnic. How many bananas did he bring?

Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for. How many bananas did he bring?
Step 3. Name what you are looking for.
Choose a variable to represent the number of bananas.
Let b=number of bananas
Step 4. Translate. Restate the problem in one sentence with all the important information.
Translate into an equation.

An illustration demonstrating how to translate the word problem 'The number of apples was 3 more than twice the number of bananas' into the algebraic equation '11 = 3 + 2b'.
Step 5. Solve the equation. A mathematical equation is displayed, 11 = 2b + 3.
Subtract 3 from each side. A mathematical equation shows '11 - 3 = 2b + 3 - 3' in a black serif font, with the number '3' highlighted in red on both sides of the equals sign, indicating subtraction from each side.
Simplify. A mathematical equation on a white background, displaying '8 = 2b' in black text. The numbers and letter are clearly visible, indicating a simple algebraic problem.
Divide each side by 2. A mathematical equation shows '8 divided by 2 equals 2b divided by 2.' The denominator '2' is highlighted in red on both sides of the equation.
Simplify. The image displays a mathematical equation, '4 = b,' indicating that the value of the variable 'b' is equal to 4, set against a plain white background.
Step 6. Check: First, is our answer reasonable? Yes, bringing four bananas to a picnic seems reasonable. The problem says the number of apples was three more than twice the number of bananas. If there are four bananas, does that make eleven apples? Twice 4 bananas is 8. Three more than 8 is 11.
Step 7. Answer the question. Yash brought 4 bananas to the picnic.

Guillermo bought textbooks and notebooks at the bookstore. The number of textbooks was 3 more than the number of notebooks. He bought 5 textbooks. How many notebooks did he buy?

Solution

2

Gerry worked Sudoku puzzles and crossword puzzles this week. The number of Sudoku puzzles he completed is seven more than the number of crossword puzzles. He completed 14 Sudoku puzzles. How many crossword puzzles did he complete?

Solution

7

In Solve Sales Tax, Commission, and Discount Applications, we learned how to translate and solve basic percent equations and used them to solve sales tax and commission applications. In the next example, we will apply our Problem Solving Strategy to more applications of percent.

Nga's car insurance premium increased by $60, which was 8% of the original cost. What was the original cost of the premium?

Solution

Solution

Step 1. Read the problem. Remember, if there are words you don't understand, look them up.
Step 2. Identify what you are looking for. the original cost of the premium
Step 3. Name. Choose a variable to represent the original cost of premium. Let c=the original cost
Step 4. Translate. Restate as one sentence. Translate into an equation.
An image demonstrating how to translate a word problem into an algebraic equation. The phrase '$60 was 8% of the original cost' is shown aligned with its mathematical translation: '60 = 0.08 * c'.
Step 5. Solve the equation. The mathematical equation '60 = 0.08c' is displayed in black text on a white background, representing a calculation where 60 is equal to 0.08 times c.
Divide both sides by 0.08. Equation: 60/0.08 = 0.08c/0.08, demonstrating both sides divided by 0.08 (highlighted in red) to solve for the variable 'c'.
Simplify. c=750
Step 6. Check: Is our answer reasonable? Yes, a $750 premium on auto insurance is reasonable. Now let's check our algebra. Is 8% of 750 equal to 60?
A mathematical equation shows '750 = c' in black text on a white background, representing the value of c as 750.
A mathematical equation shows '0.08(750) = 60' in black text against a white background.
The fundamental equality '60 = 60' is presented alongside a clear checkmark, visually confirming its undeniable truth and mathematical accuracy.
Step 7. Answer the question. The original cost of Nga's premium was $750.

Pilar's rent increased by 4%. The increase was $38. What was the original amount of Pilar's rent?

Solution

$950

Steve saves 12% of his paycheck each month. If he saved $504 last month, how much was his paycheck?

Solution

$4,200

Solve Number Problems

Now we will translate and solve number problems. In number problems, you are given some clues about one or more numbers, and you use these clues to build an equation. Number problems don't usually arise on an everyday basis, but they provide a good introduction to practicing the Problem Solving Strategy. Remember to look for clue words such as difference, of, and and.

The difference of a number and six is 13. Find the number.

Solution

Solution

Step 1. Read the problem. Do you understand all the words?
Step 2. Identify what you are looking for. the number
Step 3. Name. Choose a variable to represent the number. Let n=the number
Step 4. Translate. Restate as one sentence.
Translate into an equation.
Shows how to translate the phrase 'The difference of a number and 6 is 13' into the algebraic equation 'n - 6 = 13', breaking down the verbal statement into mathematical symbols.
Step 5. Solve the equation.
Add 6 to both sides.
Simplify.
A basic algebra equation is displayed, showing 'n - 6 = 13' in black text on a white background.
A mathematical equation 'n - 6 + 6 = 13 + 6' is displayed on a white background, demonstrating the addition property of equality with the added '6' highlighted in red.
The image displays the mathematical expression 'n = 19' centered on a white background.
Step 6. Check:
The difference of 19 and 6 is 13. It checks.
Step 7. Answer the question. The number is 19.

The difference of a number and eight is 17. Find the number.

Solution

25

The difference of a number and eleven is −7. Find the number.

Solution

4

The sum of twice a number and seven is 15. Find the number.

Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for. the number
Step 3. Name. Choose a variable to represent the number. Let n=the number
Step 4. Translate. Restate the problem as one sentence.
Translate into an equation.
The image demonstrates how to translate a word problem into an algebraic equation. The phrase 'The sum of twice a number and 7 is 15' is mapped to the equation 2 * n + 7 = 15.
Step 5. Solve the equation. The image shows a mathematical equation on a white background: '2n + 7 = 15'. The equation features the variable 'n', a multiplication operation (implied), addition, and an equality.
Subtract 7 from each side and simplify. The mathematical equation '2n = 8' is displayed in black text against a white background.
Divide each side by 2 and simplify. The mathematical expression 'n = 4' is displayed in the bottom right corner of a white background, with no other elements visible in the image.
Step 6. Check: is the sum of twice 4 and 7 equal to 15?
A simple arithmetic equation, 2 * 4 + 7 = 15, is presented on a white surface.
A mathematical equation displays '8 + 7 = 15' on a white background.
A mathematical equation '15 = 15' is displayed, followed by a checkmark, indicating its correctness.
Step 7. Answer the question. The number is 4.

The sum of four times a number and two is 14. Find the number.

Solution

3

The sum of three times a number and seven is 25. Find the number.

Solution

6

Some number word problems ask you 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. 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.

One number is five more than another. The sum of the numbers is twenty-one. Find the numbers.

Solution

Solution

Step-by-step guide demonstrating how to solve a word problem by identifying variables, translating to an equation, solving, and checking the solution.
Step 1. Read the problem.
Step 2. Identify what you are looking for. You are looking for two numbers.
Step 3. Name.
Choose a variable to represent the first number.
What do you know about the second number?
Translate.

Let n=1st number
One number is five more than another.
n+5=2nd number
Step 4. Translate.
Restate the problem as one sentence with all the important information.
Translate into an equation.
Substitute the variable expressions.

The sum of the numbers is 21.
The sum of the 1st number and the 2nd number is 21.
An algebraic equation showing the sum of two numbers, 'n' and 'n+5', equaling 21. The first number is represented by 'n', and the second number is 'n+5', together summing up to 21.
Step 5. Solve the equation. A mathematical equation is displayed on a white background, reading 'n + n + 5 = 21'.
Combine like terms. A simple linear algebraic equation, '2x + 5 = 21' is displayed, demonstrating an unknown variable multiplied by a coefficient, added to a constant, and equated to another constant.
Subtract five from both sides and simplify. The image shows the mathematical equation 2n = 16. This equation represents a linear equation where 'n' is the variable, '2' is the coefficient of 'n', and '16' is the constant term.
Divide by two and simplify. The text 'n = 8 1st number' is displayed on a white background, indicating a variable assignment and its sequence.
Find the second number too. The image displays the mathematical expression 'n + 5' followed by the text '2nd number' on a white background.
Substitute n = 8 A basic arithmetic expression, 8 + 5, is displayed on a white background. The number '8' is rendered in red, while the plus sign and the number '5' are in black.
The number '13' is displayed in a dark font against a plain white background, positioned towards the upper right side of the frame.
Step 6. Check:
Do these numbers check in the problem?
Is one number 5 more than the other?
Is thirteen, 5 more than 8? Yes.

Is the sum of the two numbers 21?
A mathematical equation asks if 13 is equal to 8 + 5, which is a true statement as both sides equal 13.
The image displays the equation '13 = 13' followed by a checkmark, indicating that the statement is correct.

A mathematical equation shows '8 + 13' followed by an equals sign with a question mark above it, then '21', posing whether 8 plus 13 equals 21.
The number 21 is equal to 21, validated with a checkmark, representing a correct mathematical statement or calculation.
Step 7. Answer the question. The numbers are 8 and 13.

One number is six more than another. The sum of the numbers is twenty-four. Find the numbers.

Solution

9, 15

The sum of two numbers is fifty-eight. One number is four more than the other. Find the numbers.

Solution

27, 31

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

Solution

Solution

A step-by-step guide demonstrating the seven-step process to solve a word problem involving finding two unknown numbers.
Step 1. Read the problem.
Step 2. Identify what you are looking for. two numbers
Step 3. Name. Choose a variable.
What do you know about the second number?
Translate.

Let n = 1st number
One number is 4 less than the other.
n - 4 = 2nd number
Step 4. Translate.
Write as one sentence.
Translate into an equation.
Substitute the variable expressions.

The sum of two numbers is negative fourteen.
This image presents an algebraic equation where the sum of two numbers, 'n' and 'n-4', equals -14. It demonstrates the translation of a word problem into a solvable mathematical expression.
Step 5. Solve the equation. A mathematical equation is displayed with a variable 'n'. The equation reads as 'n + n - 4 = -14'.
Combine like terms. The image displays the equation 2n - 4 = -14, which is an algebraic expression involving a variable 'n', constants, and arithmetic operations.
Add 4 to each side and simplify. A mathematical equation is displayed on a white background, reading '2n = -10' in black text.
Divide by 2. The image displays the mathematical expression 'n = -5 1st number' against a plain white background.
Substitute n=−5 to find the 2nd number. The image displays the algebraic expression 'n - 4' followed by the text '2nd number', suggesting it represents a second number defined in relation to a variable 'n'.
The image displays a mathematical expression '-5 - 4' horizontally on a white background. The number -5 is in red, and -4 is in black.
A white image features a small, dark numeral '-9' positioned in the upper right portion of the frame.
Step 6. Check:
Is −9 four less than −5?


Is their sum −14?
The image displays the math problem -5 - 4 with a question mark positioned directly above the equals sign, before the result -9. It asks if the equation -5 - 4 = -9 is true.
A mathematical statement showing '-9 = -9' followed by a checkmark, indicating the equality is correct.
A mathematical equation asking if -5 + (-9) equals -14.
The equation -14 = -14 is correctly shown with a checkmark, indicating it is a true statement.
Step 7. Answer the question. The numbers are −5 and −9.

The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.

Solution

−8, −15

The sum of two numbers is negative eighteen. One number is 40 more than the other. Find the numbers.

Solution

−29, 11

One number is ten more than twice another. Their sum is one. Find the numbers.

Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for. two numbers
Step 3. Name. Choose a variable.
One number is ten more than twice another.
Let x = 1st number
2x + 10 = 2nd number
Step 4. Translate. Restate as one sentence. Their sum is one.
Translate into an equation Translating a word problem into an algebraic equation: 'The sum of the two numbers is 1' becomes x + (2x + 10) = 1.
Step 5. Solve the equation. A mathematical equation is displayed against a white background: x + 2x + 10 = 1. The equation shows algebraic terms with variables and constants, combined with addition and equality signs.
Combine like terms. A mathematical equation is displayed, showing '3x + 10 = 1' in black text against a white background.
Subtract 10 from each side. A mathematical equation, '3x = -9', is displayed against a white background.
Divide each side by 3 to get the first number. The mathematical equation 'x = -3' is displayed in black text against a plain white background, presenting a simple algebraic expression.
Substitute to get the second number. The mathematical expression '2x + 10' is displayed in black text on a white background.
The mathematical expression 2(-3) + 10 is displayed on a white background, representing a calculation where two is multiplied by negative three, and then ten is added to the product.
The number 4 is displayed in black on a plain white background.
Step 6. Check.
Is 4 ten more than twice −3?



Is their sum 1?
A mathematical equation reads '2(-3) + 10 =? 4', asking if the expression on the left evaluates to 4. This challenges basic arithmetic operations including multiplication, addition, and negative numbers.


The image shows the mathematical equation -6 + 10 = 4, displaying a basic arithmetic operation involving negative and positive integers.
A mathematical equation '4 = 4' with a checkmark, indicating that the statement is correct.
An image displaying the arithmetic problem -3 + 4 with a question mark over the equals sign before the number 1, asking if the statement is true.
The equation '1=1' with a checkmark, indicating a correct and self-evident mathematical truth.
Step 7. Answer the question. The numbers are −3 and 4.

One number is eight more than twice another. Their sum is negative four. Find the numbers.

Solution

−4, 0

One number is three more than three times another. Their sum is negative five. Find the numbers.

Solution

−2, −3

Consecutive integers are integers that immediately follow each other. Some examples of consecutive integers are:

...1,2,3,4,...
...−10,−9,−8,−7,...
...150,151,152,153,...

Notice that each number is one more than the number preceding it. So 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, or n+2.

n1st integern+12nd consecutive integern+23rd consecutive integer

The sum of two consecutive integers is 47. Find the numbers.

Solution

Solution

A 7-step process for solving word problems, demonstrating the application of each step to find two consecutive integers.
Step 1. Read the problem.
Step 2. Identify what you are looking for. two consecutive integers
Step 3. Name. Let n = 1st integer
n + 1 = next consecutive integer
Step 4. Translate.
Restate as one sentence.
Translate into an equation.
An image showing the translation of the phrase 'The sum of the integers is 47' into the algebraic equation 'n + n + 1 = 47', with corresponding parts highlighted by brackets.
Step 5. Solve the equation. A mathematical equation is displayed, showing 'n + n + 1 = 47' in black text against a white background.
Combine like terms. A mathematical equation, '2n + 1 = 47', is displayed in black font against a white background, representing a linear equation to be solved for the variable 'n'.
Subtract 1 from each side. A simple mathematical equation is displayed on a white background, showing '2n = 46' in black text, representing a basic algebraic problem.
Divide each side by 2. The image displays mathematical notation 'n = 23' followed by the text '1st integer' on a white background.
Substitute to get the second number. The image shows the mathematical expression 'n + 1 2nd integer' in a horizontal layout.
The number 23 in red text is shown next to a plus sign and the number 1 in black text on a white background, forming the mathematical expression '23 + 1'.
The number 24 is displayed in black text on a white background.
Step 6. Check: A mathematical expression reads '23 + 24 =? 47', indicating a query about whether 23 plus 24 indeed equals 47.
The equation '47 = 47' is displayed with a checkmark, indicating its correctness.
Step 7. Answer the question. The two consecutive integers are 23 and 24.

The sum of two consecutive integers is 95. Find the numbers.

Solution

47, 48

The sum of two consecutive integers is −31. Find the numbers.

Solution

−15, −16

Find three consecutive integers whose sum is 42.

Solution

Solution

Illustrates the step-by-step process for solving a word problem involving finding three consecutive integers, from identification to solution verification.
Step 1. Read the problem.
Step 2. Identify what you are looking for. three consecutive integers
Step 3. Name. 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.
Translating a word problem into an algebraic equation, where 'The sum of the three integers is 42' becomes 'n + n+1 + n+2 = 42' with visual cues.
Step 5. Solve the equation. A mathematical equation is displayed on a white background, reading 'n + n + 1 + n + 2 = 42'. This equation can be simplified to 3n + 3 = 42, which implies 3n = 39, so n = 13.
Combine like terms. A mathematical equation is displayed on a white background, reading '3n + 3 = 42'.
Subtract 3 from each side. A mathematical equation is displayed against a white background, which reads '3n = 39'.
Divide each side by 3. The equation 'n = 13' is displayed next to the text '1st integer' on a white background.
Substitute to get the second number. The image shows the expression 'n + 1' alongside '2nd integer', indicating a numerical sequence or definition of an integer in relation to 'n'.
The numbers '13 + 1' are displayed on a white background, with '13' in red and '+ 1' in black, indicating a simple addition problem.
The number '14' is displayed in black text against a plain white background.
Substitute to get the third number. A mathematical expression and text on a white background, showing 'n + 2' followed by '3rd integer' in blue text.
A mathematical equation shows the number 13 in red, followed by a black plus sign and the number 2, all centered on a white background.
The number 15 is prominently displayed in a simple, clear font against a white background.
Step 6. Check: A mathematical equation displayed as 13 + 14 + 15 =? 42, posing the question of whether the sum of the three numbers equals 42.
The number 42 is shown equal to itself, followed by a checkmark, indicating a verified mathematical truth or a solved equation.
Step 7. Answer the question. The three consecutive integers are 13, 14, and 15.

Find three consecutive integers whose sum is 96.

Solution

31, 32, 33

Find three consecutive integers whose sum is −36.

Solution

−11, −12, −13

The Links to Literacy activities Math Curse, Missing Mittens and Among the Odds and Evens will provide you with another view of the topics covered in this section.

Key Concepts

  • Problem Solving Strategy
    1. Read the word problem. Make sure you understand all the words and ideas. You may need to read the problem two or more times. If there are words you don't understand, look them up in a dictionary or on the internet.
    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 first restate the problem in one sentence before translating.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem. Make sure it makes sense.
    7. Answer the question with a complete sentence.

Practice Makes Perfect

Use a Problem-solving Strategy for Word Problems

In the following exercises, use the problem-solving strategy for word problems to solve. Answer in complete sentences.

Two-thirds of the children in the fourth-grade class are girls. If there are 20 girls, what is the total number of children in the class?

Solution

There are 30 children in the class.

Three-fifths of the members of the school choir are women. If there are 24 women, what is the total number of choir members?

Zachary has 25 country music CDs, which is one-fifth of his CD collection. How many CDs does Zachary have?

Solution

Zachary has 125 CDs.

One-fourth of the candies in a bag of are red. If there are 23 red candies, how many candies are in the bag?

There are 16 girls in a school club. The number of girls is 4 more than twice the number of boys. Find the number of boys in the club.

Solution

There are 6 boys in the club.

There are 18 Cub Scouts in Troop 645. The number of scouts is 3 more than five times the number of adult leaders. Find the number of adult leaders.

Lee is emptying dishes and glasses from the dishwasher. The number of dishes is 8 less than the number of glasses. If there are 9 dishes, what is the number of glasses?

Solution

There are 17 glasses.

The number of puppies in the pet store window is twelve less than the number of dogs in the store. If there are 6 puppies in the window, what is the number of dogs in the store?

After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?

Solution

Lisa's original weight was 175 pounds.

Tricia got a 6% raise on her weekly salary. The raise was $30 per week. What was her original weekly salary?

Tim left a $9 tip for a $50 restaurant bill. What percent tip did he leave?

Solution

18%

Rashid left a $15 tip for a $75 restaurant bill. What percent tip did he leave?

Yuki bought a dress on sale for $72. The sale price was 60% of the original price. What was the original price of the dress?

Solution

The original price was $120.

Kim bought a pair of shoes on sale for $40.50. The sale price was 45% of the original price. What was the original price of the shoes?

Solve Number Problems

In the following exercises, solve each number word problem.

The sum of a number and eight is 12. Find the number.

Solution

4

The sum of a number and nine is 17. Find the number.

The difference of a number and twelve is 3. Find the number.

Solution

15

The difference of a number and eight is 4. 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 forty-two. Find the numbers.

Solution

18, 24

One number is five more than the other. Their sum is thirty-three. Find the numbers.

The sum of two numbers is twenty. One number is four less than the other. Find the numbers.

Solution

8, 12

The sum of two numbers is twenty-seven. One number is seven less than the other. Find the numbers.

A number is one more than twice another number. Their sum is negative five. Find the numbers.

Solution

−2, −3

One number is six more than five times another. Their sum is six. Find the numbers.

The sum of two numbers is fourteen. 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.

One number is fourteen less than another. If their sum is increased by seven, the result is 85. Find the numbers.

Solution

32, 46

One number is eleven less than another. If their sum is increased by eight, the result is 71. 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 two consecutive integers is −23. Find the integers.

Solution

−11, −12

The sum of two consecutive integers is −37. 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.

Everyday Math

Shopping Patty paid $35 for a purse on sale for $10 off the original price. What was the original price of the purse?

Solution

The original price was $45.

Shopping Travis bought a pair of boots on sale for $25 off the original price. He paid $60 for the boots. What was the original price of the boots?

Shopping Minh spent $6.25 on 5 sticker books to give his nephews. Find the cost of each sticker book.

Solution

Each sticker book cost $1.25.

Shopping Alicia bought a package of 8 peaches for $3.20. Find the cost of each peach.

Shopping Tom paid $1,166.40 for a new refrigerator, including $86.40 tax. What was the price of the refrigerator before tax?

Solution

The price of the refrigerator before tax was $1,080.

Shopping Kenji paid $2,279 for a new living room set, including $129 tax. What was the price of the living room set before tax?

Writing Exercises

Write a few sentences about your thoughts and opinions of word problems. Are these thoughts positive, negative, or neutral? If they are negative, how might you change your way of thinking in order to do better?

Solution

Answers will vary.

When you start to solve a word problem, how do you decide what to let the variable represent?

Self Check

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

A self-assessment grid helps users rate their math skills: approaching word problems positively, using problem-solving strategies, and solving number problems, with options for 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 Money Applications

Learning Objectives

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

  • Solve coin word problems
  • Solve ticket and stamp word problems

Before you get started, take this readiness quiz.

Multiply: 14(0.25).
If you missed this problem, review Example 5 in Decimal Operations.

Solution

3.5

Simplify: 100(0.2+0.05n).
If you missed this problem, review Example 6 in Distributive Property.

Solution

20+5n

Solve: 0.25x+0.10(x+4)=2.5
If you missed this problem, review Example 8 in Solve Equations with Fraction or Decimal Coefficients.

Solution

6

Solve Coin Word Problems

Imagine taking a handful of coins from your pocket or purse and placing them on your desk. How would you determine the value of that pile of coins?

If you can form a step-by-step plan for finding the total value of the coins, it will help you as you begin solving coin word problems.

One way to bring some order to the mess of coins would be to separate the coins into stacks according to their value. Quarters would go with quarters, dimes with dimes, nickels with nickels, and so on. To get the total value of all the coins, you would add the total value of each pile.

An image of a large stack of pennies, a large stack of nickels, a shorter stack of dimes, and a stack of quarters is shown. There are several coins in the background.
To determine the total value of a stack of nickels, multiply the number of nickels times the value of one nickel.(Credit: Darren Hester via ppdigital)

How would you determine the value of each pile? Think about the dime pile—how much is it worth? If you count the number of dimes, you'll know how many you have—the number of dimes.

But this does not tell you the value of all the dimes. Say you counted 17 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 17 dimes, multiply 17 by $0.10 to get $1.70. This is the total value of all 17 dimes.

17·$0.10=$1.70number·value=total value

Finding the Total Value for Coins of the Same Type

For coins of the same type, the total value can be found as follows:

number·value=total value

where number is the number of coins, value is the value of each coin, and total value is the total value of all the coins.

You could continue this process for each type of coin, and then you 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.

Let's look at a specific case. Suppose there are 14 quarters, 17 dimes, 21 nickels, and 39 pennies. We'll make a table to organize the information – the type of coin, the number of each, and the value.

Type Number Value ($) Total Value ($)
Quarters 14 0.25 3.50
Dimes 17 0.10 1.70
Nickels 21 0.05 1.05
Pennies 39 0.01 0.39
6.64

The total value of all the coins is $6.64. Notice how Table 1 helped us organize all the information. Let's see how this method is used to solve a coin word problem.

Adalberto has $2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?

Solution

Solution

Step 1. Read the problem. Make sure you understand all the words and ideas.

  • Determine the types of coins involved.

Think about the strategy we used to find the value of the handful of coins. The first thing you need is to notice what types of coins are involved. Adalberto has dimes and nickels.

  • 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.

We can work this problem all in cents or in dollars. Here we will do it in dollars and put in the dollar sign ($) in the table as a reminder.

The value of a dime is $0.10 and the value of a nickel is $0.05. The total value of all the coins is $2.25.

Type Number Value ($) Total Value ($)
Dimes 0.10
Nickels 0.05
2.25

Step 2. Identify what you are looking for.

  • We are asked to find the number of dimes and nickels Adalberto has.

Step 3. Name what you are looking for.

  • Use variable expressions to represent the number of each type of coin.
  • Multiply the number times the value to get the total value of each type of coin.
    In this problem you cannot count each type of coin—that is what you are looking for—but you have a clue. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.
    Letd=number of dimes.
    d+9=number of nickels
    Fill in the “number” column to help get everything organized.
Type Number Value ($) Total Value ($)
Dimes d 0.10
Nickels d+9 0.05
2.25

Now we have all the information we need from the problem!

You multiply the number times the value to get the total value of each type of coin. While you do not know the actual number, you do have an expression to represent it.

And so now multiply number·value and write the results in the Total Value column.

Type Number Value ($) Total Value ($)
Dimes d 0.10 0.10d
Nickels d+9 0.05 0.05(d+9)
2.25

Step 4. Translate into an equation. Restate the problem in one sentence. Then translate into an equation.
The sentence “Sum of the value of the dimes and value of the nickels is total value of the coins,” is written. Below “value of the dimes” is 0.10d. Below “and” is a plus sign. Below “value of the nickels” is 0.05(d plus 9). Below “is” is an equal sign. Below “total value of the coins” is 2.25.

Step 5. Solve the equation using good algebra techniques.
Write the equation. The image shows a mathematical equation: 0.10d + 0.05(d + 9) = 2.25
Distribute. A mathematical equation showing 0.10d plus 0.05d plus 0.45 equals 2.25.
Combine like terms. A mathematical equation shows '0.15d + 0.45 = 2.25' in black text on a white background.
Subtract 0.45 from each side. A mathematical equation is displayed against a white background, reading '0.15d = 1.80' in black characters.
Divide to find the number of dimes. The image displays the mathematical equation 'd = 12' in a simple and clear format on a white background.
The number of nickels is d + 9 The mathematical expression 'd+9' is displayed in black text on a white background, representing an algebraic sum of a variable 'd' and the number 9.
The mathematical expression '12 + 9' is displayed, with the number '12' in red text and '+ 9' in black text, set against a white background.
The number '21' is displayed in the upper right corner against a plain white background.

Step 6. Check.

12dimes:12(0.10)=1.2021nickels:21(0.05)=1.05_____$2.25✓

Step 7. Answer the question.

Adalberto has twelve dimes and twenty-one nickels.

If this were a homework exercise, our work might look like this:

How many of each type does he have?” Below this is a table with 4 rows and 4 columns. The first row is a header row. The headings are, “Type”, “Number”, “Value ($)”, and “Total Value ($)” Under the “Type” column are the entries dimes and nickels. Under the “Number”,column are d and d plus 9. Under the “Value”,column are the values 0.10 and 0.05. Under the “Total Value”,column are 0.10d and 0.05(d plus 9) followed by 2.25. Below the table is the word “nickels”,in bold. The equation 0.10d plus 0.05(d plus 9) equals 2.25 is shown. Below that are 2 columns. The left column says 0.10d plus 0.05d plus 0.45 equals 2.25, then 0.15d plus 0.45 equals 2.25, then 0.15d equals 1.80, then d equals 12 dimes. There is a red arrow pointing to the right column. The right column says d plus 9, then a red 12 plus 9, then 21 nickels.

Check:

12 dimes12(0.10)=1.2021 nickels21(0.05)=1.05_____$2.25

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

9 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?

Solution

17 nickels, 5 quarters

Solve a coin word problem.

  1. Read the problem. Make sure you understand all the words and ideas, and create a table to organize the information.
  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. 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.

You may find it helpful to put all the numbers into the table to make sure they check.

Type Number Value ($) Total Value

Maria has $2.43 in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each type does she have?

Solution

Solution

Step 1. Read the problem.

  • Determine the types of coins involved.
    We know that Maria has quarters and pennies.
  • 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.
Type Number Value ($) Total Value ($)
Quarters 0.25
Pennies 0.01
2.43

Step 2. Identify what you are looking for.

We are looking for the number of quarters and pennies.

Step 3. Name: Represent the number of quarters and pennies using variables.

We know Maria has twice as many pennies as quarters. The number of pennies is defined in terms of quarters.

Letqrepresent the number of quarters.

Then the number of pennies is2q.

Type Number Value ($) Total Value ($)
Quarters q 0.25
Pennies 2q 0.01
2.43

Multiply the ‘number’ and the ‘value’ to get the ‘total value’ of each type of coin.

Type Number Value ($) Total Value ($)
Quarters q 0.25 0.25q
Pennies 2q 0.01 0.01(2q)
2.43

Step 4. Translate. Write the equation by adding the 'total value’ of all the types of coins.

Step 5. Solve the equation.
Write the equation. A mathematical equation shows 0.25q + 0.01(2q) = 2.43, demonstrating an algebraic problem with decimal coefficients and a variable 'q'.
Multiply. A mathematical equation showing 0.25q plus 0.02q equals 2.43, presented in black text on a white background.
Combine like terms. An algebraic equation showing '0.27q = 2.43' in black text against a white background, representing a mathematical problem to solve for the variable 'q'.
Divide by 0.27. The image displays the equation 'q = 9 quarters' in bold, black text on a white background, representing a quantity of nine quarters.
The number of pennies is 2q. The mathematical expression '2q' is displayed in black text against a plain white background.
The numbers '2.9' are shown on a white background, with the digit '9' in red and the '2.' in black.
The image displays the text '18 pennies' in a simple, clear font against a white background.

Step 6. Check the answer in the problem.

Maria has 9 quarters and 18 pennies. Does this make $2.43?

9 quarters9(0.25)=2.2518 pennies18(0.01)=0.18_____Total$2.43✓

Step 7. Answer the question. Maria has nine quarters and eighteen pennies.

Sumanta has $4.20 in nickels and dimes in her desk drawer. She has twice as many nickels as dimes. How many coins of each type does she have?

Solution

42 nickels, 21 dimes

Alison has three times as many dimes as quarters in her purse. She has $9.35 altogether. How many coins of each type does she have?

Solution

51 dimes, 17 quarters

In the next example, we'll show only the completed table—make sure you understand how to fill it in step by step.

Danny has $2.14 worth of pennies and nickels in his piggy bank. The number of nickels is two more than ten times the number of pennies. How many nickels and how many pennies does Danny have?

Solution

Solution

A step-by-step guide demonstrating how to identify variables, assign values, and set up equations for a word problem involving coins.
Step 1: Read the problem.
Determine the types of coins involved.
Create a table.
Pennies and nickels
Write in the value of each type of coin. Pennies are worth $0.01.
Nickels are worth $0.05.
Step 2: Identify what you 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.

Let p=number of pennies
The number of nickels is two more than then times the number of pennies. 10p+2=number of nickels

Multiply the number and the value to get the total value of each type of coin.

Type Number Value ($) Total Value ($)
pennies p 0.01 0.01p
nickels 10p+2 0.05 0.05(10p+2)
$2.14

Step 4. Translate: Write the equation by adding the total value of all the types of coins.

Step 5. Solve the equation.

A mathematical equation shows 0.01p plus 0.50p plus 0.10 equals 2.14, representing an algebraic problem with a decimal coefficient and a variable 'p'.
A mathematical equation is displayed: 0.51p + 0.10 = 2.14. The equation contains decimal numbers, an unknown variable 'p', an addition operation, and an equals sign.
A mathematical equation shows '0.51p = 2.04' in black text against a white background, representing a linear equation where 'p' is an unknown variable to be solved.
The image displays the mathematical expression 'p = 4 pennies' in black text against a plain white background, indicating a variable 'p' is equal to four units of currency, specifically pennies.
How many nickels? The image displays the mathematical expression '10p + 2' in a clear, standard font, set against a plain white background.
A mathematical expression '10(4) + 2' is displayed, with the number 4 highlighted in red, indicating a calculation involving multiplication and addition.
The text '42 nickels' is displayed in a simple, clear font on a white background, indicating a quantity of money.

Step 6. Check. Is the total value of 4 pennies and 42 nickels equal to $2.14?

4(0.01)+42(0.05)=?2.142.14=2.14✓

Step 7. Answer the question. Danny has 4 pennies and 42 nickels.

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

41 nickels, 18 quarters

Elaine has $7.00 in dimes and nickels in her coin jar. The number of dimes that Elaine has is seven less than three times the number of nickels. How many of each coin does Elaine have?

Solution

22 nickels, 59 dimes

Solve Ticket and Stamp Word Problems

The strategies we used for coin problems can be easily applied to some other kinds of problems too. Problems involving tickets or stamps are very similar to coin problems, for example. Like coins, tickets and stamps have different values; so we can organize the information in tables much like we did for coin problems.

At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $6 each and adult tickets sold for $9 each. The number of adult tickets sold was 5 less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

Solution

Solution

Step 1: Read the problem.

  • Determine the types of tickets involved.
    There are student tickets and adult tickets.
  • Create a table to organize the information.
Type Number Value ($) Total Value ($)
Student 6
Adult 9
1,506

Step 2. Identify what you are looking for.

We are looking for the number of student and adult tickets.

Step 3. Name. Represent the number of each type of ticket using variables.

We know the number of adult tickets sold was5less than three times the number of student tickets sold.

Letsbe the number of student tickets.

Then3s−5is the number of adult tickets.

Multiply the number times the value to get the total value of each type of ticket.

Type Number Value ($) Total Value ($)
Student s 6 6s
Adult 3s−5 9 9(3s−5)
1,506

Step 4. Translate: Write the equation by adding the total values of each type of ticket.

6s+9(3s−5)=1506

Step 5. Solve the equation.

6s+27s−45=150633s−45=150633s=1551s=47students

Substitute to find the number of adults.

The top line says 3s minus 5 equals number of adults. The bottom line shows 3 times a red 47 minus 5 equals 136 adults.

Step 6. Check. There were 47 student tickets at $6 each and 136 adult tickets at $9 each. Is the total value $1506? We find the total value of each type of ticket by multiplying the number of tickets times its value; we then add to get the total value of all the tickets sold.

47·6=282136·9=1224_____1506✓

Step 7. Answer the question. They sold 47 student tickets and 136 adult tickets.

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?

Solution

112 adult tickets, 199 senior/child tickets

Now we'll do one where we fill in the table all at once.

Monica paid $10.44 for stamps she needed to mail the invitations to her sister's baby shower. The number of 49-cent stamps was four more than twice the number of 8-cent stamps. How many 49-cent stamps and how many 8-cent stamps did Monica buy?

Solution

Solution

The type of stamps are 49-cent stamps and 8-cent stamps. Their names also give the value.

“The number of 49 cent stamps was four more than twice the number of 8 cent stamps.”

Letx=number of 8-cent stamps2x+4=number of 49-cent stamps

Type Number Value ($) Total Value ($)
49-cent stamps 2x+4 0.49 0.49(2x+4)
8-cent stamps x 0.08 0.08x
10.44
Solution steps for an algebraic word problem, including equation formulation, solving, and verification.
Write the equation from the total values. 0.49(2x+4)+0.08x=10.44
Solve the equation. 0.98x+1.96+0.08x=10.44
1.06x+1.96=10.44
1.06x=8.48
x=8
Monica bought 8 eight-cent stamps.
Find the number of 49-cent stamps she bought by evaluating. 2x+4forx=8.
2x+4
2⋅8+4
16+4
20
Check.
8(0.08)+20(0.49)=?10.44
0.64+9.80=?10.44
10.44=10.44✓

Monica bought eight 8-cent stamps and twenty 49-cent stamps.

Eric paid $16.64 for stamps so he could mail thank you notes for his wedding gifts. The number of 49-cent stamps was eight more than twice the number of 8-cent stamps. How many 49-cent stamps and how many 8-cent stamps did Eric buy?

Solution

32 at 49 cents, 12 at 8 cents

Kailee paid $14.84 for stamps. The number of 49-cent stamps was four less than three times the number of 21-cent stamps. How many 49-cent stamps and how many 21-cent stamps did Kailee buy?

Solution

26 at 49 cents, 10 at 21 cents

Key Concepts

  • Finding the Total Value for Coins of the Same Type
    • For coins of the same type, the total value can be found as follows:
      number·value=total value
      where number is the number of coins, value is the value of each coin, and total value is the total value of all the coins.
  • Solve a Coin Word Problem
    1. Read the problem. Make sure you understand all the words and ideas, and create a table to organize the information.
    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. 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.
  • Type Number Value ($) Total Value ($)

Practice Makes Perfect

Solve Coin Word Problems

In the following exercises, solve the coin word problems.

Jaime has $2.60 in dimes and nickels. The number of dimes is 14 more than the number of nickels. How many of each coin does he have?

Solution

8 nickels, 22 dimes

Lee has $1.75 in dimes and nickels. The number of nickels is 11 more than the number of dimes. How many of each coin does he have?

Ngo has a collection of dimes and quarters with a total value of $3.50. The number of dimes is 7 more than the number of quarters. How many of each coin does he have?

Solution

15 dimes, 8 quarters

Connor has a collection of dimes and quarters with a total value of $6.30. The number of dimes is 14 more than the number of quarters. How many of each coin does he have?

Carolyn has $2.55 in her purse in nickels and dimes. The number of nickels is 9 less than three times the number of dimes. Find the number of each type of coin.

Solution

12 dimes and 27 nickels

Julio has $2.75 in his pocket in nickels and dimes. The number of dimes is 10 less than twice the number of nickels. Find the number of each type of coin.

Chi has $11.30 in dimes and quarters. The number of dimes is 3 more than three times the number of quarters. How many dimes and nickels does Chi have?

Solution

63 dimes, 20 quarters

Tyler has $9.70 in dimes and quarters. The number of quarters is 8 more than four times the number of dimes. How many of each coin does he have?

A cash box of $1 and $5 bills is worth $45. The number of $1 bills is 3 more than the number of $5 bills. How many of each bill does it contain?

Solution

10 of the $1 bills, 7 of the $5 bills

Joe's wallet contains $1 and $5 bills worth $47. The number of $1 bills is 5 more than the number of $5 bills. How many of each bill does he 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 are in the drawer?

Solution

10 of the $10 bills, 5 of the $5 bills

John 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?

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, 7 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 the ticket and stamp word problems.

The play took in $550 one night. The number of $8 adult tickets was 10 less than twice the number of $5 child tickets. How many of each ticket were sold?

Solution

30 child tickets, 50 adult tickets

If the number of $8 child tickets is seventeen less than three times the number of $12 adult tickets and the theater took in $584, how many of each ticket were sold?

The movie theater took in $1,220 one Monday night. The number of $7 child tickets was ten more than twice the number of $9 adult tickets. How many of each were sold?

Solution

110 child tickets, 50 adult tickets

The ball game took in $1,340 one Saturday. The number of $12 adult tickets was 15 more than twice the number of $5 child tickets. How many of each were sold?

Julie went to the post office and bought both $0.49 stamps and $0.34 postcards for her office's bills She spent $62.60. 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

Before he left for college out of state, Jason went to the post office and bought both $0.49 stamps and $0.34 postcards and spent $12.52. The number of stamps was 4 more than twice the number of postcards. How many of each did he buy?

Maria spent $16.80 at the post office. She bought three times as many $0.49 stamps as $0.21 stamps. How many of each did she buy?

Solution

30 at 49 cents, 10 at 21 cents

Hector spent $43.40 at the post office. He bought four times as many $0.49 stamps as $0.21 stamps. How many of each did he buy?

Hilda has $210 worth of $10 and $12 stock shares. The numbers of $10 shares is 5 more than twice the number of $12 shares. How many of each does she have?

Solution

15 at $10 shares, 5 at $12 shares

Mario invested $475 in $45 and $25 stock shares. The number of $25 shares was 5 less than three times the number of $45 shares. How many of each type of share did he buy?

Everyday Math

Parent Volunteer As the treasurer of her daughter's Girl Scout troop, Laney collected money for some girls and adults to go to a 3-day camp. Each girl paid $75 and each adult paid $30. The total amount of money collected for camp was $765. If the number of girls is three times the number of adults, how many girls and how many adults paid for camp?

Solution

9 girls, 3 adults

Parent Volunteer Laurie was completing the treasurer's report for her son's Boy Scout troop at the end of the school year. She didn't remember how many boys had paid the $24 full-year registration fee and how many had paid a $16 partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If $400 was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?

Writing Exercises

Suppose you have 6 quarters, 9 dimes, and 4 pennies. Explain how you find the total value of all the coins.

Solution

Answers will vary.

Do you find it helpful to use a table when solving coin problems? Why or why not?

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?

Solution

Answers will vary.

What similarities and differences did you see between solving the coin problems and the ticket and stamp problems?

Self Check

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

A self-assessment chart asking students to rate their ability to solve coin, ticket, and stamp word problems across three categories: 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?

Use Properties of Angles, Triangles, and the Pythagorean Theorem

Learning Objectives

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

  • Use the properties of angles
  • Use the properties of triangles
  • Use the Pythagorean Theorem

Before you get started, take this readiness quiz.

Solve: x+3+6=11.
If you missed this problem, review Example 6 in Solve Equations Using the Subtraction and Addition Properties of Equality.

Solution

x=2

Solve: a45=43.
If you missed this problem, review Example 3 in Solve Proportions and their Applications.

Solution

60

Simplify: 36+64.
If you missed this problem, review Example 4 in Simplify and Use Square Roots.

Solution

10

So far in this chapter, we have focused on solving word problems, which are similar to many real-world applications of algebra. In the next few sections, we will apply our problem-solving strategies to some common geometry problems.

Use the Properties of Angles

Are you familiar with the phrase ‘do a 180’? It means to turn so that you face the opposite direction. It comes from the fact that the measure of an angle that makes a straight line is 180 degrees. See Figure 1.

The image is a straight line with an arrow on each end. There is a dot in the center. There is an arrow pointing from one side of the dot to the other, and the angle is marked as 180 degrees.

An angle is formed by two rays that share a common endpoint. Each ray is called a side of the angle and the common endpoint is called the vertex. An angle is named by its vertex. In Figure 2, ∠A is the angle with vertex at point A. The measure of ∠A is written m∠A.

The image is an angle made up of two rays. The angle is labeled with letter A.
∠A is the angle with vertex at pointA.

We measure angles in degrees, and use the symbol ° to represent degrees. We use the abbreviation m for the measure of an angle. So if ∠A is 27°, we would write m∠A=27.

If the sum of the measures of two angles is 180°, then they are called supplementary angles. In Figure 3, each pair of angles is supplementary because their measures add to 180°. Each angle is the supplement of the other.

Part a shows a 120 degree angle next to a 60 degree angle. Together, the angles form a straight line. Below the image, it reads 120 degrees plus 60 degrees equals 180 degrees. Part b shows a 45 degree angle attached to a 135 degree angle. Together, the angles form a straight line. Below the image, it reads 45 degrees plus 135 degrees equals 180 degrees.
The sum of the measures of supplementary angles is 180°.

If the sum of the measures of two angles is 90°, then the angles are complementary angles. In Figure 4, each pair of angles is complementary, because their measures add to 90°. Each angle is the complement of the other.

Part a shows a 50 degree angle next to a 40 degree angle. Together, the angles form a right angle. Below the image, it reads 50 degrees plus 40 degrees equals 90 degrees. Part b shows a 60 degree angle attached to a 30 degree angle. Together, the angles form a right angle. Below the image, it reads 60 degrees plus 30 degrees equals 90 degrees.
The sum of the measures of complementary angles is 90°.

Supplementary and Complementary Angles

If the sum of the measures of two angles is 180°, then the angles are supplementary.

If ∠A and ∠B are supplementary, then m∠A+m∠B=180°.

If the sum of the measures of two angles is 90°, then the angles are complementary.

If ∠A and ∠B are complementary, then m∠A+m∠B=90°.

In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve geometric shapes, it will be helpful to draw a figure and then label it with the information from the problem. We will include this step in the Problem Solving Strategy for Geometry Applications.

Use a Problem Solving Strategy for Geometry Applications.

  1. Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information.
  2. Identify what you are looking for.
  3. Name what you are looking for and choose 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.

The next example will show how you can use the Problem Solving Strategy for Geometry Applications to answer questions about supplementary and complementary angles.

An angle measures 40°. Find ⓐ its supplement, and ⓑ its complement.

Solution

Solution

ⓐ
Step 1. Read the problem. Draw the figure and label it with the given information. A geometry diagram shows a straight line with two angles marked. One angle, denoted by 's°', is an obtuse angle. The other angle, denoted by '40°', is an acute angle adjacent to the obtuse angle.
Step 2. Identify what you are looking for. The text reads 'the supplement of a 40° angle.' against a white background.
Step 3. Name. Choose a variable to represent it. let s = the measure of the supplement
Step 4. Translate.
Write the appropriate formula for the situation and substitute in the given information.

A mathematical equation, m∠A + m∠B = 180, illustrating that the sum of the measures of angles A and B is 180 degrees.
A mathematical equation is displayed on a white background: s + 40 = 180. This simple linear equation involves one variable 's', a constant, and an equals sign.
Step 5. Solve the equation. The mathematical expression 's = 140' is displayed in black text against a plain white background.
Step 6. Check:
The equation '140 + 40 = 180' is displayed, with a question mark positioned directly above the equals sign, implying a query as to the truthfulness of the mathematical statement.
The image shows the equation '180 = 180' followed by a checkmark, indicating that the equality is correct or verified.
Step 7. Answer the question. The image shows the text 'The supplement of the 40° angle is 140°.' This statement defines a supplementary angle relationship where the sum of two angles is 180 degrees.
ⓑ
Step 1. Read the problem. Draw the figure and label it with the given information. A diagram with three rays originating from a common point, showing an angle labeled c degrees between the leftmost horizontal ray and a diagonal ray, and an angle of 40 degrees between the vertical ray and the diagonal ray.
Step 2. Identify what you are looking for. The text reads, 'the complement of a 40° angle,' indicating a mathematical concept. The font is a sans-serif, dark teal color against a white background.
Step 3. Name. Choose a variable to represent it. let c = the measure of the complement
Step 4. Translate.
Write the appropriate formula for the situation and substitute in the given information.

A mathematical equation is displayed against a white background, reading 'm∠A + m∠B = 90'. This indicates that the measure of angle A plus the measure of angle B equals 90 degrees.
Step 5. Solve the equation. A basic algebra equation is displayed, reading 'c + 40 = 90', suggesting a problem where one needs to solve for the variable 'c'.
The image shows a mathematical expression or variable assignment, 'c = 50', displayed in a clear, dark font against a plain white background.
Step 6. Check:
A simple arithmetic problem, 50 + 40 = 90, is displayed with a question mark above the equals sign, asking if the statement is true. The equation is correct.
The image displays the equation '90 = 90' followed by a checkmark, indicating that the equality is correct or verified.
Step 7. Answer the question. The image shows text that reads: 'The complement of the 40° angle is 50°.'

An angle measures 25°. Find its: ⓐ supplement ⓑ complement.

Solution
  1. ⓐ 155°
  2. ⓑ 65°

An angle measures 77°. Find its: ⓐ supplement ⓑ complement.

Solution
  1. ⓐ 103°
  2. ⓑ 13°

Did you notice that the words complementary and supplementary are in alphabetical order just like 90 and 180 are in numerical order?

Two angles are supplementary. The larger angle is 30° more than the smaller angle. Find the measure of both angles.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A geometry diagram shows a straight line with two adjacent angles labeled 'a' and 'a + 30', where the line appears to be broken. The angles together form a straight line, implying they are supplementary.
Step 2. Identify what you are looking for. The text in the image says 'the measures of both angles'.
Step 3. Name. Choose a variable to represent it.
The larger angle is 30° more than the smaller angle.
'let α = measure of smaller angle' in a dark teal font against a white background.
A mathematical equation states 'a + 30 = measure of larger angle'.
Step 4. Translate.
Write the appropriate formula and substitute.

The image shows the mathematical equation m∠A + m∠B = 180, indicating that the sum of the measures of angle A and angle B is 180 degrees, meaning they are supplementary angles.
Step 5. Solve the equation. A mathematical equation is displayed with black text on a white background: (a + 30) + a = 180. This equation can be used to solve for the variable 'a'.
A mathematical equation is displayed: 2a + 30 = 180. This is a linear equation with one variable, 'a', that needs to be solved.
A mathematical equation is displayed on a white background, showing '2a = 150' in black font.
The text on a white background states 'a = 75 measure of smaller angle'.
The image shows the expression 'a + 30 measure of larger angle' on a white background. The text describes a mathematical quantity for an angle.
A mathematical expression displaying '75 + 30' on a white background.
The number 105 is displayed in black text against a plain white background, centered in the frame.
Step 6. Check:
The equation m∠A + m∠B = 180 is shown, indicating that the sum of the measures of angle A and angle B is 180 degrees. This implies that angles A and B are supplementary.
A mathematical expression asks whether 75 + 105 equals 180, with a question mark placed above the equals sign.
The image displays the equation '180 = 180' followed by a checkmark, indicating that the equality is correct or verified.
Step 7. Answer the question. The measures of the angles are 75° and 105°.

Two angles are supplementary. The larger angle is 100° more than the smaller angle. Find the measures of both angles.

Solution

40°, 140°

Two angles are complementary. The larger angle is 40° more than the smaller angle. Find the measures of both angles.

Solution

25°, 65°

Use the Properties of Triangles

What do you already know about triangles? Triangle have three sides and three angles. Triangles are named by their vertices. The triangle in Figure 5 is called ΔABC, read ‘triangle ABC’. We label each side with a lower case letter to match the upper case letter of the opposite vertex.

The vertices of the triangle on the left are labeled A, B, and C. The sides are labeled a, b, and c.
ΔABC has vertices A,B,andC and sides a,b,andc.

The three angles of a triangle are related in a special way. The sum of their measures is 180°.

m∠A+m∠B+m∠C=180°

Sum of the Measures of the Angles of a Triangle

For any ΔABC, the sum of the measures of the angles is 180°.

m∠A+m∠B+m∠C=180°

The measures of two angles of a triangle are 55° and 82°. Find the measure of the third angle.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A triangle ABC is shown, with angle A measuring 82 degrees, angle B measuring 55 degrees, and angle C denoted by x.
Step 2. Identify what you are looking for. the measure of the third angle in a triangle
Step 3. Name. Choose a variable to represent it. The image shows the text 'let x = the measure of the angle' in a sans-serif font, suggesting a mathematical definition or problem statement.
Step 4. Translate.
Write the appropriate formula and substitute.

The image shows the angle sum property of a triangle, where the sum of the measures of interior angles A, B, and C is equal to 180 degrees, represented as m∠A + m∠B + m∠C = 180.
Step 5. Solve the equation. A mathematical equation is displayed on a white background: 55 + 82 + x = 180. This equation is commonly used to find the third angle of a triangle when two angles are known, as the sum of angles in a triangle is 180 degrees.
A mathematical equation is displayed, reading '137 + x = 180'.
The mathematical equation 'x = 43' is displayed in the top right corner against a plain white background.
Step 6. Check:
A mathematical equation showing 55 + 82 + 43 with a question mark above the equals sign, followed by 180, implying a verification of the sum equaling 180.
The mathematical equality 180 = 180 is displayed, followed by a checkmark indicating its correctness.
Step 7. Answer the question. The measure of the third angle is 43 degrees.

The measures of two angles of a triangle are 31° and 128°. Find the measure of the third angle.

Solution

21°

A triangle has angles of 49° and 75°. Find the measure of the third angle.

Solution

56°

Right Triangles

Some triangles have special names. We will look first at the right triangle. A right triangle has one 90° angle, which is often marked with the symbol shown in Figure 6.

A right triangle is shown. The right angle is marked with a box and labeled 90 degrees.

If we know that a triangle is a right triangle, we know that one angle measures 90° so we only need the measure of one of the other angles in order to determine the measure of the third angle.

One angle of a right triangle measures 28°. What is the measure of the third angle?

Solution
Solution
Step 1. Read the problem. Draw the figure and label it with the given information. Geometric diagram of a right-angled triangle labeled ABC, showing angles 90 degrees at A, 28 degrees at B, and x degrees at C, illustrating a problem to find angle x.
Step 2. Identify what you are looking for. The text reads, 'the measure of an angle'.
Step 3. Name. Choose a variable to represent it. The text reads 'let x = the measure of the angle' in a dark teal font on a white background, representing a mathematical definition of a variable x.
Step 4. Translate.
Write the appropriate formula and substitute.

The equation m angle A plus m angle B plus m angle C equals 180.
Step 5. Solve the equation. A mathematical equation is displayed, showing 'x + 90 + 28 = 180' in black text against a white background.
A mathematical equation is displayed with dark gray text on a white background, reading 'x + 118 = 180'.
The image displays a simple mathematical equation, 'x = 62', written in black text against a plain white background, asserting that the variable x has a value of 62.
Step 6. Check:
A mathematical equation checks if 180 is equal to the sum of 90, 28, and 62, posing '180 =? 90 + 28 + 62' on a white background. The sum is 180, so the equality is true.
The image displays the equation '180 = 180' with a checkmark, signifying a correct or verified mathematical statement.
Step 7. Answer the question. The text states, 'The measure of the third angle is 62×0.'

One angle of a right triangle measures 56°. What is the measure of the other angle?

Solution

34°

One angle of a right triangle measures 45°. What is the measure of the other angle?

Solution

45°

In the examples so far, we could draw a figure and label it directly after reading the problem. In the next example, we will have to define one angle in terms of another. So 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 20° more than the measure of the smallest angle. Find the measures of all three angles.

Solution
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.


Now draw the figure and label it with the given information.
The text
The image displays a mathematical equation written horizontally, 'a + 20 = 2nd angle', indicating a relationship where 'a plus 20' equals the second angle.
The image shows text that states '90 = 3rd angle (the right angle)', indicating that the third angle in a context, likely a geometric figure, is a right angle measuring 90 degrees.
A right-angled triangle ABC with the right angle at C. Angle A is denoted as 'a' and angle B is denoted as 'a + 20'. The sum of angles in a triangle is 180 degrees.
Step 4. Translate.
Write the appropriate formula and substitute into the formula.
Equation: m∠A + m∠B + m∠C = 180, illustrating the angle sum property of a triangle.
A mathematical equation is displayed, showing a + (a + 20) + 90 = 180, an algebraic expression typically used to solve for an unknown variable 'a' in geometry or algebra problems.
Step 5. Solve the equation. A mathematical equation is displayed, reading '2a + 110 = 180' in black text against a white background.
A mathematical equation is displayed on a white background, reading '2a = 70'.
The image displays the equation 'a = 35 first angle' in black and teal text on a white background, indicating a variable 'a' is equal to 35, referred to as the first angle.
The image shows the text 'a + 20 second angle' in a dark grey color on a white background.
A simple arithmetic expression is displayed on a white background, showing the addition problem '35 + 20.' The number 35 is rendered in red text, while the plus sign and the number 20 are in black.
The number '55' is displayed in a simple, clear black font on a plain white background.
The text '90 third angle' is displayed on a white background, indicating a numerical value and a descriptive term related to an angle or perspective.
Step 6. Check:
A mathematical equation reads '35 + 55 + 90 ?= 180', questioning if the sum of 35, 55, and 90 equals 180.
The image displays the equation '180 = 180' followed by a checkmark, indicating that the equality is correct or verified.
Step 7. Answer the question. The three angles measure 35°, 55°, 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

20°, 70°, 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

30°, 60°, 90°

Similar Triangles

When we use a map to plan a trip, a sketch to build a bookcase, or a pattern to sew a dress, we are working with similar figures. In geometry, if two figures have exactly the same shape but different sizes, we say they are similar figures. One is a scale model of the other. The corresponding sides of the two figures have the same ratio, and all their corresponding angles have the same measures.

The two triangles in Figure 7 are similar. Each side of ΔABC is four times the length of the corresponding side of ΔXYZ and their corresponding angles have equal measures.

Two triangles are shown. They appear to be the same shape, but the triangle on the right is smaller. The vertices of the triangle on the left are labeled A, B, and C. The side across from A is labeled 16, the side across from B is labeled 20, and the side across from C is labeled 12. The vertices of the triangle on the right are labeled X, Y, and Z. The side across from X is labeled 4, the side across from Y is labeled 5, and the side across from Z is labeled 3. Beside the triangles, it says that the measure of angle A equals the measure of angle X, the measure of angle B equals the measure of angle Y, and the measure of angle C equals the measure of angle Z. Below this is the proportion 16 over 4 equals 20 over 5 equals 12 over 3.
ΔABC and ΔXYZ are similar triangles. Their corresponding sides have the same ratio and the corresponding angles have the same measure.

Properties of Similar Triangles

If two triangles are similar, then their corresponding angle measures are equal and their corresponding side lengths are in the same ratio.

Two similar triangles are shown: the left larger triangle has its legs labeled lowercase c, a, b, with points labeled as uppercase A, B, and C. The right smaller triangle mirrors the left by having its legs labeled as lowercase z, x, y, and its points labeled X, Y, and Z.

The length of a side of a triangle may be referred to by its endpoints, two vertices of the triangle. For example, in ΔABC:

the lengthacan also be writtenBCthe lengthbcan also be writtenACthe lengthccan also be writtenAB

We will often use this notation when we solve similar triangles because it will help us match up the corresponding side lengths.

ΔABC and ΔXYZ are similar triangles. The lengths of two sides of each triangle are shown. Find the lengths of the third side of each triangle.

Two triangles are shown. They appear to be the same shape, but the triangle on the right is smaller. The vertices of the triangle on the left are labeled A, B, and C. The side across from A is labeled a, the side across from B is labeled 3.2, and the side across from C is labeled 4. The vertices of the triangle on the right are labeled X, Y, and Z. The side across from X is labeled 4.5, the side across from Y is labeled y, and the side across from Z is labeled 3.
Solution
Solution
Step-by-step guide demonstrating the process of finding unknown side lengths in similar triangles, from problem interpretation to solution.
Step 1. Read the problem. Draw the figure and label it with the given information. The figure is provided.
Step 2. Identify what you are looking for. The length of the sides of similar triangles
Step 3. Name. Choose a variable to represent it. Let
a = length of the third side of ΔABC
y = length of the third side ΔXYZ
Step 4. Translate.
The triangles are similar, so the corresponding sides are in the same ratio. So
ABXY=BCYZ=ACXZ

Since the side AB=4 corresponds to the side XY=3, we will use the ratio ABXY=43 to find the other sides.

Be careful to match up corresponding sides correctly.
This image displays the steps to calculate unknown side lengths 'a' and 'y' in similar triangles. It shows the ratios of corresponding sides from large and small triangles, setting up proportions like 4/3 = a/4.5 and 4/3 = 3.2/y.
Step 5. Solve the equation. Two sets of equations are solved, showing calculations for 'a' and 'y'. The first set solves 3a = 4(4.5) to find a=6. The second set solves 4y = 3(3.2) to find y=2.4.
Step 6. Check:
Two examples demonstrate how to check if two fractions are proportional by using the cross-multiplication method, showing the resulting equalities with a checkmark.
Step 7. Answer the question. The third side of ΔABC is 6 and the third side of ΔXYZ is 2.4.

ΔABC is similar to ΔXYZ. Find a.

Two triangles are shown. They appear to be the same shape, but the triangle on the right is larger The vertices of the triangle on the left are labeled A, B, and C. The side across from A is labeled a, the side across from B is labeled 15, and the side across from C is labeled 17. The vertices of the triangle on the right are labeled X, Y, and Z. The side across from X is labeled 12, the side across from Y is labeled y, and the side across from Z is labeled 25.5.
Solution

8

ΔABC is similar to ΔXYZ. Find y.

Two triangles are shown. They appear to be the same shape, but the triangle on the right is larger The vertices of the triangle on the left are labeled A, B, and C. The side across from A is labeled a, the side across from B is labeled 15, and the side across from C is labeled 17. The vertices of the triangle on the right are labeled X, Y, and Z. The side across from X is labeled 12, the side across from Y is labeled y, and the side across from Z is labeled 25.5.
Solution

22.5

Use the Pythagorean Theorem

The Pythagorean Theorem is a special property of right triangles that has been used since ancient times. It is named after the Greek philosopher and mathematician Pythagoras who lived around 500 BCE.

Remember that a right triangle has a 90° angle, which we usually mark with a small square in the corner. The side of the triangle opposite the 90° angle is called the hypotenuse, and the other two sides are called the legs. See Figure 8.

Three right triangles are shown. Each has a box representing the right angle. The first one has the right angle in the lower left corner, the next in the upper left corner, and the last one at the top. The two sides touching the right angle are labeled “leg” in each triangle. The sides across from the right angles are labeled “hypotenuse.”
In a right triangle, the side opposite the 90° angle is called the hypotenuse and each of the other sides is called a leg.

The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the two legs equals the square of the hypotenuse.

The Pythagorean Theorem

In any right triangle ΔABC,

a2+b2=c2

where c is the length of the hypotenuse a and b are the lengths of the legs.

A right triangle is shown. The right angle is marked with a box. Across from the box is side c. The sides touching the right angle are marked a and b.

To solve problems that use the Pythagorean Theorem, we will need to find square roots. In Simplify and Use Square Roots we introduced the notation m and defined it in this way:

Ifm=n2,thenm=nforn≥0

For example, we found that 25 is 5 because 52=25.

We will use this definition of square roots to solve for the length of a side in a right triangle.

Use the Pythagorean Theorem to find the length of the hypotenuse.

Right triangle with legs labeled as 3 and 4.
Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for. the length of the hypotenuse of the triangle
Step 3. Name. Choose a variable to represent it. Let c=the length of the hypotenuse
A right-angled triangle with legs of length 3 and 4, and the hypotenuse labeled 'c'. This classic 3-4-5 triangle demonstrates the Pythagorean theorem.
Step 4. Translate.
Write the appropriate formula.
Substitute.

Illustration of the Pythagorean theorem (a² + b² = c²) and a specific example (3² + 4² = c²) used to find the hypotenuse.
Step 5. Solve the equation. Solving for 'c': The image displays the calculation starting from 9 + 16 = c^2, simplifying to 25 = c^2, and then taking the square root to determine that c = 5.
Step 6. Check:
This image demonstrates the Pythagorean triple (3,4,5) with a step-by-step verification that 3^2 + 4^2 = 5^2, resulting in 9 + 16 = 25 and finally 25 = 25.
Step 7. Answer the question. The length of the hypotenuse is 5.

Use the Pythagorean Theorem to find the length of the hypotenuse.

A right triangle is shown. The right angle is marked with a box. Across from the box is side c. The sides touching the right angle are marked 6 and 8.
Solution

10

Use the Pythagorean Theorem to find the length of the hypotenuse.

A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as c. One of the sides touching the right angle is labeled as 15, the other is labeled “8”.
Solution

17

Use the Pythagorean Theorem to find the length of the longer leg.

Right triangle is shown with one leg labeled as 5 and hypotenuse labeled as 13.
Solution

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 b=the leg of the triangle
Label side b
A right-angled triangle with one vertical leg of length 5, a horizontal leg labeled 'b', and the hypotenuse of length 13. A square symbol indicates the right angle between the legs.
Step 4. Translate.
Write the appropriate formula. Substitute.
Two mathematical equations are displayed, illustrating the Pythagorean theorem. The first equation is a^2 + b^2 = c^2. The second equation shows specific values substituted: 5^2 + b^2 = 13^2.
Step 5. Solve the equation. Isolate the variable term. Use the definition of the square root.
Simplify.
Step-by-step solution for 'b' in the equation 25 + b^2 = 169. The calculation shows b^2 = 144 and concludes with b = 12, demonstrating fundamental algebraic problem-solving.
Step 6. Check:
A step-by-step mathematical verification showing that 5 squared plus 12 squared equals 13 squared, confirming a Pythagorean triple.
Step 7. Answer the question. The length of the leg is 12.

Use the Pythagorean Theorem to find the length of the leg.

A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 17. One of the sides touching the right angle is labeled as 15, the other is labeled “b”.
Solution

8

Use the Pythagorean Theorem to find the length of the leg.

A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 15. One of the sides touching the right angle is labeled as 9, the other is labeled “b”.
Solution

12

Kelvin is building a gazebo and wants to brace each corner by placing a 10-inch wooden bracket diagonally as shown. How far below the corner should he fasten the bracket if he wants the distances from the corner to each end of the bracket to be equal? Approximate to the nearest tenth of an inch.

A picture of a gazebo is shown. Beneath the roof is a rectangular shape. There are two braces from the top to each side. The brace on the left is labeled as 10 inches. From where the brace hits the side to the roof is labeled as x.
Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for. the distance from the corner that the bracket should be attached
Step 3. Name. Choose a variable to represent it. Let x = the distance from the corner
A right-angled isosceles triangle is depicted with two equal legs labeled 'x' and the hypotenuse measuring '10 inches'. This geometry problem asks for the value of x.
Step 4. Translate.
Write the appropriate formula.
Substitute.

Two mathematical equations are displayed: a squared plus b squared equals c squared, and x squared plus x squared equals 10 squared.
Step 5. Solve the equation.
Isolate the variable.
Use the definition of the square root.
Simplify. Approximate to the nearest tenth.
Mathematical calculation solving 2x^2 = 100, resulting in x = sqrt(50). The final step shows 'b' approximated as 7.1, indicating the numerical value of the square root of 50.
Step 6. Check:
The image displays two mathematical expressions related to the Pythagorean theorem. The first is the general formula a^2 + b^2 = c^2. Below it, a specific calculation is posed: (7.1)^2 + (7.1)^2 is approximately equal to 10^2, with a question mark indicating an inquiry.
Yes.
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-ft ladder 5 feet from the wall of his house. How far up the wall does the ladder reach?

A picture of a house is shown. There is a ladder leaning against the side of the house. The ladder is labeled 13 feet. The horizontal distance from the ladder's base to the house is labeled 5 feet.
Solution

12 feet

Randy wants to attach a 17-ft string of lights to the top of the 15-ft mast of his sailboat. How far from the base of the mast should he attach the end of the light string?

A picture of a boat is shown. The height of the center pole is labeled 15 feet. The string of lights is at a diagonal from the top of the pole and is labeled 17 feet.
Solution

8 feet

ACCESS ADDITIONAL ONLINE RESOURCES

  • Animation: The Sum of the Interior Angles of a Triangle
  • Similar Polygons
  • Example: Determine the Length of the Hypotenuse of a Right Triangle

Key Concepts

  • Supplementary and Complementary Angles
    • If the sum of the measures of two angles is 180°, then the angles are supplementary.
    • If ∠A and ∠B are supplementary, then m∠A+m∠B=180.
    • If the sum of the measures of two angles is 90°, then the angles are complementary.
    • If ∠A and ∠B are complementary, then m∠A+m∠B=90.
  • Solve Geometry Applications
    1. Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information.
    2. Identify what you are looking for.
    3. Name what you are looking for and choose 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.
  • Sum of the Measures of the Angles of a Triangle

    The image shows a triangle with its points labeled A, B, and C.

    • For any ΔABC, the sum of the measures is 180°
    • m∠A+m∠B+m∠C=180
  • Right Triangle

    A black and white diagram of a right-angled triangle, showing the 90-degree angle at one vertex marked with a square symbol and the label '90°' inside it.

    • A right triangle is a triangle that has one 90° angle, which is often marked with a ⦜symbol.
  • Properties of Similar Triangles
    • If two triangles are similar, then their corresponding angle measures are equal and their corresponding side lengths have the same ratio.

Practice Makes Perfect

Use the Properties of Angles

In the following exercises, find ⓐ the supplement and ⓑ the complement of the given angle.

53°

Solution
  1. ⓐ 127°
  2. ⓑ 37°

16°

29°

Solution
  1. ⓐ 151°
  2. ⓑ 61°

72°

In the following exercises, use the properties of angles to solve.

Find the supplement of a 135° angle.

Solution

45°

Find the complement of a 38° angle.

Find the complement of a 27.5° angle.

Solution

62.5°

Find the supplement of a 109.5° angle.

Two angles are supplementary. The larger angle is 56° more than the smaller angle. Find the measures of both angles.

Solution

62°, 118°

Two angles are supplementary. The smaller angle is 36° less than the larger angle. Find the measures of both angles.

Two angles are complementary. The smaller angle is 34° less than the larger angle. Find the measures of both angles.

Solution

62°, 28°

Two angles are complementary. The larger angle is 52° more than the smaller angle. Find the measures of both angles.

Use the Properties of Triangles

In the following exercises, solve using properties of triangles.

The measures of two angles of a triangle are 26° and 98°. Find the measure of the third angle.

Solution

56°

The measures of two angles of a triangle are 61° and 84°. Find the measure of the third angle.

The measures of two angles of a triangle are 105° and 31°. Find the measure of the third angle.

Solution

44°

The measures of two angles of a triangle are 47° and 72°. Find the measure of the third angle.

One angle of a right triangle measures 33°. What is the measure of the other angle?

Solution

57°

One angle of a right triangle measures 51°. What is the measure of the other angle?

One angle of a right triangle measures 22.5°. What is the measure of the other angle?

Solution

67.5°

One angle of a right triangle measures 36.5°. What is the measure of the other angle?

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 other small angle. Find the measures of all three angles.

The angles in a triangle are such that the measure of one angle is twice the measure of the smallest angle, while 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 angles in a triangle are such that the measure of one angle is 20° more than the measure of the smallest angle, while the measure of the third angle is three times the measure of the smallest angle. Find the measures of all three angles.

Find the Length of the Missing Side

In the following exercises, ΔABC is similar to ΔXYZ. Find the length of the indicated side.

Two triangles are shown. They appear to be the same shape, but the triangle on the right is smaller. The vertices of the triangle on the left are labeled A, B, and C. The side across from A is labeled 9, the side across from B is labeled b, and the side across from C is labeled 15. The vertices of the triangle on the right are labeled X, Y, and Z. The side across from X is labeled x, the side across from Y is labeled 8, and the side across from Z is labeled 10.

side b

Solution

12

side x

On a map, San Francisco, Las Vegas, and Los Angeles form a triangle whose sides are shown in the figure below. The actual distance from Los Angeles to Las Vegas is 270 miles.
A triangle is shown. The vertices are labeled San Francisco, Las Vegas, and Los Angeles. The side across from San Francisco is labeled 1 inch, the side across from Las Vegas is labeled 1.3 inches, and the side across from Los Angeles is labeled 2.1 inches.

Find the distance from Los Angeles to San Francisco.

Solution

351 miles

Find the distance from San Francisco to Las Vegas.

Use the Pythagorean Theorem

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

A right triangle is shown. The right angle is marked with a box. One of the sides touching the right angle is labeled as 9, the other as 12.
Solution

15

A right triangle is shown. The right angle is marked with a box. One of the sides touching the right angle is labeled as 16, the other as 12.
A right triangle is shown. The right angle is marked with a box. One of the sides touching the right angle is labeled as 15, the other as 20.
Solution

25

A right triangle is shown. The right angle is marked with a box. One of the sides touching the right angle is labeled as 5, the other as 12.

Find the Length of the Missing Side

In the following exercises, use the Pythagorean Theorem to find the length of the missing side. Round to the nearest tenth, if necessary.

A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 10. One of the sides touching the right angle is labeled as 6.
Solution

8

A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 17. One of the sides touching the right angle is labeled as 8.
A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 13. One of the sides touching the right angle is labeled as 5.
Solution

12

A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 20. One of the sides touching the right angle is labeled as 16.
A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 13. One of the sides touching the right angle is labeled as 8.
Solution

10.2

A right triangle is shown. The right angle is marked with a box. Both of the sides touching the right angle are labeled as 6.
A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 17. One of the sides touching the right angle is labeled as 15.
Solution

16.2

A right triangle is shown. The right angle is marked with a box. The side across from the right angle is labeled as 7. One of the sides touching the right angle is labeled as 5.

In the following exercises, solve. Approximate 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. How far from the base of the pole should the end of the string of lights be anchored?

A vertical pole is shown with a string of lights going from the top of the pole to the ground. The pole is labeled 12 feet. The string of lights is labeled 13 feet.
Solution

5 feet

Pam wants to put a banner across her garage door to congratulate her son on his college graduation. The garage door is 12 feet high and 16 feet wide. How long should the banner be to fit the garage door?

A picture of a house is shown. The rectangular garage is 12 feet high and 16 feet wide. A blue banner goes diagonally across the garage.

Chi is planning to put a path of paving stones through her flower garden. The flower garden is a square with sides of 10 feet. What will the length of the path be?

A square garden is shown. One side is labeled as 10 feet. There is a diagonal path of blue circular stones going from the lower left corner to the upper right corner.
Solution

14.1 feet

Brian borrowed a 20-foot extension ladder to paint his house. If he sets the base of the ladder 6 feet from the house, how far up will the top of the ladder reach?

A picture of a house is shown with a ladder leaning against it. The ladder is labeled 20 feet tall. The horizontal distance from the house to the base of the ladder is 6 feet.

Everyday Math

Building a scale model Joe wants to build a doll house for his daughter. He wants the doll house to look just like his house. His house is 30 feet wide and 35 feet tall at the highest point of the roof. If the dollhouse will be 2.5 feet wide, how tall will its highest point be?

Solution

2.9 feet

Measurement A city engineer plans to build a footbridge across a lake from point X to point Y, as shown in the picture below. To find the length of the footbridge, she draws a right triangle XYZ, with right angle at X. She measures the distance from X to Z,800 feet, and from Y to Z,1,000 feet. How long will the bridge be?

A lake is shown. Point Y is on one side of the lake, directly across from point X. Point Z is on the same side of the lake as point X.

Writing Exercises

Write three of the properties of triangles from this section and then explain each in your own words.

Solution

Answers will vary.

Explain how the figure below illustrates the Pythagorean Theorem for a triangle with legs of length 3 and 4.

Three squares are shown, forming a right triangle in the center. Each square is divided into smaller squares. The smallest square is divided into 9 small squares. The medium square is divided into 16 small squares. The large square is divided into 25 small squares.

Self Check

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

A self-assessment table for geometry skills, listing 'I can...' statements for properties of angles, triangles, and the Pythagorean Theorem, with options to rate understanding as 'Confidentially', '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?

angle
An angle is formed by two rays that share a common endpoint. Each ray is called a side of the angle.
complementary angles
If the sum of the measures of two angles is 90°, then they are called complementary angles.
hypotenuse
The side of the triangle opposite the 90° angle is called the hypotenuse.
legs of a right triangle
The sides of a right triangle adjacent to the right angle are called the legs.
right triangle
A right triangle is a triangle that has one 90° angle.
similar figures
In geometry, if two figures have exactly the same shape but different sizes, we say they are similar figures.
supplementary angles
If the sum of the measures of two angles is 180°, then they are called supplementary angles.
triangle
A triangle is a geometric figure with three sides and three angles.
vertex of an angle
When two rays meet to form an angle, the common endpoint is called the vertex of the angle.

Use Properties of Rectangles, Triangles, and Trapezoids

Learning Objectives

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

  • Understand linear, square, and cubic measure
  • Use properties of rectangles
  • Use properties of triangles
  • Use properties of trapezoids

Before you get started, take this readiness quiz.

The length of a rectangle is 3 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 14 in Evaluate, Simplify, and Translate Expressions.

Solution

w−3

Simplify: 12(6h).
If you missed this problem, review Example 3 in Commutative and Associative Properties.

Solution

3h

Simplify: 52(10.3−7.9).
If you missed this problem, review Example 9 in Decimals and Fractions.

Solution

6

In this section, we’ll continue working with geometry applications. We will add some more properties of triangles, and we’ll learn about the properties of rectangles and trapezoids.

Understand Linear, Square, and Cubic Measure

When you measure your height or the length of a garden hose, you use a ruler or tape measure (Figure 1). A tape measure might remind you of a line—you use it for linear measure, which measures length. Inch, foot, yard, mile, centimeter and meter are units of linear measure.

A picture of a portion of a tape measure is shown. The top shows the numbers 1 through 5. The portion from the beginning to the 1 has a red circle and an arrow to a picture from 0 to 1 inch, with 1 sixteenth, 1 eighth, 3 eighths, 1 half, and 3 fourths labeled. Above this, it is labeled “Standard Measures.” The bottom of the tape measure shows the numbers 1 through 10, then 1 and 2. The region from the edge to about 3 and a half has a red circle with an arrow pointing to a picture from 0 to 3.5. It is labeled 0, 1 cm, 1.7 cm, 2.3 cm and 3.5 cm. Above this, it is labeled “Metric (S).”
This tape measure measures inches along the top and centimeters along the bottom.

When you want to know how much tile is needed to cover a floor, or the size of a wall to be painted, you need to know the area, a measure of the region needed to cover a surface. Area is measured is square units. We often use square inches, square feet, square centimeters, or square miles to measure area. A square centimeter is a square that is one centimeter (cm) on each side. A square inch is a square that is one inch on each side (Figure 2).

Two squares are shown. The smaller one has sides labeled 1 cm and is 1 square centimeter. The larger one has sides labeled 1 inch and is 1 square inch.
Square measures have sides that are each 1 unit in length.

Figure 3 shows a rectangular rug that is 2 feet long by 3 feet wide. Each square is 1 foot wide by 1 foot long, or 1 square foot. The rug is made of 6 squares. The area of the rug is 6 square feet.

A rectangle is shown. It has 3 squares across and 2 squares down, a total of 6 squares.
The rug contains six squares of 1 square foot each, so the total area of the rug is 6 square feet.

When you measure how much it takes to fill a container, such as the amount of gasoline that can fit in a tank, or the amount of medicine in a syringe, you are measuring volume. Volume is measured in cubic units such as cubic inches or cubic centimeters. When measuring the volume of a rectangular solid, you measure how many cubes fill the container. We often use cubic centimeters, cubic inches, and cubic feet. A cubic centimeter is a cube that measures one centimeter on each side, while a cubic inch is a cube that measures one inch on each side (Figure 4).

Two cubes are shown. The smaller one has sides labeled 1 cm and is labeled as 1 cubic centimeter. The larger one has sides labeled 1 inch and is labeled as 1 cubic inch.
Cubic measures have sides that are 1 unit in length.

Suppose the cube in Figure 5 measures 3 inches on each side and is cut on the lines shown. How many little cubes does it contain? If we were to take the big cube apart, we would find 27 little cubes, with each one measuring one inch on all sides. So each little cube has a volume of 1 cubic inch, and the volume of the big cube is 27 cubic inches.

A cube is shown, comprised of smaller cubes. Each side of the cube has 3 smaller cubes across, for a total of 27 smaller cubes.
A cube that measures 3 inches on each side is made up of 27 one-inch cubes, or 27 cubic inches.
Doing the Manipulative Mathematics activity Visualizing Area and Perimeter will help you develop a better understanding of the difference between the area of a figure and its perimeter.

For each item, state whether you would use linear, square, or cubic measure:

  1. ⓐ amount of carpeting needed in a room

  2. ⓑ extension cord length

  3. ⓒ amount of sand in a sandbox

  4. ⓓ length of a curtain rod

  5. ⓔ amount of flour in a canister

  6. ⓕ size of the roof of a doghouse.

Solution

Solution

Examples of common measurements classified by their type (linear, square, or cubic measure).
ⓐ You are measuring how much surface the carpet covers, which is the area. square measure
ⓑ You are measuring how long the extension cord is, which is the length. linear measure
ⓒ You are measuring the volume of the sand. cubic measure
ⓓ You are measuring the length of the curtain rod. linear measure
ⓔ You are measuring the volume of the flour. cubic measure
ⓕ You are measuring the area of the roof. square measure

Determine whether you would use linear, square, or cubic measure for each item.

ⓐ amount of paint in a can ⓑ height of a tree ⓒ floor of your bedroom ⓓ diameter of bike wheel ⓔ size of a piece of sod ⓕ amount of water in a swimming pool

Solution
  1. ⓐ cubic
  2. ⓑ linear
  3. ⓒ square
  4. ⓓ linear
  5. ⓔ square
  6. ⓕ cubic

Determine whether you would use linear, square, or cubic measure for each item.

ⓐ volume of a packing box ⓑ size of patio ⓒ amount of medicine in a syringe ⓓ length of a piece of yarn ⓔ size of housing lot ⓕ height of a flagpole

Solution
  1. ⓐ cubic
  2. ⓑ square
  3. ⓒ cubic
  4. ⓓ linear
  5. ⓔ square
  6. ⓕ linear

Many geometry applications will involve finding the perimeter or the area of a figure. There are also many applications of perimeter and area in everyday life, so it is important to make sure you understand what they each mean.

Picture a room that needs new floor tiles. The tiles come in squares that are a foot on each side—one square foot. How many of those squares are needed to cover the floor? This is the area of the floor.

Next, think about putting new baseboard around the room, once the tiles have been laid. To figure out how many strips are needed, you must know the distance around the room. You would use a tape measure to measure the number of feet around the room. This distance is the perimeter.

Perimeter and Area

The perimeter is a measure of the distance around a figure.

The area is a measure of the surface covered by a figure.

Figure 6 shows a square tile that is 1 inch on each side. If an ant walked around the edge of the tile, it would walk 4 inches. This distance is the perimeter of the tile.

Since the tile is a square that is 1 inch on each side, its area is one square inch. The area of a shape is measured by determining how many square units cover the shape.

A 5 square by 5 square checkerboard is shown with each side labeled 1 inch. An image of an ant is shown on the top left square.
Perimeter=4inchesArea=1square inch
When the ant walks completely around the tile on its edge, it is tracing the perimeter of the tile. The area of the tile is 1 square inch.
Doing the Manipulative Mathematics activity Measuring Area and Perimeter will help you develop a better understanding of how to measure the area and perimeter of a figure.

Each of two square tiles is 1 square inch. Two tiles are shown together.

  1. ⓐ What is the perimeter of the figure?

  2. ⓑ What is the area?

    A checkerboard is shown. It has 10 squares across the top and 5 down the side.
Solution

Solution

ⓐ The perimeter is the distance around the figure. The perimeter is 6 inches.

ⓑ The area is the surface covered by the figure. There are 2 square inch tiles so the area is 2 square inches.

A checkerboard is shown. It has 10 squares across the top and 5 down the side. The top and bottom each have two adjacent 1 inch labels across, the sides have 1 inch labels.

Each box in the figure below is 1 square inch. Find the ⓐ perimeter and ⓑ area of the figure:

A rectangle is shown comprised of 3 squares.
Solution
  1. ⓐ 8 inches
  2. ⓑ 3 sq. inches

Each box in the figure below is 1 square inch. Find the ⓐ perimeter and ⓑ area of the figure:

A square is shown comprised of 4 smaller squares.
Solution
  1. ⓐ 8 centimeters
  2. ⓑ 4 sq. centimeters

Use the Properties of Rectangles

A rectangle has four sides and four right angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, L, and the adjacent side as the width, W. See Figure 7.

A rectangle is shown. Each angle is marked with a square. The top and bottom are labeled L, the sides are labeled W.
A rectangle has four sides, and four right angles. The sides are labeled L for length and W for width.

The perimeter, P, of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk L+W+L+W units, or two lengths and two widths. The perimeter then is

P=L+W+L+WorP=2L+2W

What about the area of a rectangle? Remember the rectangular rug from the beginning of this section. It was 2 feet long by 3 feet wide, and its area was 6 square feet. See Figure 8. Since A=2⋅3, we see that the area, A, is the length, L, times the width, W, so the area of a rectangle is A=L⋅W.

A rectangle is shown. It is made up of 6 squares. The bottom is 2 squares across and marked as 2, the side is 3 squares long and marked as 3.
The area of this rectangular rug is 6 square feet, its length times its width.

Properties of Rectangles

  • Rectangles have four sides and four right (90°) angles.
  • The lengths of opposite sides are equal.
  • The perimeter, P, of a rectangle is the sum of twice the length and twice the width. See Figure 7.
    P=2L+2W
  • The area, A, of a rectangle is the length times the width.
    A=L⋅W

For easy reference as we work the examples in this section, we will restate the Problem Solving Strategy for Geometry Applications here.

Use a Problem Solving Strategy for Geometry Applications

  1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
  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 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 length of a rectangle is 32 meters and the width is 20 meters. Find ⓐ the perimeter, and ⓑ the area.

Solution

Solution

ⓐ
Step 1. Read the problem. Draw the figure and label it with the given information. A rectangle is shown with sides labeled 32 m and 20 m, indicating its length and width.
Step 2. Identify what you are looking for. the perimeter of a rectangle
Step 3. Name. Choose a variable to represent it. Let P = the perimeter
Step 4. Translate.
Write the appropriate formula.
Substitute.

The formula for the perimeter of a rectangle, P = 2L + 2W, with L=32 and W=20 substituted as P = 2(32) + 2(20).
Step 5. Solve the equation. A mathematical equation showing P = 64 + 40, which is then solved to P = 104.
Step 6. Check:
A mathematical equation solving for P, showing the sum of 20 + 32 + 20 + 32 being verified as equal to 104. The final line confirms '104 = 104' with a checkmark.
Step 7. Answer the question. The perimeter of the rectangle is 104 meters.
ⓑ
Step 1. Read the problem. Draw the figure and label it with the given information. A rectangle with a length of 32 meters and a width of 20 meters, clearly indicating its dimensions.
Step 2. Identify what you are looking for. the area of a rectangle
Step 3. Name. Choose a variable to represent it. Let A = the area
Step 4. Translate.
Write the appropriate formula.
Substitute.

The image displays the formula for the area of a rectangle, A = L x W, with an example showing A = 32 m x 20 m. Brackets underneath each variable and operator indicate what each symbol represents.
Step 5. Solve the equation. The image displays the equation 'A = 640' centered on a plain white background, rendered in a standard, dark font. It appears to be a mathematical or numerical representation.
Step 6. Check:
A mathematical verification showing A equals 32 multiplied by 20, which simplifies to 640, thereby confirming the equation 640 = 640 with a checkmark.
Step 7. Answer the question. The area of the rectangle is 640 square meters.

The length of a rectangle is 120 yards and the width is 50 yards. Find ⓐ the perimeter and ⓑ the area.

Solution
  1. ⓐ 340 yd
  2. ⓑ 6000 sq. yd

The length of a rectangle is 62 feet and the width is 48 feet. Find ⓐ the perimeter and ⓑ the area.

Solution
  1. ⓐ 220 ft
  2. ⓑ 2976 sq. ft

Find the length of a rectangle with perimeter 50 inches and width 10 inches.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A rectangle is depicted, having a height of 10 inches on both vertical sides and a width labeled 'L' on both horizontal sides, indicating its dimensions.
Step 2. Identify what you are looking for. the length of the rectangle
Step 3. Name. Choose a variable to represent it. Let L = the length
Step 4. Translate.
Write the appropriate formula.
Substitute.

Mathematical equations showing the perimeter formula P = 2L + 2W. Below it, P is substituted with 50 and W with 10, leading to 50 = 2L + 2(10).
Step 5. Solve the equation. Mathematical steps showing the solution of a linear equation. The equation 50 - 20 = 2L + 20 - 20 is simplified to 30 = 2L, and then L is found to be 15 by dividing by 2.
Step 6. Check:
A mathematical equation checks if 15 + 10 + 15 + 10 equals 50, confirming the sum is indeed 50 with a checkmark, potentially demonstrating a perimeter calculation.
Step 7. Answer the question. The length is 15 inches.

Find the length of a rectangle with a perimeter of 80 inches and width of 25 inches.

Solution

15 in.

Find the length of a rectangle with a perimeter of 30 yards and width of 6 yards.

Solution

9 yd

In the next example, the width is defined in terms of the length. We’ll wait to draw the figure until we write an expression for the width so that we can label one side with that expression.

The width of a rectangle is two inches less than the length. The perimeter is 52 inches. Find the length and width.

Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for. the length and width of the rectangle
Step 3. Name. Choose a variable to represent it.

Now we can draw a figure using these expressions for the length and width.
Since the width is defined in terms of the length, we let L = length. The width is two feet less that the length, so we let L − 2 = width
A square with its top and bottom sides labeled 'L' and its left and right sides labeled 'L - 2'.
Step 4.Translate.
Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information.
Substitute in the given information.

A mathematical equation illustrating the perimeter of a rectangle, where P = 2L + 2W is shown, followed by the specific values and substitution 52 = 2L + 2(L-2).
Step 5. Solve the equation. 52=2L+2L−4
Combine like terms. 52=4L−4
Add 4 to each side. 56=4L
Divide by 4. 564=4L4
14=L
The length is 14 inches.
Now we need to find the width.
The width is L − 2. An algebraic substitution showing L-2, then substituting L with 14 (highlighted in red) to get 14-2, which equals 12.
The width is 12 inches.
Step 6. Check:
Since 14+12+14+12=52, this works!
Step 7. Answer the question. The length is 14 feet and the width is 12 feet.

The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.

Solution

18 m, 11 m

The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width.

Solution

11 ft , 19 ft

The length of a rectangle is four centimeters more than twice the width. The perimeter is 32 centimeters. Find the length and width.

Solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for. the length and width
Step 3. Name. Choose a variable to represent it. let W = width
The length is four more than twice the width.
2w + 4 = length
A rectangle is depicted with its dimensions labeled algebraically. The width is denoted by 'w' and the length is denoted by '2w + 4'.
Step 4.Translate.
Write the appropriate formula and substitute in the given information.
The image displays a two-line mathematical equation. The first line presents the formula for the perimeter of a rectangle, P = 2L + 2W. The second line shows a specific application of this formula: 32 = 2(2w + 4) + 2w.
Step 5. Solve the equation. An algebraic solution for 'w' from the equation 32 = 4w + 8 + 2w, showing w = 4 (width). The length, calculated as 2w + 4, is found to be 12 cm.
Step 6. Check:
A step-by-step mathematical verification of the perimeter formula p = 2L + 2W, using values L=12 and W=4 to show 32 = 32.
Step 7. Answer the question. The length is 12 cm and the width is 4 cm.

The length of a rectangle is eight more than twice the width. The perimeter is 64 feet. Find the length and width.

Solution

8 ft, 24 ft

The width of a rectangle is six less than twice the length. The perimeter is 18 centimeters. Find the length and width.

Solution

5 cm, 4 cm

The area of a rectangular room is 168 square feet. The length is 14 feet. What is the width?

Solution

Solution

Step 1. Read the problem. A rectangle is shown with width 'W', length '14 ft', and an area of '168 ft^2'.
Step 2. Identify what you are looking for. the width of a rectangular room
Step 3. Name. Choose a variable to represent it. Let W = width
Step 4.Translate.
Write the appropriate formula and substitute in the given information.
Two equations are displayed: A = LW (Area = Length x Width) and 168 = 14W. This represents a geometry problem where the area is 168 and the length is 14, seeking the width W.
Step 5. Solve the equation. A mathematical equation shows 168 divided by 14 equals 14W divided by 14. Below this, the solution is shown as 12 equals W, indicating the variable W has been solved for.
Step 6. Check:
A mathematical verification is shown, starting with the area formula A=LW, then questioning if 168 equals 14 multiplied by 12, and finally confirming that 168 does indeed equal 168.
Step 7. Answer the question. The width of the room is 12 feet.

The area of a rectangle is 598 square feet. The length is 23 feet. What is the width?

Solution

26 ft

The width of a rectangle is 21 meters. The area is 609 square meters. What is the length?

Solution

29 m

The perimeter of a rectangular swimming pool is 150 feet. The length is 15 feet more than the width. Find the length and width.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A rectangular body of water with width W and length W+15. The perimeter of the rectangle is given as P=150 ft.
Step 2. Identify what you are looking for. the length and width of the pool
Step 3. Name. Choose a variable to represent it.
The length is 15 feet more than the width.
Let W=width
W+15=length
Step 4.Translate.
Write the appropriate formula and substitute.
Perimeter formula P = 2L + 2W applied to a problem, substituting P=150, L=(w+15), and W=w to form the equation 150 = 2(w + 15) + 2w.
Step 5. Solve the equation. This image demonstrates the algebraic process of finding the dimensions of a pool. It solves for 'w', the width, which is 30, and then calculates the length as 'w + 15', resulting in 45.
Step 6. Check:
Mathematical steps demonstrating the calculation of a rectangle's perimeter. It shows the formula p = 2L + 2W, then substitutes L=45 and W=30 to verify if p=150, concluding with 150 = 150.
Step 7. Answer the question. The length of the pool is 45 feet and the width is 30 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

30 ft, 70 ft

The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.

Solution

60 yd, 90 yd

Use the Properties of Triangles

We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle in Figure 8, we’ve labeled the length b and the width h, so it’s area is bh.

A rectangle is shown. The side is labeled h and the bottom is labeled b. The center says A equals bh.
The area of a rectangle is the base, b, times the height, h.

We can divide this rectangle into two congruent triangles (Figure 10). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or 12bh. This example helps us see why the formula for the area of a triangle is A=12bh.

A rectangle is shown. A diagonal line is drawn from the upper left corner to the bottom right corner. The side of the rectangle is labeled h and the bottom is labeled b. Each triangle says one-half bh. To the right of the rectangle, it says “Area of each triangle,” and shows the equation A equals one-half bh.
A rectangle can be divided into two triangles of equal area. The area of each triangle is one-half the area of the rectangle.

The formula for the area of a triangle is A=12bh, where b is the base and h is the height.

To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a 90° angle with the base. Figure 11 shows three triangles with the base and height of each marked.

Three triangles are shown. The triangle on the left is a right triangle. The bottom is labeled b and the side is labeled h. The middle triangle is an acute triangle. The bottom is labeled b. There is a dotted line from the top vertex to the base of the triangle, forming a right angle with the base. That line is labeled h. The triangle on the right is an obtuse triangle. The bottom of the triangle is labeled b. The base has a dotted line extended out and forms a right angle with a dotted line to the top of the triangle. The vertical line is labeled h.
The height h of a triangle is the length of a line segment that connects the the base to the opposite vertex and makes a 90° angle with the base.

Triangle Properties

For any triangle ΔABC, the sum of the measures of the angles is 180°.

m∠A+m∠B+m∠C=180°

The perimeter of a triangle is the sum of the lengths of the sides.

P=a+b+c

The area of a triangle is one-half the base, b, times the height, h.

A=12bh
A triangle is shown. The vertices are labeled A, B, and C. The sides are labeled a, b, and c. There is a vertical dotted line from vertex B at the top of the triangle to the base of the triangle, meeting the base at a right angle. The dotted line is labeled h.

Find the area of a triangle whose base is 11 inches and whose height is 8 inches.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. An image of a triangle showing its base and height. The base is labeled as 11 inches, and the height, represented by a dashed line perpendicular to the base, is labeled as 8 inches.
Step 2. Identify what you are looking for. the area of the triangle
Step 3. Name. Choose a variable to represent it. let A = area of the triangle
Step 4.Translate.
Write the appropriate formula.
Substitute.

Two lines of an area formula: the first line shows A = 1/2 * b * h, and the second line substitutes b = 11 and h = 8 into the formula, showing A = 1/2 * 11 * 8.
Step 5. Solve the equation. The image displays the equation A = 44 square inches, indicating an area measurement.
Step 6. Check:
A mathematical calculation verifying the area of a triangle using the formula A = 1/2 bh. The steps show 44 compared to 1/2(11)8, confirming the equality 44 = 44 with a checkmark.
Step 7. Answer the question. The area is 44 square inches.

Find the area of a triangle with base 13 inches and height 2 inches.

Solution

13 sq. in.

Find the area of a triangle with base 14 inches and height 7 inches.

Solution

49 sq. in.

The perimeter of a triangular garden is 24 feet. The lengths of two sides are 4 feet and 9 feet. How long is the third side?

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A triangle with sides labeled 4 ft and 9 ft, and a third side 'c'. The perimeter of the triangle is given as P = 24 ft, implying the missing side 'c' is 11 ft.
Step 2. Identify what you are looking for. length of the third side of a triangle
Step 3. Name. Choose a variable to represent it. Let c = the third side
Step 4.Translate.
Write the appropriate formula.
Substitute in the given information.

Mathematical problem illustrating variable substitution: P = a + b + c, with 24 = 4 + 9 + c provided.
Step 5. Solve the equation. A simple algebraic equation is displayed on a white background. The first line reads '24 = 13 + c' and the second line shows its solution as '11 = c', indicating that the value of 'c' is 11.
Step 6. Check:
An image illustrating a mathematical check: P = a + b + c. The calculation verifies if 24 equals 4 + 9 + 11, concluding with a confirmed equality of 24 = 24, marked by a checkmark.
Step 7. Answer the question. The third side is 11 feet long.

The perimeter of a triangular garden is 48 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?

Solution

8 ft

The lengths of two sides of a triangular window are 7 feet and 5 feet. The perimeter is 18 feet. How long is the third side?

Solution

6 ft

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

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. An isosceles triangle shown with a base of 15 m and an indicated height 'h' perpendicular to the base.
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
Step 4.Translate.
Write the appropriate formula.
Substitute in the given information.

The area of a triangle formula A = (1/2) * b * h is shown. Below, an example calculates the height, h, with A=90 and b=15, resulting in the equation 90 = (1/2) * 15 * h.
Step 5. Solve the equation. Mathematical equations showing the calculation of 'h', where 90 equals (15/2)h, resulting in the solution h=12.
Step 6. Check:
The image illustrates the verification of a triangle's area using the formula A = (1/2)bh. It confirms that 90 is indeed the correct area when the base is 15 and height is 12, as 90 = (1/2) * 15 * 12.
Step 7. Answer the question. The height of the triangle is 12 meters.

The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

Solution

14 in.

A triangular tent door has an area of 15 square feet. The height is 5 feet. What is the base?

Solution

6 ft

Isosceles and Equilateral Triangles

Besides the right triangle, some other triangles have special names. A triangle with two sides of equal length is called an isosceles triangle. A triangle that has three sides of equal length is called an equilateral triangle. Figure 12 shows both types of triangles.

Two triangles are shown. All three sides of the triangle on the left are labeled s. It is labeled “equilateral triangle”. Two sides of the triangle on the right are labeled s. It is labeled “isosceles triangle”.
In an isosceles triangle, two sides have the same length, and the third side is the base. In an equilateral triangle, all three sides have the same length.

Isosceles and Equilateral Triangles

An isosceles triangle has two sides the same length.

An equilateral triangle has three sides of equal length.

The perimeter of an equilateral triangle is 93 inches. Find the length of each side.

Solution
Solution
Step 1. Read the problem. Draw the figure and label it with the given information. A simple black outline drawing of an equilateral triangle, with each of its three sides labeled with the letter 's', indicating that all sides have equal length.
Perimeter = 93 in.
Step 2. Identify what you are looking for. length of the sides of an equilateral triangle
Step 3. Name. Choose a variable to represent it. Let s = length of each side
Step 4.Translate.
Write the appropriate formula.
Substitute.

Mathematical equations illustrate perimeter. P=a+b+c defines the general case, while 93=s+s+s provides a specific example where the perimeter is 93 and the three sides are equal (s).
Step 5. Solve the equation. The image displays a two-step algebraic solution. The top line shows the equation 93 = 3s, and the bottom line shows the solution s = 31, indicating that both sides of the initial equation were divided by 3.
Step 6. Check:
A perfectly balanced equilateral triangle, with each of its three sides clearly marked as 31.
A math problem demonstrating 93 is equal to 31 + 31 + 31, which simplifies to 93 = 93, confirmed with a checkmark.
Step 7. Answer the question. Each side is 31 inches.

Find the length of each side of an equilateral triangle with perimeter 39 inches.

Solution

13 in.

Find the length of each side of an equilateral triangle with perimeter 51 centimeters.

Solution

17 cm

Arianna has 156 inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of
60 inches. How long can she make the two equal sides?

Solution
Solution
Step 1. Read the problem. Draw the figure and label it with the given information. A diagram of an isosceles triangle with two equal sides labeled 's' and the base labeled '60 in'.
P = 156 in.
Step 2. Identify what you are looking for. the lengths of the two equal sides
Step 3. Name. Choose a variable to represent it. Let s = the length of each side
Step 4.Translate.
Write the appropriate formula.
Substitute in the given information.

A two-row mathematical equation showing P = a + b + c in the first row, and 156 = s + 60 + s in the second row, representing a perimeter calculation with a known total and variable sides.
Step 5. Solve the equation. An image displaying the step-by-step solution to the linear equation 156 = 2s + 60, which proceeds to 96 = 2s, and finally resolves to 48 = s.
Step 6. Check:
Mathematical verification of an equation. The general formula p = a + b + c is applied to check if 156 = 48 + 60 + 48. The calculation confirms 156 = 156 with a checkmark.
Step 7. Answer the question. Arianna can make each of the two equal sides 48 inches long.

A backyard deck is in the shape of an isosceles triangle with a base of 20 feet. The perimeter of the deck is 48 feet. How long is each of the equal sides of the deck?

Solution

14 ft

A boat’s sail is an isosceles triangle with base of 8 meters. The perimeter is 22 meters. How long is each of the equal sides of the sail?

Solution

7 m

Use the Properties of Trapezoids

A trapezoid is four-sided figure, a quadrilateral, with two sides that are parallel and two sides that are not. The parallel sides are called the bases. We call the length of the smaller base b, and the length of the bigger base B. The height, h, of a trapezoid is the distance between the two bases as shown in Figure 13.

A trapezoid is shown. The top is labeled b and marked as the smaller base. The bottom is labeled B and marked as the larger base. A vertical line forms a right angle with both bases and is marked as h.
A trapezoid has a larger base, B, and a smaller base, b. The height h is the distance between the bases.

The formula for the area of a trapezoid is:

Areatrapezoid=12h(b+B)

Splitting the trapezoid into two triangles may help us understand the formula. The area of the trapezoid is the sum of the areas of the two triangles. See Figure 14.

An image of a trapezoid is shown. The top is labeled with a small b, the bottom with a big B. A diagonal is drawn in from the upper left corner to the bottom right corner.
Splitting a trapezoid into two triangles may help you understand the formula for its area.

The height of the trapezoid is also the height of each of the two triangles. See Figure 15.

An image of a trapezoid is shown. The top is labeled with a small b, the bottom with a big B. A diagonal is drawn in from the upper left corner to the bottom right corner. There is an arrow pointing to a second trapezoid. The upper right-hand side of the trapezoid forms a blue triangle, with the height of the trapezoid drawn in as a dotted line. The lower left-hand side of the trapezoid forms a red triangle, with the height of the trapezoid drawn in as a dotted line.

The formula for the area of a trapezoid is

This image shows the formula for the area of a trapezoid and says “area of trapezoid equals one-half h times smaller base b plus larger base B).

If we distribute, we get,

The top line says area of trapezoid equals one-half times blue little b times h plus one-half times red big B times h. Below this is area of trapezoid equals A sub blue triangle plus A sub red triangle.

Properties of Trapezoids

  • A trapezoid has four sides. See Figure 13.
  • Two of its sides are parallel and two sides are not.
  • The area, A, of a trapezoid is A=12h(b+B).

Find the area of a trapezoid whose height is 6 inches and whose bases are 14 and 11 inches.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A trapezoid is displayed with a top base of 14 inches, a bottom base of 11 inches, and a height of 6 inches, indicated by a dashed line and right angle symbols.
Step 2. Identify what you are looking for. the area of the trapezoid
Step 3. Name. Choose a variable to represent it. Let A=the area
Step 4.Translate.
Write the appropriate formula.
Substitute.

Illustrates the area formula for a trapezoid, A = 1/2 * h * (b + B), and its application with h=6, b=11, B=14.
Step 5. Solve the equation. This image demonstrates the calculation of area A. Starting with A = 1/2 * 6 * 25, the equation is simplified to A = 3 * 25, which ultimately results in an area of A = 75 square inches.
Step 6. Check: Is this answer reasonable?

If we draw a rectangle around the trapezoid that has the same big base B and a height h, its area should be greater than that of the trapezoid.

If we draw a rectangle inside the trapezoid that has the same little base b and a height h, its area should be smaller than that of the trapezoid.

A table is shown with 3 columns and 4 rows. The first column has an image of a trapezoid with a rectangle drawn around it in red. The larger base of the trapezoid is labeled 14 and is the same as the base of the rectangle. The height of the trapezoid is labeled 6 and is the same as the height of the rectangle. The smaller base of the trapezoid is labeled 11. Below this is A sub rectangle equals b times h. Below is A sub rectangle equals 14 times 6. Below is A sub rectangle equals 84 square inches. The second column has an image of a trapezoid. The larger base is labeled 14, the smaller base is labeled 11, and the height is labeled 6. Below this is A sub trapezoid equals one-half times h times parentheses little b plus big B. Below this is A sub trapezoid equals one-half times 6 times parentheses 11 plus 14. Below this is A sub trapezoid equals 75 square inches. The third column has an image of a trapezoid with a red rectangle drawn inside of it. The height is labeled 6. Below this is A sub rectangle equals b times h. Below is A sub rectangle equals 11 times 6. Below is A sub rectangle equals 66 square inches.

The area of the larger rectangle is 84 square inches and the area of the smaller rectangle is 66 square inches. So it makes sense that the area of the trapezoid is between 84 and 66 square inches

Step 7. Answer the question. The area of the trapezoid is 75 square inches.

The height of a trapezoid is 14 yards and the bases are 7 and 16 yards. What is the area?

Solution

161 sq. yd

The height of a trapezoid is 18 centimeters and the bases are 17 and 8 centimeters. What is the area?

Solution

225 sq. cm

Find the area of a trapezoid whose height is 5 feet and whose bases are 10.3 and 13.7 feet.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A white background shows a trapezoid with its top base measuring 10.3 ft, its bottom base measuring 13.7 ft, and its height measuring 5 ft. A dashed vertical line indicates the height, with right angle symbols.
Step 2. Identify what you are looking for. the area of the trapezoid
Step 3. Name. Choose a variable to represent it. Let A = the area
Step 4.Translate.
Write the appropriate formula.
Substitute.

Formula for the area of a trapezoid (A = 1/2 * h * (b + B)) and its application with specific values: h=5, b=10.3, B=13.7.
Step 5. Solve the equation. A mathematical calculation is shown where the area (A) is determined. Starting with A = (1/2) * 5 * 24, the steps lead to A = 12 * 5, and finally A = 60 square feet, presented in three lines on a white background.
Step 6. Check: Is this answer reasonable?
The area of the trapezoid should be less than the area of a rectangle with base 13.7 and height 5, but more than the area of a rectangle with base 10.3 and height 5.
An image of a trapezoid is shown with a red rectangle drawn around it. The larger base of the trapezoid is labeled 13.7 ft. and is the same as the base of the rectangle. The height of both the trapezoid and the rectangle is 5 ft. Next to this is an image of a trapezoid with a black rectangle drawn inside it. The smaller base of the trapezoid is labeled 10.3 ft. and is the same as the base of the rectangle. Below the images is A sub red rectangle is greater than A sub trapezoid is greater than A sub rectangle. Below this is 68.5, 60, and 51.5.
Step 7. Answer the question. The area of the trapezoid is 60 square feet.

The height of a trapezoid is 7 centimeters and the bases are 4.6 and 7.4 centimeters. What is the area?

Solution

42 sq. cm

The height of a trapezoid is 9 meters and the bases are 6.2 and 7.8 meters. What is the area?

Solution

63 sq. m

Vinny has a garden that is shaped like a trapezoid. The trapezoid has a height of 3.4 yards and the bases are 8.2 and 5.6 yards. How many square yards will be available to plant?

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A right trapezoid is shown with a top base of 5.6 yd, a bottom base of 8.2 yd, and a height of 3.4 yd, indicated by right angle markings on the vertical side.
Step 2. Identify what you are looking for. the area of a trapezoid
Step 3. Name. Choose a variable to represent it. Let A = the area
Step 4.Translate.
Write the appropriate formula.
Substitute.

Formula for the area of a trapezoid, A = (1/2) * h * (b + B), with h=3.4, b=5.6, and B=8.2 substituted in the second line.
Step 5. Solve the equation. A mathematical calculation showing A equals one-half times 3.4 times 13.8, resulting in A equals 23.46 square yards.
Step 6. Check: Is this answer reasonable?
Yes. The area of the trapezoid is less than the area of a rectangle with a base of 8.2 yd and height 3.4 yd, but more than the area of a rectangle with base 5.6 yd and height 3.4 yd.

This image is a table with two rows. the first row is split into three columns. The first column is the formula Area of a rectangle equals base times height. On the next line under this it has numbers plugged into the formula; the base, 8.2 in parentheses times the height 3.4 in parentheses. Under this is it has “equals 27.88 yards squared”. The center column includes the formula of a trapezoid and says Area of a trapezoid equals one half times 3.5 yards in parentheses times 5.8 plus 8.2 in parentheses. Under this it has “equals 23.46 yards squared”. In the third column it it has the formula the area of a rectangle equals base times height. Under this it has equals 5.6 in parentheses times 3.4 in parentheses. Under this it has “equals 19.04 yards squared.” In the second row, centered from left to right it has “Area of a rectangle” and a “greater than” sign, “Area of a trapezoid” and a greater than sign and “area of a rectangle”. Under Area of a rectangle it has 27.88, then 23.46 under “area of a trapezoid”, then 19.04 under “area of a rectangle”.
Step 7. Answer the question. Vinny has 23.46 square yards in which he can plant.

Lin wants to sod his lawn, which is shaped like a trapezoid. The bases are 10.8 yards and 6.7 yards, and the height is 4.6 yards. How many square yards of sod does he need?

Solution

40.25 sq. yd

Kira wants cover his patio with concrete pavers. If the patio is shaped like a trapezoid whose bases are 18 feet and 14 feet and whose height is 15 feet, how many square feet of pavers will he need?

Solution

240 sq. ft

The Links to Literacy activity Spaghetti and Meatballs for All will provide you with another view of the topics covered in this section."

ACCESS ADDITIONAL ONLINE RESOURCES

  • Perimeter of a Rectangle
  • Area of a Rectangle
  • Perimeter and Area Formulas
  • Area of a Triangle
  • Area of a Triangle with Fractions
  • Area of a Trapezoid

Key Concepts

  • Properties of Rectangles
    • Rectangles have four sides and four right (90°) angles.
    • The lengths of opposite sides are equal.
    • The perimeter, P, of a rectangle is the sum of twice the length and twice the width.
      • P=2L+2W
    • The area, A, of a rectangle is the length times the width.
      • A=L⋅W
  • Triangle Properties
    • For any triangle ΔABC, the sum of the measures of the angles is 180°.
      • m∠A+m∠B+m∠C=180°
    • The perimeter of a triangle is the sum of the lengths of the sides.
      • P=a+b+c
    • The area of a triangle is one-half the base, b, times the height, h.
      • A=12bh

Practice Makes Perfect

Understand Linear, Square, and Cubic Measure

In the following exercises, determine whether you would measure each item using linear, square, or cubic units.

amount of water in a fish tank

Solution

cubic

length of dental floss

living area of an apartment

Solution

square

floor space of a bathroom tile

height of a doorway

Solution

linear

capacity of a truck trailer

In the following exercises, find the ⓐ perimeter and ⓑ area of each figure. Assume each side of the square is 1 cm.

A rectangle is shown comprised of 4 squares forming a horizontal line.
Solution
  1. ⓐ 10 cm
  2. ⓑ 4 sq. cm
A rectangle is shown comprised of 3 squares forming a vertical line.
Three squares are shown. There is one on the bottom left, one on the bottom right, and one on the top right.
Solution
  1. ⓐ 8 cm
  2. ⓑ 3 sq. cm
Four squares are shown. Three form a horizontal line, and there is one above the center square.
Five squares are shown. There are three forming a horizontal line across the top and two underneath the two on the right.
Solution
  1. ⓐ 10 cm
  2. ⓑ 5 sq. cm
A square is shown. It is comprised of nine smaller squares.

Use the Properties of Rectangles

In the following exercises, find the ⓐ perimeter and ⓑ area of each rectangle.

The length of a rectangle is 85 feet and the width is 45 feet.

Solution
  1. ⓐ 260 ft
  2. ⓑ 3825 sq. ft

The length of a rectangle is 26 inches and the width is 58 inches.

A rectangular room is 15 feet wide by 14 feet long.

Solution
  1. ⓐ 58 ft
  2. ⓑ 210 sq. ft

A driveway is in the shape of a rectangle 20 feet wide by 35 feet long.

In the following exercises, solve.

Find the length of a rectangle with perimeter 124 inches and width 38 inches.

Solution

24 inches

Find the length of a rectangle with perimeter 20.2 yards and width of 7.8 yards.

Find the width of a rectangle with perimeter 92 meters and length 19 meters.

Solution

27 meters

Find the width of a rectangle with perimeter 16.2 meters and length 3.2 meters.

The area of a rectangle is 414 square meters. The length is 18 meters. What is the width?

Solution

23 m

The area of a rectangle is 782 square centimeters. The width is 17 centimeters. What is the length?

The length of a rectangle is 9 inches more than the width. The perimeter is 46 inches. Find the length and the width.

Solution

7 in., 16 in.

The width of a rectangle is 8 inches more than the length. The perimeter is 52 inches. Find the length and the width.

The perimeter of a rectangle is 58 meters. The width of the rectangle is 5 meters less than the length. Find the length and the width of the rectangle.

Solution

17 m, 12 m

The perimeter of a rectangle is 62 feet. The width is 7 feet less than the length. Find the length and the 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 a rectangle is three times the width. The perimeter is 72 feet. Find the length and width of the rectangle.

The length of a rectangle is 3 meters less than twice the width. The perimeter is 36 meters. Find the length and width.

Solution

l = 11 m, w = 7 m

The length of a rectangle is 5 inches more than twice the width. The perimeter is 34 inches. Find the length and width.

The width of a rectangular window is 24 inches. The area is 624 square inches. What is the length?

Solution

26 in.

The length of a rectangular poster is 28 inches. The area is 1316 square inches. What is the width?

The area of a rectangular roof is 2310 square meters. The length is 42 meters. What is the width?

Solution

55 m

The area of a rectangular tarp is 132 square feet. The width is 12 feet. What is the length?

The perimeter of a rectangular courtyard is 160 feet. The length is 10 feet more than the width. Find the length and the width.

Solution

35 ft, 45 ft

The perimeter of a rectangular painting is 306 centimeters. The length is 17 centimeters more than the width. Find the length and the width.

The width of a rectangular window is 40 inches less than the height. The perimeter of the doorway is 224 inches. Find the length and the width.

Solution

76 in., 36 in.

The width of a rectangular playground is 7 meters less than the length. The perimeter of the playground is 46 meters. Find the length and the width.

Use the Properties of Triangles

In the following exercises, solve using the properties of triangles.

Find the area of a triangle with base 12 inches and height 5 inches.

Solution

30 sq. in.

Find the area of a triangle with base 45 centimeters and height 30 centimeters.

Find the area of a triangle with base 8.3 meters and height 6.1 meters.

Solution

25.315 sq. m

Find the area of a triangle with base 24.2 feet and height 20.5 feet.

A triangular flag has base of 1 foot and height of 1.5 feet. What is its area?

Solution

0.75 sq. ft

A triangular window has base of 8 feet and height of 6 feet. What is its area?

If a triangle has sides of 6 feet and 9 feet and the perimeter is 23 feet, how long is the third side?

Solution

8 ft

If a triangle has sides of 14 centimeters and 18 centimeters and the perimeter is 49 centimeters, how long is the third side?

What is the base of a triangle with an area of 207 square inches and height of 18 inches?

Solution

23 in.

What is the height of a triangle with an area of 893 square inches and base of 38 inches?

The perimeter of a triangular reflecting pool is 36 yards. The lengths of two sides are 10 yards and 15 yards. How long is the third side?

Solution

11 yd

A triangular courtyard has perimeter of 120 meters. The lengths of two sides are 30 meters and 50 meters. How long is the third side?

An isosceles triangle has a base of 20 centimeters. If the perimeter is 76 centimeters, find the length of each of the other sides.

Solution

28 cm

An isosceles triangle has a base of 25 inches. If the perimeter is 95 inches, find the length of each of the other sides.

Find the length of each side of an equilateral triangle with a perimeter of 51 yards.

Solution

17 yd

Find the length of each side of an equilateral triangle with a perimeter of 54 meters.

The perimeter of an equilateral triangle is 18 meters. Find the length of each side.

Solution

6 m

The perimeter of an equilateral triangle is 42 miles. Find the length of each side.

The perimeter of an isosceles triangle is 42 feet. The length of the shortest side is 12 feet. Find the length of the other two sides.

Solution

15 ft

The perimeter of an isosceles triangle is 83 inches. The length of the shortest side is 24 inches. Find the length of the other two sides.

A dish is in the shape of an equilateral triangle. Each side is 8 inches long. Find the perimeter.

Solution

24 in.

A floor tile is in the shape of an equilateral triangle. Each side is 1.5 feet long. Find the perimeter.

A road sign in the shape of an isosceles triangle has a base of 36 inches. If the perimeter is 91 inches, find the length of each of the other sides.

Solution

27.5 in.

A scarf in the shape of an isosceles triangle has a base of 0.75 meters. If the perimeter is 2 meters, find the length of each of the other sides.

The perimeter of a triangle is 39 feet. One side of the triangle is 1 foot longer than the second side. The third side is 2 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 5 feet longer than the second side. The third side is 3 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 5 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 3 feet more than the shortest side. The perimeter is 13 feet. Find the lengths of all three sides.

Use the Properties of Trapezoids

In the following exercises, solve using the properties of trapezoids.

The height of a trapezoid is 12 feet and the bases are 9 and 15 feet. What is the area?

Solution

144 sq. ft

The height of a trapezoid is 24 yards and the bases are 18 and 30 yards. What is the area?

Find the area of a trapezoid with a height of 51 meters and bases of 43 and 67 meters.

Solution

2805 sq. m

Find the area of a trapezoid with a height of 62 inches and bases of 58 and 75 inches.

The height of a trapezoid is 15 centimeters and the bases are 12.5 and 18.3 centimeters. What is the area?

Solution

231 sq. cm

The height of a trapezoid is 48 feet and the bases are 38.6 and 60.2 feet. What is the area?

Find the area of a trapezoid with a height of 4.2 meters and bases of 8.1 and 5.5 meters.

Solution

28.56 sq. m

Find the area of a trapezoid with a height of 32.5 centimeters and bases of 54.6 and 41.4 centimeters.

Laurel is making a banner shaped like a trapezoid. The height of the banner is 3 feet and the bases are 4 and 5 feet. What is the area of the banner?

Solution

13.5 sq. ft

Niko wants to tile the floor of his bathroom. The floor is shaped like a trapezoid with width 5 feet and lengths 5 feet and 8 feet. What is the area of the floor?

Theresa needs a new top for her kitchen counter. The counter is shaped like a trapezoid with width 18.5 inches and lengths 62 and 50 inches. What is the area of the counter?

Solution

1036 sq. in.

Elena is knitting a scarf. The scarf will be shaped like a trapezoid with width 8 inches and lengths 48.2 inches and 56.2 inches. What is the area of the scarf?

Everyday Math

Fence Jose just removed the children’s playset from his back yard to make room for a rectangular garden. He wants to put a fence around the garden to keep out the dog. 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 if he wants to use the entire roll of fence?

Solution

15 ft

Gardening Lupita wants to fence in her tomato garden. The garden is rectangular and the length is twice the width. It will take 48 feet of fencing to enclose the garden. Find the length and width of her garden.

Fence Christa wants to put a fence around her triangular flowerbed. The sides of the flowerbed are 6 feet, 8 feet, and 10 feet. The fence costs $10 per foot. How much will it cost for Christa to fence in her flowerbed?

Solution

$240

Painting Caleb wants to paint one wall of his attic. The wall is shaped like a trapezoid with height 8 feet and bases 20 feet and 12 feet. The cost of the painting one square foot of wall is about $0.05. About how much will it cost for Caleb to paint the attic wall?

A right trapezoid is shown.

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.

A rectangle is shown on the left. It is labeled as 2 by 8. A square is shown on the right. It is labeled as 4 by 4.

ⓐ 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?

Solution

Answers will vary.

The length of a rectangle is 5 feet more than the width. The area is 50 square feet. Find the length and the width.

ⓐ Write the equation you would use to solve the problem.

ⓑ Why can’t you solve this equation with the methods you learned in the previous chapter?

Self Check

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

A self-assessment table with geometry skills. The rows list skills like understanding measures and using properties of rectangles, triangles, and trapezoids. Columns are for 'Confidently', 'With some help', and '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?

area
The area is a measure of the surface covered by a figure.
equilateral triangle
A triangle with all three sides of equal length is called an equilateral triangle.
isosceles triangle
A triangle with two sides of equal length is called an isosceles triangle.
perimeter
The perimeter is a measure of the distance around a figure.
rectangle
A rectangle is a geometric figure that has four sides and four right angles.
trapezoid
A trapezoid is four-sided figure, a quadrilateral, with two sides that are parallel and two sides that are not.

Solve Geometry Applications: Circles and Irregular Figures

Learning Objectives

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

  • Use the properties of circles
  • Find the area of irregular figures

Before you get started, take this readiness quiz.

Evaluate x2 when x=5.
If you missed this problem, review Example 3 in Evaluate, Simplify, and Translate Expressions.

Solution

25

Using 3.14 for π, approximate the (a) circumference and (b) the area of a circle with radius 8 inches.
If you missed this problem, review Example 12 in Decimals and Fractions.

Solution

(a) 50.24in.; (b) 200.96sq.in.

Simplify 227(0.25)2 and round to the nearest thousandth.
If you missed this problem, review Example 9 in Decimals and Fractions.

Solution

0.196

In this section, we’ll continue working with geometry applications. We will add several new formulas to our collection of formulas. To help you as you do the examples and exercises in this section, we will show the Problem Solving Strategy for Geometry Applications here.

Problem Solving Strategy for Geometry Applications
  1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
  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 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.

Use the Properties of Circles

Do you remember the properties of circles from Decimals and Fractions Together? We’ll show them here again to refer to as we use them to solve applications.

Properties of Circles

An image of a circle is shown. There is a line drawn through the widest part at the center of the circle with a red dot indicating the center of the circle. The line is labeled d. The two segments from the center of the circle to the outside of the circle are each labeled r.
  • r is the length of the radius
  • d is the length of the diameter
  • d=2r
  • Circumference is the perimeter of a circle. The formula for circumference is
    C=2πr
  • The formula for area of a circle is
    A=πr2

Remember, that we approximate π with 3.14 or 227 depending on whether the radius of the circle is given as a decimal or a fraction. If you use the π key on your calculator to do the calculations in this section, your answers will be slightly different from the answers shown. That is because the π key uses more than two decimal places.

A circular sandbox has a radius of 2.5 feet. Find the ⓐ circumference and ⓑ area of the sandbox.

Solution

Solution

ⓐ
Step 1. Read the problem. Draw the figure and label it with the given information.
An image showing a black outline of a circle. A horizontal line from the center to the edge is labeled 'r = 2.5 ft', indicating the circle's radius is 2.5 feet.
Step 2. Identify what you are looking for. the circumference of the circle
Step 3. Name. Choose a variable to represent it. Let c = circumference of the circle
Step 4. Translate.
Write the appropriate formula
Substitute

C=2πr
C=2π(2.5)
Step 5. Solve the equation. C≈2(3.14)(2.5)
C≈15ft
Step 6. Check. Does this answer make sense?
Yes. If we draw a square around the circle, its sides would be 5 ft (twice the radius), so its perimeter would be 20 ft. This is slightly more than the circle's circumference, 15.7 ft.
A geometric diagram showing a circle with a radius of 2.5 ft perfectly inscribed within a square with 5 ft sides.
Step 7. Answer the question. The circumference of the sandbox is 15.7 feet.
ⓑ
Step 1. Read the problem. Draw the figure and label it with the given information.
An image showing a black outline of a circle. A horizontal line from the center to the edge is labeled 'r = 2.5 ft', indicating the circle's radius is 2.5 feet.
Step 2. Identify what you are looking for. the area of the circle
Step 3. Name. Choose a variable to represent it. Let A = the area of the circle
Step 4. Translate.
Write the appropriate formula
Substitute

A=πr2
A=π(2.5)2
Step 5. Solve the equation. A≈(3.14)(2.5)2
A≈19.625sq. ft
Step 6. Check.
Yes. If we draw a square around the circle, its sides would be 5 ft, as shown in part ⓐ. So the area of the square would be 25 sq. ft. This is slightly more than the circle's area, 19.625 sq. ft.
Step 7. Answer the question. The area of the circle is 19.625 square feet.


A circular mirror has radius of 5 inches. Find the ⓐ circumference and ⓑ area of the mirror.

Solution
  1. ⓐ 31.4 in.
  2. ⓑ 78.5 sq. in.

A circular spa has radius of 4.5 feet. Find the ⓐ circumference and ⓑ area of the spa.

Solution
  1. ⓐ 28.26 ft
  2. ⓑ 63.585 sq. ft

We usually see the formula for circumference in terms of the radius r of the circle:

C=2πr

But since the diameter of a circle is two times the radius, we could write the formula for the circumference in terms ofd.

C=2πrUsing the commutative property, we getC=π·2rThen substitutingd=2rC=π·dSoC=πd

We will use this form of the circumference when we’re given the length of the diameter instead of the radius.

A circular table has a diameter of four feet. What is the circumference of the table?

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A circular wooden surface or object with a diameter labeled as 4 feet.
Step 2. Identify what you are looking for. the circumference of the table
Step 3. Name. Choose a variable to represent it. Let c = the circumference of the table
Step 4. Translate.
Write the appropriate formula for the situation.
Substitute.

C=πd
C=π(4)
Step 5. Solve the equation, using 3.14 for π. C≈(3.14)(4)
C≈12.56feet
Step 6. Check: If we put a square around the circle, its side would be 4.
The perimeter would be 16. It makes sense that the circumference of the circle, 12.56, is a little less than 16.
A circle with a diameter of 4 ft is inscribed within a square, where each side of the square also measures 4 ft.
Step 7. Answer the question. The diameter of the table is 12.56 feet.

Find the circumference of a circular fire pit whose diameter is 5.5 feet.

Solution

17.27 ft

If the diameter of a circular trampoline is 12 feet, what is its circumference?

Solution

37.68 ft

Find the diameter of a circle with a circumference of 47.1 centimeters.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. A circle is shown with its diameter 'd' and its circumference 'C = 47.1 cm' labeled, representing a geometry problem to calculate the diameter from the given circumference.
Step 2. Identify what you are looking for. the diameter of the circle
Step 3. Name. Choose a variable to represent it. Let d = the diameter of the circle
Step 4. Translate.
Write the formula.
Substitute, using 3.14 to approximate π.
A mathematical formula is displayed on a white background: C = πd. This formula represents the circumference of a circle (C) as the product of pi (π) and its diameter (d).
A mathematical equation shows '47.1 ≈ 3.14d' on a white background, representing an approximation where 47.1 is approximately equal to 3.14 times d.
Step 5. Solve. A mathematical expression shows the fraction 47.1 over 3.14 approximately equal to the fraction 3.14d over 3.14.
The mathematical expression '15 is approximately equal to d' is displayed on a white background, signifying an estimated value or approximation in an equation.
Step 6. Check:
The mathematical formula for the circumference of a circle, C = πd, where C is the circumference, π (pi) is a mathematical constant approximately equal to 3.14159, and d is the diameter of the circle.
47.1=?(3.14)(15)
47.1=47.1✓
Step 7. Answer the question. The diameter of the circle is approximately 15 centimeters.

Find the diameter of a circle with circumference of 94.2 centimeters.

Solution

30 cm

Find the diameter of a circle with circumference of 345.4 feet.

Solution

110 ft

Find the Area of Irregular Figures

So far, we have found area for rectangles, triangles, trapezoids, and circles. An irregular figure is a figure that is not a standard geometric shape. Its area cannot be calculated using any of the standard area formulas. But some irregular figures are made up of two or more standard geometric shapes. To find the area of one of these irregular figures, we can split it into figures whose formulas we know and then add the areas of the figures.

Find the area of the shaded region.

An image of an attached horizontal rectangle and a vertical rectangle is shown. The top is labeled 12, the side of the horizontal rectangle is labeled 4. The side is labeled 10, the width of the vertical rectangle is labeled 2.
Solution

Solution

The given figure is irregular, but we can break it into two rectangles. The area of the shaded region will be the sum of the areas of both rectangles.

An image of an attached horizontal rectangle and a vertical rectangle is shown. The top is labeled 12, the side of the horizontal rectangle is labeled 4. The side is labeled 10, the width of the vertical rectangle is labeled 2.

The blue rectangle has a width of 12 and a length of 4. The red rectangle has a width of 2, but its length is not labeled. The right side of the figure is the length of the red rectangle plus the length of the blue rectangle. Since the right side of the blue rectangle is 4 units long, the length of the red rectangle must be 6 units.

An image of a blue horizontal rectangle attached to a red vertical rectangle is shown. The top is labeled 12, the side of the blue rectangle is labeled 4. The whole side is labeled 10, the blue portion is labeled 4 and the red portion is labeled 6. The width of the red rectangle is labeled 2. The first line says A sub figure equals A sub rectangle plus A sub red rectangle. Below this is A sub figure equals bh plus red bh. Below this is A sub figure equals 12 times 4 plus red 2 times 6. Below this is A sub figure equals 48 plus red 12. Below this is A sub figure equals 60.

The area of the figure is 60 square units.

Is there another way to split this figure into two rectangles? Try it, and make sure you get the same area.

Find the area of each shaded region:

A blue geometric shape is shown. It looks like a horizontal rectangle attached to a vertical rectangle. The top is labeled as 8, the width of the horizontal rectangle is labeled as 2. The side is labeled as 6, the width of the vertical rectangle is labeled as 3.
Solution

28 sq. units

Find the area of each shaded region:

A blue geometric shape is shown. It looks like a horizontal rectangle attached to a vertical rectangle. The top is labeled as 14, the width of the horizontal rectangle is labeled as 5. The side is labeled as 10, the width of the missing space is labeled as 6.
Solution

110 sq. units

Find the area of the shaded region.

A blue geometric shape is shown. It looks like a rectangle with a triangle attached to the top on the right side. The left side is labeled 4, the top 5, the bottom 8, the right side 7.
Solution

Solution

We can break this irregular figure into a triangle and rectangle. The area of the figure will be the sum of the areas of triangle and rectangle.

The rectangle has a length of 8 units and a width of 4 units.

We need to find the base and height of the triangle.

Since both sides of the rectangle are 4, the vertical side of the triangle is 3, which is 7−4.

The length of the rectangle is 8, so the base of the triangle will be 3, which is 8−5.

A geometric shape is shown. It is a blue rectangle with a red triangle attached to the top on the right side. The left side is labeled 4, the top 5, the bottom 8, the right side 7. The right side of the rectangle is labeled 4. The right side and bottom of the triangle are labeled 3.

Now we can add the areas to find the area of the irregular figure.
The top line reads A sub figure equals A sub rectangle plus A sub red triangle. The second line reads A sub figure equals lw plus one-half red bh. The next line says A sub figure equals 8 times 4 plus one-half times red 3 times red 3. The next line reads A sub figure equals 32 plus red 4.5. The last line says A sub figure equals 36.5 sq. units.

The area of the figure is 36.5 square units.

Find the area of each shaded region.

A blue geometric shape is shown. It looks like a rectangle with a triangle attached to the lower right side. The base of the rectangle is labeled 8, the height of the rectangle is labeled 4. The distance from the top of the rectangle to where the triangle begins is labeled 3, the top of the triangle is labeled 3.
Solution

36.5 sq. units

Find the area of each shaded region.

A blue geometric shape is shown. It looks like a rectangle with an equilateral triangle attached to the top. The base of the rectangle is labeled 12, each side is labeled 5. The base of the triangle is split into two pieces, each labeled 2.5.
Solution

70 sq. units

A high school track is shaped like a rectangle with a semi-circle (half a circle) on each end. The rectangle has length 105 meters and width 68 meters. Find the area enclosed by the track. Round your answer to the nearest hundredth.

A track is shown, shaped like a rectangle with a semi-circle attached to each side.
Solution

Solution

We will break the figure into a rectangle and two semi-circles. The area of the figure will be the sum of the areas of the rectangle and the semicircles.

A blue geometric shape is shown. It looks like a rectangle with a semi-circle attached to each side. The base of the rectangle is labeled 105 m. The height of the rectangle and diameter of the circle on the left is labeled 68 m.

The rectangle has a length of 105 m and a width of 68 m. The semi-circles have a diameter of 68 m, so each has a radius of 34 m.
The top line reads A sub figure equals A sub rectangle plus A sub semicircles. The second line reads A sub figure equals bh plus red 2 times (in parentheses) red 1/2pi times r squared. The next line says A sub figure approximately equals 105 times 68 plus red 2 times (in parentheses) red 1/2 times 3.14 times 34 squared. The next line reads A sub figure approximately equals 7140 plus red 3629.84. The last line says A sub figure approximately equals 10,769.84 square meters.

Find the area:

A shape is shown. It is a blue rectangle with a portion of the rectangle missing. There is a red circle the same height as the rectangle attached to the missing side of the rectangle. The top of the rectangle is labeled 15, the height is labeled 9.
Solution

103.2 sq. units

Find the area:

A blue geometric shape is shown. It appears to be two trapezoids with a semicircle at the top. The base of the semicircle is labeled 5.2. The height of the trapezoids is labeled 6.5. The combined base of the trapezoids is labeled 3.3.
Solution

38.24 sq. units

ACCESS ADDITIONAL ONLINE RESOURCES

  • Circumference of a Circle
  • Area of a Circle
  • Area of an L-shaped polygon
  • Area of an L-shaped polygon with Decimals
  • Perimeter Involving a Rectangle and Circle
  • Area Involving a Rectangle and Circle

Key Concepts

  • Problem Solving Strategy for Geometry Applications
    1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
    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 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.
  • Properties of Circles

    A circle with its diameter (d) and two radii (r) labeled, illustrating that the diameter is twice the radius.

    • d=2r
    • Circumference: C=2πr or C=πd
    • Area: A=πr2

Practice Makes Perfect

Use the Properties of Circles

In the following exercises, solve using the properties of circles.

The lid of a paint bucket is a circle with radius 7 inches. Find the ⓐ circumference and ⓑ area of the lid.

Solution
  1. ⓐ 43.96 in.
  2. ⓑ 153.86 sq. in.

An extra-large pizza is a circle with radius 8 inches. Find the ⓐ circumference and ⓑ area of the pizza.

A farm sprinkler spreads water in a circle with radius of 8.5 feet. Find the ⓐ circumference and ⓑ area of the watered circle.

Solution
  1. ⓐ 53.38 ft
  2. ⓑ 226.865 sq. ft

A circular rug has radius of 3.5 feet. Find the ⓐ circumference and ⓑ area of the rug.

A reflecting pool is in the shape of a circle with diameter of 20 feet. What is the circumference of the pool?

Solution

62.8 ft

A turntable is a circle with diameter of 10 inches. What is the circumference of the turntable?

A circular saw has a diameter of 12 inches. What is the circumference of the saw?

Solution

37.68 in.

A round coin has a diameter of 3 centimeters. What is the circumference of the coin?

A barbecue grill is a circle with a diameter of 2.2 feet. What is the circumference of the grill?

Solution

6.908 ft

The top of a pie tin is a circle with a diameter of 9.5 inches. What is the circumference of the top?

A circle has a circumference of 163.28 inches. Find the diameter.

Solution

52 in.

A circle has a circumference of 59.66 feet. Find the diameter.

A circle has a circumference of 17.27 meters. Find the diameter.

Solution

5.5 m

A circle has a circumference of 80.07 centimeters. Find the diameter.

In the following exercises, find the radius of the circle with given circumference.

A circle has a circumference of 150.72 feet.

Solution

24 ft

A circle has a circumference of 251.2 centimeters.

A circle has a circumference of 40.82 miles.

Solution

6.5 mi

A circle has a circumference of 78.5 inches.

Find the Area of Irregular Figures

In the following exercises, find the area of the irregular figure. Round your answers to the nearest hundredth.

A geometric shape is shown. It is a horizontal rectangle attached to a vertical rectangle. The top is labeled 6, the height of the horizontal rectangle is labeled 2, the distance from the edge of the horizontal rectangle to the start of the vertical rectangle is 4, the base of the vertical rectangle is 2, the right side of the shape is 4.
Solution

16 sq. units

A geometric shape is shown. It is an L-shape. The base is labeled 10, the right side 1, the top and left side are each labeled 4.
A geometric shape is shown. It is a sideways U-shape. The top is labeled 6, the left side is labeled 6. An inside horizontal piece is labeled 3. Each of the vertical pieces on the right are labeled 2.
Solution

30 sq. units

A geometric shape is shown. It is a U-shape. The base is labeled 7. The right side is labeled 5. The two horizontal lines at the top and the vertical line on the inside are all labeled 3.
A geometric shape is shown. It is a rectangle with a triangle attached to the bottom left side. The top is labeled 4. The right side is labeled 10. The base is labeled 9. The vertical line from the top of the triangle to the top of the rectangle is labeled 3.
Solution

57.5 sq. units

A trapezoid is shown. The bases are labeled 5 and 10, the height is 5.
Two triangles are shown. They appear to be right triangles. The bases are labeled 3, the heights 4, and the longest sides 5.
Solution

12 sq. units

A geometric shape is shown. It appears to be composed of two triangles. The shared base of both triangles is 8, the heights are both labeled 6.
A geometric shape is shown. It is composed of two trapezoids. The base is labeled 10. The height of one trapezoid is 2. The horizontal and vertical sides are all labeled 5.
Solution

67.5 sq. units

A geometric shape is shown. It is a trapezoid attached to a triangle. The base of the triangle is labeled 6, the height is labeled 5. The height of the trapezoid is 6, one base is 3.
A geometric shape is shown. It is a rectangle with a triangle and another rectangle attached. The left side is labeled 8, the bottom is 8, the right side is 13, and the width of the smaller rectangle is 2.
Solution

89 sq. units

A geometric shape is shown. It is a rectangle with a triangle and another rectangle attached. The left side is labeled 12, the right side 7, the base 6. The width of the smaller rectangle is labeled 1.
A geometric shape is shown. It is a rectangle attached to a semi-circle. The base of the rectangle is labeled 5, the height is 7.
Solution

44.81 sq. units

A geometric shape is shown. It is a rectangle attached to a semi-circle. The base of the rectangle is labeled 10, the height is 6. The portion of the rectangle on the left of the semi-circle is labeled 5, the portion on the right is labeled 2.
A geometric shape is shown. A triangle is attached to a semi-circle. The base of the triangle is labeled 4. The height of the triangle and the diameter of the circle are 8.
Solution

41.12 sq. units

A geometric shape is shown. A triangle is attached to a semi-circle. The height of the triangle is labeled 4. The base of the triangle, also the diameter of the semi-circle, is labeled 4.
A geometric shape is shown. It is a rectangle attached to a semi-circle. The base of the rectangle is labeled 5, the height is 7.
Solution

35.13 sq. units

A geometric shape is shown. A trapezoid is shown with a semi-circle attached to the top. The diameter of the circle, which is also the top of the trapezoid, is labeled 8. The height of the trapezoid is 6. The bottom of the trapezoid is 13.
A geometric shape is shown. It is a rectangle with a triangle attached to the top on the left side and a circle attached to the top right corner. The diameter of the circle is labeled 5. The height of the triangle is labeled 5, the base is labeled 4. The height of the rectangle is labeled 6, the base 11.
Solution

95.625 sq. units

A geometric shape is shown. It is a trapezoid with a triangle attached to the top, and a circle attached to the triangle. The diameter of the circle is 4. The height of the triangle is 5, the base of the triangle, which is also the top of the trapezoid, is 6. The bottom of the trapezoid is 9. The height of the trapezoid is 7.

In the following exercises, solve.

A city park covers one block plus parts of four more blocks, as shown. The block is a square with sides 250 feet long, and the triangles are isosceles right triangles. Find the area of the park.

A square is shown with four triangles coming off each side.
Solution

187,500 sq. ft

A gift box will be made from a rectangular piece of cardboard measuring 12 inches by 20 inches, with squares cut out of the corners of the sides, as shown. The sides of the squares are 3 inches. Find the area of the cardboard after the corners are cut out.

A rectangle is shown. Each corner has a gray shaded square. There are dotted lines drawn across the side of each square attached to the next square.

Perry needs to put in a new lawn. His lot is a rectangle with a length of 120 feet and a width of 100 feet. The house is rectangular and measures 50 feet by 40 feet. His driveway is rectangular and measures 20 feet by 30 feet, as shown. Find the area of Perry’s lawn.

A rectangular lot is shown. In it is a home shaped like a rectangle attached to a rectangular driveway.
Solution

9400 sq. ft

Denise is planning to put a deck in her back yard. The deck will be a 20-ft by 12-ft rectangle with a semicircle of diameter 6 feet, as shown below. Find the area of the deck.

A picture of a deck is shown. It is shaped like a rectangle with a semi-circle attached to the top on the left side.

Everyday Math

Area of a Tabletop Yuki bought a drop-leaf kitchen table. The rectangular part of the table is a 1-ft by 3-ft rectangle with a semicircle at each end, as shown. ⓐ Find the area of the table with one leaf up. ⓑ Find the area of the table with both leaves up.

An image of a table is shown. There is a rectangular portion attached to a semi-circular portion. There is another semi-circular leaf folded down on the other side of the rectangle.
Solution
  1. ⓐ 6.5325 sq. ft
  2. ⓑ 10.065 sq. ft

Painting Leora wants to paint the nursery in her house. The nursery is an 8-ft by 10-ft rectangle, and the ceiling is 8 feet tall. There is a 3-ft by 6.5-ft door on one wall, a 3-ft by 6.5-ft closet door on another wall, and one 4-ft by 3.5-ft window on the third wall. The fourth wall has no doors or windows. If she will only paint the four walls, and not the ceiling or doors, how many square feet will she need to paint?

Writing Exercises

Describe two different ways to find the area of this figure, and then show your work to make sure both ways give the same area.

A geometric shape is shown. It is a vertical rectangle attached to a horizontal rectangle. The width of the vertical rectangle is 3, the left side is labeled 6, the bottom is labeled 9, and the width of the horizontal rectangle is labeled 3. The top of the horizontal rectangle is labeled 6, and the distance from the top of that rectangle to the top of the other rectangle is labeled 3.
Solution

Answers will vary.

A circle has a diameter of 14 feet. Find the area of the circle ⓐ using 3.14 forπ ⓑ using 227 for π. ⓒ Which calculation to do prefer? Why?

Self Check

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

A self-assessment table for students to rate their understanding of using properties of circles and finding areas of irregular figures. Options are 'Confidently', 'With some help', and '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?

irregular figure
An irregular figure is a figure that is not a standard geometric shape. Its area cannot be calculated using any of the standard area formulas.

Solve Geometry Applications: Volume and Surface Area

Learning Objectives

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

  • Find volume and surface area of rectangular solids
  • Find volume and surface area of spheres
  • Find volume and surface area of cylinders
  • Find volume of cones

Before you get started, take this readiness quiz.

Evaluate x3 when x=5.
If you missed this problem, review Example 3 in Evaluate, Simplify, and Translate Expressions.

Solution

125

Evaluate 2x when x=5.
If you missed this problem, review Example 4 in Evaluate, Simplify, and Translate Expressions.

Solution

32

Find the area of a circle with radius 72.
If you missed this problem, review Example 12 in Decimals and Fractions.

Solution

772

In this section, we will finish our study of geometry applications. We find the volume and surface area of some three-dimensional figures. Since we will be solving applications, we will once again show our Problem-Solving Strategy for Geometry Applications.

    Problem Solving Strategy for Geometry Applications

  1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
  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 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.

Find Volume and Surface Area of Rectangular Solids

A cheerleading coach is having the squad paint wooden crates with the school colors to stand on at the games. (See Figure 1). The amount of paint needed to cover the outside of each box is the surface area, a square measure of the total area of all the sides. The amount of space inside the crate is the volume, a cubic measure.

This is an image of a wooden crate.
This wooden crate is in the shape of a rectangular solid.

Each crate is in the shape of a rectangular solid. Its dimensions are the length, width, and height. The rectangular solid shown in Figure 2 has length 4 units, width 2 units, and height 3 units. Can you tell how many cubic units there are altogether? Let’s look layer by layer.

A rectangular solid is shown. Each layer is composed of 8 cubes, measuring 2 by 4. The top layer is pink. The middle layer is orange. The bottom layer is green. Beside this is an image of the top layer that says “The top layer has 8 cubic units.” The orange layer is shown and says “The middle layer has 8 cubic units.” The green layer is shown and says, “The bottom layer has 8 cubic units.”
Breaking a rectangular solid into layers makes it easier to visualize the number of cubic units it contains. This 4 by 2 by 3 rectangular solid has 24 cubic units.

Altogether there are 24 cubic units. Notice that 24 is the length×width×height.

The top line says V equals L times W times H. Beneath the V is 24, beneath the equal sign is another equal sign, beneath the L is a 4, beneath the W is a 2, beneath the H is a 3.

The volume, V, of any rectangular solid is the product of the length, width, and height.

V=LWH

We could also write the formula for volume of a rectangular solid in terms of the area of the base. The area of the base, B, is equal to length×width.

B=L·W

We can substitute B for L·W in the volume formula to get another form of the volume formula.

The top line says V equals red L times red W times H. Below this is V equals red parentheses L times W times H. Below this is V equals red capital B times h.

We now have another version of the volume formula for rectangular solids. Let’s see how this works with the 4×2×3 rectangular solid we started with. See Figure 2.

An image of a rectangular solid is shown. It is made up of cubes. It is labeled as 2 by 4 by 3. Beside the solid is V equals Bh. Below this is V equals Base times height. Below Base is parentheses 4 times 2. The next line says V equals parentheses 4 times 2 times 3. Below that is V equals 8 times 3, then V equals 24 cubic units.

To find the surface area of a rectangular solid, think about finding the area of each of its faces. How many faces does the rectangular solid above have? You can see three of them.

Afront=L×WAside=L×WAtop=L×WAfront=4·3Aside=2·3Atop=4·2Afront=12Aside=6Atop=8

Notice for each of the three faces you see, there is an identical opposite face that does not show.

S=(front+back)+(left side+right side)+(top+bottom)S=(2·front)+(2·left side)+(2·top)S=2·12+2·6+2·8S=24+12+16S=52sq. units

The surface area S of the rectangular solid shown in Figure 3 is 52 square units.

In general, to find the surface area of a rectangular solid, remember that each face is a rectangle, so its area is the product of its two dimensions, either length and width, length and height, or width and height (see Figure 4). Find the area of each face that you see and then multiply each area by two to account for the face on the opposite side.

S=2LH+2LW+2WH
A rectangular solid is shown. The sides are labeled L, W, and H. One face is labeled LW and another is labeled WH.
For each face of the rectangular solid facing you, there is another face on the opposite side. There are 6 faces in all.

Volume and Surface Area of a Rectangular Solid

For a rectangular solid with length L, width W, and height H:

A rectangular solid is shown. The sides are labeled L, W, and H. Beside it is Volume: V equals LWH equals BH. Below that is Surface Area: S equals 2LH plus 2LW plus 2WH.
Doing the Manipulative Mathematics activity “Painted Cube” will help you develop a better understanding of volume and surface area.

For a rectangular solid with length 14 cm, height 17 cm, and width 9 cm, find the ⓐ volume and ⓑ surface area.

Solution

Solution

Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

Step 1. Read the problem. Draw the figure and
label it with the given information.
A cuboid shown in 3D perspective with dimensions: length 14, width 9, and height 17. Hidden edges are indicated by dashed lines.
This table outlines the systematic steps for solving a problem, exemplified by calculating the volume of a rectangular solid.
ⓐ
Step 2. Identify what you are looking for. the volume of the rectangular solid
Step 3. Name. Choose a variable to represent it. Let V= volume
Step 4. Translate.
Write the appropriate formula.
Substitute.

V=LWH
V=14⋅9⋅17
Step 5. Solve the equation. V=2,142
Step 6. Check
We leave it to you to check your calculations.
Step 7. Answer the question. The volume is 2,142 cubic centimeters.
Step-by-step guide detailing the process and results for calculating the surface area of a solid.
ⓑ
Step 2. Identify what you are looking for. the surface area of the solid
Step 3. Name. Choose a variable to represent it. Let S= surface area
Step 4. Translate.
Write the appropriate formula.
Substitute.

S=2LH+2LW+2WH
S=2(14⋅17)+2(14⋅9)+2(9⋅17)
Step 5. Solve the equation. S=1,034
Step 6. Check: Double-check with a calculator.
Step 7. Answer the question. The surface area is 1,034 square centimeters.

Find the ⓐ volume and ⓑ surface area of rectangular solid with the: length 8 feet, width 9 feet, and height 11 feet.

Solution
  1. ⓐ 792 cu. ft
  2. ⓑ 518 sq. ft

Find the ⓐ volume and ⓑ surface area of rectangular solid with the: length 15 feet, width 12 feet, and height 8 feet.

Solution
  1. ⓐ 1,440 cu. ft
  2. ⓑ 792 sq. ft

A rectangular crate has a length of 30 inches, width of 25 inches, and height of 20 inches. Find its ⓐ volume and ⓑ surface area.

Solution

Solution

Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

Step 1. Read the problem. Draw the figure and
label it with the given information.
A cuboid with dimensions 30 (length), 25 (width), and 20 (height) units. The front and top faces are visible, with dashed lines indicating hidden edges. It illustrates a basic 3D geometric shape with measurements.
Outlines a step-by-step method for solving problems, specifically demonstrating the calculation of a crate's volume from identification to the final solution.
ⓐ
Step 2. Identify what you are looking for. the volume of the crate
Step 3. Name. Choose a variable to represent it. let V= volume
Step 4. Translate.
Write the appropriate formula.
Substitute.

V=LWH
V=30⋅25⋅20
Step 5. Solve the equation. V=15,000
Step 6. Check: Double check your math.
Step 7. Answer the question. The volume is 15,000 cubic inches.
Step-by-step guide to solving a problem, exemplified by calculating the surface area of a crate.
ⓑ
Step 2. Identify what you are looking for. the surface area of the crate
Step 3. Name. Choose a variable to represent it. let S= surface area
Step 4. Translate.
Write the appropriate formula.
Substitute.

S=2LH+2LW+2WH
S=2(30⋅20)+2(30⋅25)+2(25⋅20)
Step 5. Solve the equation. S=3,700
Step 6. Check: Check it yourself!
Step 7. Answer the question. The surface area is 3,700 square inches.

A rectangular box has length 9 feet, width 4 feet, and height 6 feet. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 216 cu. ft
  2. ⓑ 228 sq. ft

A rectangular suitcase has length 22 inches, width 14 inches, and height 9 inches. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 2,772 cu. in.
  2. ⓑ 1,264 sq. in.

Volume and Surface Area of a Cube

A cube is a rectangular solid whose length, width, and height are equal. See Volume and Surface Area of a Cube, below. Substituting, s for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get:

V=LWHS=2LH+2LW+2WHV=s·s·sS=2s·s+2s·s+2s·sV=s3S=2s2+2s2+2s2S=6s2

So for a cube, the formulas for volume and surface area are V=s3 and S=6s2.

Volume and Surface Area of a Cube

For any cube with sides of length s,

An image of a cube is shown. Each side is labeled s. Beside this is Volume: V equals s cubed. Below that is Surface Area: S equals 6 times s squared.

A cube is 2.5 inches on each side. Find its ⓐ volume and ⓑ surface area.

Solution
Solution

Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

Step 1. Read the problem. Draw the figure and
label it with the given information.
A simple line drawing of a cube, with each visible side dimension labeled as 2.5, indicating equal length, width, and height. The cube is rendered in an isometric projection.
Demonstrates a seven-step problem-solving method, including identifying, naming variables, translating to a formula, solving, checking, and answering, using a cube's volume as an example.
ⓐ
Step 2. Identify what you are looking for. the volume of the cube
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate.
Write the appropriate formula.

V=s3
Step 5. Solve. Substitute and solve. V=(2.5)3
V=15.625
Step 6. Check: Check your work.
Step 7. Answer the question. The volume is 15.625 cubic inches.
Outlines a multi-step problem-solving method, demonstrated by calculating the surface area of a cube, including identifying, naming, translating, solving, and answering.
ⓑ
Step 2. Identify what you are looking for. the surface area of the cube
Step 3. Name. Choose a variable to represent it. let S = surface area
Step 4. Translate.
Write the appropriate formula.

S=6s2
Step 5. Solve. Substitute and solve. S=6⋅(2.5)2
S=37.5
Step 6. Check: The check is left to you.
Step 7. Answer the question. The surface area is 37.5 square inches.

For a cube with side 4.5 meters, find the ⓐ volume and ⓑ surface area of the cube.

Solution
  1. ⓐ 91.125 cu. m
  2. ⓑ 121.5 sq. m

For a cube with side 7.3 yards, find the ⓐ volume and ⓑ surface area of the cube.

Solution
  1. ⓐ 389.017 cu. yd.
  2. ⓑ 319.74 sq. yd.

A notepad cube measures 2 inches on each side. Find its ⓐ volume and ⓑ surface area.

Solution
Solution
Step 1. Read the problem. Draw the figure and
label it with the given information.
A wireframe drawing of a cube, indicating its dimensions are 2 units on each side. The three visible edges are labeled with the number '2', denoting its length, width, and height.
This table outlines the sequential steps for solving a mathematical problem, specifically demonstrating how to find the volume of a cube.
ⓐ
Step 2. Identify what you are looking for. the volume of the cube
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate.
Write the appropriate formula.

V=s3
Step 5. Solve the equation. V=23
V=8
Step 6. Check: Check that you did the calculations
correctly.
Step 7. Answer the question. The volume is 8 cubic inches.
A demonstration of problem-solving steps applied to calculate the surface area of a cube.
ⓑ
Step 2. Identify what you are looking for. the surface area of the cube
Step 3. Name. Choose a variable to represent it. let S = surface area
Step 4. Translate.
Write the appropriate formula.

S=6s2
Step 5. Solve the equation. S=6⋅22
S=24
Step 6. Check: The check is left to you.
Step 7. Answer the question. The surface area is 24 square inches.

A packing box is a cube measuring 4 feet on each side. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 64 cu. ft
  2. ⓑ 96 sq. ft

A wall is made up of cube-shaped bricks. Each cube is 16 inches on each side. Find the ⓐ volume and ⓑ surface area of each cube.

Solution
  1. ⓐ 4,096 cu. in.
  2. ⓑ 1536 sq. in.

Find the Volume and Surface Area of Spheres

A sphere is the shape of a basketball, like a three-dimensional circle. Just like a circle, the size of a sphere is determined by its radius, which is the distance from the center of the sphere to any point on its surface. The formulas for the volume and surface area of a sphere are given below.

Showing where these formulas come from, like we did for a rectangular solid, is beyond the scope of this course. We will approximate π with 3.14.

Volume and Surface Area of a Sphere

For a sphere with radius r:

An image of a sphere is shown. The radius is labeled r. Beside this is Volume: V equals four-thirds times pi times r cubed. Below that is Surface Area: S equals 4 times pi times r squared.

A sphere has a radius 6 inches. Find its ⓐ volume and ⓑ surface area.

Solution

Solution

Step 1 is the same for both ⓐ and ⓑ , so we will show it just once.

Step 1. Read the problem. Draw the figure and label
it with the given information.
A 3D illustration of a sphere, indicating a radius of 6 from its center to the perimeter. A dashed line represents the hidden part of the equatorial plane.
Step-by-step method for calculating the volume of a sphere.
ⓐ
Step 2. Identify what you are looking for. the volume of the sphere
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate.
Write the appropriate formula.

V=43πr3
Step 5. Solve. V≈43(3.14)63
V≈904.32cubic inches
Step 6. Check: Double-check your math on a calculator.
Step 7. Answer the question. The volume is approximately 904.32 cubic inches.
This table outlines the steps for solving a problem, specifically demonstrating how to calculate the surface area of a sphere with corresponding actions and results.
ⓑ
Step 2. Identify what you are looking for. the surface area of the sphere
Step 3. Name. Choose a variable to represent it. let S = surface area
Step 4. Translate.
Write the appropriate formula.

S=4πr2
Step 5. Solve. S≈4(3.14)62
S≈452.16
Step 6. Check: Double-check your math on a calculator
Step 7. Answer the question. The surface area is approximately 452.16 square inches.

Find the ⓐ volume and ⓑ surface area of a sphere with radius 3 centimeters.

Solution
  1. ⓐ 113.04 cu. cm
  2. ⓑ 113.04 sq. cm

Find the ⓐ volume and ⓑ surface area of each sphere with a radius of 1 foot

Solution
  1. ⓐ 4.19 cu. ft
  2. ⓑ 12.56 sq. ft

A globe of Earth is in the shape of a sphere with radius 14 inches. Find its ⓐ volume and ⓑ surface area. Round the answer to the nearest hundredth.

Solution

Solution

Step 1. Read the problem. Draw a figure with the
given information and label it.
A stylized globe showing North and South America, Europe, and Africa. A dashed line represents the equator, and a horizontal line segment with the number '14' indicates a measurement from the center.
This table illustrates a seven-step problem-solving process, from identifying the unknown to providing the final answer, exemplified by calculating the volume of a sphere.
ⓐ
Step 2. Identify what you are looking for. the volume of the sphere
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate.
Write the appropriate formula.
Substitute. (Use 3.14 for π)

V=43πr3
V≈43(3.14)143
Step 5. Solve. V≈11,488.21
Step 6. Check: We leave it to you to check your calculations.
Step 7. Answer the question. The volume is approximately 11,488.21 cubic inches.
A step-by-step guide demonstrating the calculation of the surface area of a sphere using a structured problem-solving approach.
ⓑ
Step 2. Identify what you are looking for. the surface area of the sphere
Step 3. Name. Choose a variable to represent it. let S = surface area
Step 4. Translate.
Write the appropriate formula.
Substitute. (Use 3.14 for π)

S=4πr2
S≈4(3.14)142
Step 5. Solve. S≈2461.76
Step 6. Check: We leave it to you to check your calculations.
Step 7. Answer the question. The surface area is approximately 2461.76 square inches.

A beach ball is in the shape of a sphere with radius of 9 inches. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 3052.08 cu. in.
  2. ⓑ 1017.36 sq. in.

A Roman statue depicts Atlas holding a globe with radius of 1.5 feet. Find the ⓐ volume and ⓑ surface area of the globe.

Solution
  1. ⓐ 14.13 cu. ft
  2. ⓑ 28.26 sq. ft

Find the Volume and Surface Area of a Cylinder

If you have ever seen a can of soda, you know what a cylinder looks like. A cylinder is a solid figure with two parallel circles of the same size at the top and bottom. The top and bottom of a cylinder are called the bases. The height h of a cylinder is the distance between the two bases. For all the cylinders we will work with here, the sides and the height, h , will be perpendicular to the bases.

An image of a cylinder is shown. There is a red arrow pointing to the radius of the top labeling it r, radius. There is a red arrow pointing to the height of the cylinder labeling it h, height.
A cylinder has two circular bases of equal size. The height is the distance between the bases.

Rectangular solids and cylinders are somewhat similar because they both have two bases and a height. The formula for the volume of a rectangular solid, V=Bh , can also be used to find the volume of a cylinder.

For the rectangular solid, the area of the base, B , is the area of the rectangular base, length × width. For a cylinder, the area of the base, B, is the area of its circular base, πr2. Figure 6 compares how the formula V=Bh is used for rectangular solids and cylinders.

In (a), a rectangular solid is shown. The sides are labeled L, W, and H. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses lw times h, then V equals lwh. In (b), a cylinder is shown. The radius of the top is labeled r, the height is labeled h. Below this is V equals capital Bh, then V equals Base times h, then V equals parentheses pi r squared times h, then V equals pi times r squared times h.
Seeing how a cylinder is similar to a rectangular solid may make it easier to understand the formula for the volume of a cylinder.

To understand the formula for the surface area of a cylinder, think of a can of vegetables. It has three surfaces: the top, the bottom, and the piece that forms the sides of the can. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle. See Figure 7.

A cylindrical can of green beans is shown. The height is labeled h. Beside this are pictures of circles for the top and bottom of the can and a rectangle for the other portion of the can. Above the circles is C equals 2 times pi times r. The top of the rectangle says l equals 2 times pi times r. The left side of the rectangle is labeled h, the right side is labeled w.
By cutting and unrolling the label of a can of vegetables, we can see that the surface of a cylinder is a rectangle. The length of the rectangle is the circumference of the cylinder’s base, and the width is the height of the cylinder.

The distance around the edge of the can is the circumference of the cylinder’s base it is also the length L of the rectangular label. The height of the cylinder is the width W of the rectangular label. So the area of the label can be represented as

The top line says A equals l times red w. Below the l is 2 times pi times r. Below the w is a red h.

To find the total surface area of the cylinder, we add the areas of the two circles to the area of the rectangle.

A rectangle is shown with circles coming off the top and bottom.

The surface area of a cylinder with radius r and height h, is

S=2πr2+2πrh

Volume and Surface Area of a Cylinder

For a cylinder with radius r and height h:

A cylinder is shown. The height is labeled h and the radius of the top is labeled r. Beside it is Volume: V equals pi times r squared times h or V equals capital B times h. Below this is Surface Area: S equals 2 times pi times r squared plus 2 times pi times r times h.

A cylinder has height 5 inches and radius 3 inches. Find the ⓐ volume and ⓑ surface area.

Solution

Solution

Step 1. Read the problem. Draw the figure and label
it with the given information.
A simple black and white line drawing of a cylinder with its dimensions labeled: a radius of 3 units and a height of 5 units.
Steps for calculating the volume of a cylinder, detailing each problem-solving action and its mathematical result.
ⓐ
Step 2. Identify what you are looking for. the volume of the cylinder
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate.
Write the appropriate formula.
Substitute. (Use 3.14 for π)

V=πr2h
V≈(3.14)32⋅5
Step 5. Solve. V≈141.3
Step 6. Check: We leave it to you to check your calculations.
Step 7. Answer the question. The volume is approximately 141.3 cubic inches.
Illustrates the step-by-step calculation of a cylinder's surface area, detailing problem-solving phases like identification, translation, and solution.
ⓑ
Step 2. Identify what you are looking for. the surface area of the cylinder
Step 3. Name. Choose a variable to represent it. let S = surface area
Step 4. Translate.
Write the appropriate formula.
Substitute. (Use 3.14 for π)

S=2πr2+2πrh
S≈2(3.14)32+2(3.14)(3)5
Step 5. Solve. S≈150.72
Step 6. Check: We leave it to you to check your calculations.
Step 7. Answer the question. The surface area is approximately 150.72 square inches.

Find the ⓐ volume and ⓑ surface area of the cylinder with radius 4 cm and height 7cm.

Solution
  1. ⓐ 351.68 cu. cm
  2. ⓑ 276.32 sq. cm

Find the ⓐ volume and ⓑ surface area of the cylinder with given radius 2 ft and height 8 ft.

Solution
  1. ⓐ 100.48 cu. ft
  2. ⓑ 125.6 sq. ft

Find the ⓐ volume and ⓑ surface area of a can of soda. The radius of the base is 4 centimeters and the height is 13 centimeters. Assume the can is shaped exactly like a cylinder.

Solution

Solution

Step 1. Read the problem. Draw the figure and
label it with the given information.
An illustration of a soda can with a blue and pink abstract design. The can's height is labeled as 13 and its diameter as 4, indicating its geometric dimensions.
This table outlines the systematic steps for calculating the volume of a cylinder, from identifying the unknown to providing the final answer.
ⓐ
Step 2. Identify what you are looking for. the volume of the cylinder
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate.
Write the appropriate formula.
Substitute. (Use 3.14 for π)

V=πr2h
V≈(3.14)42⋅13
Step 5. Solve. V≈653.12
Step 6. Check: We leave it to you to check.
Step 7. Answer the question. The volume is approximately 653.12 cubic centimeters.
This table outlines the systematic steps for calculating the surface area of a cylinder, from problem identification to the final solution.
ⓑ
Step 2. Identify what you are looking for. the surface area of the cylinder
Step 3. Name. Choose a variable to represent it. let S = surface area
Step 4. Translate.
Write the appropriate formula.
Substitute. (Use 3.14 for π)

S=2πr2+2πrh
S≈2(3.14)42+2(3.14)(4)13
Step 5. Solve. S≈427.04
Step 6. Check: We leave it to you to check your calculations.
Step 7. Answer the question. The surface area is approximately 427.04 square centimeters.

Find the ⓐ volume and ⓑ surface area of a can of paint with radius 8 centimeters and height 19 centimeters. Assume the can is shaped exactly like a cylinder.

Solution
  1. ⓐ 3,818.24 cu. cm
  2. ⓑ 1,356.48 sq. cm

Find the ⓐ volume and ⓑ surface area of a cylindrical drum with radius 2.7 feet and height 4 feet. Assume the drum is shaped exactly like a cylinder.

Solution
  1. ⓐ 91.5624 cu. ft
  2. ⓑ 113.6052 sq. ft

Find the Volume of Cones

The first image that many of us have when we hear the word ‘cone’ is an ice cream cone. There are many other applications of cones (but most are not as tasty as ice cream cones). In this section, we will see how to find the volume of a cone.

In geometry, a cone is a solid figure with one circular base and a vertex. The height of a cone is the distance between its base and the vertex.The cones that we will look at in this section will always have the height perpendicular to the base. See Figure 8.

An image of a cone is shown. The top is labeled vertex. The height is labeled h. The radius of the base is labeled r.
The height of a cone is the distance between its base and the vertex.

Earlier in this section, we saw that the volume of a cylinder is V=πr2h. We can think of a cone as part of a cylinder. Figure 9 shows a cone placed inside a cylinder with the same height and same base. If we compare the volume of the cone and the cylinder, we can see that the volume of the cone is less than that of the cylinder.

An image of a cone is shown. There is a cylinder drawn around it.
The volume of a cone is less than the volume of a cylinder with the same base and height.

In fact, the volume of a cone is exactly one-third of the volume of a cylinder with the same base and height. The volume of a cone is

The formula V equals one-third times capital B times h is shown.

Since the base of a cone is a circle, we can substitute the formula of area of a circle, πr2 , for B to get the formula for volume of a cone.

The formula V equals one-third times pi times r squared times h is shown.

In this book, we will only find the volume of a cone, and not its surface area.

Volume of a Cone

For a cone with radius r and height h.

An image of a cone is shown. The height is labeled h, the radius of the base is labeled r. Beside this is Volume: V equals one-third times pi times r squared times h.

Find the volume of a cone with height 6 inches and radius of its base 2 inches.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it
with the given information.
A 3D image of a cone with its height shown as 6 and its base radius as 2. The circular base is shaded in purple.
Step 2. Identify what you are looking for. the volume of the cone
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate.
Write the appropriate formula.
Substitute. (Use 3.14 for π)

V=13πr2h
V≈133.14(2)2(6)
Step 5. Solve. V≈25.12
Step 6. Check: We leave it to you to check your
calculations.
Step 7. Answer the question. The volume is approximately 25.12 cubic inches.

Find the volume of a cone with height 7 inches and radius 3 inches

Solution

65.94 cu. in.

Find the volume of a cone with height 9 centimeters and radius 5 centimeters

Solution

235.5 cu. cm

Marty’s favorite gastro pub serves french fries in a paper wrap shaped like a cone. What is the volume of a conic wrap that is 8 inches tall and 5 inches in diameter? Round the answer to the nearest hundredth.

Solution

Solution

Step 1. Read the problem. Draw the figure and label it with the given information. Notice here that the base is the circle at the top of the cone. An inverted cone is depicted with a radius of 5 units and a height of 8 units. The circular top is shaded to indicate its open face.
Step 2. Identify what you are looking for. the volume of the cone
Step 3. Name. Choose a variable to represent it. let V = volume
Step 4. Translate. Write the appropriate formula. Substitute. (Use 3.14 for π, and notice that we were given the distance across the circle, which is its diameter. The radius is 2.5 inches.)
V=13πr2h
V≈133.14(2.5)2(8)
Step 5. Solve. V≈52.33
Step 6. Check: We leave it to you to check your calculations.
Step 7. Answer the question. The volume of the wrap is approximately 52.33 cubic inches.

How many cubic inches of candy will fit in a cone-shaped piñata that is 18 inches long and 12 inches across its base? Round the answer to the nearest hundredth.

Solution

678.24 cu. in.

What is the volume of a cone-shaped party hat that is 10 inches tall and 7 inches across at the base? Round the answer to the nearest hundredth.

Solution

128.2 cu. in.

Summary of Geometry Formulas

The following charts summarize all of the formulas covered in this chapter.

A table is shown that summarizes all of the formulas in the chapter. The first cell is for Supplementary and Complementary Angles, and says that the measure of angle A plus the measure of angle B equals 180 degrees for supplementary angles A and B and the measure of angle C plus the measure of angle D equals 90 degrees for complementary angles C and D. There is an image of two angles A and B that together form a straight line and two angles C and D that together form a right angle. The next cell says Rectangular Solid and shows the formulas Volume equals LWH and Surface Area equals 2LH plus 2LW plus 2WH. An image of a rectangular solid with sides L, W, and H is shown. The next cell says Triangle. An image of a triangle is shown with sides a, b, and c, vertices A, B, and C, and height h. It says, “For triangle ABC, angle measures measure of angle A plus measure of angle B plus measure of angle C equal 180 degrees. Below this is Perimeter, P equals a plus b plus c. Below this is Area, A equals one-half bh. The next cell says Cube and shows an image of a cube with sides s. It says Volume V equals s cubed and Surface Area S equals 6 times s squared. The next cell says Similar Triangles. It shows two similar triangles ABC and XYZ. It says if triangle ABC is similar to triangle XYZ, then measure of angle A equals measure of angle X, measure of angle B equals measure of angle Y, and measure of angle C equals measure of angle Z. It then says a over x equals b over y equal c over z. The next cell says Sphere and shows an image of a sphere with radius r. It says volume V equals four-thirds times pi times r and Surface Area S equals 4 times pi times r squared. The next cell says Circle. There is an image with two radii labeled r and the diameter labeled d. It says Circumference C equals 2 pi times r and C equals pi times d. It says Area equals pi times r squared. The next cell says Cylinder and shows an image of a cylinder with height h and radius of the base r. It says Volume V equals pi times r squared times h. Below this is V equals Bh. Below that is Surface Area S equals 2 times pi times r squared plus 2 times pi times rh. The next cell says Rectangle and shows an image of a rectangle with sides W and L. It says Perimeter P equals 2L plus 2W, then Area A equals LW. The next cell says Cone and shows an image of a cone with height h and radius of the base r. It says Volume V equals one-third times pi times r squared times h. The last cell says Trapezoid and shows an image of a trapezoid with bases little b and capital B, and height h. It says Area A equals one-half times h times parentheses little b plus capital B. This image shows a row with three columns. The first column says Rectangular solid with the formula below that says volume: V equals LWH. Under this, it says Surface Area: S equals 2LH plus 2LW plus 2WH. An image shows an image of a rectangular solid with the sides labeled L , W and H. The middle column says Rectangle. Under this it says Perimeter P equals 2L plus 2W, then Area A equals LW.  An image of a rectangle with sides W and L. The right column says Cube. Under this it says “Volume: V equals s to the third power.” Under this is says “Surface area: S equals 6 times s squared. Below it is an image of a cube with three sides labeled “s”.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Volume of a Cone

Key Concepts

  • Volume and Surface Area of a Rectangular Solid
    • V=LWH
    • S=2LH+2LW+2WH
  • Volume and Surface Area of a Cube
    • V=s3
    • S=6s2
  • Volume and Surface Area of a Sphere
    • V=43πr3
    • S=4πr2
  • Volume and Surface Area of a Cylinder
    • V=πr2h
    • S=2πr2+2πrh
  • Volume of a Cone
    • For a cone with radius r and height h:
      Volume: V=13πr2h

Practice Makes Perfect

Find Volume and Surface Area of Rectangular Solids

In the following exercises, find ⓐ the volume and ⓑ the surface area of the rectangular solid with the given dimensions.

length 2 meters, width 1.5 meters, height 3 meters

Solution
  1. ⓐ 9 cu. m
  2. ⓑ 27 sq. m

length 5 feet, width 8 feet, height 2.5 feet

length 3.5 yards, width 2.1 yards, height 2.4 yards

Solution
  1. ⓐ 17.64 cu. yd.
  2. ⓑ 41.58 sq. yd.

length 8.8 centimeters, width 6.5 centimeters, height 4.2 centimeters

In the following exercises, solve.

Moving van A rectangular moving van has length 16 feet, width 8 feet, and height 8 feet. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 1,024 cu. ft
  2. ⓑ 640 sq. ft

Gift box A rectangular gift box has length 26 inches, width 16 inches, and height 4 inches. Find its ⓐ volume and ⓑ surface area.

Carton A rectangular carton has length 21.3 cm, width 24.2 cm, and height 6.5 cm. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 3,350.49 cu. cm
  2. ⓑ 1,622.42 sq. cm

Shipping container A rectangular shipping container has length 22.8 feet, width 8.5 feet, and height 8.2 feet. Find its ⓐ volume and ⓑ surface area.

In the following exercises, find ⓐ the volume and ⓑ the surface area of the cube with the given side length.

5 centimeters

Solution
  1. ⓐ 125 cu. cm
  2. ⓑ 150 sq. cm

6 inches

10.4 feet

Solution
  1. ⓐ 1124.864 cu. ft.
  2. ⓑ 648.96 sq. ft

12.5 meters

In the following exercises, solve.

Science center Each side of the cube at the Discovery Science Center in Santa Ana is 64 feet long. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 262,144 cu. ft
  2. ⓑ 24,576 sq. ft

Museum A cube-shaped museum has sides 45 meters long. Find its ⓐ volume and ⓑ surface area.

Base of statue The base of a statue is a cube with sides 2.8 meters long. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 21.952 cu. m
  2. ⓑ 47.04 sq. m

Tissue box A box of tissues is a cube with sides 4.5 inches long. Find its ⓐ volume and ⓑ surface area.

Find the Volume and Surface Area of Spheres

In the following exercises, find ⓐ the volume and ⓑ the surface area of the sphere with the given radius. Round answers to the nearest hundredth.

3 centimeters

Solution
  1. ⓐ 113.04 cu. cm
  2. ⓑ 113.04 sq. cm

9 inches

7.5 feet

Solution
  1. ⓐ 1,766.25 cu. ft
  2. ⓑ 706.5 sq. ft

2.1 yards

In the following exercises, solve. Round answers to the nearest hundredth.

Exercise ball An exercise ball has a radius of 15 inches. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 14,130 cu. in.
  2. ⓑ 2,826 sq. in.

Balloon ride The Great Park Balloon is a big orange sphere with a radius of 36 feet . Find its ⓐ volume and ⓑ surface area.

Golf ball A golf ball has a radius of 4.5 centimeters. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 381.51 cu. cm
  2. ⓑ 254.34 sq. cm

Baseball A baseball has a radius of 2.9 inches. Find its ⓐ volume and ⓑ surface area.

Find the Volume and Surface Area of a Cylinder

In the following exercises, find ⓐ the volume and ⓑ the surface area of the cylinder with the given radius and height. Round answers to the nearest hundredth.

radius 3 feet, height 9 feet

Solution
  1. ⓐ 254.34 cu. ft
  2. ⓑ 226.08 sq. ft

radius 5 centimeters, height 15 centimeters

radius 1.5 meters, height 4.2 meters

Solution
  1. ⓐ 29.673 cu. m
  2. ⓑ 53.694 sq. m

radius 1.3 yards, height 2.8 yards

In the following exercises, solve. Round answers to the nearest hundredth.

Coffee can A can of coffee has a radius of 5 cm and a height of 13 cm. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 1,020.5 cu. cm
  2. ⓑ 565.2 sq. cm

Snack pack A snack pack of cookies is shaped like a cylinder with radius 4 cm and height 3 cm. Find its ⓐ volume and ⓑ surface area.

Barber shop pole A cylindrical barber shop pole has a diameter of 6 inches and height of 24 inches. Find its ⓐ volume and ⓑ surface area.

Solution
  1. ⓐ 678.24 cu. in.
  2. ⓑ 508.68 sq. in.

Architecture A cylindrical column has a diameter of 8 feet and a height of 28 feet. Find its ⓐ volume and ⓑ surface area.

Find the Volume of Cones

In the following exercises, find the volume of the cone with the given dimensions. Round answers to the nearest hundredth.

height 9 feet and radius 2 feet

Solution

37.68 cu. ft

height 8 inches and radius 6 inches

height 12.4 centimeters and radius 5 cm

Solution

324.47 cu. cm

height 15.2 meters and radius 4 meters

In the following exercises, solve. Round answers to the nearest hundredth.

Teepee What is the volume of a cone-shaped teepee tent that is 10 feet tall and 10 feet across at the base?

Solution

261.67 cu. ft

Popcorn cup What is the volume of a cone-shaped popcorn cup that is 8 inches tall and 6 inches across at the base?

Silo What is the volume of a cone-shaped silo that is 50 feet tall and 70 feet across at the base?

Solution

64,108.33 cu. ft

Sand pile What is the volume of a cone-shaped pile of sand that is 12 meters tall and 30 meters across at the base?

Everyday Math

Street light post The post of a street light is shaped like a truncated cone, as shown in the picture below. It is a large cone minus a smaller top cone. The large cone is 30 feet tall with base radius 1 foot. The smaller cone is 10 feet tall with base radius of 0.5 feet. To the nearest tenth,

  1. ⓐ find the volume of the large cone.

  2. ⓑ find the volume of the small cone.

  3. ⓒ find the volume of the post by subtracting the volume of the small cone from the volume of the large cone.

    An image of a cone is shown. There is a dark dotted line at the top indicating a smaller cone.
Solution
  1. ⓐ 31.4 cu. ft
  2. ⓑ 2.6 cu. ft
  3. ⓒ 28.8 cu. ft

Ice cream cones A regular ice cream cone is 4 inches tall and has a diameter of 2.5 inches. A waffle cone is 7 inches tall and has a diameter of 3.25 inches. To the nearest hundredth,

  1. ⓐ find the volume of the regular ice cream cone.

  2. ⓑ find the volume of the waffle cone.

  3. ⓒ how much more ice cream fits in the waffle cone compared to the regular cone?

Writing Exercises

The formulas for the volume of a cylinder and a cone are similar. Explain how you can remember which formula goes with which shape.

Solution

Answers will vary.

Which has a larger volume, a cube of sides of 8 feet or a sphere with a diameter of 8 feet? Explain your reasoning.

Self Check

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

A self-assessment table for students to rate their understanding of finding volume and surface area for rectangular solids, spheres, cylinders, and volume of cones. The options are Confidently, With some help, and No-I don't get it!

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

cone
A cone is a solid figure with one circular base and a vertex.
cube
A cube is a rectangular solid whose length, width, and height are equal.
cylinder
A cylinder is a solid figure with two parallel circles of the same size at the top and bottom.

Solve a Formula for a Specific Variable

Learning Objectives

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

  • Use the distance, rate, and time formula
  • Solve a formula for a specific variable

Before you get started, take this readiness quiz.

Write 35 miles per gallon as a unit rate.
If you missed this problem, review Example 8 in Ratios and Rate.

Solution

35 miles1 gallon

Solve 6x+24=96.
If you missed this problem, review Example 1 in Solve Equations with Variables and Constants on Both Sides.

Solution

12

Find the simple interest earned after 5 years on $1,000 at an interest rate of 4%.
If you missed this problem, review Example 1 in Solve Simple Interest Applications.

Solution

$200

Use the Distance, Rate, and Time Formula

One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you travel if you drove at a steady rate of 60 miles per hour for 2 hours? (This might happen if you use your car’s cruise control while driving on the Interstate.) If you said 120 miles, you already know how to use this formula!

The math to calculate the distance might look like this:

distance=(60miles1hour)(2hours)distance=120miles

In general, the formula relating distance, rate, and time is

distance=rate·time

Distance, Rate and Time

For an object moving in at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula

d=rt

where d= distance, r= rate, and t= time.

Notice that the units we used above for the rate were miles per hour, which we can write as a ratio mileshour. Then when we multiplied by the time, in hours, the common units ‘hour’ divided out. The answer was in miles.

Jamal rides his bike at a uniform rate of 12 miles per hour for 312 hours. How much distance has he traveled?

Solution

Solution

Step 1. Read the problem.
You may want to create a mini-chart to summarize the
information in the problem.
d=?
r=12mph
t=312hours
Step 2. Identify what you are looking for. distance traveled
Step 3. Name. Choose a variable to represent it. let d = distance
Step 4. Translate.
Write the appropriate formula for the situation.
Substitute in the given information.
d=rt

d=12⋅312
Step 5. Solve the equation. d=42miles
Step 6. Check: Does 42 miles make sense?
A list shows Jamal rides 12, 24, 36, and 48 miles in 1, 2, 3, and 4 hours respectively. An arrow points to the statement that 42 miles in 3 1/2 hours is reasonable.
Step 7. Answer the question with a complete sentence. Jamal rode 42 miles.

Lindsay drove for 512 hours at 60 miles per hour. How much distance did she travel?

Solution

330 mi

Trinh walked for 213 hours at 3 miles per hour. How far did she walk?

Solution

7 mi

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of 520 miles. If he can drive at a steady rate of 65 miles per hour, how many hours will the trip take?

Solution

Solution

A seven-step guide to solving a distance, rate, and time problem.
Step 1. Read the problem.
Summarize the information in the problem.
d=520miles
r=65mph
t=?
Step 2. Identify what you are looking for. how many hours (time)
Step 3. Name:
Choose a variable to represent it.
let t = time
Step 4. Translate.
Write the appropriate formula.
Substitute in the given information.
d=rt
520=65t
Step 5. Solve the equation. t=8
Step 6. Check:
Substitute the numbers into the formula and make sure
the result is a true statement.
d=rt
520=?65⋅8
520=520>✓
Step 7. Answer the question with a complete sentence.
We know the units of time will be hours because
we divided miles by miles per hour.
Rey's trip will take 8 hours.

Lee wants to drive from Phoenix to his brother’s apartment in San Francisco, a distance of 770 miles. If he drives at a steady rate of 70 miles per hour, how many hours will the trip take?

Solution

11 hours

Yesenia is 168 miles from Chicago. If she needs to be in Chicago in 3 hours, at what rate does she need to drive?

Solution

56 mph

Solve a Formula for a Specific Variable

In this chapter, you became familiar with some formulas used in geometry. Formulas are also very useful in the sciences and social sciences—fields such as chemistry, physics, biology, psychology, sociology, and criminal justice. Healthcare workers use formulas, too, even for something as routine as dispensing medicine. The widely used spreadsheet program Microsoft ExcelTM relies on formulas to do its calculations. Many teachers use spreadsheets to apply formulas to compute student grades. It is important to be familiar with formulas and be able to manipulate them easily.

In Example 1 and Example 2, we used the formula d=rt. This formula gives the value of d when you substitute in the values of r and t. But in Example 2, we had to find the value of t. We substituted in values of d and r and then used algebra to solve to t. If you had to do this often, you might wonder why there isn’t a formula that gives the value of t when you substitute in the values of d and r. We can get a formula like this by solving the formula d=rt for t.

To solve a formula for a specific variable means to get that variable by itself with a coefficient of 1 on one side of the equation and all the other variables and constants on the other side. We will call this solving an equation for a specific variable in general. This process is also called solving a literal equation. The result is another formula, made up only of variables. The formula contains letters, or literals.

Let’s try a few examples, starting with the distance, rate, and time formula we used above.

Solve the formula d=rt for t:
  1. ⓐ when d=520 and r=65
  2. ⓑ in general.
Solution

Solution

We’ll write the solutions side-by-side so you can see that solving a formula in general uses the same steps as when we have numbers to substitute.

Steps to solve d=rt for time (t), showing both a specific numerical example (d=520, r=65) and the general algebraic solution.
ⓐ when d = 520 and r = 65 ⓑ in general
Write the forumla. The mathematical formula d = rt, representing distance equals rate times time, is displayed in bold black text on a plain white background. The mathematical formula for distance, rate, and time: d = rt, where 'd' is distance, 'r' is rate (or speed), and 't' is time. This fundamental equation is used in physics and everyday calculations.
Substitute any given values. A mathematical equation is displayed on a white background, reading '520 = 65t'.
Divide to isolate t. A mathematical equation showing 520 divided by 65 equals 65t divided by 65. Illustrating the division step to solve for 't' in the equation d = rt, resulting in t = d/r.
Simplify. The mathematical expressions '8=t' and 't=8' are shown, illustrating that equality is symmetrical. The formula for time (t) derived from distance (d) and rate (r), expressed as d/r = t and t = d/r, representing the same fundamental relationship.

Notice that the solution for ⓐ is the same as that in Example 2. We say the formula t=dr is solved for t. We can use this version of the formula anytime we are given the distance and rate and need to find the time.

Solve the formula d=rt for r:
  1. ⓐ when d=180 and t=4
  2. ⓐ in general
Solution
  1. ⓐ r=45
  2. ⓑ r=dt
Solve the formula d=rt for r:
  1. ⓐ when d=780 and t=12
  2. ⓑ in general
Solution
  1. ⓐ r=65
  2. ⓑ r=dt

We used the formula A=12bh in Use Properties of Rectangles, Triangles, and Trapezoids to find the area of a triangle when we were given the base and height. In the next example, we will solve this formula for the height.

The formula for area of a triangle is A=12bh. Solve this formula for h:
  1. ⓐ when A=90 and b=15
  2. ⓑ in general
Solution

Solution

Comparison of step-by-step solutions for a mathematical formula, showing both a specific case (A=90, b=15) and a general derivation.
ⓐ when A = 90 and b = 15 ⓑ in general
Write the forumla. The mathematical formula A = 1/2bh, which represents the area of a triangle where A is the area, b is the base, and h is the height. The formula for the area of a triangle, A = (1/2)bh, where A is the area, b is the base, and h is the height, is displayed on a white background.
Substitute any given values. A mathematical equation reads '90 = 1/2 * 15 * h' displayed in black font against a white background.
Clear the fractions. A mathematical equation reads 2 multiplied by 90 equals 2 multiplied by 1/2 multiplied by 15 multiplied by h. The number 2 on both sides and the fraction 1/2 are highlighted in red. A mathematical equation illustrating the formula for the area of a triangle, where both sides of the equation, 2 * A and 2 * (1/2) * b * h, are multiplied by 2, with the number 2 highlighted in red.
Simplify. A mathematical equation is displayed on a white background: 180 = 15h. The image shows the mathematical equation 2A = bh, representing a formula for calculating area in geometry, where A is area, b is base, and h is height.
Solve for h. The equation '12 = h' is displayed on a white background, representing a mathematical statement where the variable h is equal to the number 12. A mathematical equation shows '2A/b = h' with '2A' over 'b' on the left side, an equals sign in the middle, and 'h' on the right side. The characters are black on a white background.

We can now find the height of a triangle, if we know the area and the base, by using the formula

h=2Ab
Use the formula A=12bh to solve for h:
  1. ⓐ when A=170 and b=17
  2. ⓑ in general
Solution

ⓐ h=20 ⓑ h=2Ab

Use the formula A=12bh to solve for b:
  1. ⓐ when A=62 and h=31
  2. ⓑ in general
Solution
  1. ⓐ b=4
  2. ⓑ b=2Ah

In Solve Simple Interest Applications, we used the formula I=Prt to calculate simple interest, where I is interest, P is principal, r is rate as a decimal, and t is time in years.

Solve the formula I=Prt to find the principal, P:
  1. ⓐ when I=$5,600,r=4%,t=7years
  2. ⓑ in general
Solution

Solution

This table illustrates the step-by-step process of calculating the principal (P) from simple interest, providing both a specific numerical example and the general formula derivation.
I = $5600, r = 4%, t = 7 years in general
Write the forumla. The simple interest formula is displayed, showing I = Prt. This equation represents how to calculate simple interest (I) based on the principal amount (P), the annual interest rate (r), and the time (t) in years. The simple interest formula I=Prt is displayed in black text on a white background. 'I' represents interest, 'P' is the principal amount, 'r' is the annual interest rate, and 't' is the time in years.
Substitute any given values. A mathematical equation showing 5600 equals P multiplied by 0.04 and then by 7. This represents a simple interest calculation where 5600 is the interest, P is the principal, 0.04 is the rate, and 7 is the time. The simple interest formula I=Prt is displayed in black text on a white background. 'I' represents interest, 'P' is the principal amount, 'r' is the annual interest rate, and 't' is the time in years.
Multiply r ⋅ t. A mathematical equation is displayed, reading '5600 = P(0.28)', where P likely represents a variable being multiplied by 0.28 to equal 5600. The image shows the simple interest formula, I = P(rt), where I is interest, P is principal, r is the interest rate, and t is time.
Divide to isolate P. A mathematical equation shows '5600 over 0.28 equals P(0.28) over 0.28'. The numerators are 5600 and P(0.28), while the denominators are 0.28 for both sides, with the 0.28 in red. Demonstration of isolating 'P' in the simple interest formula by dividing both sides of I = Prt by 'rt', resulting in I/rt = P(rt)/rt.
Simplify. An equation displays '20,000 = P' in black bold font against a white background. A mathematical equation displays I over rt equals P, written as I/(rt) = P, on a white background. The variables are in italic font.
State the answer. The principal is $20,000. A mathematical formula displaying P = I / (rt). This equation relates a principal amount (P) to interest (I), rate (r), and time (t), often seen in financial or physics contexts.

Use the formula I=Prt.

Find t: ⓐ when I=$2,160,r=6%,P=$12,000; ⓑ in general

Solution
  1. ⓐ t=3 years
  2. ⓑ t=IPr

Use the formula I=Prt.

Find r: ⓐ when I=$5,400,P=$9,000,t=5years ⓑ in general

Solution
  1. ⓐ r=0.12=12%
  2. ⓑ r=IPt

Later in this class, and in future algebra classes, you’ll encounter equations that relate two variables, usually x and y. You 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. To do this, we will follow the same steps that we used to solve a formula for a specific variable.

Solve the formula 3x+2y=18 for y:
  1. ⓐ when x=4
  2. ⓑ in general
Solution

Solution

when x = 4 in general
Write the equation. A mathematical equation is displayed, showing '3x + 2y = 18' in black text against a white background. A mathematical equation is displayed on a white background, reading '3x + 2y = 18'. The equation is presented in a clear, standard mathematical font.
Substitute any given values. The image displays the algebraic equation '3(4) + 2y = 18' centered on a white background. A mathematical equation is displayed on a white background, reading '3x + 2y = 18'. The equation is presented in a clear, standard mathematical font.
Simplify if possible. A mathematical equation is displayed on a white background, which reads '12 + 2y = 18'. A mathematical equation is displayed on a white background, reading '3x + 2y = 18'. The equation is presented in a clear, standard mathematical font.
Subtract to isolate the y-term. An algebraic equation showing a step in solving for 'y', specifically: 12 - 12 + 2y = 18 - 12. The number 12 is highlighted in red where it is being subtracted on both sides. An algebraic equation is shown, with 3x - 3x + 2y on the left side and 18 - 3x on the right. The terms -3x on the left and -3x on the right are highlighted in red, indicating an operation or change.
Simplify. The image displays the algebraic equation '2y = 6' in a clear, standard mathematical notation, presented on a white background. A mathematical equation is displayed on a white background: 2y = 18 - 3x.
Divide. The equation 2y/2 = 6/2 is displayed, with the number 2 in the denominator of both fractions highlighted in red to show division. A mathematical equation showing 2y divided by 2 equals the expression (18 minus 3x) divided by 2, with the number 2 in the denominators highlighted in red.
Simplify. The mathematical equation 'y = 3' is displayed in a black serif font against a plain white background, centered in the frame. A mathematical equation shows y equals a fraction. The numerator is 18 minus 3x, and the denominator is 2.
Solve the formula 3x+4y=10 for y:
  1. ⓐ when x=2
  2. ⓑ in general
Solution
  1. ⓐ y=1
  2. ⓑ y=10−3x4
Solve the formula 5x+2y=18 for y:
  1. ⓐ when x=4
  2. ⓑ in general
Solution
  1. ⓐ y=−1
  2. ⓑ y=18−5x2

In the previous examples, we used the numbers in part (a) as a guide to solving in general in part (b). Do you think you’re ready to solve a formula in general without using numbers as a guide?

Solve the formula P=a+b+c for a.

Solution

Solution

We will isolate a on one side of the equation.
We will isolate a on one side of the equation.
Write the equation. P=a+b+c
Subtract b and c from both sides to isolate a. A mathematical equation is displayed: P - b - c = a + b + c - b - c. The variables 'b' and 'c' are highlighted in red on both sides of the equation, suggesting simplification or cancellation.
Simplify. P−b−c=a

So, a=P−b−c

Solve the formula P=a+b+c for b.

Solution

b = P − a − c

Solve the formula P=a+b+c for c.

Solution

c = P − a − b

Solve the equation 3x+y=10 for y.

Solution

Solution

We will isolate y on one side of the equation.
We will isolate y on one side of the equation.
Write the equation. 3x+y=10
Subtract 3x from both sides to isolate y. A mathematical equation showing the subtraction of 3x from both sides: 3x - 3x + y = 10 - 3x. This is a step to isolate the variable y.
Simplify. y=10−3x

Solve the formula 7x+y=11 for y.

Solution

y = 11 − 7x

Solve the formula 11x+y=8 for y.

Solution

y = 8 − 11x

Solve the equation 6x+5y=13 for y.

Solution

Solution

We will isolate y on one side of the equation.
We will isolate y on one side of the equation.
Write the equation. A linear equation in two variables, 6x + 5y = 13, is displayed in black text on a white background.
Subtract to isolate the term with y. A math equation: 6x + 5y - 6x = 13 - 6x. The 6x terms being subtracted from both sides are highlighted in red, illustrating a step to simplify the equation and isolate the 5y term.
Simplify. A mathematical equation is displayed on a white background, reading '5y = 13 - 6x' in black text.
Divide 5 to make the coefficient 1. A mathematical equation shows 5y/5 = (13 - 6x)/5. The number 5, as the denominator on both sides, is highlighted in red, indicating a division operation.
Simplify. A mathematical equation is displayed: y = (13 - 6x) / 5.

Solve the formula 4x+7y=9 for y.

Solution

y=9−4x7

Solve the formula 5x+8y=1 for y.

Solution

y=1−5x8

The Links to Literacy activity What's Faster than a Speeding Cheetah? will provide you with another view of the topics covered in this section.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Distance=RatexTime
  • Distance, Rate, Time
  • Simple Interest
  • Solving a Formula for a Specific Variable
  • Solving a Formula for a Specific Variable

Key Concepts

  • Distance, Rate, and Time
    • d=rt

Section Exercises

Practice Makes Perfect

Use the Distance, Rate, and Time Formula

In the following exercises, solve.

Steve drove for 812 hours at 72 miles per hour. How much distance did he travel?

Solution

612 mi

Socorro drove for 456 hours at 60 miles per hour. How much distance did she travel?

Yuki walked for 134 hours at 4 miles per hour. How far did she walk?

Solution

7 mi

Francie rode her bike for 212 hours at 12 miles per hour. How far did she ride?

Connor wants to drive from Tucson to the Grand Canyon, a distance of 338 miles. If he drives at a steady rate of 52 miles per hour, how many hours will the trip take?

Solution

6.5 hours

Megan is taking the bus from New York City to Montreal. The distance is 384 miles and the bus travels at a steady rate of 64 miles per hour. How long will the bus ride be?

Aurelia is driving from Miami to Orlando at a rate of 65 miles per hour. The distance is 235 miles. To the nearest tenth of an hour, how long will the trip take?

Solution

3.6 hours

Kareem wants to ride his bike from St. Louis, Missouri to Champaign, Illinois. The distance is 180 miles. If he rides at a steady rate of 16 miles per hour, how many hours will the trip take?

Javier is driving to Bangor, Maine, which is 240 miles away from his current location. If he needs to be in Bangor in 4 hours, at what rate does he need to drive?

Solution

60 mph

Alejandra is driving to Cincinnati, Ohio, 450 miles away. If she wants to be there in 6 hours, at what rate does she need to drive?

Aisha took the train from Spokane to Seattle. The distance is 280 miles, and the trip took 3.5 hours. What was the speed of the train?

Solution

80 mph

Philip got a ride with a friend from Denver to Las Vegas, a distance of 750 miles. If the trip took 10 hours, how fast was the friend driving?

Solve a Formula for a Specific Variable

In the following exercises, use the formula. d=rt.

Solve for t:
  1. ⓐ when d=350 and r=70
  2. ⓑ in general
Solution
  1. ⓐ t=5
  2. ⓑ t=dr
Solve for t:
  1. ⓐ when d=240 and r=60
  2. ⓑ in general
Solve for t:
  1. ⓐ when d=510 and r=60
  2. ⓑ in general
Solution
  1. ⓐ t=8.5
  2. ⓑ t=dr
Solve for t:
  1. ⓐ when d=175 and r=50
  2. ⓑ in general
Solve for r:
  1. ⓐ when d=204 and t=3
  2. ⓑ in general
Solution
  1. ⓐ r=68
  2. ⓑ r=dt
Solve for r:
  1. ⓐ when d=420 and t=6
  2. ⓑ in general
Solve for r:
  1. ⓐ when d=160 and t=2.5
  2. ⓑ in general
Solution
  1. ⓐ r=64
  2. ⓑ r=dt
Solve for r:
  1. ⓐ when d=180 and t=4.5
  2. ⓑ in general.

In the following exercises, use the formula A=12bh.

Solve for b:
  1. ⓐ when A=126 and h=18
  2. ⓑ in general
Solution
  1. ⓐ b=14
  2. ⓑ b=2Ah
Solve for h:
  1. ⓐ when A=176 and b=22
  2. ⓑ in general
Solve for h:
  1. ⓐ when A=375 and b=25
  2. ⓑ in general
Solution
  1. ⓐ h=30
  2. ⓑ h=2Ab
Solve for b:
  1. ⓐ when A=65 and h=13
  2. ⓑ in general

In the following exercises, use the formula I=Prt.

Solve for the principal, P for:
  1. ⓐ I=$5,480, r=4%, t=7years
  2. ⓑ in general
Solution
  1. ⓐ P=$19,571.43
  2. ⓑ P=Irt
Solve for the principal, P for:
  1. ⓐ I=$3,950, r=6%, t=5years
  2. ⓑ in general
Solve for the time, t for:
  1. ⓐ I=$2,376, P=$9,000, r=4.4%
  2. ⓑ in general
Solution
  1. ⓐ t=6years
  2. ⓑ t=IPr
Solve for the time, t for:
  1. ⓐ I=$624, P=$6,000, r=5.2%
  2. ⓑ in general

In the following exercises, solve.

Solve the formula 2x+3y=12 for y:
  1. ⓐ when x=3
  2. ⓑ in general
Solution
  1. ⓐ y=2
  2. ⓑ y=12−2x3
Solve the formula 5x+2y=10 for y:
  1. ⓐ when x=4
  2. ⓑ in general
Solve the formula 3x+y=7 for y:
  1. ⓐ when x=−2
  2. ⓑ in general
Solution
  1. ⓐ y = 13
  2. ⓑ y = 7 − 3x
Solve the formula 4x+y=5 for y:
  1. ⓐ when x=−3
  2. ⓑ in general

Solve a+b=90 for b.

Solution
  1. b = 90 − a

Solve a+b=90 for a.

Solve 180=a+b+c for a.

Solution

a = 180 − b − c

Solve 180=a+b+c for c.

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 4x+3y=7 for y.

Solution

y=7−4x3

Solve the formula 3x+2y=11 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 P=2L+2W for L.

Solution

L=P−2W2

Solve the formula P=2L+2W for W.

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.

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 temperature in Fahrenheit.

Solution

104° F

Converting temperature Yon was visiting the United States and he saw that the temperature in Seattle was 50° Fahrenheit. Solve for C in the formula F=95C+32 to find the temperature in Celsius.

Writing Exercises

Solve the equation 2x+3y=6 for y:
  1. ⓐ when x=−3
  2. ⓑ in general
  3. ⓒ Which solution is easier for you? Explain why.
Solution

Answers will vary

Solve the equation 5x−2y=10 for x:
  1. ⓐ when y=10
  2. ⓑ in general
  3. ⓒ Which solution is easier for you? Explain why.

Self Check

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

A self-assessment rubric for math, where students evaluate their proficiency in using the distance, rate, and time formula, and solving equations for variables, using options like 'Confidently' or 'No-I don't get it!'

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Use a Problem Solving Strategy

Approach Word Problems with a Positive Attitude

In the following exercises, solve.

How has your attitude towards solving word problems changed as a result of working through this chapter? Explain.

Solution

Answers will vary.

Did the Problem Solving Strategy help you solve word problems in this chapter? Explain.

Use a Problem Solving Strategy for Word Problems

In the following exercises, solve using the problem-solving strategy for word problems. Remember to write a complete sentence to answer each question.

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 at the concert.

There are 9 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.

Reza was very sick and lost 15% of his original weight. He lost 27 pounds. What was his original weight?

Solution

His original weight was 180 pounds.

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?

Solve Number 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

Twice the difference of a number and ten is fifty-four. Find the number.

One number is nine less than another. Their sum is twenty-seven. Find the numbers.

Solution

18, 9

The sum of two consecutive integers is −135. Find the numbers.

Solve Money Applications

Solve Coin Word Problems

In the following exercises, solve each coin word problem.

Francie has $4.35 in dimes and quarters. The number of dimes is 5 more than the number of quarters. How many of each coin does she have?

Solution

16 dimes, 11 quarters

Scott has $0.39 in pennies and nickels. The number of pennies is 8 times the number of nickels. How many of each coin does he have?

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?

Solution

6 of $5 bills, 11 of $10 bills

Lenny has $3.69 in pennies, dimes, and quarters. The number of pennies is 3 more than the number of dimes. The number of quarters is twice the number of dimes. How many of each coin does he have?

Solve Ticket and Stamp Word Problems

In the following exercises, solve each ticket or stamp word problem.

A church luncheon made $842. Adult tickets cost $10 each and children’s tickets cost $6 each. The number of children was 12 more than twice the number of adults. How many of each ticket were sold?

Solution

35 adults, 82 children

Tickets for a basketball game cost $2 for students and $5 for adults. The number of students was 3 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?

Ana spent $4.06 buying stamps. The number of $0.41 stamps she bought was 5 more than the number of $0.26 stamps. How many of each did she buy?

Solution

3 of 26 -cent stamps, 8 of 41 -cent stamps

Yumi spent $34.15 buying stamps. The number of $0.56 stamps she bought was 10 less than 4 times the number of $0.41 stamps. How many of each did she buy?

Use Properties of Angles, Triangles, and the Pythagorean Theorem

Use Properties of Angles

In the following exercises, solve using properties of angles.

What is the supplement of a 48° angle?

Solution

132°

What is the complement of a 61° angle?

Two angles are complementary. The smaller angle is 24° less than the larger angle. Find the measures of both angles.

Solution

33°, 57°

Two angles are supplementary. The larger angle is 45° more than the smaller angle. Find the measures of both angles.

Use Properties of Triangles

In the following exercises, solve using properties of triangles.

The measures of two angles of a triangle are 22 and 85 degrees. Find the measure of the third angle.

Solution

73°

One angle of a right triangle measures 41.5 degrees. What is the measure of the other small angle?

One angle of a triangle is 30° more than the smallest angle. The largest angle is the sum of the other angles. Find the measures of all three angles.

Solution

30°, 60°, 90°

One angle of a triangle is twice the measure of the smallest angle. The third angle is 60° more than the measure of the smallest angle. Find the measures of all three angles.

In the following exercises, ΔABC is similar to ΔXYZ. Find the length of the indicated side.

Two triangles are shown. Triangle ABC is on the left. The side across from A is labeled 21, across from B is b, and across from C is 11.2. Triangle XYZ is on the right. The side across from X is labeled x, across from Y is 10, and across from Z is 8.

side x

Solution

15

side b

Use the Pythagorean Theorem

In the following exercises, use the Pythagorean Theorem to find the length of the missing side. Round to the nearest tenth, if necessary.

A right triangle is shown. The base is labeled 10, the height is labeled 24.
Solution

26

A right triangle is shown. The base is labeled 6, the height is labeled 8.
A right triangle is shown. The height is labeled 15, the hypotenuse is labeled 17.
Solution

8

A right triangle is shown. The height is labeled 15, the hypotenuse is labeled 25.
A right triangle is shown. The height is labeled 7, the base is labeled 4.
Solution

8.1

A right triangle is shown. The height is labeled 11, the base is labeled 10.

In the following exercises, solve. Approximate to the nearest tenth, if necessary.

Sergio needs to attach a wire to hold the antenna to the roof of his house, as shown in the figure. The antenna is 8 feet tall and Sergio has 10 feet of wire. How far from the base of the antenna can he attach the wire?

An image of a house is shown. A 10-foot wire is going from the roof of the house to the ground. The wire hits the house at a height of 8 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?

A rectangular shelf is shown, with a diagonal drawn in from the lower left corner to the upper right corner. The side is labeled 15 inches, the top is labeled 36 inches.

Use Properties of Rectangles, Triangles, and Trapezoids

Understand Linear, Square, Cubic Measure

In the following exercises, would you measure each item using linear, square, or cubic measure?

amount of sand in a sandbag

Solution

cubic

height of a tree

size of a patio

Solution

square

length of a highway

In the following exercises, find
  1. ⓐ the perimeter
  2. ⓑ the area of each figure
Three squares are shown, in a sideways L shape.
Solution
  1. ⓐ 8 units
  2. ⓑ 3 sq. units
Five squares are shown, in a T-shape. There are three squares across the top and three squares down.

Use Properties of Rectangles

In the following exercises, find the ⓐ perimeter ⓑ area of each rectangle

The length of a rectangle is 42 meters and the width is 28 meters.

Solution
  1. ⓐ 140 m
  2. ⓑ 1176 sq. m

The length of a rectangle is 36 feet and the width is 19 feet.

A sidewalk in front of Kathy’s house is in the shape of a rectangle 4 feet wide by 45 feet long.

Solution
  1. ⓐ 98 ft.
  2. ⓑ 180 sq. ft.

A rectangular room is 16 feet wide by 12 feet long.

In the following exercises, solve.

Find the length of a rectangle with perimeter of 220 centimeters and width of 85 centimeters.

Solution

25 cm

Find the width of a rectangle with perimeter 39 and length 11.

The area of a rectangle is 2356 square meters. The length is 38 meters. What is the width?

Solution

62 m

The width of a rectangle is 45 centimeters. The area is 2700 square centimeters. What is the length?

The length of a rectangle is 12 centimeters more than the width. The perimeter is 74 centimeters. Find the length and the width.

Solution

24.5 cm., 12.5 cm.

The width of a rectangle is 3 more than twice the length. The perimeter is 96 inches. Find the length and the width.

Use Properties of Triangles

In the following exercises, solve using the properties of triangles.

Find the area of a triangle with base 18 inches and height 15 inches.

Solution

135 sq. in.

Find the area of a triangle with base 33 centimeters and height 21 centimeters.

A triangular road sign has base 30 inches and height 40 inches. What is its area?

Solution

600 sq. in.

If a triangular courtyard has sides 9 feet and 12 feet and the perimeter is 32 feet, how long is the third side?

A tile in the shape of an isosceles triangle has a base of 6 inches. If the perimeter is 20 inches, find the length of each of the other sides.

Solution

7 in., 7 in.

Find the length of each side of an equilateral triangle with perimeter of 81 yards.

The perimeter of a triangle is 59 feet. One side of the triangle is 3 feet longer than the shortest side. The third side is 5 feet longer than the shortest side. Find the length of each side.

Solution

17 ft., 20 ft., 22 ft.

One side of a triangle is three times the smallest side. The third side is 9 feet more than the shortest side. The perimeter is 39 feet. Find the lengths of all three sides.

Use Properties of Trapezoids

In the following exercises, solve using the properties of trapezoids.

The height of a trapezoid is 8 feet and the bases are 11 and 14 feet. What is the area?

Solution

100 sq. ft.

The height of a trapezoid is 5 yards and the bases are 7 and 10 yards. What is the area?

Find the area of the trapezoid with height 25 meters and bases 32.5 and 21.5 meters.

Solution

675 sq. m

A flag is shaped like a trapezoid with height 62 centimeters and the bases are 91.5 and 78.1 centimeters. What is the area of the flag?

Solve Geometry Applications: Circles and Irregular Figures

Use Properties of Circles

In the following exercises, solve using the properties of circles. Round answers to the nearest hundredth.

A circular mosaic has radius 3 meters. Find the
  1. ⓐ circumference
  2. ⓑ area of the mosaic
Solution
  1. ⓐ 18.84 m
  2. ⓑ 28.26 sq. m
A circular fountain has radius 8 feet. Find the
  1. ⓐ circumference
  2. ⓑ area of the fountain

Find the diameter of a circle with circumference 150.72 inches.

Solution

48 in.

Find the radius of a circle with circumference 345.4 centimeters

Find the Area of Irregular Figures

In the following exercises, find the area of each shaded region.

A geometric shape is shown, formed by two rectangles. The top is labeled 8. The width of the top rectangle is labeled 3. The right side of the figure is labeled 5. The width of the bottom rectangle is labeled 3.
Solution

30 sq. units

A geometric shape is shown. It is a U-shape. The base is labeled 5, the height 6. The horizontal and vertical lines at the top are labeled 2.
A geometric shape is shown. It is formed by two triangles. The shared base of the two triangles is labeled 20. The height of each triangle is labeled 15.
Solution

300 sq. units

A geometric shape is shown. It is a trapezoid with a triangle attached to the top on the right side.  The height of the trapezoid is labeled 8, the bottom base is labeled 12, and the top is labeled 9. The height of the triangle is labeled 8.
A geometric shape is shown. It is a rectangle with a semi-circle attached to the top. The base of the rectangle, also the diameter of the semi-circle, is labeled 10. The height of the rectangle is labeled 16.
Solution

199.25 sq. units

A geometric shape is shown. It is a triangle with a semicircle attached. The base of the triangle, also the diameter of the semi-circle, is labeled 5. The height of the triangle is also labeled 5.

Solve Geometry Applications: Volume and Surface Area

Find Volume and Surface Area of Rectangular Solids

In the following exercises, find the
  1. ⓐ volume
  2. ⓑ surface area of the rectangular solid

a rectangular solid with length 14 centimeters, width 4.5 centimeters, and height 10 centimeters

Solution
  1. ⓐ 630 cu. cm
  2. ⓑ 496 sq. cm

a cube with sides that are 3 feet long

a cube of tofu with sides 2.5 inches

Solution
  1. ⓐ 15.625 cu. in.
  2. ⓑ 37.5 sq. in.

a rectangular carton with length 32 inches, width 18 inches, and height 10 inches

Find Volume and Surface Area of Spheres

In the following exercises, find the
  1. ⓐ volume
  2. ⓑ surface area of the sphere.

a sphere with radius 4 yards

Solution
  1. ⓐ 267.95 cu. yd.
  2. ⓑ 200.96 sq. yd.

a sphere with radius 12 meters

a baseball with radius 1.45 inches

Solution
  1. ⓐ 12.76 cu. in.
  2. ⓑ 26.41 sq. in.

a soccer ball with radius 22 centimeters

Find Volume and Surface Area of Cylinders

In the following exercises, find the
  1. ⓐ volume
  2. ⓑ surface area of the cylinder

a cylinder with radius 2 yards and height 6 yards

Solution
  1. ⓐ 75.36 cu. yd.
  2. ⓑ 100.48 sq. yd.

a cylinder with diameter 18 inches and height 40 inches

a juice can with diameter 8 centimeters and height 15 centimeters

Solution
  1. ⓐ 753.6 cu. cm
  2. ⓑ 477.28 sq. cm

a cylindrical pylon with diameter 0.8 feet and height 2.5 feet

Find Volume of Cones

In the following exercises, find the volume of the cone.

a cone with height 5 meters and radius 1 meter

Solution

5.233 cu. m

a cone with height 24 feet and radius 8 feet

a cone-shaped water cup with diameter 2.6 inches and height 2.6 inches

Solution

4.599 cu. in.

a cone-shaped pile of gravel with diameter 6 yards and height 5 yards

Solve a Formula for a Specific Variable

Use the Distance, Rate, and Time Formula

In the following exercises, solve using the formula for distance, rate, and time.

A plane flew 4 hours at 380 miles per hour. What distance was covered?

Solution

1520 miles

Gus rode his bike for 112 hours at 8 miles per hour. How far did he ride?

Jack is driving from Bangor to Portland at a rate of 68 miles per hour. The distance is 107 miles. To the nearest tenth of an hour, how long will the trip take?

Solution

1.6 hours

Jasmine took the bus from Pittsburgh to Philadelphia. The distance is 305 miles and the trip took 5 hours. What was the speed of the bus?

Solve a Formula for a Specific Variable

In the following exercises, use the formula d=rt.

Solve for t:

  1. ⓐ when d=403 and r=65
  2. ⓑ in general
Solution
  1. ⓐ t=6.2
  2. ⓑ t=dr

Solve for r:

  1. ⓐ when d=750 and t=15
  2. ⓑ in general

In the following exercises, use the formula A=12bh.

Solve for b:

  1. ⓐ when A=416 and h=32
  2. ⓑ in general
Solution
  1. ⓐ b=26
  2. ⓑ b=2Ah

Solve for h:

  1. ⓐ when A=48 and b=8
  2. ⓑ in general

In the following exercises, use the formula I=Prt.

Solve for the principal, P, for:

  1. ⓐ I=$720, r=4%, t=3years
  2. ⓑ in general
Solution
  1. ⓐ P=$6000
  2. ⓑ P=I(r⋅t)
Solve for the time, t for:
  1. ⓐ I=$3630, P=$11,000, r=5.5%
  2. ⓑ in general

In the following exercises, solve.

Solve the formula 6x+5y=20 for y:
  1. ⓐ when x=0
  2. ⓑ in general
Solution
  1. ⓐ y=4
  2. ⓑ y=20−6x5
Solve the formula 2x+y=15 for y:
  1. ⓐ when x=−5
  2. ⓑ in general

Solve a+b=90 for a.

Solution

a = 90 − b

Solve 180=a+b+c for a.

Solve the formula 4x+y=17 for y.

Solution

y = 17 − 4x

Solve the formula −3x+y=−6 for y.

Solve the formula P=2L+2W for W.

Solution

W=P−2L2

Solve the formula V=LWH for H.

Describe how you have used two topics from this chapter in your life outside of math class during the past month.

Chapter Practice Test

Four-fifths of the people on a hike are children. If there are 12 children, what is the total number of people on the hike?

The sum of 13 and twice a number is −19. Find the number.

Solution

−16

One number is 3 less than another number. Their sum is 65. Find the numbers.

Bonita has $2.95 in dimes and quarters in her pocket. If she has 5 more dimes than quarters, how many of each coin does she have?

Solution

7 quarters, 12 dimes

At a concert, $1600 in tickets were sold. Adult tickets were $9 each and children’s tickets were $4 each. If the number of adult tickets was 30 fewer than twice the number of children’s tickets, how many of each kind were sold?

Find the complement of a 52° angle.

Solution

38°

The measure of one angle of a triangle is twice the measure of the smallest angle. The measure of the third angle is 14 more than the measure of the smallest angle. Find the measures of all three angles.

The perimeter of an equilateral triangle is 145 feet. Find the length of each side.

Solution

48.3

ΔABC is similar to ΔXYZ. Find the length of side c.

Two triangles are shown. Triangle XYZ is on the left. The side across from X is labeled 5, the side across from Y is labeled 10, the side across from Z is labeled 7. Triangle ABC is on the right. The side across from A is labeled 6, the side across from B is labeled 12, and the side across from C is labeled c.

Find the length of the missing side. Round to the nearest tenth, if necessary.

A right triangle is shown. The height is labeled 24 and the hypotenuse is labeled 26.
Solution

10

Find the length of the missing side. Round to the nearest tenth, if necessary.

A right triangle is shown. The base is labeled 6 and the height is labeled 9.

A baseball diamond is shaped like a square with sides 90 feet long. How far is it from home plate to second base, as shown?

A baseball diamond is shown. It is in the shape of a sideways square. The bottom corner is labeled Home and there is a dotted line to the top corner, labeled 2nd base. The right corner is labeled 1st base and the left corner is labeled 3rd base.
Solution

127.3 ft

The length of a rectangle is 2 feet more than five times the width. The perimeter is 40 feet. Find the dimensions of the rectangle.

A triangular poster has base 80 centimeters and height 55 centimeters. Find the area of the poster.

Solution

2200 square centimeters

A trapezoid has height 14 inches and bases 20 inches and 23 inches. Find the area of the trapezoid.

A circular pool has diameter 90 inches. What is its circumference? Round to the nearest tenth.

Solution

282.6 inches

Find the area of the shaded region. Round to the nearest tenth.

A geometric shape is shown. It is a rectangle with a semi-circle attached on the left and a triangle attached on the right. The height of the rectangle, also the height of the triangle and the diameter of the semi-circle, is labeled 4. The base of the figure is labeled 10. The top of the rectangle is labeled 7.

Find the volume of a rectangular room with width 12 feet, length 15 feet, and height 8 feet.

Solution

1440

A coffee can is shaped like a cylinder with height 7 inches and radius 5 inches. Find (a) the surface area and (b) the volume of the can. Round to the nearest tenth.

A traffic cone has height 75 centimeters. The radius of the base is 20 centimeters. Find the volume of the cone. Round to the nearest tenth.

Solution

31,400 cubic inches

Leon drove from his house in Cincinnati to his sister’s house in Cleveland. He drove at a uniform rate of 63 miles per hour and the trip took 4 hours. What was the distance?

The Catalina Express takes 112 hours to travel from Long Beach to Catalina Island, a distance of 22 miles. To the nearest tenth, what is the speed of the boat?

Solution

14.7 miles per hour

Use the formula I=Prt to solve for the principal, P, for:
  1. ⓐ I=$1380,r=5%,t=3 years
  2. ⓑ in general
Solve the formula A=12bh for h:
  1. ⓐ when A=1716 and b=66
  2. ⓑ in general
Solution
  1. ⓐ height=52
  2. ⓑ h=2Ab

Solve x+5y=14 for y.

Introduction to Polynomials

This is an image of a space shuttle blasting off into space.
The paths of rockets are calculated using polynomials. (credit: NASA, Public Domain)

Expressions known as polynomials are used widely in algebra. Applications of these expressions are essential to many careers, including economists, engineers, and scientists. In this chapter, we will find out what polynomials are and how to manipulate them through basic mathematical operations.

Add and Subtract Polynomials

Learning Objectives

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

  • Identify polynomials, monomials, binomials, and trinomials
  • Determine the degree of polynomials
  • Add and subtract monomials
  • Add and subtract polynomials
  • Evaluate a polynomial for a given value

Before you get started, take this readiness quiz.

Simplify: 8x+3x.
If you missed this problem, review Example 10 in Evaluate, Simplify, and Translate Expressions.

Solution

11x

Subtract: (5n+8)−(2n−1).
If you missed this problem, review Example 13 in Distributive Property.

Solution

3n+9

Evaluate: 4y2 when y=5
If you missed this problem, review Example 6 in Evaluate, Simplify, and Translate Expressions.

Solution

100

Identify Polynomials, Monomials, Binomials, and Trinomials

In Evaluate, Simplify, and Translate Expressions, you learned that a term is a constant or the product of a constant and one or more variables. The constant is called a coefficient. When it is of the form axm, where a is a constant and m is a whole number, it is called a monomial. A monomial, or a sum and/or difference of monomials, is called a polynomial.

Polynomials

polynomial—A monomial, or two or more monomials, combined by addition or subtraction

monomial—A polynomial with exactly one term

binomial— A polynomial with exactly two terms

trinomial—A polynomial with exactly three terms

Notice the roots:

  • poly- means many
  • mono- means one
  • bi- means two
  • tri- means three

Here are some examples of polynomials:

Polynomial b+1 4y2−7y+2 5x5−4x4+x3+8x2−9x+1
Monomial 5 4b2 −9x3
Binomial 3a−7 y2−9 17x3+14x2
Trinomial x2−5x+6 4y2−7y+2 5a4−3a3+a

Notice that every monomial, binomial, and trinomial is also a polynomial. They are 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’.

Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial:

  1. ⓐ 8x2−7x−9
  2. ⓑ −5a4
  3. ⓒ x4−7x3−6x2+5x+2
  4. ⓓ 11−4y3
  5. ⓔ n
Solution

Solution

Polynomial Number of terms Type
ⓐ 8x2−7x−9 3 Trinomial
ⓑ −5a4 1 Monomial
ⓒ x4−7x3−6x2+5x+2 5 Polynomial
ⓓ 11−4y3 2 Binomial
ⓔ n 1 Monomial
Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial.
  1. ⓐ z
  2. ⓑ 2x3−4x2−x−8
  3. ⓒ 6x2−4x+1
  4. ⓓ 9−4y2
  5. ⓔ 3x7
Solution
  1. ⓐ monomial
  2. ⓑ polynomial
  3. ⓒ trinomial
  4. ⓓ binomial
  5. ⓔ monomial
Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial.
  1. ⓐ y3−8
  2. ⓑ 9x3−5x2−x
  3. ⓒ x4−3x2−4x−7
  4. ⓓ −y4
  5. ⓔ w
Solution
  1. ⓐ binomial
  2. ⓑ trinomial
  3. ⓒ polynomial
  4. ⓓ monomial
  5. ⓒ monomial

Determine the Degree of Polynomials

In this section, we will work with polynomials that have only one variable in each term. 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—it has no variable.

Degree of a Polynomial

The degree of a term is the exponent of its variable.

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.

Remember: Any base written without an exponent has an implied exponent of 1.

A table is shown. The top row is titled “Monomials” and lists the following monomials: 5, 4 b squared, negative 9 x cubed, negative 18. The next row is titled “Degree” and lists, in blue, 0, 2, 3, and 0. The next row is titled “Binomial” and lists the following binomials: b plus 1, 3a minus 7, y squared minus 9, 17 x cubed plus 14 x squared. The next row is titled “Degree of each term,” with “term” written in blue. This row lists 1, 0, 1, 0, 2, 0, 3, 2 in blue. The next row is titled “Degree of polynomial,” with “polynomial” written in red. This row lists 1, 1, 2, 3 in red. The next row is titled “Trinomial” and lists the following trinomials: x squared minus 5x plus 6, 4 y squared minus 7y plus 2, 5 a to the fourth minus 3 a cubed plus a, and x to the fourth plus 2 x squared minus 5. The next row is titled “Degree of each term,” with “term” written in blue. This row lists 2, 1, 0, 2, 1, 0, 4, 3, 1, 4, 2, 0 in blue. The next row is titled “Degree of polynomial,” with “polynomial” written in red. This row lists 2, 2, 4, 4 in red. The next row is titled “Polynomial” and lists the following polynomials: b plus 1, 4 y squared minus 7y plus 2, and 4 x to the fourth plus x cubed plus 8 x squared minus 9x plus 1. The next row is titled “Degree of each term,” with “term” written in blue. This row lists 1, 0, 2, 1, 0, 4, 3, 2, 1, 0 in blue. The next row is titled “Degree of polynomial,” with “polynomial” written in red. This row lists 1, 2, 4 in red.

Find the degree of the following polynomials:

  1. ⓐ 4x
  2. ⓑ 3x3−5x+7
  3. ⓒ −11
  4. ⓓ −6x2+9x−3
  5. ⓔ 8x+2
Solution

Solution

This table illustrates how to determine the degree of various mathematical expressions, including monomials, polynomials, and constants, by identifying the highest exponent of the variable.
ⓐ 4x
The exponent of x is one. x=x1 The degree is 1.
ⓑ 3x3−5x+7
The highest degree of all the terms is 3. The degree is 3
ⓒ −11
The degree of a constant is 0. The degree is 0.
ⓓ −6x2+9x−3
The highest degree of all the terms is 2. The degree is 2.
ⓔ 8x+2
The highest degree of all the terms is 1. The degree is 1.
Find the degree of the following polynomials:
  1. ⓐ −6y
  2. ⓑ 4x−1
  3. ⓒ 3x4+4x2−8
  4. ⓓ 2y2+3y+9
  5. ⓔ −18
Solution
  1. ⓐ 1
  2. ⓑ 1
  3. ⓒ 4
  4. ⓓ 2
  5. ⓔ 0
Find the degree of the following polynomials:
  1. ⓐ 47
  2. ⓑ 2x2−8x+2
  3. ⓒ x4−16
  4. ⓓ y5−5y3+y
  5. ⓔ 9a3
Solution
  1. ⓐ 0
  2. ⓑ 2
  3. ⓒ 4
  4. ⓓ 5
  5. ⓔ 3

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. Look back at the polynomials in Example 2. Notice that they are all written in standard form. Get in the habit of writing the term with the highest degree first.

Add and Subtract Monomials

In The Language of Algebra, you simplified expressions by combining like terms. Adding and subtracting monomials is the same as combining like terms. Like terms must have the same variable with the same exponent. Recall that when combining like terms only the coefficients are combined, never the exponents.

Add: 17x2+6x2.

Solution

Solution

Steps to simplify an algebraic expression by combining like terms.
17x2+6x2
Combine like terms. 23x2

Add: 12x2+5x2.

Solution

17x2

Add: −11y2+8y2.

Solution

−3y2

Subtract: 11n−(−8n).

Solution

Solution

Illustrates the simplification of the algebraic expression 11n - (-8n) by combining like terms to 19n.
11n−(−8n)
Combine like terms. 19n

Subtract: 9n−(−5n).

Solution

14n

Subtract: −7a3−(−5a3).

Solution

−2a3

Simplify: a2+4b2−7a2.

Solution

Solution

Example demonstrating the simplification of an algebraic expression by combining like terms.
a2+4b2−7a2
Combine like terms. −6a2+4b2

Remember, −6a2 and 4b2 are not like terms. The variables are not the same.

Add: 3x2+3y2−5x2.

Solution

−2x2 + 3y2

Add: 2a2+b2−4a2.

Solution

−2a2 + b2

Add and Subtract Polynomials

Adding and subtracting polynomials can be thought of as just adding and subtracting like terms. Look for like terms—those with the same variables with the same exponent. The Commutative Property allows us to rearrange the terms to put like terms together. It may also be helpful to underline, circle, or box like terms.

Find the sum: (4x2−5x+1)+(3x2−8x−9).

Solution

Solution

A mathematical expression showing the addition of two quadratic polynomials: (4x^2 - 5x + 1) + (3x^2 - 8x - 9).
Identify like terms. A mathematical expression shows: 4x^2 - 5x + [1] + 3x^2 - 8x - [9]. Terms with 'x' and 'x^2' are underlined with single and double lines, respectively, and the constants '1' and '9' are boxed.
Rearrange to get the like terms together. An algebraic expression demonstrating the combination of like terms: 4x^2 + 3x^2 - 5x - 8x + 1 - 9. Terms with x squared, x, and constants are visually grouped by underlines.
Combine like terms. A mathematical expression displaying a quadratic polynomial: 7x^2 - 13x - 8.

Find the sum: (3x2−2x+8)+(x2−6x+2).

Solution

4x2 − 8x + 10

Find the sum: (7y2+4y−6)+(4y2+5y+1).

Solution

11y2 + 9y − 5

Parentheses are grouping symbols. When we add polynomials as we did in Example 6, we can rewrite the expression without parentheses and then combine like terms. But when we subtract polynomials, we must be very careful with the signs.

Find the difference: (7u2−5u+3)−(4u2−2).

Solution

Solution

A mathematical expression showing the subtraction of two polynomials: (7u^2 - 5u + 3) - (4u^2 - 2).
Distribute and identify like terms. An algebraic expression: 7u^2 - 5u + [3] - 4u^2 + [2]. Terms 7u^2 and 4u^2 are double-underlined, 5u is single-underlined, and 3 and 2 are boxed.
Rearrange the terms. A mathematical expression is displayed, showing '7u^2 - 4u^2 - 5u + 3 + 2'. The terms '7u^2' and '4u^2' are double-underlined, '5u' is single-underlined, and the final number '2' is enclosed in a box.
Combine like terms. A mathematical expression displays three terms: '3u^2 - 5u + 5' on a white background. This quadratic expression features a squared variable, a linear variable, and a constant term.

Find the difference: (6y2+3y−1)−(3y2−4).

Solution

3y2 + 3y + 3

Find the difference: (8u2−7u−2)−(5u2−6u−4).

Solution

3u2 − u + 2

Subtract: (m2−3m+8) from (9m2−7m+4).

Solution

Solution

A mathematical problem asking to subtract the polynomial (m^2 - 3m + 8) from (9m^2 - 7m + 4).
Distribute and identify like terms. An algebraic expression is displayed: 9m^2 - 7m + [4 - m^2 + 3m - 8]. Some terms are underlined with single or double lines, and constants 4 and 8 are enclosed in square boxes.
Rearrange the terms. A mathematical expression showing 9m^2 - m^2 - 7m + 3m + 4 - 8, featuring like terms for simplification.
Combine like terms. The image displays the mathematical expression 8m^2 - 4m - 4, which is a quadratic trinomial. The expression involves the variable 'm' raised to the power of two and one, along with constant terms.

Subtract: (4n2−7n−3) from (8n2+5n−3).

Solution

4n2 + 12n

Subtract: (a2−4a−9) from (6a2+4a−1).

Solution

5a2 + 8a + 8

Evaluate a Polynomial for a Given Value

In The Language of Algebra we evaluated expressions. Since polynomials are expressions, we'll follow the same procedures to evaluate polynomials—substitute the given value for the variable into the polynomial, and then simplify.

Evaluate 3x2−9x+7 when
  1. ⓐ x=3
  2. ⓑ x=−1
Solution

Solution

Step-by-step evaluation of the algebraic expression 3x^2 - 9x + 7 for x = 3.
ⓐ x=3
3x2−9x+7
Substitute 3 for x 3(3)2−9(3)+7
Simplify the expression with the exponent. 3·9−9(3)+7
Multiply. 27−27+7
Simplify. 7
Step-by-step evaluation of the algebraic expression 3x^2 - 9x + 7 when x = -1, showing each stage of simplification.
ⓑ x=−1
3x2−9x+7
Substitute −1 for x 3(−1)2−9(−1)+7
Simplify the expression with the exponent. 3·1−9(−1)+7
Multiply. 3+9+7
Simplify. 19
Evaluate: 2x2+4x−3 when
  1. ⓐ x=2
  2. ⓑ x=−3
Solution
  1. ⓐ 13
  2. ⓑ 3
Evaluate: 7y2−y−2 when
  1. ⓐ y=−4
  2. ⓑ y=0
Solution
  1. ⓐ 114
  2. ⓑ −2

The polynomial −16t2+300 gives the height of an object t seconds after it is dropped from a 300 foot tall bridge. Find the height after t=3 seconds.

Solution

Solution

The image shows the mathematical expression -16t^2 + 300, presented in a clear, typeset format against a white background.
Substitute 3 for t The image shows the mathematical expression -16(3)^2 + 300, which evaluates to -16 * 9 + 300 = -144 + 300 = 156.
Simplify the expression with the exponent. A mathematical expression showing -16 multiplied by 9, then added to 300.
Multiply. The mathematical expression -144 + 300 is displayed in black font on a white background.
Simplify. The number 156 is displayed in black text against a plain white background.

The polynomial −8t2+24t+4 gives the height, in feet, of a ball t seconds after it is tossed into the air, from an initial height of 4 feet. Find the height after t=3 seconds.

Solution

4 feet

The polynomial −8t2+24t+4 gives the height, in feet, of a ball x seconds after it is tossed into the air, from an initial height of 4 feet. Find the height after t=2 seconds.

Solution

20 feet

ACCESS ADDITIONAL ONLINE RESOURCES

  • Adding Polynomials
  • Subtracting Polynomials

Practice Makes Perfect

Identify Polynomials, Monomials, Binomials and Trinomials

In the following exercises, determine if each of the polynomials is a monomial, binomial, trinomial, or other polynomial.

5x+2

Solution

binomial

z2−5z−6

a2+9a+18

Solution

trinomial

−12p4

y3−8y2+2y−16

Solution

polynomial

10−9x

23y2

Solution

monomial

m4+4m3+6m2+4m+1

Determine the Degree of Polynomials

In the following exercises, determine the degree of each polynomial.

8a5−2a3+1

Solution

5

5c3+11c2−c−8

3x−12

Solution

1

4y+17

−13

Solution

0

−22

Add and Subtract Monomials

In the following exercises, add or subtract the monomials.

6x2+9x2

Solution

15x2

4y3+6y3

−12u+4u

Solution

−8u

−3m+9m

5a+7b

Solution

5a + 7b

8y+6z

Add: 4a,−3b,−8a

Solution

−4a −3b

Add: 4x,3y,−3x

18x−2x

Solution

16x

13a−3a

Subtract 5x6from−12x6

Solution

−17x6

Subtract 2p4from−7p4

Add and Subtract Polynomials

In the following exercises, add or subtract the polynomials.

(4y2+10y+3)+(8y2−6y+5)

Solution

12y2 + 4y + 8

(7x2−9x+2)+(6x2−4x+3)

(x2+6x+8)+(−4x2+11x−9)

Solution

−3x2 + 17x − 1

(y2+9y+4)+(−2y2−5y−1)

(3a2+7)+(a2−7a−18)

Solution

4a2 − 7a − 11

(p2−5p−11)+(3p2+9)

(6m2−9m−3)−(2m2+m−5)

Solution

4m2 − 10m + 2

(3n2−4n+1)−(4n2−n−2)

(z2+8z+9)−(z2−3z+1)

Solution

11z + 8

(z2−7z+5)−(z2−8z+6)

(12s2−15s)−(s−9)

Solution

12s2 − 16s + 9

(10r2−20r)−(r−8)

Find the sum of (2p3−8) and (p2+9p+18)

Solution

2p3 + p2 + 9p + 10

Find the sum of (q2+4q+13) and (7q3−3)

Subtract (7x2−4x+2) from (8x2−x+6)

Solution

x2 + 3x + 4

Subtract (5x2−x+12) from (9x2−6x−20)

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)

Evaluate a Polynomial for a Given Value

In the following exercises, evaluate each polynomial for the given value.

Evaluate8y2−3y+2
  1. ⓐ y=5
  2. ⓑ y=−2
  3. ⓒ y=0
Solution
  1. ⓐ 187
  2. ⓑ 40
  3. ⓒ 2
Evaluate5y2−y−7when:
  1. ⓐ y=−4
  2. ⓑ y=1
  3. ⓒ y=0
Evaluate4−36xwhen:
  1. ⓐ x=3
  2. ⓑ x=0
  3. ⓒ x=−1
Solution
  1. ⓐ −104
  2. ⓑ 4
  3. ⓒ 40
Evaluate16−36x2when:
  1. ⓐ x=−1
  2. ⓑ x=0
  3. ⓒ x=2

A window washer drops a squeegee from a platform 275 feet high. The polynomial −16t2+275 gives the height of the squeegee t seconds after it was dropped. Find the height after t=4 seconds.

Solution

19 feet

A manufacturer of microwave ovens has found that the revenue received from selling microwaves at a cost of p dollars each is given by the polynomial −5p2+350p. Find the revenue received when p=50 dollars.

Everyday Math

Fuel Efficiency The fuel efficiency (in miles per gallon) of a bus going at a speed of x miles per hour is given by the polynomial −1160x2+12x. Find the fuel efficiency when x=40mph.

Solution

10 mpg

Stopping Distance The number of feet it takes for a car traveling at x miles per hour to stop on dry, level concrete is given by the polynomial 0.06x2+1.1x. Find the stopping distance when x=60mph.

Writing Exercises

Using your own words, explain the difference between a monomial, a binomial, and a trinomial.

Solution

Answers will vary.

Eloise thinks the sum 5x2+3x4 is 8x6. What is wrong with her reasoning?

Self Check

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

Math self-assessment grid for polynomial skills: identifying, determining degree, adding/subtracting monomials & polynomials, and evaluating expressions.

ⓑ 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 a 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 of a polynomial is the exponent of its variable.
monomial
A term of the form axm, where a is a constant and m is a whole number, is called a monomial.
polynomial
A polynomial is a monomial, or two or more monomials, combined by addition or subtraction.
trinomial
A trinomial is a polynomial with exactly three terms.

Use Multiplication Properties of Exponents

Learning Objectives

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

  • Simplify expressions with exponents
  • Simplify expressions using the Product Property of Exponents
  • Simplify expressions using the Power Property of Exponents
  • Simplify expressions using the Product to a Power Property
  • Simplify expressions by applying several properties
  • Multiply monomials

Before you get started, take this readiness quiz.

Simplify: 34·34.
If you missed the problem, review Example 7 in Multiply and Divide Fractions.

Solution

916

Simplify: (−2)(−2)(−2).
If you missed the problem, review Example 6 in Multiply and Divide Integers.

Solution

−8

Simplify Expressions with Exponents

Remember that an exponent indicates repeated multiplication of the same quantity. For example, 24 means to multiply four factors of 2, so 24 means 2·2·2·2. This format is known as exponential notation.

Exponential Notation

On the left side, a raised to the m is shown. The m is labeled in blue as an exponent. The a is labeled in red as the base. On the right, it says a to the m means multiply m factors of a. Below this, it says a to the m equals a times a times a times a, with m factors written below in blue.

This is read a to the mth power.

In the expression am, the exponent tells us how many times we use the base a as a factor.

On the left side, 7 to the 3rd power is shown. Below is 7 times 7 times 7, with 3 factors written below. On the right side, parentheses negative 8 to the 5th power is shown. Below is negative 8 times negative 8 times negative 8 times negative 8 times negative 8, with 5 factors written below.

Before we begin working with variable expressions containing exponents, let’s simplify a few expressions involving only numbers.

Simplify:
  1. ⓐ 53
  2. ⓑ 91
Solution

Solution

This table details the step-by-step process of evaluating the exponential expression 5^3, illustrating its expansion and final simplification.
ⓐ
53
Multiply 3 factors of 5. 5·5·5
Simplify. 125
This table illustrates the definition and calculation of 9 raised to the power of 1, showing the expression and its numerical value.
ⓑ
91
Multiply 1 factor of 9. 9
Simplify:
  1. ⓐ 43
  2. ⓑ 111
Solution
  1. ⓐ 64
  2. ⓑ 11
Simplify:
  1. ⓐ 34
  2. ⓑ 211
Solution
  1. ⓐ 81
  2. ⓐ 21
Simplify:
  1. ⓐ (78)2
  2. ⓑ (0.74)2
Solution

Solution

Steps illustrating the evaluation of the square of a fraction (7/8) through multiplication and simplification.
ⓐ
(78)2
Multiply two factors. (78)(78)
Simplify. 4964
Illustrates the steps to calculate the square of 0.74, from the initial power expression to the final simplified decimal result.
ⓑ
(0.74)2
Multiply two factors. (0.74)(0.74)
Simplify. 0.5476
Simplify:
  1. ⓐ (58)2
  2. ⓑ (0.67)2
Solution
  1. ⓐ 2564
  2. ⓑ 0.4489
Simplify:
  1. ⓐ (25)3
  2. ⓑ (0.127)2
Solution
  1. ⓐ 8125
  2. ⓑ 0.016129
Simplify:
  1. ⓐ (−3)4
  2. ⓑ −34
Solution

Solution

Step-by-step evaluation of the exponential expression (-3)^4.
ⓐ
(−3)4
Multiply four factors of −3. (−3)(−3)(−3)(−3)
Simplify. 81
Step-by-step simplification of the mathematical expression -3^4, demonstrating the calculation process and final result.
ⓑ
−34
Multiply two factors. −(3·3·3·3)
Simplify. −81

Notice the similarities and differences in parts ⓐ and ⓑ. Why are the answers different? In part ⓐ the parentheses tell us to raise the (−3) to the 4th power. In part ⓑ we raise only the 3 to the 4th power and then find the opposite.

Simplify:
  1. ⓐ (−2)4
  2. ⓑ −24
Solution
  1. ⓐ 16
  2. ⓑ −16
Simplify:
  1. ⓐ (−8)2
  2. ⓑ −82
Solution
  1. ⓐ 64
  2. ⓑ −64

Simplify Expressions Using the Product Property of Exponents

You have seen that when you combine like terms by adding and subtracting, you 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. We’ll derive the properties of exponents by looking for patterns in several examples. All the exponent properties hold true for any real numbers, but right now we will only use whole number exponents.

First, we will look at an example that leads to the Product Property.

A mathematical expression shows 'x' raised to the power of 2, followed by a multiplication dot, and then 'x' raised to the power of 3. This can be simplified to x to the power of 5.
What does this mean?

How many factors altogether?
Multiplying terms with the same base: (x*x) has 2 factors, and (x*x*x) has 3 factors. Their product (x*x*x*x*x) has a total of 5 factors, demonstrating the addition of exponents.
So, we have The mathematical expression x raised to the power of 5 (x^5) is centered on a plain white background.
Notice that 5 is the sum of the exponents, 2 and 3. The image shows the mathematical expression x^2 multiplied by x^3, demonstrating that it equals x^(2+3), or x^5, illustrating the rule of adding exponents when multiplying powers with the same base.
We write: x2⋅x3
x2+3
x5

The base stayed the same and we added the exponents. This leads to the Product Property for Exponents.

Product Property of Exponents

If a is a real number and m,n are counting numbers, then

am·an=am+n

To multiply with like bases, add the exponents.

An example with numbers helps to verify this property.

22·23=?22+34·8=?2532=32✓

Simplify: x5·x7.

Solution

Solution

x5·x7
Use the product property, am·an=am+n. A mathematical expression showing 'x' raised to the power of '5+7', with the exponent '5+7' in red.
Simplify. x12

Simplify: x7·x8.

Solution

x15

Simplify: x5·x11.

Solution

x16

Simplify: b4·b.

Solution

Solution

b4·b
Rewrite, b=b1. b4·b1
Use the product property, am·an=am+n. A black lowercase letter 'b' with a red superscript '4+1', representing the mathematical expression b^(4+1).
Simplify. b5

Simplify: p9·p.

Solution

p10

Simplify: m·m7.

Solution

m8

Simplify: 27·29.

Solution

Solution

27·29
Use the product property, am·an=am+n. The mathematical expression 2 with an exponent of 7 plus 9, where the exponent 7+9 is rendered in red, indicating a potential highlight or distinction from the base number 2 which is in black.
Simplify. 216

Simplify: 6·69.

Solution

610

Simplify: 96·99.

Solution

915

Simplify: y17·y23.

Solution

Solution

y17·y23
Notice, the bases are the same, so add the exponents. The mathematical expression y^(17+23) is displayed on a white background, featuring a black 'y' with a red exponent that includes the numbers 17 and 23 separated by a plus sign.
Simplify. y40

Simplify: y24·y19.

Solution

y43

Simplify: z15·z24.

Solution

z39

We can extend the Product Property of Exponents to more than two factors.

Simplify: x3·x4·x2.

Solution

Solution

x3·x4·x2
Add the exponents, since the bases are the same. A mathematical expression shows 'x' raised to the power of '3+4+2', with the exponent partially in red.
Simplify. x9

Simplify: x7·x5·x9.

Solution

x21

Simplify: y3·y8·y4.

Solution

y15

Simplify Expressions Using the Power Property of Exponents

Now let’s look at an exponential expression that contains a power raised to a power. See if you can discover a general property.

A mathematical expression showing x squared raised to the power of 3, written as (x^2)^3.
The mathematical expression 'x^2 . x^2 . x^2' is displayed, illustrating the multiplication of x squared by itself three times.
What does this mean?

How many factors altogether?
Illustration of factors: Three sets of 'x * x', each labeled '2 factors', combine to show a total of '6 factors' in an algebraic expression.
So, we have The mathematical expression x^6 is displayed in black text on a plain white background, centered within the frame.
Notice that 6 is the product of the exponents, 2 and 3. A mathematical expression states that (x^2)^3 is equal to x^(2*3) or x^6, demonstrating the power of a power rule in exponents.
We write: (x2)3
x2⋅3
x6

We multiplied the exponents. This leads to the Power Property for Exponents.

Power Property of Exponents

If a is a real number and m,n are whole numbers, then

(am)n=am·n

To raise a power to a power, multiply the exponents.

An example with numbers helps to verify this property.

(52)3=?52·3(25)3=?5615,625=15,625✓
Simplify:
  1. ⓐ (x5)7
  2. ⓐ (36)8
Solution

Solution

ⓐ
(x5)7
Use the Power Property, (am)n=am·n. A mathematical expression showing the variable 'x' raised to the power of '5.7', where the exponent '5.7' is distinctively colored in red.
Simplify. x35
ⓑ
(36)8
Use the Power Property, (am)n=am·n. The mathematical expression 3 to the power of 6.8.
Simplify. 348
Simplify:
  1. ⓐ (x7)4
  2. ⓑ (74)8
Solution
  1. ⓐ x28
  2. ⓑ 732
Simplify:
  1. ⓐ (x6)9
  2. ⓑ (86)7
Solution
  1. ⓐ x54
  2. ⓑ 842

Simplify Expressions Using the Product to a Power Property

We will now look at an expression containing a product that is raised to a power. Look for a pattern.

This table demonstrates the Power of a Product Rule by showing the step-by-step expansion and simplification of (2x)^3 into 2^3 * x^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. (2x)3is23·x3
We write: (2x)3
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 of 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.

An example with numbers helps to verify this property:

(2·3)2=?22·3262=?4·936=36✓

Simplify: (−11x)2.

Solution

Solution

(−11x)2
Use the Power of a Product Property, (ab)m=ambm. A mathematical expression is shown, consisting of an opening parenthesis, a minus sign, the number 11, a closing parenthesis, a superscript 2, the letter x, and another superscript 2. The superscripts are in red.
Simplify. 121x2

Simplify: (−14x)2.

Solution

196x2

Simplify: (−12a)2.

Solution

144a2

Simplify: (3xy)3.

Solution

Solution

(3xy)3
Raise each factor to the third power. A mathematical expression showing '3x^3y^3', where the exponents for 'x' and 'y' are red threes.
Simplify. 27x3y3

Simplify: (−4xy)4.

Solution

256x4y4

Simplify: (6xy)3.

Solution

216x3y3

Simplify Expressions by Applying Several Properties

We now have three properties for multiplying expressions with exponents. Let’s summarize them and then we’ll do some examples that use more than one of the properties.

Properties of Exponents

If a,b are real numbers and m,n are whole numbers, then

Product Propertyam·an=am+nPower Property(am)n=am·nProduct to a Power Property(ab)m=ambm

Simplify: (x2)6(x5)4.

Solution

Solution

Step-by-step simplification of an exponential expression, applying the power property and adding exponents.
(x2)6(x5)4
Use the Power Property. x12·x20
Add the exponents. x32

Simplify: (x4)3(x7)4.

Solution

x40

Simplify: (y9)2(y8)3.

Solution

y42

Simplify: (−7x3y4)2.

Solution

Solution

This table demonstrates the step-by-step simplification of the algebraic expression (-7x^3y^4)^2 using the Power Property of exponents.
(−7x3y4)2
Take each factor to the second power. (−7)2(x3)2(y4)2
Use the Power Property. 49x6y8

Simplify: (−8x4y7)3.

Solution

−512x12y21

Simplify: (−3a5b6)4.

Solution

81a20b24

Simplify: (6n)2(4n3).

Solution

Solution

Step-by-step simplification of an algebraic expression, demonstrating exponent rules and properties.
(6n)2(4n3)
Raise 6n to the second power. 62n2·4n3
Simplify. 36n2·4n3
Use the Commutative Property. 36·4·n2·n3
Multiply the constants and add the exponents. 144n5

Notice that in the first monomial, the exponent was outside the parentheses and it applied to both factors inside. In the second monomial, the exponent was inside the parentheses and so it only applied to the n.

Simplify: (7n)2(2n12).

Solution

98n14

Simplify: (4m)2(3m3).

Solution

48m5

Simplify: (3p2q)4(2pq2)3.

Solution

Solution

Step-by-step simplification of the algebraic expression (3p^2q)^4(2pq^2)^3 to its simplified form, demonstrating exponent properties.
(3p2q)4(2pq2)3
Use the Power of a Product Property. 34(p2)4q4·23p3(q2)3
Use the Power Property. 81p8q4·8p3q6
Use the Commutative Property. 81·8·p8·p3·q4·q6
Multiply the constants and add the exponents for
each variable.
648p11q10

Simplify: (u3v2)5(4uv4)3.

Solution

64u18v22

Simplify: (5x2y3)2(3xy4)3.

Solution

675x7y18

Multiply Monomials

Since a monomial is an algebraic expression, we can use the properties for simplifying expressions with exponents to multiply the monomials.

Multiply: (4x2)(−5x3).

Solution

Solution

Steps to multiply two monomials, illustrating the rearrangement of factors using the Commutative Property to simplify the expression.
(4x2)(−5x3)
Use the Commutative Property to rearrange the factors. 4·(−5)·x2·x3
Multiply. −20x5

Multiply: (7x7)(−8x4).

Solution

−56x11

Multiply: (−9y4)(−6y5).

Solution

54y9

Multiply: (34c3d)(12cd2).

Solution

Solution

Step-by-step simplification of an algebraic expression, showing the application of the Commutative Property to rearrange factors before multiplication.
(34c3d)(12cd2)
Use the Commutative Property to rearrange
the factors.
34·12·c3·c·d·d2
Multiply. 9c4d3

Multiply: (45m4n3)(15mn3).

Solution

12m5n6

Multiply: (23p5q)(18p6q7).

Solution

12p11q8

ACCESS ADDITIONAL ONLINE RESOURCES

  • Exponent Properties
  • Exponent Properties 2

Key Concepts

  • Exponential Notation On the left side, a raised to the m is shown. The m is labeled in blue as an exponent. The a is labeled in red as the base. On the right, it says a to the m means multiply m factors of a. Below this, it says a to the m equals a times a times a times a, with m factors written below in blue.

    This is read a to the mth power.

  • Product Property of Exponents
    • If a is a real number and m,n are counting numbers, then
      am·an=am+n
    • To multiply with like bases, add the exponents.
  • Power Property for Exponents
    • If a is a real number and m,n are counting numbers, then
      (am)n =am⋅n
  • Product to a Power Property for Exponents
    • If a and b are real numbers and m is a whole number, then
      (ab)m=ambm

Practice Makes Perfect

Simplify Expressions with Exponents

In the following exercises, simplify each expression with exponents.

45

Solution

1,024

103

(12)2

Solution

14

(35)2

(0.2)3

Solution

0.008

(0.4)3

(−5)4

Solution

625

(−3)5

−54

Solution

−625

−35

−104

Solution

−10,000

−26

(−23)3

Solution

−827

(−14)4

−0.52

Solution

−0.25

−0.14

Simplify Expressions Using the Product Property of Exponents

In the following exercises, simplify each expression using the Product Property of Exponents.

x3·x6

Solution

x9

m4·m2

a·a4

Solution

a5

y12·y

35·39

Solution

314

510·56

z·z2·z3

Solution

z6

a·a3·a5

xa·x2

Solution

xa+2

yp·y3

ya·yb

Solution

ya+b

xp·xq

Simplify Expressions Using the Power Property of Exponents

In the following exercises, simplify each expression using the Power Property of Exponents.

(u4)2

Solution

u8

(x2)7

(y5)4

Solution

y20

(a3)2

(102)6

Solution

1012

(28)3

(x15)6

Solution

x90

(y12)8

(x2)y

Solution

x2y

(y3)x

(5x)y

Solution

5xy

(7a)b

Simplify Expressions Using the Product to a Power Property

In the following exercises, simplify each expression using the Product to a Power Property.

(5a)2

Solution

25a2

(7x)2

(−6m)3

Solution

−216m3

(−9n)3

(4rs)2

Solution

16r2s2

(5ab)3

(4xyz)4

Solution

256x4y4z4

(−5abc)3

Simplify Expressions by Applying Several Properties

In the following exercises, simplify each expression.

(x2)4·(x3)2

Solution

x14

(y4)3·(y5)2

(a2)6·(a3)8

Solution

a36

(b7)5·(b2)6

(3x)2(5x)

Solution

45x3

(2y)3(6y)

(5a)2(2a)3

Solution

200a5

(4b)2(3b)3

(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

Multiply Monomials

In the following exercises, multiply the following monomials.

(12x2)(−5x4)

Solution

−60x6

(−10y3)(7y2)

(−8u6)(−9u)

Solution

72u7

(−6c4)(−12c)

(15r8)(20r3)

Solution

4r11

(14a5)(36a2)

(4a3b)(9a2b6)

Solution

36a5b7

(6m4n3)(7mn5)

(47xy2)(14xy3)

Solution

8x2y5

(58u3v)(24u5v)

(23x2y)(34xy2)

Solution

12x3y3

(35m3n2)(59m2n3)

Everyday Math

Email Janet emails a joke to six of her friends and tells them to forward it to six of their friends, who forward it to six of their friends, and so on. The number of people who receive the email on the second round is 62, on the third round is 63, as shown in the table. How many people will receive the email on the eighth round? Simplify the expression to show the number of people who receive the email.

Round Number of people
1 6
2 62
3 63
… …
8 ?
Solution

1,679,616

Salary Raul’s boss gives him a 5% raise every year on his birthday. This means that each year, Raul’s salary is 1.05 times his last year’s salary. If his original salary was $40,000, his salary after 1 year was $40,000(1.05), after 2 years was $40,000(1.05)2, after 3 years was $40,000(1.05)3, as shown in the table below. What will Raul’s salary be after 10 years? Simplify the expression, to show Raul’s salary in dollars.

Year Salary
1 $40,000(1.05)
2 $40,000(1.05)2
3 $40,000(1.05)3
… …
10 ?

Writing Exercises

Use the Product Property for Exponents to explain why x·x=x2.

Solution

Answers will vary.

Explain why −53=(−5)3 but −54≠(−5)4.

Jorge thinks (12)2 is 1. What is wrong with his reasoning?

Solution

Answers will vary.

Explain why x3·x5 is x8, and not x15.

Self Check

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

A self-assessment checklist for students to rate their understanding of simplifying expressions with exponents and multiplying monomials, using categories like 'Confidently' and 'No-I don't get it!'

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

Multiply Polynomials

Learning Objectives

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

  • Multiply a polynomial by a monomial
  • Multiply a binomial by a binomial
  • Multiply a trinomial by a binomial

Before you get started, take this readiness quiz.

Distribute: 2(x+3).
If you missed the problem, review Example 1 in Distributive Property.

Solution

2x+6

Distribute: −11(4−3a).
If you missed the problem, review Example 10 in Distributive Property.

Solution

−44+33a

Combine like terms: x2+9x+7x+63.
If you missed the problem, review Example 9 in Evaluate, Simplify, and Translate Expressions.

Solution

x2+16x+63

Multiply a Polynomial by a Monomial

In Distributive Property you learned to use the Distributive Property to simplify expressions such as 2(x−3). You multiplied both terms in the parentheses, xand3, by 2, to get 2x−6. With this chapter's new vocabulary, you can say you were multiplying a binomial, x−3, by a monomial, 2. Multiplying a binomial by a monomial is nothing new for you!

Multiply: 3(x+7).

Solution

Solution

3(x+7)
Distribute. An image illustrating the distributive property in algebra, showing red arrows indicating how to multiply 3 by both 'x' and '7' in the expression 3(x + 7).
3·x+3·7
Simplify. 3x+21

Multiply: 6(x+8).

Solution

6x + 48

Multiply: 2(y+12).

Solution

2y + 24

Multiply: x(x−8).

Solution

Solution

A mathematical expression reads x(x-8) in black text on a white background.
Distribute. Red arrows indicate the distributive property for the expression x(x - 8), showing that the outer 'x' multiplies each term within the parentheses.
The mathematical expression x squared minus 8x, written in black font on a white background.
Simplify. The mathematical expression x squared minus 8x, written in black font on a white background.

Multiply: y(y−9).

Solution

y2 − 9y

Multiply: p(p−13).

Solution

p2 − 13p

Multiply: 10x(4x+y).

Solution

Solution

The image shows the algebraic expression 10x(4x + y).
Distribute. An algebraic expression 10x(4x + y) is shown with red curved arrows illustrating the distributive property, pointing from 10x to both 4x and y inside the parentheses.
The image shows the mathematical expression 10x times 4x plus 10x times y.
Simplify. The image shows the mathematical expression 40x^2 + 10xy, rendered in a black serif font against a plain white background.

Multiply: 8x(x+3y).

Solution

8x2 + 24xy

Multiply: 3r(6r+s).

Solution

18r2 + 3rs

Multiplying a monomial by a trinomial works in much the same way.

Multiply: −2x(5x2+7x−3).

Solution

Solution

−2x(5x2+7x−3)
Distribute. The image shows the distributive property being applied to the expression -2x(5x^2 + 7x - 3), with red arrows illustrating how -2x is multiplied by each term inside the parentheses.
−2x⋅5x2+(−2x)⋅7x−(−2x)⋅3
Simplify. −10x3−14x2+6x

Multiply: −4y(8y2+5y−9).

Solution

−32y3 − 20y2 + 36y

Multiply: −6x(9x2+x−1).

Solution

−54x3 − 6x2 + 6x

Multiply: 4y3(y2−8y+1).

Solution

Solution

4y3(y2−8y+1)
Distribute. An algebraic expression shows the distributive property. The term 4y^3 is multiplied by each term inside the parentheses (y^2 - 8y + 1), indicated by red curved arrows.
4y3⋅y2−4y3⋅8y+4y3⋅1
Simplify. 4y5−32y4+4y3

Multiply: 3x2(4x2−3x+9).

Solution

12x4 − 9x3 + 27x2

Multiply: 8y2(3y2−2y−4).

Solution

24y4 − 16y3 − 32y2

Now we will have the monomial as the second factor.

Multiply: (x+3)p.

Solution

Solution

(x+3)p
Distribute. Illustrating the distributive property in algebra, showing p being multiplied by x and by 3 in the expression (x+3)p.
x⋅p+3⋅p
Simplify. xp+3p

Multiply: (x+8)p.

Solution

xp + 8p

Multiply: (a+4)p.

Solution

ap + 4p

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.

Using the Distributive Property

We will start by using the Distributive Property. Look again at Example 6.

An algebraic expression (x+3)p, demonstrating the distributive property with curved red arrows indicating that 'p' multiplies both 'x' and '3' within the parentheses.
We distributed the p to get The mathematical expression xp + 3p is shown, featuring the variables 'x' and 'p' with 'p' highlighted in red, indicating a common factor.
What if we have (x+7) instead of p?
The image displays the text 'Think of the (x + 7) as the p above.' in a teal-like font, with the letter 'p' highlighted in red.
The distributive property (FOIL method) applied to the binomial multiplication (x+3)(x+7), with red arrows showing the term distribution.
Distribute (x+7). An algebraic expression showing x(x + 7) + 3(x + 7) with red arrows illustrating the distributive property applied to both terms.
Distribute again. x2+7x+3x+21
Combine like terms. x2+10x+21

Notice that before combining like terms, we had four terms. We multiplied the two terms of the first binomial by the two terms of the second binomial—four multiplications.

Be careful to distinguish between a sum and a product.

SumProductx+xx·x2xx2combine like termsadd exponents of like bases

Multiply: (x+6)(x+8).

Solution
Solution
(x+6)(x+8)
This image shows the initial steps of multiplying two binomials, (x+6)(x+8), using the distributive property. Red arrows demonstrate 'x' from the first term multiplying 'x' and '8' in the second term.
Distribute (x+8). A mathematical expression shows x multiplied by the quantity (x plus 8), added to 6 multiplied by the quantity (x plus 8).
Distribute again. x2+8x+6x+48
Simplify. x2+14x+48

Multiply: (x+8)(x+9).

Solution

x2 + 17x + 72

Multiply: (a+4)(a+5).

Solution

a2 + 9a + 20

Now we'll see how to multiply binomials where the variable has a coefficient.

Multiply: (2x+9)(3x+4).

Solution
Solution
(2x+9)(3x+4)
Distribute. (3x+4) A mathematical expression shows 2x multiplied by the quantity 3x plus 4, added to 9 multiplied by the quantity 3x plus 4. The terms (3x + 4) are highlighted in red, indicating a common factor.
Distribute again. 6x2+8x+27x+36
Simplify. 6x2+35x+36

Multiply: (5x+9)(4x+3).

Solution

20x2 + 51x + 27

Multiply: (10m+9)(8m+7).

Solution

80m2 + 142m + 63

In the previous examples, the binomials were sums. When there are differences, we pay special attention to make sure the signs of the product are correct.

Multiply: (4y+3)(6y−5).

Solution
Solution
(4y+3)(6y−5)
Distribute. The image shows the mathematical expression 4y(6y - 5) + 3(6y - 5), which can be simplified by factoring out the common term (6y - 5).
Distribute again. 24y2−20y+18y−15
Simplify. 24y2−2y−15

Multiply: (7y+1)(8y−3).

Solution

56y2 − 13y − 3

Multiply: (3x+2)(5x−8).

Solution

15x2 − 14x − 16

Up to this point, the product of two binomials has been a trinomial. This is not always the case.

Multiply: (x+2)(x−y).

Solution
Solution
A mathematical expression in black text on a white background, showing the product of two binomials: (x + 2)(x - y).
Distribute. The image displays the mathematical expression x(x - y) + 2(x - y), which involves variables x and y, parentheses, multiplication, addition, and subtraction.
Distribute again. The image displays the algebraic expression x^2 - xy + 2x - 2y, which is a polynomial with four terms involving variables x and y, and constant coefficients.
Simplify. There are no like terms to combine.

Multiply: (x+5)(x−y).

Solution

x2 − xy + 5x − 5y

Multiply: (x+2y)(x−1).

Solution

x2 − x + 2xy − 2y

Using the FOIL Method

Remember that when you multiply a binomial by a binomial you get four terms. Sometimes you can combine like terms to get a trinomial, but sometimes there are no like terms to combine. Let's look at the last example again and pay particular attention to how we got the four terms.

(x+2)(x−y)
x2−xy+2x−2y

Where did the first term, x2, come from?

It is the product of xandx, the first terms in (x+2)and(x−y).

Parentheses x plus 2 times parentheses x minus y is shown. There is a red arrow from the first x to the second. Beside this, “First” is written in red.

The next term, −xy, is the product of xand−y, the two outer terms.

Parentheses x plus 2 times parentheses x minus y is shown. There is a black arrow from the first x to the second x. There is a red arrow from the first x to the y. Beside this, “Outer” is written in red.

The third term, +2x, is the product of 2andx, the two inner terms.

Parentheses x plus 2 times parentheses x minus y is shown. There is a black arrow from the first x to the second x. There is a black arrow from the first x to the y. There is a red arrow from the 2 to the x. Below that, “Inner” is written in red.

And the last term, −2y, came from multiplying the two last terms.

Parentheses x plus 2 times parentheses x minus y is shown. There is a black arrow from the first x to the second x. There is a black arrow from the first x to the y. There is a black arrow from the 2 to the x. There is a red arrow from the 2 to the y. Above that, “Last” is written in red.

We abbreviate “First, Outer, Inner, Last” as FOIL. The letters stand for ‘First, Outer, Inner, Last’. The word FOIL is easy to remember and ensures we find all four products. We might say we use the FOIL method to multiply two binomials.

Parentheses a plus b times parentheses c plus d is shown. Above a is first, above b is last, above c is first, above d is last. There is a brace connecting a and d that says outer. There is a brace connecting b and c that says inner.

Let's look at (x+3)(x+7) again. Now we will work through an example where we use the FOIL pattern to multiply two binomials.

Comparing the Distributive Property and FOIL method for multiplying binomials, like (x+3)(x+7), both resulting in x^2 + 10x + 21.

Multiply using the FOIL method: (x+6)(x+9).

Solution
Solution
Step 1: Multiply the First terms. An algebraic expression (x + 6)(x + 9) being expanded using the FOIL method, with the first term x² explicitly shown and placeholders for the Outer, Inner, and Last terms.
Step 2: Multiply the Outer terms. This image demonstrates the 'First' step of the FOIL method for multiplying binomials (x+6)(x+9). The red arrow points from the 'x' in the first term to the 'x' in the second, showing x * x = x^2. The partial result includes x^2 and the 'Outer' term, 9x, with spaces for the 'Inner' and 'Last' terms.
Step 3: Multiply the Inner terms. This image illustrates the FOIL method for multiplying binomials. It shows (x+6)(x+9) expanding to x^2 + 9x + 6x + ___, with the red arrow emphasizing the 'Inner' term multiplication of 6 and x.
Step 4: Multiply the Last terms. The FOIL method for multiplying binomials is shown with (x+6)(x+9) expanding to x^2 + 9x + 6x + 54, clearly marking the First, Outer, Inner, and Last products.
Step 5: Combine like terms, when possible. The image displays the quadratic expression x^2 + 15x + 54.

Multiply using the FOIL method: (x+7)(x+8).

Solution

x2 + 15x + 56

Multiply using the FOIL method: (y+14)(y+2).

Solution

y2 + 16y + 28

We summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!

Use the FOIL method for multiplying two binomials.

  1. Multiply the First terms.
  2. Multiply the Outer terms.
  3. Multiply the Inner terms.
  4. Multiply the Last terms.
  5. Combine like terms, when possible.
Parentheses a plus b times parentheses c plus d is shown. Above a is first, above b is last, above c is first, above d is last. There is a brace connecting a and d that says outer. There is a brace connecting b and c that says inner.

Multiply: (y−8)(y+6).

Solution
Solution
Step 1: Multiply the First terms. This image demonstrates the FOIL method for multiplying binomials. It shows (y-8)(y+6) expanding to y^2 + _ + _ + _ with F, O, I, L terms indicated below, where y^2 is the 'First' term.
Step 2: Multiply the Outer terms. The image illustrates the FOIL method for multiplying binomials (y-8)(y+6). Arrows highlight the First (y*y), Outer (y*6), Inner (-8*y), and Last (-8*6) terms. The partial expansion y^2 + 6y + _ + _ is labeled with F, O, I, L.
Step 3: Multiply the Inner terms. An image illustrating the FOIL method for multiplying two binomials, (y - 8)(y + 6). It shows the 'First', 'Outer', and 'Inner' terms of the expansion y^2 + 6y - 8y, with a blank for the 'Last' term.
Step 4: Multiply the Last terms. This image illustrates the FOIL method for multiplying two binomials: (y-8)(y+6). It shows how to expand it to y^2 + 6y - 8y - 48, indicating the 'First', 'Outer', 'Inner', and 'Last' terms with arrows and labels.
Step 5: Combine like terms A mathematical expression, y squared minus 2y minus 48, is centered on a white background. It represents a quadratic polynomial.

Multiply: (y−3)(y+8).

Solution

y2 + 5y − 24

Multiply: (q−4)(q+5).

Solution

q2 + q − 20

Multiply: (2a+3)(3a−1).

Solution
Solution
The image displays the mathematical expression (2a + 3)(3a - 1).
The FOIL method is illustrated for multiplying the binomials (2a + 3) and (3a - 1), showing the distributive steps with curved arrows.
Multiply the First terms. A mathematics diagram illustrating the start of the FOIL method, showing the product of the first terms '2a * 3a = 6a^2' with placeholders for Outer, Inner, and Last terms denoted by F, O, I, L.
Multiply the Outer terms. This image presents a math exercise on the FOIL method, displaying the First (6a^2) and Outer (-2a) terms of a binomial product. Blanks are provided for the Inner and Last terms, accompanied by a hint: 2a * (-1).
Multiply the Inner terms. Image showing an algebraic expression 6a^2 - 2a + 9a + __ with F, O, I, L labels below, illustrating the FOIL method for multiplying binomials. A separate term '3 * 3a' is also visible.
Multiply the Last terms. Mathematical notation showing the expansion of a binomial using the FOIL method, with terms 6a^2, -2a, +9a, and -3 labeled F, O, I, L respectively. A multiplication 3 * (-1) is also visible.
Combine like terms. The image displays the quadratic expression 6a^2 + 7a - 3 in black text on a white background.

Multiply: (4a+9)(5a−2).

Solution

20a2 + 37a − 18

Multiply: (7x+4)(7x−8).

Solution

49x2 − 28x − 32

Multiply: (5x−y)(2x−7).

Solution
Solution
A mathematical expression showing the product of two binomials: (5x - y) and (2x - 7).
An algebraic expression showing the multiplication of two binomials, (5x - y) and (2x - 7), with curved arrows illustrating the FOIL method for distribution.
Multiply the First terms. An algebra example demonstrating the FOIL method for multiplying binomials, with '10x^2' as the 'F' (First) term and blanks for the 'O', 'I', and 'L' terms.
Multiply the Outer terms. An algebraic expression demonstrating the FOIL method, showing the 'First' term as 10x^2 and the 'Outer' term as -35x, with blanks representing the 'Inner' and 'Last' terms.
Multiply the Inner terms. A partial algebraic expression 10x^2 - 35x - 2xy + ___ with the letters F, O, I, L annotated below, illustrating the FOIL method where the 'L' (Last) term is missing.
Multiply the Last terms. The polynomial expression 10x² - 35x - 2xy + 7y is displayed, with the letters F, O, I, L beneath its terms, referencing the FOIL method. The last term '+7y' is notably highlighted in red.
Combine like terms. There are none. A mathematical expression reads 10x^2 - 35x - 2xy + 7y. The numbers and variables are black except for the '+ 7y' at the end, which is rendered in red.

Multiply: (12x−y)(x−5).

Solution

12x2 − 60x − xy + 5y

Multiply: (6a−b)(2a−9).

Solution

12a2 − 54a − 2ab + 9b

Using the Vertical Method

The FOIL method is usually the quickest method for multiplying two binomials, but it works only 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.

A vertical multiplication problem is shown. 23 times 46 is written with a line underneath. Beneath the line is 138. Beside 138 is written “partial product.” Beneath 138 is 92. Beside 92 is written “partial product.” Beneath 92 is a line and 1058. Beside 1058 is written “product.”

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: (5x−1)(2x−7).

Solution
Solution

It does not matter which binomial goes on the top. Line up the columns when you multiply as we did when we multiplied 23(46).

A vertical setup for multiplying two binomials, (2x - 7) by (5x - 1), using a format similar to long multiplication, with the multiplication symbol and an underline indicating the operation.
Multiply 2x−7 by −1. The image shows the expression '-2x + 7' in red, followed by the words 'partial product' in teal.
Multiply 2x−7 by 5x. The expression 10x^2 - 35x is presented as a 'partial product,' representing an intermediate step in a larger polynomial multiplication or division calculation.
Add like terms. A quadratic expression 10x^2 - 37x + 7 is shown next to the word 'product'.

Notice the partial products are the same as the terms in the FOIL method.

On the left, 5x minus 1 times 2x minus 7 is shown. Below that is 10 x squared minus 35x minus 2x plus 7. The first two terms are in blue, the second two in red. Beneath that is 10 x squared minus 37x plus 7. On the right, a vertical multiplication problem is shown. 2xx minus 7 times 5x minus 1 is written with a line underneath. Beneath the line is a red negative 2x plus 7. Beneath that is 10 x squared minus 35 x in blue. Beneath that, there is another line. Beneath that line is 10 x squared minus 37x plus 7.

Multiply using the vertical method: (4m−9)(3m−7).

Solution

12m2 − 55m + 63

Multiply using the vertical method: (6n−5)(7n−2).

Solution

42n2 − 47n + 10

We have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The three methods are listed here to help you remember them.

Multiplying Two Binomials

To multiply binomials, use the:
  • Distributive Property
  • FOIL Method
  • Vertical Method
Remember, FOIL only works when multiplying two binomials.

Multiply a Trinomial by a Binomial

We have multiplied monomials by monomials, monomials by polynomials, and binomials by binomials. Now we're ready to multiply a trinomial by a binomial. Remember, the FOIL method will not work in this case, but we can use either the Distributive Property or the Vertical Method. We first look at an example using the Distributive Property.

Multiply using the Distributive Property: (x+3)(2x2−5x+8).

Solution

Solution

Diagram depicting the initial steps of multiplying two polynomials: (x+3) and (2x^2-5x+8), highlighting the distribution of 'x' to the first two terms.
Distribute. An algebraic expression showing x(2x^2 - 5x + 8) + 3(2x^2 - 5x + 8), which represents a sum of terms suitable for factoring or expansion.
Multiply. 2x3−5x2+8x+6x2−15x+24
Combine like terms. 2x3+x2−7x+24

Multiply using the Distributive Property: (y−1)(y2−7y+2).

Solution

y3 − 8y2 + 9y − 2

Multiply using the Distributive Property: (x+2)(3x2−4x+5).

Solution

3x3 + 2x2 − 3x + 10

Now let's do this same multiplication using the Vertical Method.

Multiply using the Vertical Method: (x+3)(2x2−5x+8).

Solution

Solution

It is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products this way.

A vertical long multiplication setup for polynomials. The top polynomial is 2x^2 - 5x + 8, and the bottom polynomial is x + 3, with a multiplication symbol on the left.
Multiply (2x2−5x+8) by 3. The image shows the mathematical expression 6x^2 - 15x + 24.
Multiply (2x2−5x+8) by x. A mathematical expression, 2x^3 - 5x^2 + 8x, is displayed above a horizontal line, representing a fraction or a numerator.
Add like terms. The mathematical expression 2x^3 + x^2 - 7x + 24 is displayed in black font against a white background.

Multiply using the Vertical Method: (y−1)(y2−7y+2).

Solution

y3 − 8y2 + 9y − 2

Multiply using the Vertical Method: (x+2)(3x2−4x+5).

Solution

3x3 + 2x2 − 3x + 10

ACCESS ADDITIONAL ONLINE RESOURCES

  • Multiply Monomials
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  • Multiply Polynomials Review
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  • Multiply Binomials

Key Concepts

  • Use the FOIL method for multiplying two binomials.
    Illustrative steps for applying the FOIL method for multiplying binomials.
    Step 1. Multiply the First terms. A diagram illustrating the 'first', 'outer', 'inner', and 'last' (FOIL) terms when multiplying two binomials, (a+b)(c+d).
    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
    • Vertical Method
  • Multiplying a Trinomial by a Binomial: To multiply a trinomial by a binomial, use the:
    • Distributive Property
    • Vertical Method

Practice Makes Perfect

Multiply a Polynomial by a Monomial

In the following exercises, multiply.

4(x+10)

Solution

4x + 40

6(y+8)

15(r−24)

Solution

15r − 360

12(v−30)

−3(m+11)

Solution

−3m − 33

−4(p+15)

−8(z−5)

Solution

−8z + 40

−3(x−9)

u(u+5)

Solution

u2 + 5u

q(q+7)

n(n2−3n)

Solution

n3 − 3n2

s(s2−6s)

12x(x−10)

Solution

12x2 − 120x

9m(m−11)

−9a(3a+5)

Solution

−27a2 − 45a

−4p(2p+7)

6x(4x+y)

Solution

24x2 + 6xy

5a(9a+b)

5p(11p−5q)

Solution

55p2 − 25pq

12u(3u−4v)

3(v2+10v+25)

Solution

3v2 + 30v + 75

6(x2+8x+16)

2n(4n2−4n+1)

Solution

8n3 − 8n2 + 2n

3r(2r2−6r+2)

−8y(y2+2y−15)

Solution

−8y3 − 16y2 + 120y

−5m(m2+3m−18)

5q3(q2−2q+6)

Solution

5q5 − 10q4 + 30q3

9r3(r2−3r+5)

−4z2(3z2+12z−1)

Solution

−12z4 − 48z3 + 4z2

−3x2(7x2+10x−1)

(2y−9)y

Solution

2y2 − 9y

(8b−1)b

(w−6)·8

Solution

8w − 48

(k−4)·5

Multiply a Binomial by a Binomial

In the following exercises, multiply the following binomials using: ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical method

(x+4)(x+6)

Solution

x2 + 10x + 24

(u+8)(u+2)

(n+12)(n−3)

Solution

n2 + 9n − 36

(y+3)(y−9)

In the following exercises, multiply the following binomials. Use any method.

(y+8)(y+3)

Solution

y2 + 11y + 24

(x+5)(x+9)

(a+6)(a+16)

Solution

a2 + 22a + 96

(q+8)(q+12)

(u−5)(u−9)

Solution

u2 − 14u + 45

(r−6)(r−2)

(z−10)(z−22)

Solution

z2 − 32z + 220

(b−5)(b−24)

(x−4)(x+7)

Solution

x2 + 3x − 28

(s−3)(s+8)

(v+12)(v−5)

Solution

v2 + 7v − 60

(d+15)(d−4)

(6n+5)(n+1)

Solution

6n2 + 11n + 5

(7y+1)(y+3)

(2m−9)(10m+1)

Solution

20m2 − 88m − 9

(5r−4)(12r+1)

(4c−1)(4c+1)

Solution

16c2 − 1

(8n−1)(8n+1)

(3u−8)(5u−14)

Solution

15u2 − 82u + 112

(2q−5)(7q−11)

(a+b)(2a+3b)

Solution

2a2 + 5ab + 3b2

(r+s)(3r+2s)

(5x−y)(x−4)

Solution

5x2 − 20x − xy + 4y

(4z−y)(z−6)

Multiply a Trinomial by a Binomial

In the following exercises, multiply using ⓐ the Distributive Property and ⓑ the Vertical Method.

(u+4)(u2+3u+2)

Solution

u3 + 7u2 + 14u + 8

(x+5)(x2+8x+3)

(a+10)(3a2+a−5)

Solution

3a3 + 31a2 + 5a − 50

(n+8)(4n2+n−7)

In the following exercises, multiply. Use either method.

(y−6)(y2−10y+9)

Solution

y3 − 16y2 + 69y − 54

(k−3)(k2−8k+7)

(2x+1)(x2−5x−6)

Solution

2x3 − 9x2 − 17x − 6

(5v+1)(v2−6v−10)

Everyday Math

Mental math You can use binomial multiplication to multiply numbers without a calculator. Say you need to multiply 13 times 15. Think of 13 as 10+3 and 15 as 10+5.

  1. ⓐ Multiply (10+3)(10+5) by the FOIL method.
  2. ⓑ Multiply 13·15 without using a calculator.
  3. ⓒ Which way is easier for you? Why?
Solution
  1. ⓐ 195
  2. ⓑ 195
  3. ⓒ Answers will vary.

Mental math You can use binomial multiplication to multiply numbers without a calculator. Say you need to multiply 18 times 17. Think of 18 as 20−2 and 17 as 20−3.

  1. ⓐ Multiply (20−2)(20−3) by the FOIL method.
  2. ⓑ Multiply 18·17 without using a calculator.
  3. ⓒ Which way is easier for you? Why?

Writing Exercises

Which method do you prefer to use when multiplying two binomials—the Distributive Property, the FOIL method, or the Vertical Method? Why?

Solution

Answers will vary.

Which method do you prefer to use when multiplying a trinomial by a binomial—the Distributive Property or the Vertical Method? Why?

Self Check

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

A self-assessment table for students to evaluate their understanding of multiplying polynomials, binomials, and trinomials, with options to rate their skill level as '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?

Divide Monomials

Learning Objectives

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

  • Simplify expressions using the Quotient Property of Exponents
  • Simplify expressions with zero exponents
  • Simplify expressions using the Quotient to a Power Property
  • Simplify expressions by applying several properties
  • Divide monomials

Before you get started, take this readiness quiz.

Simplify: 824.
If you missed the problem, review Example 1 in Multiply and Divide Fractions.

Solution

13

Simplify: (2m3)5.
If you missed the problem, review Example 13 in Use Multiplication Properties of Exponents.

Solution

32m15

Simplify: 12x12y.
If you missed the problem, review Example 5 in Multiply and Divide Fractions.

Solution

xy

Simplify Expressions Using the Quotient Property of Exponents

Earlier in this chapter, we developed the properties of exponents for multiplication. We summarize these properties here.

Summary of Exponent Properties for Multiplication

If a,b are real numbers and m,n are whole numbers, then

Product Propertyam⋅an=am+nPower Property(am)n=am⋅nProduct to a Power(ab)m=ambm

Now we will look at the exponent properties for division. A quick memory refresher may help before we get started. In Fractions you learned that fractions may be simplified by dividing out common factors from the numerator and denominator using the Equivalent Fractions Property. This property will also help us work with algebraic fractions—which are also quotients.

Equivalent Fractions Property

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

ab=a·cb·canda·cb·c=ab

As before, we'll try to discover a property by looking at some examples.

Considerx5x2andx2x3What do they mean?x⋅x⋅x⋅x⋅xx⋅xx⋅xx⋅x⋅xUse the Equivalent Fractions Property.x⋅x⋅x⋅x⋅xx⋅x⋅1x⋅x⋅1x⋅x⋅xSimplify.x31x

Notice that in each case the bases were the same and we subtracted the exponents.

  • When the larger exponent was in the numerator, we were left with factors in the numerator and 1 in the denominator, which we simplified.
  • When the larger exponent was in the denominator, we were left with factors in the denominator, and 1 in the numerator, which could not be simplified.

We write:

x5x2x2x3x5−21x3−2x31x

Quotient Property of Exponents

If a is a real number, a≠0, and m,n are whole numbers, then

aman=am−n,m>nandaman=1an−m,n>m

A couple of examples with numbers may help to verify this property.

3432=?34−25253=?153−2819=?3225125=?1519=9✓15=15✓

When we work with numbers and the exponent is less than or equal to 3, we will apply the exponent. When the exponent is greater than 3, we leave the answer in exponential form.

Simplify:
  1. ⓐ x10x8
  2. ⓑ 2922
Solution

Solution

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

ⓐ
Since 10 > 8, there are more factors of x in the numerator. x10x8
Use the quotient property with m>n,aman=am−n. The mathematical expression x^(10-8) is displayed, with the exponent '10-8' rendered in red atop a black 'x' on a white background.
Simplify. x2
ⓑ
Since 9 > 2, there are more factors of 2 in the numerator. 2922
Use the quotient property with m>n,aman=am−n. A mathematical expression showing the number 2 raised to the power of 9 minus 2 (2^(9-2)).
Simplify. 27

Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.

Simplify:
  1. ⓐ x12x9
  2. ⓑ 71475
Solution
  1. ⓐ x3
  2. ⓑ 79
Simplify:
  1. ⓐ y23y17
  2. ⓑ 81587
Solution
  1. ⓐ y6
  2. ⓑ 88
Simplify:
  1. ⓐ b10b15
  2. ⓑ 3335
Solution

Solution

To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.

ⓐ
Since 15 > 10, there are more factors of b in the denominator. b10b15
Use the quotient property with n>m,aman=1an−m. A mathematical expression showing the fraction 1 over b raised to the power of 15 minus 10.
Simplify. 1b5
ⓑ
Since 5 > 3, there are more factors of 3 in the denominator. 3335
Use the quotient property with n>m,aman=1an−m. The fraction 1 over 3 raised to the power of (5-3). The numerator '1' and the exponent '5-3' are red, with the base '3' in black.
Simplify. 132
Apply the exponent. 19

Notice that when the larger exponent is in the denominator, we are left with factors in the denominator and 1 in the numerator.

Simplify:
  1. ⓐ x8x15
  2. ⓑ 12111221
Solution
  1. ⓐ 1x7
  2. ⓑ 11210
Simplify:
  1. ⓐ m17m26
  2. ⓑ 78714
Solution
  1. ⓐ 1m9
  2. ⓑ 176
Simplify:
  1. ⓐ a5a9
  2. ⓑ x11x7
Solution

Solution

ⓐ
Since 9 > 5, there are more a's in the denominator and so we will end up with factors in the denominator. a5a9
Use the Quotient Property for n>m,aman=1an−m. The fraction 1 over 'a' raised to the power of (9 minus 5). The numerator '1' and the exponent '9-5' are highlighted in red, contrasting with the black 'a' and fraction bar.
Simplify. 1a4
ⓑ
Notice there are more factors of x in the numerator, since 11 > 7. So we will end up with factors in the numerator. x11x7
Use the Quotient Property for m>n,aman=an−m. A mathematical expression showing 'x' raised to the power of '11-7', with the exponent in red color.
Simplify. x4
Simplify:
  1. ⓐ b19b11
  2. ⓑ z5z11
Solution
  1. ⓐ b8
  2. ⓑ 1z6
Simplify:
  1. ⓐ p9p17
  2. ⓑ w13w9
Solution
  1. ⓐ 1p8
  2. ⓑ w4

Simplify Expressions with Zero Exponents

A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like amam. From earlier work with fractions, we know that

22=11717=1−43−43=1

In words, a number divided by itself is 1. So xx=1, for any x (x≠0), since any number divided by itself is 1.

The Quotient Property of Exponents shows us how to simplify aman when m>n and when n<m by subtracting exponents. What if m=n?

Now we will simplify amam in two ways to lead us to the definition of the zero exponent.

Consider first 88, which we know is 1.

88=1
Write 8 as 23. 2323=1
Subtract exponents. 23−3=1
Simplify. 20=1
This image demonstrates the proof that any non-zero number raised to the power of zero equals one (a^0 = 1), using both exponent rules and the cancellation of factors.

We see aman simplifies to a a0 and to 1. So a0=1.

Zero Exponent

If a is a non-zero number, then a0=1.

Any nonzero 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:
  1. ⓐ 120
  2. ⓑ y0
Solution

Solution

The definition says any non-zero number raised to the zero power is 1.

Illustrates the evaluation of 12^0 by applying the definition of the zero exponent, resulting in 1.
ⓐ
120
Use the definition of the zero exponent. 1
This table illustrates the zero exponent rule, showing that y^0 simplifies to 1 based on its definition.
ⓑ
y0
Use the definition of the zero exponent. 1
Simplify:
  1. ⓐ 170
  2. ⓑ m0
Solution
  1. ⓐ 1
  2. ⓑ 1
Simplify:
  1. ⓐ k0
  2. ⓑ 290
Solution
  1. ⓐ 1
  2. ⓑ 1

Now that we have defined the zero exponent, we can expand all the Properties of Exponents to include whole number exponents.

What about raising an expression to the zero power? Let's look at (2x)0. We can use the product to a power rule to rewrite this expression.

Step-by-step simplification of the expression (2x)^0, demonstrating the application of product to a power and zero exponent rules to arrive at the final value of 1.
(2x)0
Use the Product to a Power Rule. 20x0
Use the Zero Exponent Property. 1⋅1
Simplify. 1

This tells us that any non-zero expression raised to the zero power is one.

Simplify: (7z)0.

Solution

Solution

This table demonstrates the zero exponent rule, showing (7z)^0 evaluates to 1.
(7z)0
Use the definition of the zero exponent. 1

Simplify: (−4y)0.

Solution

1

Simplify: (23x)0.

Solution

1

Simplify:
  1. ⓐ (−3x2y)0
  2. ⓑ −3x2y0
Solution

Solution

This table illustrates the simplification of an expression raised to the zero power, demonstrating the zero exponent rule.
ⓐ
The product is raised to the zero power. (−3x2y)0
Use the definition of the zero exponent. 1
Steps demonstrating the simplification of the algebraic expression -3x^2y^0 using the zero exponent rule.
ⓑ
Notice that only the variable y is being raised to the zero power. −3x2y0
Use the definition of the zero exponent. −3x2⋅1
Simplify. −3x2
Simplify:
  1. ⓐ (7x2y)0
  2. ⓑ 7x2y0
Solution
  1. ⓐ 1
  2. ⓑ 7x2
Simplify:
  1. ⓐ −23x2y0
  2. ⓑ (−23x2y)0
Solution
  1. ⓐ −23x2
  2. ⓑ 1

Simplify Expressions Using the Quotient to a Power Property

Now we will look at an example that will lead us to the Quotient to a Power Property.

Demonstration of simplifying a quotient raised to a power, showing (x/y)^3 expands to x^3/y^3.
(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.

We write:(xy)3x3y3

This leads to the Quotient to a Power Property for Exponents.

Quotient to a Power Property of Exponents

If a and b are real numbers, b≠0, and m is a counting number, then

(ab)m=ambm

To raise a fraction to a power, raise the numerator and denominator to that power.

An example with numbers may help you understand this property:

(23)3=?233323⋅23⋅23=?827827=827✓
Simplify:
  1. ⓐ (58)2
  2. ⓑ (x3)4
  3. ⓒ (ym)3
Solution

Solution

ⓐ
The mathematical expression (5/8) squared, representing the fraction five-eighths raised to the power of two.
Use the Quotient to a Power Property, (ab)m=ambm. A mathematical fraction showing 5 squared divided by 8 squared, with the exponents '2' in red, highlighting the power to which both the numerator and denominator are raised.
Simplify. A mathematical fraction displays 25 over 64, with the number 25 positioned above a horizontal division line and the number 64 below it, representing the ratio or division of 25 by 64.
ⓑ
The mathematical expression (x/3) raised to the power of 4.
Use the Quotient to a Power Property, (ab)m=ambm. A mathematical fraction featuring 'x' raised to the 4th power divided by '3' raised to the 4th power, with the exponent '4' colored red in both the numerator and denominator.
Simplify. The mathematical expression x to the power of 4 divided by 81.
ⓒ
A mathematical expression showing the fraction y over m, enclosed in parentheses, all raised to the power of 3.
Raise the numerator and denominator to the third power. A mathematical fraction displaying y cubed divided by m cubed, with the exponents in red.
Simplify:
  1. ⓐ (79)2
  2. ⓑ (y8)3
  3. ⓒ (pq)6
Solution
  1. ⓐ 4981
  2. ⓑ y3512
  3. ⓒ p6q6
Simplify:
  1. ⓐ (18)2
  2. ⓑ (−5m)3
  3. ⓒ (rs)4
Solution
  1. ⓐ 164
  2. ⓑ −125m3
  3. ⓒ r4s4

Simplify Expressions by Applying Several Properties

We'll now summarize all the properties of exponents so they are all together to refer to as we simplify expressions using several properties. Notice that they are now defined for whole number exponents.

Summary of Exponent Properties

If a,b are real numbers and m,n are whole numbers, then

Product Propertyam⋅an=am+nPower Property(am)n=am⋅nProduct to a Power Property(ab)m=ambmQuotient Propertyaman=am−n,a≠0,m>naman=1an−m,a≠0,n>mZero Exponent Definitiona0=1,a≠0Quotient to a Power Property(ab)m=ambm,b≠0

Simplify: (x2)3x5.

Solution

Solution

Step-by-step simplification of a rational expression using exponent properties.
(x2)3x5
Multiply the exponents in the numerator, using the
Power Property.
x6x5
Subtract the exponents. x

Simplify: (a4)5a9.

Solution

a11

Simplify: (b5)6b11.

Solution

b19

Simplify: m8(m2)4.

Solution

Solution

This table demonstrates the step-by-step simplification of the exponential expression m^8 / (m^2)^4 to 1, applying exponent properties.
m8(m2)4
Multiply the exponents in the numerator, using the
Power Property.
m8m8
Subtract the exponents. m0
Zero power property 1

Simplify: k11(k3)3.

Solution

k2

Simplify: d23(d4)6.

Solution

1d

Simplify: (x7x3)2.

Solution

Solution

Step-by-step simplification of the algebraic expression (x^7/x^3)^2.
(x7x3)2
Remember parentheses come before exponents, and the
bases are the same so we can simplify inside the
parentheses. Subtract the exponents.
(x7−3)2
Simplify. (x4)2
Multiply the exponents. x8

Simplify: (f14f8)2.

Solution

f12

Simplify: (b6b11)2.

Solution

1b10

Simplify: (p2q5)3.

Solution

Solution

Here we cannot simplify inside the parentheses first, since the bases are not the same.

Steps to simplify a power of a quotient using exponent properties.
(p2q5)3
Raise the numerator and denominator to the third power
using the Quotient to a Power Property, (ab)m=ambm
(p2)3(q5)3
Use the Power Property, (am)n=am⋅n. p6q15

Simplify: (m3n8)5.

Solution

m15n40

Simplify: (t10u7)2.

Solution

t20u14

Simplify: (2x33y)4.

Solution

Solution

Step-by-step simplification of the algebraic expression (2x^3 / 3y)^4, demonstrating exponent properties.
(2x33y)4
Raise the numerator and denominator to the fourth
power using the Quotient to a Power Property.
(2x3)4(3y)4
Raise each factor to the fourth power, using the Power
to a Power Property.
24(x3)434y4
Use the Power Property and simplify. 16x1281y4

Simplify: (5b9c3)2.

Solution

25b281c6

Simplify: (4p47q5)3.

Solution

64p12343q15

Simplify: (y2)3(y2)4(y5)4.

Solution

Solution

Illustrates the step-by-step simplification of an algebraic expression by applying various properties of exponents.
(y2)3(y2)4(y5)4
Use the Power Property. (y6)(y8)y20
Add the exponents in the numerator, using the Product Property. y14y20
Use the Quotient Property. 1y6

Simplify: (y4)4(y3)5(y7)6.

Solution

1y11

Simplify: (3x4)2(x3)4(x5)3.

Solution

9x5

Divide Monomials

We have now seen all the properties of exponents. We'll use them to divide monomials. Later, you'll use them to divide polynomials.

Find the quotient: 56x5÷7x2.

Solution

Solution

Steps for dividing the monomials 56x^5 by 7x^2, illustrating rewriting as a fraction, separating terms, and applying the Quotient Property.
56x5÷7x2
Rewrite as a fraction. 56x57x2
Use fraction multiplication to separate the number
part from the variable part.
567⋅x5x2
Use the Quotient Property. 8x3

Find the quotient: 63x8÷9x4.

Solution

7x4

Find the quotient: 96y11÷6y8.

Solution

16y3

When we divide monomials with more than one variable, we write one fraction for each variable.

Find the quotient: 42x2y3−7xy5.

Solution

Solution

Step-by-step simplification of a rational algebraic expression demonstrating the use of fraction multiplication and quotient property.
42x2y3−7xy5
Use fraction multiplication. 42−7⋅x2x⋅y3y5
Simplify and use the Quotient Property. −6⋅x⋅1y2
Multiply. −6xy2

Find the quotient: −84x8y37x10y2.

Solution

−12yx2

Find the quotient: −72a4b5−8a9b5.

Solution

9a5

Find the quotient: 24a5b348ab4.

Solution

Solution

Steps to simplify a rational algebraic expression using fraction multiplication and the quotient property.
24a5b348ab4
Use fraction multiplication. 2448⋅a5a⋅b3b4
Simplify and use the Quotient Property. 12⋅a4⋅1b
Multiply. a42b

Find the quotient: 16a7b624ab8.

Solution

2a63b2

Find the quotient: 27p4q7−45p12q.

Solution

−3q65p8

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

Solution

Simplification of a rational algebraic expression using the Quotient Property.
14x7y1221x11y6
Simplify and use the Quotient Property. 2y63x4

Be very careful to simplify 1421 by dividing out a common factor, and to simplify the variables by subtracting their exponents.

Find the quotient: 28x5y1449x9y12.

Solution

4y27x4

Find the quotient: 30m5n1148m10n14.

Solution

58m5n3

In all examples so far, there was no work to do in the numerator or denominator before simplifying the fraction. In the next example, we'll first find the product of two monomials in the numerator before we simplify the fraction.

Find the quotient: (3x3y2)(10x2y3)6x4y5.

Solution

Solution

Remember, the fraction bar is a grouping symbol. We will simplify the numerator first.

Step-by-step simplification of a rational algebraic expression using exponent rules.
(3x3y2)(10x2y3)6x4y5
Simplify the numerator. 30x5y56x4y5
Simplify, using the Quotient Rule. 5x

Find the quotient: (3x4y5)(8x2y5)12x5y8.

Solution

2xy2

Find the quotient: (−6a6b9)(−8a5b8)−12a10b12.

Solution

−4ab5

ACCESS ADDITIONAL ONLINE RESOURCES

  • Simplify a Quotient
  • Zero Exponent
  • Quotient Rule
  • Polynomial Division
  • Polynomial Division 2

Key Concepts

  • Equivalent Fractions Property
    • If a,b,c are whole numbers where b≠0,c≠0, then
      ab=a·cb·canda·cb·c=ab
  • Zero Exponent
    • If a is a non-zero number, then a0=1.
    • Any nonzero number raised to the zero power is 1.
  • Quotient Property for Exponents
    • If a is a real number, a≠0, and m,n are whole numbers, then
      aman=am−n,m>nandaman=1an−m,n>m
  • Quotient to a Power Property for Exponents
    • If a and b are real numbers, b≠0, and m is a counting number, then
      (ab)m=ambm
    • To raise a fraction to a power, raise the numerator and denominator to that power.

Practice Makes Perfect

Simplify Expressions Using the Quotient Property of Exponents

In the following exercises, simplify.

4842

Solution

46

31234

x12x3

Solution

x9

u9u3

r5r

Solution

r4

y4y

y4y20

Solution

1y16

x10x30

1031015

Solution

11012

r2r8

aa9

Solution

1a8

225

Simplify Expressions with Zero Exponents

In the following exercises, simplify.

50

Solution

1

100

a0

Solution

1

x0

−70

Solution

−1

−40

  1. ⓐ (10p)0
  2. ⓑ 10p0
Solution
  1. ⓐ 1
  2. ⓑ 10
  1. ⓐ (3a)0
  2. ⓑ 3a0
  1. ⓐ (−27x5y)0
  2. ⓑ −27x5y0
Solution
  1. ⓐ 1
  2. ⓑ −27x5
  1. ⓐ (−92y8z)0
  2. ⓑ −92y8z0
  1. ⓐ 150
  2. ⓑ 151
Solution
  1. ⓐ 1
  2. ⓑ 15
  1. ⓐ −60
  2. ⓑ −61

2·x0+5·y0

Solution

7

8·m0−4·n0

Simplify Expressions Using the Quotient to a Power Property

In the following exercises, simplify.

(32)5

Solution

24332

(45)3

(m6)3

Solution

m3216

(p2)5

(xy)10

Solution

x10y10

(ab)8

(a3b)2

Solution

a29b2

(2xy)4

Simplify Expressions by Applying Several Properties

In the following exercises, simplify.

(x2)4x5

Solution

x3

(y4)3y7

(u3)4u10

Solution

u2

(y2)5y6

y8(y5)2

Solution

1y2

p11(p5)3

r5r4·r

Solution

1

a3·a4a7

(x2x8)3

Solution

1x18

(uu10)2

(a4·a6a3)2

Solution

a14

(x3·x8x4)3

(y3)5(y4)3

Solution

y3

(z6)2(z2)4

(x3)6(x4)7

Solution

1x10

(x4)8(x5)7

(2r35s)4

Solution

16r12625s4

(3m24n)3

(3y2·y5y15·y8)0

Solution

1

(15z4·z90.3z2)0

(r2)5(r4)2(r3)7

Solution

1r3

(p4)2(p3)5(p2)9

(3x4)3(2x3)2(6x5)2

Solution

3x8

(−2y3)4(3y4)2(−6y3)2

Divide Monomials

In the following exercises, divide the monomials.

48b8÷6b2

Solution

8b6

42a14÷6a2

36x3÷(−2x9)

Solution

−18x6

20u8÷(−4u6)

18x39x2

Solution

2x

36y94y7

−35x7−42x13

Solution

56x6

18x5−27x9

18r5s3r3s9

Solution

6r2s8

24p7q6p2q5

8mn1064mn4

Solution

n68

10a4b50a2b6

−12x4y915x6y3

Solution

−4y65x2

48x11y9z336x6y8z5

64x5y9z748x7y12z6

Solution

4z3x2y3

(10u2v)(4u3v6)5u9v2

(6m2n)(5m4n3)3m10n2

Solution

10n2m4

(6a4b3)(4ab5)(12a8b)(a3b)

(4u5v4)(15u8v)(12u3v)(u6v)

Solution

5u4v3

Mixed Practice

  1. ⓐ 24a5+2a5
  2. ⓑ 24a5−2a5
  3. ⓒ 24a5⋅2a5
  4. ⓓ 24a5÷2a5
  1. ⓐ 15n10+3n10
  2. ⓑ 15n10−3n10
  3. ⓒ 15n10⋅3n10
  4. ⓓ 15n10÷3n10
Solution
  1. ⓐ 18n10
  2. ⓑ 12n10
  3. ⓒ 45n20
  4. ⓓ 5
  1. ⓐ p4⋅p6
  2. ⓑ (p4)6
  1. ⓐ q5⋅q3
  2. ⓑ (q5)3
Solution
  1. ⓐ q8
  2. ⓑ q15
  1. ⓐ y3y
  2. ⓑ yy3
  1. ⓐ z6z5
  2. ⓑ z5z6
Solution
  1. ⓐ z
  2. ⓑ 1z

(8x5)(9x)÷6x3

(4y)(12y7)÷8y2

Solution

6y6

27a73a3+54a99a5

32c114c5+42c96c3

Solution

15c6

32y58y2−60y105y7

48x66x4−35x97x7

Solution

3x2

63r6s39r4s2−72r2s26s

56y4z57y3z3−45y2z25y

Solution

−yz2

Everyday Math

Memory One megabyte is approximately 106 bytes. One gigabyte is approximately 109 bytes. How many megabytes are in one gigabyte?

Memory One megabyte is approximately 106 bytes. One terabyte is approximately 1012 bytes. How many megabytes are in one terabyte?

Solution

1,000,000

Writing Exercises

Vic thinks the quotient x20x4 simplifies to x5. What is wrong with his reasoning?

Mai simplifies the quotient y3y by writing y3y=3. What is wrong with her reasoning?

Solution

Answers will vary.

When Dimple simplified −30 and (−3)0 she got the same answer. Explain how using the Order of Operations correctly gives different answers.

Roxie thinks n0 simplifies to 0. What would you say to convince Roxie she is wrong?

Solution

Answers will vary.

Self Check

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

A math self-assessment sheet listing various exponent properties and monomial division skills, with columns for students to rate their proficiency: 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?

zero exponent
If a is a non-zero number, then a0=1. Any nonzero number raised to the zero power is 1.

Integer Exponents and Scientific Notation

Learning Objectives

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

  • Use the definition of a negative exponent
  • Simplify expressions with integer exponents
  • Convert from decimal notation to scientific notation
  • Convert scientific notation to decimal form
  • Multiply and divide using scientific notation

Before you get started, take this readiness quiz.

What is the place value of the 6 in the number 64,891?
If you missed this problem, review Example 3 in Introduction to Whole Numbers.

Solution

ten thousand

Name the decimal 0.0012.
If you missed this problem, review Example 1 in Decimals.

Solution

twelve ten-thousandths

Subtract: 5−(−3).
If you missed this problem, review Example 8 in Subtract Integers.

Solution

8

Use the Definition of a Negative Exponent

The Quotient Property of Exponents, introduced in Divide Monomials, had two forms depending on whether the exponent in the numerator or denominator was larger.

Quotient Property of Exponents

If a is a real number, a≠0, and m,n are whole numbers, then

aman=am−n,m>nandaman=1an−m,n>m

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.

x2x5
x2−5
x−3

We can also simplify x2x5 by dividing out common factors: x2x5.

A fraction is shown. The numerator is x times x, the denominator is x times x times x times x times x. Two x's are crossed out in red on the top and on the bottom. Below that, the fraction 1 over x cubed is shown.

This implies that x−3=1x3 and it leads us to the definition of a negative exponent.

Negative Exponent

If n is a positive integer and a≠0, then a−n=1an.

The negative exponent tells us to re-write 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 an expression with only positive exponents.

Simplify:
  1. ⓐ 4−2
  2. ⓑ 10−3
Solution

Solution

Steps to simplify an expression with a negative exponent, demonstrating the conversion and calculation of 4^-2 to 1/16.
ⓐ
4−2
Use the definition of a negative exponent, a−n=1an. 142
Simplify. 116
Step-by-step simplification of 10 to the power of -3, demonstrating the use of negative exponent definition.
ⓑ
10−3
Use the definition of a negative exponent, a−n=1an. 1103
Simplify. 11000
Simplify:
  1. ⓐ 2−3
  2. ⓑ 10−2
Solution
  1. ⓐ 18
  2. ⓑ 1100
Simplify:
  1. ⓐ 3−2
  2. ⓑ 10−4
Solution
  1. ⓐ 19
  2. ⓑ 110,000

When simplifying any expression with exponents, we must be careful to correctly identify the base that is raised to each exponent.

Simplify:
  1. ⓐ (−3)−2
  2. ⓑ −3−2
Solution

Solution

The negative in the exponent does not affect the sign of the base.

Steps to simplify an expression with a negative exponent, detailing each mathematical operation.
ⓐ
The exponent applies to the base, −3. (−3)−2
Take the reciprocal of the base and change the sign of the exponent. 1(−3)2
Simplify. 19
Step-by-step evaluation of -3^(-2), demonstrating rules for negative exponents and bases.
ⓑ
The expression −3−2 means "find the opposite of 3−2".
The exponent applies only to the base, 3.
−3−2
Rewrite as a product with −1. −1·3−2
Take the reciprocal of the base and change the sign of the exponent. −1·132
Simplify. −19
Simplify:
  1. ⓐ (−5)−2
  2. ⓑ −5−2
Solution
  1. ⓐ 125
  2. ⓑ −125
Simplify:
  1. ⓐ (−2)−2
  2. ⓑ −2−2
Solution
  1. ⓐ 14
  2. ⓑ −14

We must be careful to follow the order of operations. In the next example, parts ⓐ and ⓑ look similar, but we get different results.

Simplify:
  1. ⓐ 4·2−1
  2. ⓑ (4·2)−1
Solution

Solution

Remember to always follow the order of operations.

Step-by-step evaluation of 4 * 2^-1, demonstrating exponent rules and order of operations.
ⓐ
Do exponents before multiplication. 4·2−1
Use a−n=1an. 4·121
Simplify. 2
Demonstrates the step-by-step simplification of (4 * 2)^-1 using negative exponent rules.
ⓑ (4·2)−1
Simplify inside the parentheses first. (8)−1
Use a−n=1an. 181
Simplify. 18
Simplify:
  1. ⓐ 6·3−1
  2. ⓑ (6·3)−1
Solution
  1. ⓐ 2
  2. ⓑ 118
Simplify:
  1. ⓐ 8·2−2
  2. ⓑ (8·2)−2
Solution
  1. ⓐ 2
  2. ⓑ 1256

When a variable is raised to a negative exponent, we apply the definition the same way we did with numbers.

Simplify: x−6.

Solution

Solution

Demonstrates how to apply the definition of a negative exponent to simplify a mathematical expression.
x−6
Use the definition of a negative exponent, a−n=1an. 1x6

Simplify: y−7.

Solution

1y7

Simplify: z−8.

Solution

1z8

When there is a product and an exponent we have to be careful to apply the exponent to the correct quantity. According to the order of operations, expressions in parentheses are simplified before exponents are applied. We’ll see how this works in the next example.

Simplify:
  1. ⓐ 5y−1
  2. ⓑ (5y)−1
  3. ⓒ (−5y)−1
Solution

Solution

This table illustrates the step-by-step simplification of the expression 5y^-1, demonstrating how to handle negative exponents.
ⓐ
Notice the exponent applies to just the base y. 5y−1
Take the reciprocal of y and change the sign of the exponent. 5·1y1
Simplify. 5y
Steps demonstrating the simplification of the algebraic expression (5y)^-1 using rules for negative exponents.
ⓑ
Here the parentheses make the exponent apply to the base 5y. (5y)−1
Take the reciprocal of 5y and change the sign of the exponent. 1(5y)1
Simplify. 15y
Step-by-step simplification of the algebraic expression (-5y)^-1, demonstrating the rule for negative exponents.
ⓒ
(−5y)−1
The base is −5y. Take the reciprocal of −5y and change the sign of the exponent. 1(−5y)1
Simplify. 1−5y
Use a−b=−ab. −15y
Simplify:
  1. ⓐ 8p−1
  2. ⓑ (8p)−1
  3. ⓒ (−8p)−1
Solution
  1. ⓐ 8p
  2. ⓑ 18p
  3. ⓒ −18p
Simplify:
  1. ⓐ 11q−1
  2. ⓑ (11q)−1
  3. ⓒ (−11q)−1
Solution
  1. ⓐ 11q
  2. ⓑ 111q
  3. ⓒ −111q

Now that we have defined negative exponents, the Quotient Property of Exponents needs only one form, aman=am−n, where a≠0 and m and n are integers.

When the exponent in the denominator is larger than the exponent in the numerator, the exponent of the quotient will be negative. If the result gives us a negative exponent, we will rewrite it by using the definition of negative exponents, a−n=1an.

Simplify Expressions with Integer Exponents

All the exponent properties we developed earlier in this chapter with whole number exponents apply to integer exponents, too. We restate them here for reference.

Summary of Exponent Properties

If a,b are real numbers and m,n are integers, then

Product Propertyam·an=am+nPower Property(am)n=am·nProduct to a Power Property(ab)m=ambmQuotient Propertyaman=am−n,a≠0Zero Exponent Propertya0=1,a≠0Quotient to a Power Property(ab)m=ambm,b≠0Definition of Negative Exponenta−n=1an
Simplify:
  1. ⓐ x−4·x6
  2. ⓑ y−6·y4
  3. ⓒ z−5·z−3
Solution

Solution

Steps for simplifying the exponential expression x^(-4) * x^6 using the product property, resulting in x^2.
ⓐ
x−4·x6
Use the Product Property, am·an=am+n. x−4+6
Simplify. x2
Step-by-step simplification of an exponential expression involving the product rule and negative exponents.
ⓑ
y−6·y4
The bases are the same, so add the exponents. y−6+4
Simplify. y−2
Use the definition of a negative exponent, a−n=1an. 1y2
Steps for simplifying the exponential expression z^(-5) * z^(-3) by applying the product rule and negative exponent definition.
ⓒ
z−5·z−3
The bases are the same, so add the exponents. z−5−3
Simplify. z−8
Use the definition of a negative exponent, a−n=1an. 1z8
Simplify:
  1. ⓐ x−3·x7
  2. ⓑ y−7·y2
  3. ⓒ z−4·z−5
Solution
  1. ⓐ x4
  2. ⓑ 1y5
  3. ⓒ 1z9
Simplify:
  1. ⓐ a−1·a6
  2. ⓑ b−8·b4
  3. ⓒ c−8·c−7
Solution
  1. ⓐ a5
  2. ⓑ 1b4
  3. ⓒ 1c15

In the next two examples, we’ll start by using the Commutative Property to group the same variables together. This makes it easier to identify the like bases before using the Product Property of Exponents.

Simplify: (m4n−3)(m−5n−2).

Solution

Solution

Demonstrates the step-by-step simplification of an algebraic expression involving 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

Simplify: (p6q−2)(p−9q−1).

Solution

1p3q3

Simplify: (r5s−3)(r−7s−5).

Solution

1r2s8

If the monomials have numerical coefficients, we multiply the coefficients, just as we did in Use Multiplication Properties of Exponents.

Simplify: (2x−6y8)(−5x5y−3).

Solution

Solution

Step-by-step simplification of an algebraic expression involving exponents.
(2x−6y8)(−5x5y−3)
Rewrite with the like bases together. 2(−5)·(x−6x5)·(y8y−3)
Simplify. −10·x−1·y5
Use the definition of a negative exponent, a−n=1an. −10·1x1·y5
Simplify. −10y5x

Simplify: (3u−5v7)(−4u4v−2).

Solution

−12v5u

Simplify: (−6c−6d4)(−5c−2d−1).

Solution

30d3c8

In the next two examples, we’ll use the Power Property and the Product to a Power Property.

Simplify: (k3)−2.

Solution

Solution

Illustrates the step-by-step simplification of the exponential expression (k^3)^(-2) using exponent rules, yielding 1/k^6.
(k3)−2
Use the Product to a Power Property, (ab)m=ambm. k3(−2)
Simplify. k−6
Rewrite with a positive exponent. 1k6

Simplify: (x4)−1.

Solution

1x4

Simplify: (y2)−2.

Solution

1y4

Simplify: (5x−3)2.

Solution

Solution

Step-by-step simplification of 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 52 and multiply the exponents of x using the
Power Property, (am)n=am·n.
25x−6
Rewrite x−6 by using the definition of a negative
exponent, a−n=1an.
25·1x6
Simplify 25x6

Simplify: (8a−4)2.

Solution

64a8

Simplify: (2c−4)3.

Solution

8c12

To simplify a fraction, we use the Quotient Property.

Simplify: r5r−4.

Solution

Solution

The exponent expression r^5 over r^-4.
Use the Quotient Property, aman=am−n. A mathematical expression reads 'r to the power of 5 minus negative 4,' with '-4' highlighted in red.
The text reads 'Be careful to subtract 5 - (-4).', highlighting the subtraction of a negative number.
Simplify. The expression r^9 is displayed on a white background.

Simplify: x8x−3.

Solution

x11

Simplify: y7y−6.

Solution

y13

Convert from Decimal Notation to Scientific Notation

Remember working with place value for whole numbers and decimals? Our number system is based on powers of 10. We use tens, hundreds, thousands, and so on. Our decimal numbers are also based on powers of tens—tenths, hundredths, thousandths, and so on.

Consider the numbers 4000 and 0.004. We know that 4000 means 4×1000 and 0.004 means 4×11000. If we write the 1000 as a power of ten in exponential form, we can rewrite these numbers in this way:

40000.0044×10004×110004×1034×11034×10−3

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 10, 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×10n

where a≥1 and a<10 and n is an integer.

It is customary in scientific notation to use × as the multiplication sign, even though we avoid using this sign elsewhere in algebra.

Scientific notation is a useful way of writing very large or very small numbers. It is used often in the sciences to make calculations easier.

If we look at what happened to the decimal point, we can see a method to easily convert from decimal notation to scientific notation.

On the left, we see 4000 equals 4 times 10 cubed. Beneath that is the same thing, but there is an arrow from after the last 0 in 4000 to between the 4 and the first 0. Beneath, it says, “Moved the decimal point 3 places to the left.” On the right, we see 0.004 equals 4 times 10 to the negative 3. Beneath that is the same thing, but there is an arrow from the decimal point to after the 4. Beneath, it says, “Moved the decimal point 3 places to the right.”

In both cases, the decimal was moved 3 places to get the first factor, 4, by itself.

  • The power of 10 is positive when the number is larger than 1:4000=4×103.
  • The power of 10 is negative when the number is between 0 and 1:0.004=4×10−3.

Write 37,000 in scientific notation.

Solution

Solution

Step 1: Move the decimal point so that the first factor is greater than or equal to 1 but less than 10. The number 37000. is displayed in black text. Underneath the last three zeros of the number, there are three small, light blue wavy arrows pointing downwards, indicating a numerical operation or a specific feature of the zeros.
Step 2: Count the number of decimal places, n, that the decimal point was moved. 3.70000
4 places
Step 3: Write the number as a product with a power of 10. 3.7×104
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
Step 4: Check.
104 is 10,000 and 10,000 times 3.7 will be 37,000.
37,000=3.7×104

Write in scientific notation: 96,000.

Solution

9.6 × 104

Write in scientific notation: 48,300.

Solution

4.83 × 104

Convert from decimal notation 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: 0.0052.

Solution

Solution

0.0052
Move the decimal point to get 5.2, a number between 1 and 10. The number 0.0052 is shown with three blue wavy arrows underneath, indicating a shift of the decimal point three places to the right.
Count the number of decimal places the point was moved. 3 places
Write as a product with a power of 10. 5.2 ×10−3
Check your answer:
5.2×10−35.2×11035.2×110005.2×0.0010.0052
0.0052=5.2×10−3

Write in scientific notation: 0.0078.

Solution

7.8 × 10−3

Write in scientific notation: 0.0129.

Solution

1.29 × 10−2

Convert Scientific Notation to Decimal Form

How can we convert from scientific notation to decimal form? Let’s look at two numbers written in scientific notation and see.

9.12×1049.12×10−49.12×10,0009.12×0.000191,2000.000912

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.

On the left, we see 9.12 times 10 to the 4th equals 91,200. Beneath that is 9.12 followed by 2 spaces, with an arrow from the decimal to after the second space, times 10 to the 4th equals 91,200.  On the right, we see 9.12 times 10 to the negative 4 equals 0.000912. Beneath that is three spaces followed by 9.12 with an arrow from the decimal to after the first space, times 10 to the negative 4 equals 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 to decimal form: 6.2×103.

Solution

Solution

Step 1: Determine the exponent, n, on the factor 10. 6.2×103
Step 2: Move the decimal point n places, adding zeros if needed. The number 6.200 is shown with three blue curved arrows pointing left under the digits '200', visually representing a three-place decimal shift or counting of digits.
  • 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.
6,200
Step 3: Check to see if your answer makes sense.
103 is 1000 and 1000 times 6.2 will be 6,200. 6.2 ×103=6,200

Convert to decimal form: 1.3×103.

Solution

1,300

Convert to decimal form: 9.25×104.

Solution

92,500

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: 8.9×10−2.

Solution

Solution

8.9 ×10−2
Determine the exponent n, on the factor 10. The exponent is −2.
Move the decimal point 2 places to the left. The number -8.9 is displayed in black text on a white background, with light blue wavy lines underneath the -8 portion of the number, suggesting a focus or grouping on the integer part.
Add zeros as needed for placeholders. 0.089
8.9×10−2=0.089
The Check is left to you.

Convert to decimal form: 1.2×10−4.

Solution

0.00012

Convert to decimal form: 7.5×10−2.

Solution

0.075

Multiply and Divide Using Scientific Notation

We use the Properties of Exponents to multiply and divide numbers in scientific notation.

Multiply. Write answers in decimal form: (4×105)(2×10−7).

Solution

Solution

Steps to multiply numbers in scientific notation, showing rearrangement, multiplication, and conversion to decimal form.
(4×105)(2×10−7)
Use the Commutative Property to rearrange the factors. 4·2·105·10−7
Multiply 4 by 2 and use the Product Property to multiply 105 by 10−7. 8×10−2
Change to decimal form by moving the decimal two places left. 0.08

Multiply. Write answers in decimal form: (3×106)(2×10−8).

Solution

0.06

Multiply. Write answers in decimal form: (3×10−2)(3×10−1).

Solution

0.009

Divide. Write answers in decimal form: 9×1033×10−2.

Solution

Solution

Steps for simplifying a fraction with scientific notation, demonstrating the process from separation of factors to final decimal form.
9×1033×10−2
Separate the factors. 93×10310−2
Divide 9 by 3 and use the Quotient Property to divide 103 by 10−2. 3×105
Change to decimal form by moving the decimal five places right. 300,000

Divide. Write answers in decimal form: 8×1042×10−1.

Solution

400,000

Divide. Write answers in decimal form: 8×1024×10−2.

Solution

20,000

ACCESS ADDITIONAL ONLINE RESOURCES

  • Negative Exponents
  • Examples of Simplifying Expressions with Negative Exponents
  • Scientific Notation

Key Concepts

  • Summary of Exponent Properties
    • If a,b are real numbers and m,n are integers, then
      Product Propertyam·an=am+nPower Property(am)n=am·nProduct to a Power Property(ab)m=ambmQuotient Propertyaman=am−n,a≠0Zero Exponent Propertya0=1,a≠0Quotient to a Power Property(ab)m=ambm,b≠0Definition of Negative Exponenta−n=1an
  • Convert from Decimal Notation to Scientific Notation: 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.
      • If the original number is between 0 and 1, the power of 10 will be 10-n.
    4. Check.
  • Convert Scientific Notation to Decimal Form: 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

Use the Definition of a Negative Exponent

In the following exercises, simplify.

5−3

8−2

Solution

164

3−4

2−5

Solution

132

7−1

10−1

Solution

110

2−3+2−2

3−2+3−1

Solution

49

3−1+4−1

10−1+2−1

Solution

35

100−10−1+10−2

20−2−1+2−2

Solution

34

  1. ⓐ (−6)−2
  2. ⓑ −6−2
  1. ⓐ (−8)−2
  2. ⓑ −8−2
Solution
  1. ⓐ 164
  2. ⓑ −164
  1. ⓐ (−10)−4
  2. ⓑ −10−4
  1. ⓐ (−4)−6
  2. ⓑ −4−6
Solution
  1. ⓐ 14096
  2. ⓑ −14096
  1. ⓐ 5·2−1
  2. ⓑ (5·2)−1
  1. ⓐ 10·3−1
  2. ⓑ (10·3)−1
Solution
  1. ⓐ 103
  2. ⓑ 130
  1. ⓐ 4·10−3
  2. ⓑ (4·10)−3
  1. ⓐ 3·5−2
  2. ⓑ (3·5)−2
Solution
  1. ⓐ 325
  2. ⓑ 1225

n−4

p−3

Solution

1p3

c−10

m−5

Solution

1m5

  1. ⓐ 4x−1
  2. ⓑ (4x)−1
  3. ⓒ (−4x)−1
  1. ⓐ 3q−1
  2. ⓑ (3q)−1
  3. ⓒ (−3q)−1
Solution
  1. ⓐ 3q
  2. ⓑ 13q
  3. ⓒ −13q
  1. ⓐ 6m−1
  2. ⓑ (6m)−1
  3. ⓒ (−6m)−1
  1. ⓐ 10k−1
  2. ⓑ (10k)−1
  3. ⓒ (−10k)−1
Solution
  1. ⓐ 10k
  2. ⓑ 110k
  3. ⓒ −110k

Simplify Expressions with Integer Exponents

In the following exercises, simplify.

p−4·p8

r−2·r5

Solution

r3

n−10·n2

q−8·q3

Solution

1q5

k−3·k−2

z−6·z−2

Solution

1z8

a·a−4

m·m−2

Solution

1m

p5·p−2·p−4

x4·x−2·x−3

Solution

1x

a3b−3

u2v−2

Solution

u2v2

(x5y−1)(x−10y−3)

(a3b−3)(a−5b−1)

Solution

1a2b4

(uv−2)(u−5v−4)

(pq−4)(p−6q−3)

Solution

1p5q7

(−2r−3s9)(6r4s−5)

(−3p−5q8)(7p2q−3)

Solution

−21q5p3

(−6m−8n−5)(−9m4n2)

(−8a−5b−4)(−4a2b3)

Solution

32a3b

(a3)−3

(q10)−10

Solution

1q100

(n2)−1

(x4)−1

Solution

1x4

(y−5)4

(p−3)2

Solution

1p6

(q−5)−2

(m−2)−3

Solution

m6

(4y−3)2

(3q−5)2

Solution

9q10

(10p−2)−5

(2n−3)−6

Solution

n1864

u9u−2

b5b−3

Solution

b8

x−6x4

m5m−2

Solution

m7

q3q12

r6r9

Solution

1r3

n−4n−10

p−3p−6

Solution

p3

Convert from Decimal Notation to Scientific Notation

In the following exercises, write each number in scientific notation.

45,000

280,000

Solution

2.8 × 105

8,750,000

1,290,000

Solution

1.29 × 106

0.036

0.041

Solution

4.1 × 10−2

0.00000924

0.0000103

Solution

1.03 × 10−5

The population of the United States on July 4, 2010 was almost 310,000,000.

The population of the world on July 4, 2010 was more than 6,850,000,000.

Solution

6.85 × 109

The average width of a human hair is 0.0018 centimeters.

The probability of winning the 2010 Megamillions lottery is about 0.0000000057.

Solution

5.7 × 10−9

Convert Scientific Notation to Decimal Form

In the following exercises, convert each number to decimal form.

4.1×102

8.3×102

Solution

830

5.5×108

1.6×1010

Solution

16,000,000,000

3.5×10−2

2.8×10−2

Solution

0.028

1.93×10−5

6.15×10−8

Solution

0.0000000615

In 2010, the number of Facebook users each day who changed their status to ‘engaged’ was 2×104.

At the start of 2012, the US federal budget had a deficit of more than $1.5×1013.

Solution

$15,000,000,000,000

The concentration of carbon dioxide in the atmosphere is 3.9×10−4.

The width of a proton is 1×10−5 of the width of an atom.

Solution

0.00001

Multiply and Divide Using Scientific Notation

In the following exercises, multiply or divide and write your answer in decimal form.

(2×105)(2×10−9)

(3×102)(1×10−5)

Solution

0.003

(1.6×10−2)(5.2×10−6)

(2.1×10−4)(3.5×10−2)

Solution

0.00000735

6×1043×10−2

8×1064×10−1

Solution

20,000,000

7×10−21×10−8

5×10−31×10−10

Solution

50,000,000

Everyday Math

Calories In May 2010 the Food and Beverage Manufacturers pledged to reduce their products by 1.5 trillion calories by the end of 2015.

  1. ⓐ Write 1.5 trillion in decimal notation.
  2. ⓑ Write 1.5 trillion in scientific notation.

Length of a year The difference between the calendar year and the astronomical year is 0.000125 day.

  1. ⓐ Write this number in scientific notation.
  2. ⓑ How many years does it take for the difference to become 1 day?
Solution
  1. ⓐ 1.25 × 10−4
  2. ⓐ 8,000

Calculator display Many calculators automatically show answers in scientific notation if there are more digits than can fit in the calculator’s display. To find the probability of getting a particular 5-card hand from a deck of cards, Mario divided 1 by 2,598,960 and saw the answer 3.848×10−7. Write the number in decimal notation.

Calculator display Many calculators automatically show answers in scientific notation if there are more digits than can fit in the calculator’s display. To find the number of ways Barbara could make a collage with 6 of her 50 favorite photographs, she multiplied 50·49·48·47·46·45. Her calculator gave the answer 1.1441304×1010. Write the number in decimal notation.

Solution

11,441,304,000

Writing Exercises

  1. ⓐ Explain the meaning of the exponent in the expression 23.
  2. ⓑ Explain the meaning of the exponent in the expression 2−3

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.

A self-assessment checklist for math skills, including exponents and scientific notation, with options to mark '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?

negative exponent
If n is a positive integer and a≠0, then a−n=1an.
scientific notation
A number expressed in scientific notation when it is of the form a×10n, where a≥1 and a<10, and n is an integer.

Introduction to Factoring Polynomials

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

Before you get started, take this readiness quiz.

Factor 56 into primes.
If you missed this problem, review Example 1 in Prime Factorization and the Least Common Multiple.

Solution

2⋅2⋅2⋅7

Multiply: −3(6a+11).
If you missed this problem, review Example 9 in Distributive Property.

Solution

−18a−33

Multiply: 4x2(x2+3x−1).
If you missed this problem, review Example 5 in Multiply Polynomials.

Solution

4x4+12x3−4x2

Find the Greatest Common Factor of Two or More Expressions

Earlier we multiplied factors together to get a product. Now, we will be reversing this process; we will start with a product and then break it down into its factors. Splitting a product into factors is called factoring.

On the left, the equation 8 times 7 equals 56 is shown. 8 and 7 are labeled factors, 56 is labeled product. On the right, the equation 2x times parentheses x plus 3 equals 2 x squared plus 6x is shown. 2x and x plus 3 are labeled factors, 2 x squared plus 6x is labeled product. There is an arrow on top pointing to the right that says “multiply” in red. There is an arrow on the bottom pointing to the left that says “factor” in red.

In The Language of Algebra we factored 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.

First we will find the greatest common factor of two numbers.

Find the greatest common factor of 24 and 36.

Solution

Solution

Step 1: Factor each coefficient into primes. Write all variables with exponents in expanded form. Factor 24 and 36. Factor trees demonstrating the prime factorization of 24 and 36.
Step 2: List all factors--matching common factors in a column. Two lines of mathematical equations demonstrating the prime factorization of 24 and 36. 24 is shown as 2 x 2 x 2 x 3, and 36 as 2 x 2 x 3 x 3, with an underline beneath the latter.
In each column, circle the common factors. Circle the 2, 2, and 3 that are shared by both numbers. A step-by-step example of finding the Greatest Common Factor (GCF) of 24 and 36 using prime factorization, illustrating how common prime factors (2, 2, and 3) are multiplied to get the GCF, which is 12.
Step 3: Bring down the common factors that all expressions share. Bring down the 2, 2, 3 and then multiply.
Step 4: Multiply the factors. The GCF of 24 and 36 is 12.

Notice that since the GCF is a factor of both numbers, 24 and 36 can be written as multiples of 12.

24=12·236=12·3

Find the greatest common factor: 54,36.

Solution

18

Find the greatest common factor: 48,80.

Solution

16

In the previous example, we found the greatest common factor of constants. The greatest common factor of an algebraic expression can contain variables raised to powers along with coefficients. We summarize the steps we use to find the greatest common factor.

Find the greatest common factor.

  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.

Find the greatest common factor of 5xand15.

Solution

Solution

Factor each number into primes.
Circle the common factors in each column.
Bring down the common factors.
An image shows the calculation for the Greatest Common Factor (GCF) of 5x and 15. It factors 5x as 5*x and 15 as 3*5. The common factor '5' is circled, resulting in GCF = 5.
The GCF of 5x and 15 is 5.

Find the greatest common factor: 7y,14.

Solution

7

Find the greatest common factor: 22,11m.

Solution

11

In the examples so far, the greatest common factor was a constant. In the next two examples we will get variables in the greatest common factor.

Find the greatest common factor of 12x2 and 18x3.

Solution

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.
Multiply the factors.
An algebraic problem illustrating the calculation of the Greatest Common Factor (GCF) for 12x^2 and 18x^3. The image factors both expressions, using pink circles to highlight common factors that combine to form the GCF, 6x^2.
The GCF of12x2and18x3is6x2

Find the greatest common factor: 16x2,24x3.

Solution

8x2

Find the greatest common factor: 27y3,18y4.

Solution

9y3

Find the greatest common factor of 14x3,8x2,10x.

Solution

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.
Multiply the factors.
An image illustrating the process of finding the Greatest Common Factor (GCF) of 14x^3, 8x^2, and 10x. The prime factorization of each term is listed, with common factors (2 and x) circled, leading to a GCF of 2x.
The GCF of14x3and8x2, and10xis2x

Find the greatest common factor: 21x3,9x2,15x.

Solution

3x

Find the greatest common factor: 25m4,35m3,20m2.

Solution

5m2

Factor the Greatest Common Factor from a Polynomial

Just like in arithmetic, where it is sometimes useful to represent a number in factored form (for example, 12 as 2·6or3·4), in algebra it can be useful to represent a polynomial in factored form. One way to do this is by finding the greatest common factor of all the terms. Remember that you can multiply a polynomial by a monomial as follows:

2(x + 7)factors 2·x + 2·7 2x + 14product

Here, we will start with a product, like 2x+14, and end with its factors, 2(x+7). To do this we apply the Distributive Property “in reverse”.

Distributive Property

If a,b,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 we use the Distributive Property to factor a polynomial? We find the GCF of all the terms and write the polynomial as a product!

Factor: 2x+14.

Solution

Solution

Step 1: Find the GCF of all the terms of the polynomial. Find the GCF of 2x and 14. A math problem demonstrating how to find the Greatest Common Factor (GCF) of 2x and 14. The number 2 is circled as the common factor, yielding GCF = 2.
Step 2: Rewrite each term as a product using the GCF. Rewrite 2x and 14 as products of their GCF, 2.
2x=2⋅x
14=2⋅7
A mathematical expression showing the process of factoring out a common number. The first line is 2x + 14, and the second line breaks down 14 into 2 * 7, highlighting the common factor of 2 in red as 2 * x + 2 * 7.
Step 3: Use the Distributive Property 'in reverse' to factor the expression. 2(x+7)
Step 4: Check by multiplying the factors. Check:
An image illustrating the distributive property in algebra, showing the expansion of 2(x+7) into 2x+14. The steps demonstrate multiplying 2 by both x and 7, resulting in 2x + 14, confirmed with a checkmark.

Factor: 4x+12.

Solution

4(x + 3)

Factor: 6a+24.

Solution

6(a + 4)

Notice that in Example 5, we used the word factor as both a noun and a verb:

Noun7is a factor of14Verbfactor2from2x+14

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 Distributive Property ‘in reverse’ to factor the expression.
  4. Check by multiplying the factors.

Factor: 3a+3.

Solution

Solution

A math problem illustrating how to find the Greatest Common Factor (GCF) of 3a and 3. The solution shows 3a factored as 3 * a and 3 as 3, highlighting 3 as the common factor, resulting in GCF = 3.
The mathematical expression '3a + 3' is displayed in a bold, dark gray font against a plain white background.
Rewrite each term as a product using the GCF. A mathematical expression showing 3 multiplied by 'a' plus 3 multiplied by 1, which can be factored as 3(a+1).
Use the Distributive Property 'in reverse' to factor the GCF. A mathematical expression: 3(a + 1).
Check by multiplying the factors to get the original polynomial.
Illustration of the distributive property, showing 3(a+1) expanding to 3a + 3 with a checkmark for correctness.

Factor: 9a+9.

Solution

9(a + 1)

Factor: 11x+11.

Solution

11(x + 1)

The expressions in the next example have several factors in common. Remember to write the GCF as the product of all the common factors.

Factor: 12x−60.

Solution

Solution

Steps to find the Greatest Common Factor (GCF) of 12x and 60 using prime factorization, showing the common factors circled and the final GCF as 12.
The mathematical expression '12x - 60' is displayed in a dark gray font on a white background.
Rewrite each term as a product using the GCF. The mathematical expression 12 multiplied by x minus 12 multiplied by 5 is shown.
Factor the GCF. A mathematical expression shows twelve multiplied by the quantity x minus five, written as 12(x - 5).
Check by multiplying the factors.
Applying the distributive property to simplify the algebraic expression 12(x-5) step-by-step, resulting in 12x-60.

Factor: 11x−44.

Solution

11(x − 4)

Factor: 13y−52.

Solution

13(y − 4)

Now we’ll factor the greatest common factor from a trinomial. We start by finding the GCF of all three terms.

Factor: 3y2+6y+9.

Solution

Solution

Finding the GCF of 3y^2, 6y, and 9 using prime factorization. Each term's factors are listed, and the common factor, 3, is circled, showing GCF = 3.
A mathematical expression displaying 3y squared plus 6y plus 9.
Rewrite each term as a product using the GCF. A mathematical expression displaying 3 multiplied by y squared, plus 3 multiplied by 2y, plus 3 multiplied by 3. The number 3 is highlighted in red in each term of the expression.
Factor the GCF. A mathematical expression showing 3 multiplied by the quantity (y-squared plus 2y plus 3), as 3(y^2 + 2y + 3).
Check by multiplying.
An algebraic expression 3(y^2 + 2y + 3) is expanded using the distributive property, showing the steps to arrive at the simplified form 3y^2 + 6y + 9, which is marked as correct.

Factor: 4y2+8y+12.

Solution

4(y2 + 2y + 3)

Factor: 6x2+42x−12.

Solution

6(x2 + 7x − 2)

In the next example, we factor a variable from a binomial.

Factor: 6x2+5x.

Solution

Solution

6x2+5x
Find the GCF of 6x2 and 5x and the math that goes with it. This image illustrates the process of finding the Greatest Common Factor (GCF) of 6x^2 and 5x by factoring. The common factor 'x' is highlighted, leading to a GCF of x.
Rewrite each term as a product. A mathematical expression showing 'x times 6x plus x times 5'. The variable 'x' is highlighted in red for both instances, emphasizing its presence in the terms.
Factor the GCF. x(6x+5)
Check by multiplying.
x(6x+5)
x⋅6x+x⋅5
6x2+5x✓

Factor: 9x2+7x.

Solution

x(9x + 7)

Factor: 5a2−12a.

Solution

a(5a − 12)

When there are several common factors, as we’ll see in the next two examples, good organization and neat work helps!

Factor: 4x3−20x2.

Solution

Solution

An algebraic example showing how to find the Greatest Common Factor (GCF) of 4x^3 and 20x^2. Prime factors are listed and common factors are circled in pink, resulting in a GCF of 4x^2.
A mathematical expression reads 4x^3 - 20x^2 in black text on a white background, formatted with standard mathematical notation for exponents and subtraction.
Rewrite each term. The mathematical expression '4x^2 * x - 4x^2 * 5' is displayed, showing a subtraction operation between two terms, both involving 4x^2 multiplied by another factor.
Factor the GCF. The image displays the algebraic expression 4x^2(x-5) in black text on a white background, representing a polynomial in factored form. The terms are clearly visible, indicating multiplication.
Check. Algebraic expansion of 4x^2(x-5) to 4x^3 - 20x^2, illustrating the distributive property applied correctly and step-by-step.

Factor: 2x3+12x2.

Solution

2x2(x + 6)

Factor: 6y3−15y2.

Solution

3y2(2y − 5)

Factor: 21y2+35y.

Solution

Solution

Find the GCF of 21y2 and 35y Finding the Greatest Common Factor (GCF) of 21y^2 and 35y. The image demonstrates the prime factorization of both terms, highlighting common factors (7 and y) to calculate the GCF, which is 7y.
The image displays the mathematical expression '21y^2 + 35y' in black text against a white background.
Rewrite each term. A mathematical expression displaying the equation 7y multiplied by 3y, added to 7y multiplied by 5, with '7y' highlighted in red.
Factor the GCF. A mathematical expression showing the term 7y multiplied by the binomial (3y + 5).

Factor: 18y2+63y.

Solution

9y(2y + 7)

Factor: 32k2+56k.

Solution

8k(4k + 7)

Factor: 14x3+8x2−10x.

Solution

Solution

Previously, we found the GCF of 14x3,8x2,and10x to be 2x.

14x3+8x2−10x
Rewrite each term using the GCF, 2x. A mathematical expression: 2x * 7x^2 + 2x * 4x - 2x * 5, showing a common factor of '2x' highlighted in red across three terms for potential factoring.
Factor the GCF. 2x(7x2+4x−5)
Algebraic check using the distributive property: 2x(7x^2 + 4x - 5) = 2x*7x^2 + 2x*4x - 2x*5 = 14x^3 + 8x^2 - 10x, confirmed with a checkmark.

Factor: 18y3−6y2−24y.

Solution

6y(3y2 − y − 4)

Factor: 16x3+8x2−12x.

Solution

4x(4x2 + 2x − 3)

When the leading coefficient, the coefficient of the first term, is negative, we factor the negative out as part of the GCF.

Factor: −9y−27.

Solution

Solution

When the leading coefficient is negative, the GCF will be negative. Ignoring the signs of the terms, we first find the GCF of 9y and 27 is 9. The image demonstrates finding the Greatest Common Factor (GCF) of 9y and 27. It shows the prime factorization of 9y as 3*3*y and 27 as 3*3*3, with common factors 3 and 3 circled. The GCF is calculated as 9.
Since the expression −9y−27 has a negative leading coefficient, we use −9 as the GCF.
−9y − 27
Rewrite each term using the GCF. The image shows the mathematical expression -9 * y + (-9) * 3, which demonstrates the distributive property in an algebraic context. The number -9 is a common factor in both terms of the expression.
Factor the GCF. −9(y+3)
A mathematical check demonstrates the distributive property, simplifying -9(y + 3) to -9y - 27, confirmed correct with a checkmark.

Factor: −5y−35.

Solution

−5(y + 7)

Factor: −16z−56.

Solution

−8(2z + 7)

Pay close attention to the signs of the terms in the next example.

Factor: −4a2+16a.

Solution

Solution

The leading coefficient is negative, so the GCF will be negative.
The Greatest Common Factor (GCF) of 4a^2 and 16a is determined by prime factorization. Common factors (2, 2, a) are circled, and their product yields the GCF, which is 4a.
Since the leading coefficient is negative, the GCF is negative, −4a.
−4a2+16a
Rewrite each term. -4a * a - (-4a) * 4
Factor the GCF. −4a(a−4)
Check on your own by multiplying.

Factor: −7a2+21a.

Solution

−7a(a − 3)

Factor: −6x2+x.

Solution

−x(6x − 1)

ACCESS ADDITIONAL ONLINE RESOURCES

  • Factor GCF
  • Factor a Binomial
  • Identify GCF

Key Concepts

  • Find the greatest common factor.
    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, c are real numbers, then
      a(b+c)=ab+ac and ab+ac=a(b+c)
  • 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 Distributive Property ‘in reverse’ to factor the expression.
    4. Check by multiplying the factors.

Section Exercises

Practice Makes Perfect

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

40,56

45,75

Solution

15

72,162

150,275

Solution

25

3x,12

4y,28

Solution

4

10a,50

5b,30

Solution

5

16y,24y2

9x,15x2

Solution

3x

18m3,36m2

12p4,48p3

Solution

12p3

10x,25x2,15x3

18a,6a2,22a3

Solution

2a

24u,6u2,30u3

40y,10y2,90y3

Solution

10y

15a4,9a5,21a6

35x3,10x4,5x5

Solution

5x3

27y2,45y3,9y4

14b2,35b3,63b4

Solution

7b2

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

2x+8

5y+15

Solution

5(y + 3)

3a−24

4b−20

Solution

4(b − 5)

9y−9

7x−7

Solution

7(x − 1)

5m2+20m+35

3n2+21n+12

Solution

3(n2 + 7n + 4)

8p2+32p+48

6q2+30q+42

Solution

6(q2 + 5q + 7)

8q2+15q

9c2+22c

Solution

c(9c + 22)

13k2+5k

17x2+7x

Solution

x(17x + 7)

5c2+9c

4q2+7q

Solution

q(4q + 7)

5p2+25p

3r2+27r

Solution

3r(r + 9)

24q2−12q

30u2−10u

Solution

10u(3u − 1)

yz+4z

ab+8b

Solution

b(a + 8)

60x−6x3

55y−11y4

Solution

11y(5 − y3)

48r4−12r3

45c3−15c2

Solution

15c2(3c − 1)

4a3−4ab2

6c3−6cd2

Solution

6c(c2 − d2)

30u3+80u2

48x3+72x2

Solution

24x2(2x + 3)

120y6+48y4

144a6+90a3

Solution

18a3(8a3 + 5)

4q2+24q+28

10y2+50y+40

Solution

10(y2 + 5y + 4)

15z2−30z−90

12u2−36u−108

Solution

12(u2 − 3u − 9)

3a4−24a3+18a2

5p4−20p3−15p2

Solution

5p2(p2 − 4p − 3)

11x6+44x5−121x4

8c5+40c4−56c3

Solution

8c3(c2 + 5c − 7)

−3n−24

−7p−84

Solution

−7(p + 12)

−15a2−40a

−18b2−66b

Solution

−6b(3b + 11)

−10y3+60y2

−8a3+32a2

Solution

−8a2(a − 4)

−4u5+56u3

−9b5+63b3

Solution

−9b3(b2 − 7)

Everyday Math

Revenue A manufacturer of microwave ovens has found that the revenue received from selling microwaves a cost of p dollars each is given by the polynomial −5p2+150p. Factor the greatest common factor from this polynomial.

Height of a baseball The height of a baseball hit with velocity 80 feet/second at 4 feet above ground level is −16t2+80t+4, with t= the number of seconds since it was hit. Factor the greatest common factor from this polynomial.

Solution

−4(4t2 − 20t − 1)

Writing Exercises

The greatest common factor of 36 and 60 is 12. Explain what this means.

What is the GCF of y4, y5, and y10? Write a general rule that tells how to find the GCF of ya, yb, and yc.

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 table for math skills, with columns for 'Confidently,' 'With some help,' and 'No-I don't get it!' The skills listed are finding the greatest common factor and factoring it from a polynomial.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Add and Subtract Polynomials

Identify Polynomials, Monomials, Binomials and Trinomials

In the following exercises, determine if each of the following polynomials is a monomial, binomial, trinomial, or other polynomial.

y2+8y−20

Solution

trinomial

−6a4

9x3−1

Solution

binomial

n3−3n2+3n−1

Determine the Degree of Polynomials

In the following exercises, determine the degree of each polynomial.

16x2−40x−25

Solution

2

5m+9

−15

Solution

0

y2+6y3+9y4

Add and Subtract Monomials

In the following exercises, add or subtract the monomials.

4p+11p

Solution

15p

−8y3−5y3

Add 4n5,−n5,−6n5

Solution

−3n5

Subtract 10x2 from 3x2

Add and Subtract Polynomials

In the following exercises, add or subtract the polynomials.

(4a2+9a−11)+(6a2−5a+10)

Solution

10a2 + 4a − 1

(8m2+12m−5)−(2m2−7m−1)

(y2−3y+12)+(5y2−9)

Solution

6y2 − 3y + 3

(5u2+8u)−(4u−7)

Find the sum of 8q3−27 and q2+6q−2

Solution

8q3 + q2 + 6q − 29

Find the difference of x2+6x+8 and x2−8x+15

Evaluate a Polynomial for a Given Value of the Variable

In the following exercises, evaluate each polynomial for the given value.

200x−15x2 when x=5

Solution

995

200x−15x2 when x=0

200x−15x2 when x=15

Solution

2,955

5+40x−12x2 when x=10

5+40x−12x2 when x=−4

Solution

−163

5+40x−12x2 when x=0

A pair of glasses is dropped off a bridge 640 feet above a river. The polynomial −16t2+640 gives the height of the glasses t seconds after they were dropped. Find the height of the glasses when t=6.

Solution

64 feet

The fuel efficiency (in miles per gallon) of a bus going at a speed of x miles per hour is given by the polynomial −1160x2+12x. Find the fuel efficiency when x=20 mph.

Use Multiplication Properties of Exponents

Simplify Expressions with Exponents

In the following exercises, simplify.

63

Solution

216

(12)4

(−0.5)2

Solution

0.25

−32

Simplify Expressions Using the Product Property of Exponents

In the following exercises, simplify each expression.

p3·p10

Solution

p13

2·26

a·a2·a3

Solution

a6

x·x8

Simplify Expressions Using the Power Property of Exponents

In the following exercises, simplify each expression.

(y4)3

Solution

y12

(r3)2

(32)5

Solution

310

(a10)y

Simplify Expressions Using the Product to a Power Property

In the following exercises, simplify each expression.

(8n)2

Solution

64n2

(−5x)3

(2ab)8

Solution

256a8b8

(−10mnp)4

Simplify Expressions by Applying Several Properties

In the following exercises, simplify each expression.

(3a5)3

Solution

27a15

(4y)2(8y)

(x3)5(x2)3

Solution

x21

(5st2)3(2s3t4)2

Multiply Monomials

In the following exercises, multiply the monomials.

(−6p4)(9p)

Solution

−54p5

(13c2)(30c8)

(8x2y5)(7xy6)

Solution

56x3y11

(23m3n6)(16m4n4)

Multiply Polynomials

Multiply a Polynomial by a Monomial

In the following exercises, multiply.

7(10−x)

Solution

70 − 7x

a2(a2−9a−36)

−5y(125y3−1)

Solution

−625y4 + 5y

(4n−5)(2n3)

Multiply a Binomial by a Binomial

In the following exercises, multiply the binomials using various methods.

(a+5)(a+2)

Solution

a2 + 7a + 10

(y−4)(y+12)

(3x+1)(2x−7)

Solution

6x2 − 19x − 7

(6p−11)(3p−10)

(n+8)(n+1)

Solution

n2 + 9n + 8

(k+6)(k−9)

(5u−3)(u+8)

Solution

5u2 + 37u − 24

(2y−9)(5y−7)

(p+4)(p+7)

Solution

p2 + 11p + 28

(x−8)(x+9)

(3c+1)(9c−4)

Solution

27c2 − 3c − 4

(10a−1)(3a−3)

Multiply a Trinomial by a Binomial

In the following exercises, multiply using any method.

(x+1)(x2−3x−21)

Solution

x3 − 2x2 − 24x − 21

(5b−2)(3b2+b−9)

(m+6)(m2−7m−30)

Solution

m3 − m2 − 72m − 180

(4y−1)(6y2−12y+5)

Divide Monomials

Simplify Expressions Using the Quotient Property of Exponents

In the following exercises, simplify.

2822

Solution

26 or 64

a6a

n3n12

Solution

1n9

xx5

Simplify Expressions with Zero Exponents

In the following exercises, simplify.

30

Solution

1

y0

(14t)0

Solution

1

12a0−15b0

Simplify Expressions Using the Quotient to a Power Property

In the following exercises, simplify.

(35)2

Solution

925

(x2)5

(5mn)3

Solution

125m3n3

(s10t)2

Simplify Expressions by Applying Several Properties

In the following exercises, simplify.

(a3)2a4

Solution

a2

u3u2·u4

(xx9)5

Solution

1x40

(p4·p5p3)2

(n5)3(n2)8

Solution

1n

(5s24t)3

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

42r2s46rs3−54rs29s

Integer Exponents and Scientific Notation

Use the Definition of a Negative Exponent

In the following exercises, simplify.

6−2

Solution

136

(−10)−3

5·2−4

Solution

516

(8n)−1

Simplify Expressions with Integer Exponents

In the following exercises, simplify.

x−3·x9

Solution

x6

r−5·r−4

(uv−3)(u−4v−2)

Solution

1u3v5

(m5)−1

(k−2)−3

Solution

k6

q4q20

b8b−2

Solution

b10

n−3n−5

Convert from Decimal Notation to Scientific Notation

In the following exercises, write each number in scientific notation.

5,300,000

Solution

5.3 × 106

0.00814

The thickness of a piece of paper is about 0.097 millimeter.

Solution

9.7 × 10−2 millimeter

According to www.cleanair.com, U.S. businesses use about 21,000,000 tons of paper per year.

Convert Scientific Notation to Decimal Form

In the following exercises, convert each number to decimal form.

2.9×104

Solution

29,000

1.5×108

3.75×10−1

Solution

0.375

9.413×10−5

Multiply and Divide Using Scientific Notation

In the following exercises, multiply and write your answer in decimal form.

(3×107)(2×10−4)

Solution

6,000

(1.5×10−3)(4.8×10−1)

6×1092×10−1

Solution

30,000,000,000

9×10−31×10−6

Introduction to Factoring Polynomials

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

5n,45

Solution

5

8a,72

12x2,20x3,36x4

Solution

4x2

9y4,21y5,15y6

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

16u−24

Solution

8(2u − 3)

15r+35

6p2+6p

Solution

6p(p + 1)

10c2−10c

−9a5−9a3

Solution

−9a3(a2 + 1)

−7x8−28x3

5y2−55y+45

Solution

5(y2 − 11y + 9)

2q5−16q3+30q2

Chapter Practice Test

For the polynomial 8y4−3y2+1

  1. ⓐ Is it a monomial, binomial, or trinomial?
  2. ⓑ What is its degree?
Solution
  1. ⓐ trinomial
  2. ⓑ 4

In the following exercises, simplify each expression.

(5a2+2a−12)+(9a2+8a−4)

(10x2−3x+5)−(4x2−6)

Solution

6x2 − 3x + 11

(−34)3

n·n4

Solution

n5

(10p3q5)2

(8xy3)(−6x4y6)

Solution

−48x5y9

4u(u2−9u+1)

(s+8)(s+9)

Solution

s2 + 17s + 72

(m+3)(7m−2)

(11a−6)(5a−1)

Solution

55a2 − 41a + 6

(n−8)(n2−4n+11)

(4a+9b)(6a−5b)

Solution

24a2 + 34ab − 45b2

5658

(x3·x9x5)2

Solution

x14

(47a18b23c5)0

24r3s6r2s7

Solution

4rs6

8y2−16y+204y

(15xy3−35x2y)÷5xy

Solution

3y2 − 7x

4−1

(2y)−3

Solution

18y3

p−3·p−8

x4x−5

Solution

x9

In the following exercises, factor the greatest common factor from each polynomial.

80a3+120a2+40a

−6x2−30x

Solution

−6x(x + 5)

According to www.cleanair.org, the amount of trash generated in the US in one year averages out to 112,000 pounds of trash per person. Write this number in scientific notation.

Convert 5.25×10−4 to decimal form.

Solution

0.000525

In the following exercises, simplify, and write your answer in decimal form.

(2.4×108)(2×10−5)

9×1043×10−1

Solution

300,000

A hiker drops a pebble from a bridge 240 feet above a canyon. The polynomial −16t2+240 gives the height of the pebble t seconds a after it was dropped. Find the height when t=3.

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.

Graphs

This photo shows a pack of cyclists in a road race.
Cyclists speed toward the finish line. (credit: ewan traveler, Flickr)

Which cyclist will win the race? What will the winning time be? How many seconds will separate the winner from the runner-up? One way to summarize the information from the race is by creating a graph. In this chapter, we will discuss the basic concepts of graphing. The applications of graphing go far beyond races. They are used to present information in almost every field, including healthcare, business, and entertainment.

Use the Rectangular Coordinate System

Learning Objectives

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

  • Plot points on a rectangular coordinate system
  • Identify points on a graph
  • Verify solutions to an equation in two variables
  • Complete a table of solutions to a linear equation
  • Find solutions to linear equations in two variables

Before you get started, take this readiness quiz.

Evaluate: x+3 when x=−1.
If you missed this problem, review Example 10 in Add Integers.

Solution

2

Evaluate: 2x−5y when x=3,y=−2.
If you missed this problem, review Example 10 in Multiply and Divide Integers.

Solution

16

Solve for y:40−4y=20.
If you missed this problem, review Example 1 in Solve Equations with Variables and Constants on Both Sides.

Solution

5

Plot Points on a Rectangular Coordinate System

Many maps, such as the Campus Map shown in Figure 1, use a grid system to identify locations. Do you see the numbers 1,2,3, and 4 across the top and bottom of the map and the letters A, B, C, and D along the sides? Every location on the map can be identified by a number and a letter.

For example, the Student Center is in section 2B. It is located in the grid section above the number 2 and next to the letter B. In which grid section is the Stadium? The Stadium is in section 4D.

The figure shows a labeled grid representing the Campus Map. The columns are labeled 1 through 4 and the rows are labeled A through D. At position A-1 is the title Parking Garage. At position A-4 is a rectangle labeled Residence Halls. At position B-2 is a rectangle labeled Student Center. At position B-3 is a rectangle labeled Engineering Building. At position C-1 is a rectangle labeled Taylor Hall. At position C-2 is a rectangle labeled Library.  At position C-4 is a rectangle labeled Tiger Field. At position D-4 is a rectangle labeled Stadium.

Use the map in Figure 1.

  1. ⓐ Find the grid section of the Residence Halls.
  2. ⓑ What is located in grid section 4C?
Solution

Solution

  1. ⓐ Read the number below the Residence Halls, 4, and the letter to the side, A. So the Residence Halls are in grid section 4A.
  2. ⓑ Find 4 across the bottom of the map and C along the side. Look below the 4 and next to the C. Tiger Field is in grid section 4C.

Use the map in Figure 1.

  1. ⓐ Find the grid section of Taylor Hall.
  2. ⓑ What is located in section 3B?
Solution
  1. ⓐ 1C
  2. ⓑ Engineering Building

Use the map in Figure 1.

  1. ⓐ Find the grid section of the Parking Garage.
  2. ⓑ What is located in section 2C?
Solution
  1. ⓐ 1A
  2. ⓑ Library

Just as 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. To create a rectangular coordinate system, start with a horizontal number line. Show both positive and negative numbers as you did before, using a convenient scale unit. This horizontal number line is called the x-axis.

The figure shows a number line with integer values labeled from -5 to 5.

Now, make a vertical number line passing through the x-axis at 0. Put the positive numbers above 0 and the negative numbers below 0. See Figure 2. This vertical line is called the y-axis.

Vertical grid lines pass through the integers marked on the x-axis. Horizontal grid lines pass through the integers marked on the y-axis. The resulting grid is the rectangular coordinate system.

The rectangular coordinate system is also called the x-y plane, the coordinate plane, or the Cartesian coordinate system (since it was developed by a mathematician named René Descartes.)

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7.  An arrow points to the horizontal axis with the label “x-axis”. An arrow points to the vertical axis with label “y-axis”. An arrow points to the intersection of the axes with label “origin”.
The rectangular coordinate system.

The x-axis and the y-axis form the rectangular coordinate system. These axes divide a plane into four areas, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See Figure 3.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The top-right portion of the plane is labeled “I”, the top-left portion of the plane is labeled “II”, the bottom-left portion of the plane is labelled “III” and the bottom-right portion of the plane is labeled “IV”
The four quadrants of the rectangular coordinate system

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.

Ordered Pair

An ordered pair, (x,y) gives the coordinates of a point in a rectangular coordinate system.

The first number is thex-coordinate.The second number is they-coordinate.
The ordered pair x y is labeled with the first coordinate x labeled as “x-coordinate” and the second coordinate y labeled as “y-coordinate”

So how do the coordinates of a point help you locate a point on the x-y plane?

Let’s try locating the point (2,5). In this ordered pair, the x-coordinate is 2 and the y-coordinate is 5.

We start by locating the x value, 2, on the x-axis. Then we lightly sketch a vertical line through x=2, as shown in Figure 4.

The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. There is a vertical dotted line passing through 2 on the x-axis.

Now we locate the y value, 5, on the y-axis and sketch a horizontal line through y=5. The point where these two lines meet is the point with coordinates (2,5). We plot the point there, as shown in Figure 5.

The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. An arrow starts at the origin and extends right to the number 2 on the x-axis. An arrow starts at the end of the first arrow at 2 on the x-axis and goes vertically 5 units to a point labeled “2, 5” in parentheses.

Plot (1,3) and (3,1) in the same rectangular coordinate system.

Solution

Solution

The coordinate values are the same for both points, but the x and y values are reversed. Let’s begin with point (1,3). The x-coordinate is 1 so find 1 on the x-axis and sketch a vertical line through x=1. The y-coordinate is 3 so we find 3 on the y-axis and sketch a horizontal line through y=3. Where the two lines meet, we plot the point (1,3).
The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. A horizontal dotted line passes through 3 on the y axis. A vertical dotted line passes through 1 on the x axis. The dotted lines intersect at a point labeled “ordered pair 1, 3”.

To plot the point (3,1), we start by locating 3 on the x-axis and sketch a vertical line through x=3. Then we find 1 on the y-axis and sketch a horizontal line through y=1. Where the two lines meet, we plot the point (3,1).
The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. A horizontal dotted line passes  through 1 on the y-axis. A vertical dotted line passes through 3 on the x axis. The dotted line intersects at a point labeled “ordered pair 3, 1”.

Notice that the order of the coordinates does matter, so, (1,3) is not the same point as (3,1).

Plot each point on the same rectangular coordinate system: (5,2),(2,5).

Solution


This answer graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. There are two labeled points: the first is ordered pair (5, 2), and the second is (2, 5)

Plot each point on the same rectangular coordinate system: (4,2),(2,4).

Solution


This answer graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. There are two labeled points: the first is ordered pair (2, 4), and the second is (4, 2)

Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:

  1. ⓐ (−1,3)
  2. ⓑ (−3,−4)
  3. ⓒ (2,−3)
  4. ⓓ (3,52)
Solution

Solution

The first number of the coordinate pair is the x-coordinate, and the second number is the y-coordinate.

ⓐ Since x=−1,y=3, the point (−1,3) is in Quadrant II.

ⓑ Since x=−3,y=−4, the point (−3,−4) is in Quadrant III.

ⓒ Since x=2,y=−1, the point (2,−1) is in Quadrant lV.

ⓓ Since x=3,y=52, the point (3,52) is in Quadrant I. It may be helpful to write 52 as the mixed number, 212, or decimal, 2.5. Then we know that the point is halfway between 2 and 3 on the y-axis.
The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 3, 5 over 2” is labeled “ordered pair “3,5 over 2”. The point “ordered pair -1, 3” is labeled “ordered pair -1, 3”. The point “ordered pair -3, -4” is labeled “ordered pair -3, -4”. The point “ordered pair 2, -1” is labeled “ordered pair 2, -1”.

Plot each point on a rectangular coordinate system and identify the quadrant in which the point is located.
  1. ⓐ (−2,1)
  2. ⓑ (−3,−1)
  3. ⓒ (4,−4)
  4. ⓓ (−4,32)
Solution

(a) Quadrant II, (b) Quadrant III, (c) Quadrant IV, (d) Quadrant II


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -2, 1” is labeled “a”. The point “ordered pair -3,  1” is labeled “b”.  The point “ordered pair 4, -4 is labeled “c”. The point “ordered pair -4, 3/2” is labeled “d”.

Plot each point on a rectangular coordinate system and identify the quadrant in which the point is located.
  1. ⓐ (−4,1)
  2. ⓑ (−2,3)
  3. ⓒ (2,−5)
  4. ⓓ (−3,52)
Solution

(a) Quadrant II, (b) Quadrant II, (c) Quadrant IV, (d) Quadrant II


This image is an answer graph and shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -4, 1” is labeled “a”. The point “ordered pair -2,  3” is labeled “b”. The point “ordered pair 2, -5” is labeled “c”. The point “ordered pair -3, 5/2” is labeled “d”.

How do the signs affect the location of the points?

Plot each point:
  1. ⓐ (−5,2)
  2. ⓑ (−5,−2)
  3. ⓒ (5,2)
  4. ⓓ (5,−2)
Solution

Solution

As we locate the x-coordinate and the y-coordinate, we must be careful with the signs.
The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 5, 2” is labeled “ordered pair 5, 2”. The point “ordered pair -5, 2” is labeled “ordered pair -5, 2”. The point “ordered pair -5, -2” is labeled “ordered pair -5, -2”. The point “ordered pair 5, -2” is labeled “ordered pair 5, -2”.

Plot each point:
  1. ⓐ (4,−3)
  2. ⓑ (4,3)
  3. ⓒ (−4,−3)
  4. ⓓ (−4,3)
Solution


This image is an answer graph and  shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair 4, -3” is labeled “a”. The point “ordered pair 4, 3” is labeled “b”. The point “ordered pair -4, -3” is labeled “c”. The point “ordered pair -4, 3” is labeled “d”.

Plot each point:
  1. ⓐ (−1,4)
  2. ⓑ (1,4)
  3. ⓒ (1,−4)
  4. ⓓ (−1,−4)
Solution


This image is an answer graph and  shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -1, -4” is labeled “a”. The point “ordered pair 1, 4” is labeled “b”. The point “ordered pair 1, -4” is labeled “c”. The point “ordered pair -1, -4” is labeled “d”.

You may have noticed some patterns as you graphed the points in the two previous examples.

For each point in Quadrant IV, what do you notice about the signs of the coordinates?

What about the signs of the coordinates of the points in the third quadrant? The second quadrant? The first quadrant?

Can you tell just by looking at the coordinates in which quadrant the point (−2, 5) is located? In which quadrant is (2, −5) located?

A Cartesian coordinate system shows points (-2, 5) in Quadrant II and (2, -5) in Quadrant IV. The axes range from -7 to 7, with grid lines every unit.

We can summarize sign patterns of the quadrants as follows. Also see Figure 7.

Quadrant I Quadrant II Quadrant III Quadrant IV
(x,y) (x,y) (x,y) (x,y)
(+,+) (−,+) (−,−) (+,−)
The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The top-right portion of the plane is labeled “I” and “ordered pair +, +”, the top-left portion of the plane is labeled “II” and “ordered pair -, +”, the bottom-left portion of the plane is labelled “III”  “ordered pair -, -” and the bottom-right portion of the plane is labeled “IV” and “ordered pair +, -”.

What if one coordinate is zero? Where is the point (0,4) located? Where is the point (−2,0) located? The point (0,4) is on the y-axis and the point (−2,0) is on the x-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).

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 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.
Plot each point on a coordinate grid:
  1. ⓐ (0,5)
  2. ⓑ (4,0)
  3. ⓒ (−3,0)
  4. ⓓ (0,0)
  5. ⓔ (0,−1)
Solution

Solution

  1. ⓐ Since x=0, the point whose coordinates are (0,5) is on the y-axis.
  2. ⓑ Since y=0, the point whose coordinates are (4,0) is on the x-axis.
  3. ⓒ Since y=0, the point whose coordinates are (−3,0) is on the x-axis.
  4. ⓓ Since x=0 and y=0, the point whose coordinates are (0,0) is the origin.
  5. ⓔ Since x=0, the point whose coordinates are (0,−1) is on the y-axis. The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 0, 0” is labeled “0, 0” in parentheses. The point “ordered pair 4, 0” is labeled “4, 0” in parentheses. The point “ordered pair 0, 5” is labeled “0, 5” in parentheses. The point “ordered pair 0, -1” is labeled “ 0, -1” in parentheses.
Plot each point on a coordinate grid:
  1. ⓐ (4,0)
  2. ⓑ (−2,0)
  3. ⓒ (0,0)
  4. ⓓ (0,2)
  5. ⓔ (0,−3)
Solution


This image is an answer graph and  shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The  point for ordered pair 4, 0 is plotted.  The point for ordered pair -2, 0 is plotted. The point for ordered pair 0,0 is plotted. The point for ordered pair 0, 2 is plotted. The point for ordered pair 0,-3 is plotted.

Plot each point on a coordinate grid:
  1. ⓐ (−5,0)
  2. ⓑ (3,0)
  3. ⓒ (0,0)
  4. ⓓ (0,−1)
  5. ⓔ (0,4)
Solution


This image is an answer graph and  shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The  point for ordered pair -5, 0 is plotted.  The point for ordered pair 3, 0 is plotted. The point for ordered pair 0,0 is plotted. The point for ordered pair 0, -1 is plotted. The point for ordered pair 0,4 is plotted.

Identify Points on a Graph

In algebra, being able to identify the coordinates of a point shown on a graph is just as important as being able to plot points. To identify the x-coordinate of a point on a graph, read the number on the x-axis directly above or below the point. To identify the y-coordinate of a point, read the number on the y-axis directly to the left or right of the point. Remember, to write the ordered pair using the correct order (x,y).

Name the ordered pair of each point shown:

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 2, 4” is labeled C. The point “ordered pair -3, 3” is labeled A.  The point “ordered pair -1, -3” is labeled B. The point “ordered pair 4, -4” is labeled D.
Solution

Solution

Point A is above −3 on the x-axis, so the x-coordinate of the point is −3. The point is to the left of 3 on the y-axis, so the y-coordinate of the point is 3. The coordinates of the point are (−3,3).

Point B is below −1 on the x-axis, so the x-coordinate of the point is −1. The point is to the left of −3 on the y-axis, so the y-coordinate of the point is −3. The coordinates of the point are (−1,−3).

Point C is above 2 on the x-axis, so the x-coordinate of the point is 2. The point is to the right of 4 on the y-axis, so the y-coordinate of the point is 4. The coordinates of the point are (2,4).

Point D is below 4 on the x-axis, so the x-coordinate of the point is 4. The point is to the right of −4 on the y-axis, so the y-coordinate of the point is −4. The coordinates of the point are (4,−4).

Name the ordered pair of each point shown:

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 5, 1” is labeled A. The point “ordered pair -2, 4” is labeled B.  The point “ordered pair -5, -1” is labeled C. The point “ordered pair 3, -2” is labeled D.
Solution
  1. A: (5,1)
  2. B: (−2,4)
  3. C: (−5,−1)
  4. D: (3,−2)

Name the ordered pair of each point shown:

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 4, 2” is labeled A. The point “ordered pair -2, 3” is labeled B.  The point “ordered pair -4, -4” is labeled C. The point “ordered pair 3, -5” is labeled D.
Solution
  1. A: (4,2)
  2. B: (−2,3)
  3. C: (−4,−4)
  4. D: (3,−5)

Name the ordered pair of each point shown:

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 3, 0” is labeled C. The point “ordered pair 0, 1” is labeled D.  The point “ordered pair -4, 0” is labeled A. The point “ordered pair 0, -2” is labeled B.
Solution

Solution

Points defined by their axis location and their corresponding Cartesian coordinates.
Point A is on the x-axis at x=−4. The coordinates of point A are (−4,0).
Point B is on the y-axis at y=−2 The coordinates of point B are (0,−2).
Point C is on the x-axis at x=3. The coordinates of point C are (3,0).
Point D is on the y-axis at y=1. The coordinates of point D are (0,1).

Name the ordered pair of each point shown:

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 4, 0” is labeled A. The point “ordered pair 0, 3” is labeled B.  The point “ordered pair -3, 0” is labeled C. The point “ordered pair 0, -5” is labeled D.
Solution
  1. A: (4,0)
  2. B: (0,3)
  3. C: (−3,0)
  4. D: (0,−5)

Name the ordered pair of each point shown:

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The point “ordered pair 5, 0” is labeled C. The point “ordered pair 0, 2” is labeled D.  The point “ordered pair -3, 0” is labeled A. The point “ordered pair 0,-3” is labeled B.
Solution
  1. A: (−3,0)
  2. B: (0,−3)
  3. C: (5,0)
  4. D: (0,2)

Verify Solutions to an Equation in Two Variables

All the equations we solved so far have been equations with one variable. In almost every case, when we solved the equation we got exactly one solution. The process of solving an equation ended with a statement such as x=4. Then we checked the solution by substituting back into the equation.

Here’s an example of a linear equation in one variable, and its one solution.

3x+5=173x=12x=4

But equations can have more than one variable. Equations with two variables can be written in the general 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 AandB are not both zero, is called a linear equation in two variables.

Notice that the word “line” is in linear.

Here is an example of a linear equation in two variables, x and y:

A series of equations is shown. The first line shows A x + B x = C. The “A” is red, the “B” is blue, and the “C” is turquoise. The second line shows x + 4 y = 8. The “4” is blue and the “8” is turquoise. The last line shows A =1 in red, B = 4 in blue, and C =8 in turquoise.

Is y=−5x+1 a linear equation? It does not appear to be in the form Ax+By=C. But we could rewrite it in this form.

The image displays the linear equation y = -5x + 1, written in a standard mathematical notation on a white background. This represents a line with a negative slope and a positive y-intercept.
Add 5x to both sides. The image shows the algebraic equation y + 5x = -5x + 1 + 5x.
Simplify. A linear equation is displayed as y + 5x = 1, featuring variables y and x, constants 5 and 1, and arithmetic operators addition and equality. The text is in a clear, dark font against a white background.
Use the Commutative Property to put it in Ax+By=C. Two linear equations are displayed: Ax + By = C (where A is red and B is light blue) and 5x + y = 1. The first equation shows a general form, while the second is a specific example.

By rewriting y=−5x+1 as 5x+y=1, we can see that it is a linear equation in two variables because it can be written in the form Ax+By=C.

Linear equations in two variables 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 to a Linear Equation in Two Variables

An ordered pair (x,y) is a solution to the linear equation Ax+By=C, if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.

Determine which ordered pairs are solutions of the equation x+4y=8:
  1. ⓐ (0,2)
  2. ⓑ (2,−4)
  3. ⓒ (−4,3)
Solution

Solution

Substitute the x- andy-values from each ordered pair into the equation and determine if the result is a true statement.

ⓐ (0,2) ⓑ (2,−4) ⓒ (−4,3)
This image demonstrates how to verify that x=0 and y=2 is a solution to the equation x+4y=8 by substituting the values, which results in the true statement 8=8. The image demonstrates checking if x=2 and y=-4 satisfy the equation x + 4y = 8, showing that -14 does not equal 8, thus proving they are not solutions. A step-by-step verification demonstrating that the values x = -4 and y = 3 satisfy the equation x + 4y = 8, leading to 8 = 8, confirmed with a checkmark.
(0,2) is a solution. (2,−4) is not a solution. (−4,3) is a solution.
Determine which ordered pairs are solutions to the given equation: 2x+3y=6
  1. ⓐ (3,0)
  2. ⓑ (2,0)
  3. ⓒ (6,−2)
Solution

ⓐ , ⓒ

Determine which ordered pairs are solutions to the given equation: 4x−y=8
  1. ⓐ (0,8)
  2. ⓑ (2,0)
  3. ⓒ (1,−4)
Solution

ⓑ , ⓒ

Determine which ordered pairs are solutions of the equation. y=5x−1:
  1. ⓐ (0,−1)
  2. ⓑ (1,4)
  3. ⓒ (−2,−7)
Solution

Solution

Substitute the x- and y-values from each ordered pair into the equation and determine if it results in a true statement.

ⓐ (0,−1) ⓑ (1,4) ⓒ (−2,−7)
Mathematical steps verify if x=0, y=-1 satisfies y=5x-1. Substitution leads to -1 ?= 5(0)-1, simplifying to -1 = -1, marked with a check. The point is a solution. A step-by-step mathematical verification is shown, starting with x=1 and y=4, and the equation y=5x-1. The values are substituted into the equation, leading to the confirmation 4=4 with a checkmark. A step-by-step verification of whether the point (-2, -7) satisfies the equation y = 5x - 1, demonstrating that -7 is not equal to -11, thus the point is not on the line.
(0,−1) is a solution. (1,4) is a solution. (−2,−7) is not a solution.
Determine which ordered pairs are solutions of the given equation: y=4x−3
  1. ⓐ (0,3)
  2. ⓑ (1,1)
  3. ⓒ (1,0)
Solution

ⓑ

Determine which ordered pairs are solutions of the given equation: y=−2x+6
  1. ⓐ (0,6)
  2. ⓑ (1,4)
  3. ⓒ (−2,−2)
Solution

ⓐ , ⓑ

Complete a Table of Solutions to a Linear Equation

In the previous examples, we substituted the x- andy-values of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do we find the ordered pairs if they are not given? One way is to choose a value for x and then solve the equation for y. Or, choose a value for y and then solve for x.

We’ll start by looking at the solutions to the equation y=5x−1 we found in Example 9. We can summarize this information in a table of solutions.

y=5x−1
x y (x,y)
0 −1 (0,−1)
1 4 (1,4)

To find a third solution, we’ll let x=2 and solve for y.

y=5x−1
The image shows the text 'Substitute x = 2.' in a blue-green gradient font against a white background. A mathematical equation shows 'y = 5(2) - 1', where the number 2 is highlighted in blue.
Multiply. y=10−1
Simplify. y=9

The ordered pair is a solution to y=5x-1. We will add it to the table.

y=5x−1
x y (x,y)
0 −1 (0,−1)
1 4 (1,4)
2 9 (2,9)

We can find more solutions to the equation by substituting any value of x or any value of y and solving the resulting equation to get another ordered pair that is a solution. There are an infinite number of solutions for this equation.

Complete the table to find three solutions to the equation y=4x−2:

y=4x−2
x y (x,y)
0
−1
2
Solution

Solution

Substitute x=0,x=−1, and x=2 into y=4x−2.

A mathematical equation displaying 'x = 0' with the number 0 highlighted in a light blue color, set against a plain white background. A mathematical equation on a white background, displaying 'x = -1'. The 'x' and '=' symbols are in a dark gray tone, while the '-1' is depicted in a lighter blue color, indicating a numerical value. A simple mathematical equation shows the variable 'x' is equal to the number 2, displayed with 'x' in black and '2' in light blue.
y=4x−2 y=4x−2 y=4x−2
A mathematical equation: y = 4 * 0 - 2, with the number 0 highlighted in blue. The equation evaluates to y = -2. A mathematical equation, y = 4(-1) - 2, is displayed. A mathematical equation 'y = 4 * 2 - 2' is displayed, with the second '2' highlighted in a light blue color.
y=0−2 y=−4−2 y=8−2
y=−2 y=−6 y=6
(0,−2) (−1,−6) (2,6)

The results are summarized in the table.

y=4x−2
x y (x,y)
0 −2 (0,−2)
−1 −6 (−1,−6)
2 6 (2,6)

Complete the table to find three solutions to the equation: y=3x−1.

y=3x−1
x y (x,y)
0
−1
2
Solution
y=3x−1
x y (x,y)
0 −1 (0,−1)
−1 −4 (−1,−4)
2 5 (2,5)

Complete the table to find three solutions to the equation: y=6x+1

y=6x+1
x y (x,y)
0
1
−2
Solution
y=6x+1
x y (x,y)
0 1 (0,1)
1 7 (1,7)
−2 −11 (−2,−11)

Complete the table to find three solutions to the equation 5x−4y=20:

5x−4y=20
x y (x,y)
0
0
5
Solution

Solution

The figure shows three algebraic substitutions into an equation. The first substitution is x = 0, with 0 shown in blue. The next line is 5 x- 4 y = 20.  The next line is 5 times 0, shown in blue - 4 y = 20.  The next line is 0 - 4 y = 20.  The next line is - 4 y = 20. The next line is y = -5.   The last line is “ordered pair 0, -5”. The second substitution is y = 0, with 0 shown in red. The next line is 5 x- 4 y = 20.  The next line is 5 x - 4 times 0, with 0 shown in red. The next line is 5 x  - 0 = 20.  The next line is 5 x = 20. The next line is x = 4.   The last line is “ordered pair 4, 0”. The third substitution is  y = 5, with 5 shown in red.  The next line is 5 x- 4 y = 20.  The next line is 5 x - 4 times 5, with 5 shown in blue. The next line is 5 x  - 20 = 20.  The next line is 5 x = 40. The next line is x = 8.   The last line is “ordered pair 8, 5”.

The results are summarized in the table.

5x−4y=20
x y (x,y)
0 −5 (0,−5)
4 0 (4,0)
8 5 (8,5)

Complete the table to find three solutions to the equation: 2x−5y=20.

2x−5y=20
x y (x,y)
0
0
−5
Solution
2x−5y=20
x y (x,y)
0 −4 (0,−4)
10 0 (10,0)
−5 −6 (−5,−6)

Complete the table to find three solutions to the equation: 3x−4y=12.

3x−4y=12
x y (x,y)
0
0
−4
Solution
3x−4y=12
x y (x,y)
0 −3 (0,−3)
4 0 (4,0)
−4 −6 (−4,−6)

Find Solutions to Linear Equations in Two Variables

To find a solution to a linear equation, we can choose any number we want to substitute into the equation for either x or y. We could choose 1,100,1,000, or any other value we want. But it’s a good idea to choose a number that’s easy to work with. We’ll usually choose 0 as one of our values.

Find a solution to the equation 3x+2y=6.

Solution

Solution

Step 1: Choose any value for one of the variables in the equation. We can substitute any value we want for x or any value for y.
Let's pick x=0.
What is the value of y if x=0?
Step 2: Substitute that value into the equation.
Solve for the other variable.

Substitute 0 for x.
Simplify.

Divide both sides by 2.
Step-by-step calculation showing how to find the y-intercept (y=3) of the equation 3x + 2y = 6 by setting x to 0.
Step 3: Write the solution as an ordered pair. So, when x=0,y=3. This solution is represented by the ordered pair (0,3).
Step 4: Check. Instruction to substitute x=0 and y=3 into the equation 3x + 2y = 6.
Is the result a true equation?
Yes!
Verifying if the point (0, 3) satisfies the equation 3x + 2y = 6. Substituting x=0 and y=3 proves the equality 6=6, confirming it's a solution.

Find a solution to the equation: 4x+3y=12.

Solution

Answers will vary.

Find a solution to the equation: 2x+4y=8.

Solution

Answers will vary.

We said that linear equations in two variables have infinitely many solutions, and we’ve just found one of them. Let’s find some other solutions to the equation 3x+2y=6.

Find three more solutions to the equation 3x+2y=6.

Solution

Solution

To find solutions to 3x+2y=6, choose a value for x or y. Remember, we can choose any value we want for x or y. Here we chose 1 for x, and 0 and −3 for y.
Substitute it into the equation. Algebraic substitution showing y=0 being substituted into the equation 3x+2y=6 to simplify it to 3x+2(0)=6. An image depicting a step in solving a system of equations, where the value x = 1 is substituted into the equation 3x + 2y = 6, leading to 3(1) + 2y = 6. The number 1 is highlighted. A step-by-step display of algebraic substitution, showing the value y = -3 being substituted into the equation 3x + 2y = 6, resulting in 3x + 2(-3) = 6. The -3 is highlighted in red in both its initial definition and its substituted form.
Simplify.
Solve.
A mathematical equation shown in two steps: first, 3x + 0 = 6, and then simplified to 3x = 6. This illustrates the identity property of addition where adding zero does not change the value. Mathematical problem showing the simplification of 3 + 2y = 6 to 2y = 3, illustrating a step in solving an algebraic equation. An algebra problem showing the steps to solve for x, with the initial equation 3x - 6 = 6 followed by 3x = 12.
The mathematical equation 'x = 2' is displayed in black text on a white background, representing a simple algebraic solution or assignment. The equation y = 3/2 is displayed in black text on a white background. The image displays a simple algebraic equation, 'x = 4,' centered on a white background. The variable 'x' is set equal to the number '4', representing a straightforward mathematical statement.
Write the ordered pair. (2,0) (1,32) (4,−3)
Check your answers.
(2,0) (1,32) (4,−3)
Checking if (2,0) is a solution to the equation 3x + 2y = 6. Substituting x=2 and y=0, the equation simplifies to 6 + 0 = 6, confirming that (2,0) is indeed a valid solution. Solution verification for 3x + 2y = 6: substituting x=1 and y=3/2 correctly yields 6=6, confirming the values satisfy the equation. This image demonstrates how to check if the point (4, -3) is a solution to the equation 3x + 2y = 6. Substituting x=4 and y=-3 into the equation simplifies it to 6=6, confirming it's a valid solution.

So (2,0),(1,32) and (4,−3) are all solutions to the equation 3x+2y=6. In the previous example, we found that (0,3) is a solution, too. We can list these solutions in a table.

3x+2y=6
x y (x,y)
0 3 (0,3)
2 0 (2,0)
1 32 (1,32)
4 −3 (4,−3)

Find three solutions to the equation: 2x+3y=6.

Solution

Answers will vary.

Find three solutions to the equation: 4x+2y=8.

Solution

Answers will vary.

Let’s find some solutions to another equation now.

Find three solutions to the equation x−4y=8.

Solution

Solution

The image displays the linear equation x - 4y = 8, written in a standard algebraic format. The image displays the equation x - 4y = 8. A mathematical equation displays 'x - 4y = 8' in a bold, black font against a white background.
Choose a value for x or y. A close-up image displaying the mathematical equation 'x = 0' in black and teal typography against a white background. The equation y = 0 is displayed, with the number 0 highlighted in red. The equation y = 3 is displayed with the number 3 highlighted in red, indicating a constant value for y.
Substitute it into the equation. The image displays the mathematical equation 0 - 4y = 8. The '0' is rendered in a light blue color, while the rest of the equation, including the minus sign, '4y', equals sign, and '8', is in black. A mathematical equation shows 'x - 4 * 0 = 8', where the number 0 is highlighted in red. The mathematical equation x - 4 ⋅ 3 = 8, with the number 3 highlighted in red.
Solve. Solving the linear equation -4y = 8, which results in y = -2. A straightforward algebraic problem showing x - 0 = 8, simplifying to x = 8. A clear example of the additive identity property. An algebraic equation x - 12 = 8 is shown, with its solution x = 20 presented below it. The image displays a simple mathematical problem and its answer.
Write the ordered pair. (0,−2) (8,0) (20,3)

So (0,−2),(8,0), and (20,3) are three solutions to the equation x−4y=8.

x−4y=8
x y (x,y)
0 −2 (0,−2)
8 0 (8,0)
20 3 (20,3)

Remember, there are an infinite number of solutions to each linear equation. Any point you find is a solution if it makes the equation true.

Find three solutions to the equation: 4x+y=8.

Solution

Answers will vary.

Find three solutions to the equation: x+5y=10.

Solution

Answers will vary.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Plotting Points
  • Identifying Quadrants
  • Verifying Solution to Linear Equation

Key Concepts

  • Sign Patterns of the Quadrants
    Quadrant I Quadrant II Quadrant III Quadrant IV
    (x,y) (x,y) (x,y) (x,y)
    (+,+) (−,+) (−,−) (+,−)
  • Coordinates of Zero
    • Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates ( a, 0).
    • Points with a x-coordinate equal to 0 are on the y-axis, and have coordinates ( 0, b).
    • The point (0, 0) is called the origin. It is the point where the x-axis and y-axis intersect.

Practice Makes Perfect

Plot Points on a Rectangular Coordinate System

In the following exercises, plot each point on a coordinate grid.

(3,2)

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair 3, 2” is labeled

(4,1)

(1,5)

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair 1, 5” is labeled

(3,4)

(4,1),(1,4)

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair 1, 4” is labeled. The point “ordered pair 4, 1” is labeled.

(3,2),(2,3)

(3,4),(4,3)

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair 3, 4” is labeled. The point “ordered pair 4, 3” is labeled.

In the following exercises, plot each point on a coordinate grid and identify the quadrant in which the point is located.

  1. ⓐ (−4,2)
  2. ⓑ (−1,−2)
  3. ⓒ (3,−5)
  4. ⓓ (2,52)
  1. ⓐ (−2,−3)
  2. ⓑ (3,−3)
  3. ⓒ (−4,1)
  4. ⓓ (1,32)
Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The quadrants are labeled I, II, III, and IV. The point (-1, 1) is labeled a, the point (-2, -1) is labeled b. The point (1, -4) is labeled c, and the point (3, 7/2) is labeled d.

  1. ⓐ (−1,1)
  2. ⓑ (−2,−1)
  3. ⓒ (1,−4)
  4. ⓓ (3,72)

In the following exercises, plot each point on a coordinate grid.

  1. ⓐ (3,−2)
  2. ⓑ (−3,2)
  3. ⓒ (−3,−2)
  4. ⓓ (3,2)
Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point (3, -2) is labeled a, the point (-3, 2) is labeled b. The point (-3, -2) is labeled c, and the point (3, 2) is labeled d.

  1. ⓐ (4,−1)
  2. ⓑ (−4,1)
  3. ⓒ (−4,−1)
  4. ⓓ (4,1)
  1. ⓐ (−2,0)
  2. ⓑ (−3,0)
  3. ⓒ (0,4)
  4. ⓓ (0,2)
Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point (-2, 0) is labeled a, the point (-3, 0) is labeled b. The point (0, 4) is labeled c, and the point (0, 2) is labeled d.

Identify Points on a Graph

In the following exercises, name the ordered pair of each point shown.

The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -4, 1” is labeled “A”. The point “ordered pair -3, -4” is labeled “B”.
The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair 4, 3” is labeled “D”. The point “ordered pair 1, -3” is labeled “C”.
Solution

C(1, -3) D(4, 3)

The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -3, -2” is labeled “X”. The point “ordered pair 5, -1” is labeled “Y”.
The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -2, 4” is labeled “S”. The point “ordered pair -4, -2” is labeled “T”.
Solution

S(-2, 4) T(-4, -2)

The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -2, 0” is labeled “B”. The point “ordered pair 0, -2” is labeled “A”.
The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair -1, 0” is labeled “D”. The point “ordered pair 0,  -1” is labeled “C”.
Solution

C(0, -1) D(-1, 0)

The graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The point “ordered pair 3, 0” is labeled “T”. The point “ordered pair -4,  0” is labeled “S”.

Verify Solutions to an Equation in Two Variables

In the following exercises, determine which ordered pairs are solutions to the given equation.

2x+y=6
  1. ⓐ (1,4)
  2. ⓑ (3,0)
  3. ⓒ (2,3)
Solution

ⓐ , ⓑ

x+3y=9

  1. ⓐ (0,3)
  2. ⓑ (6,1)
  3. ⓒ (−3,−3)
4x−2y=8
  1. ⓐ (3,2)
  2. ⓑ (1,4)
  3. ⓒ (0,−4)
Solution

ⓐ , ⓒ

3x−2y=12
  1. ⓐ (4,0)
  2. ⓑ (2,−3)
  3. ⓒ (1,6)
y=4x+3
  1. ⓐ (4,3)
  2. ⓑ (−1,−1)
  3. ⓒ (12,5)
Solution

ⓑ , ⓒ

y=2x−5
  1. ⓐ (0,−5)
  2. ⓑ (2,1)
  3. ⓒ (12,−4)
y=12x−1
  1. ⓐ (2,0)
  2. ⓑ (−6,−4)
  3. ⓒ (−4,−1)
Solution

ⓐ , ⓑ

y=13x+1
  1. ⓐ (−3,0)
  2. ⓑ (9,4)
  3. ⓒ (−6,−1)

Find Solutions to Linear Equations in Two Variables

In the following exercises, complete the table to find solutions to each linear equation.

y=2x−4

x y (x,y)
−1
0
2
Solution
x y (x,y)
−1 −6 (−1,−6)
0 −4 (0,−4)
2 0 (2,0)

y=3x−1

x y (x,y)
−1
0
2

y=−x+5

x y (x,y)
−2
0
3
Solution
x y (x,y)
−2 7 (−2,7)
0 5 (0,5)
3 2 (3,2)

y=13x+1

x y (x,y)
0
3
6

y=−32x−2

x y (x,y)
−2
0
2
Solution
x y (x,y)
−2 1 (−2,1)
0 −2 (0,−2)
2 −5 (2,−5)

x+2y=8

x y (x,y)
0
4
0

Everyday Math

Weight of a baby Mackenzie recorded her baby’s weight every two months. The baby’s age, in months, and weight, in pounds, are listed in the table, and shown as an ordered pair in the third column.

ⓐ Plot the points on a coordinate grid.

Age Weight (x,y)
0 7 (0,7)
2 11 (2,11)
4 15 (4,15)
6 16 (6,16)
8 19 (8,19)
10 20 (10,20)
12 21 (12,21)

ⓑ Why is only Quadrant I needed?

Solution
  1. ⓐ This figure shows points plot on the x y coordinate plane. There are 7 points graphed without labeled at approximately the points “ordered pair 0, 7”, “ordered pair 2, 11”, “ordered pair 4, 15”, “ordered pair 6, 16”, “ordered pair 8, 19”, “ordered pair 10, 20”, “ordered pair 12, 21”.
  2. ⓑ Age and weight are only positive.

Weight of a child Latresha recorded her son’s height and weight every year. His height, in inches, and weight, in pounds, are listed in the table, and shown as an ordered pair in the third column.

ⓐ Plot the points on a coordinate grid.

Heightx Weighty (x,y)
28 22 (28,22)
31 27 (31,27)
33 33 (33,33)
37 35 (37,35)
40 41 (40,41)
42 45 (42,45)

ⓑ Why is only Quadrant I needed?

Writing Exercises

Have you ever used a map with a rectangular coordinate system? Describe the map and how you used it.

Solution

Answers may vary.

How do you determine if an ordered pair is a solution to a given equation?

Self Check

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

An empty self-assessment checklist for students to rate their confidence in math skills like plotting points, identifying points on graphs, and solving linear equations with categories: 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.

linear equation
An equation of the form Ax+By=C, where AandB 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.
(x,y)x-coordinate,y-coordinate
origin
The point (0,0) is called the origin. It is the point where the x-axis and y-axis intersect.
quadrants
The x-axis and y-axis divide a rectangular coordinate system into four areas, called quadrants.
solution to a linear equation in two variables
An ordered pair (x,y) is a solution to the linear equation Ax+By=C, if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.
x-axis
The x-axis is the horizontal axis in a rectangular coordinate system.
y-axis
The y-axis is the vertical axis on a rectangular coordinate system.

Graphing Linear Equations

Learning Objectives

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

  • Recognize the relation between the solutions of an equation and its graph
  • Graph a linear equation by plotting points
  • Graph vertical and horizontal lines

Before you get started, take this readiness quiz.

Evaluate: 3x+2 when x=−1.
If you missed this problem, review Example 10 in Multiply and Divide Integers.

Solution

−1

Solve the formula: 5x+2y=20 for y.
If you missed this problem, review Example 6 in Solve a Formula for a Specific Variable.

Solution

y=20−5x2

Simplify: 38(−24).
If you missed this problem, review Example 10 in Multiply and Divide Fractions.

Solution

−9

Recognize the Relation Between the Solutions of an Equation and its Graph

In Use the Rectangular Coordinate System, we found a few solutions to the equation 3x+2y=6. They are listed in the table below. So, the ordered pairs (0,3), (2,0), (1,32), (4,−3), are some solutions to the equation3x+2y=6. We can plot these solutions in the rectangular coordinate system as shown on the graph at right.

The image shows a table breaking up points by their x components and y components. These points are then mapped onto a graph. Points include (0, 3), (2, 0), (1, 3/2), and (4, –3).

Notice how the points line up perfectly? We connect the points with a straight line to get the graph of the equation 3x+2y=6. Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

A graph displays the linear equation 3x + 2y = 6. The line passes through points such as (0, 3), (2, 0), and (3, -1.5), demonstrating its downward slope on a Cartesian coordinate system.

Every point on the line is a solution of the equation. Also, every solution of this equation is a point on this line. Points not on the line are not solutions!

Notice that the point whose coordinates are (−2,6) is on the line shown in Figure 1. If you substitute x=−2 and y=6 into the equation, you find that it is a solution to the equation.

The image shows the process of testing point (–2, 6) as a solution to the given equation. Graphing it out, the line properly intercepts this point but does not intercept point (4, 1) which was also tested but failed with the equation.

So (4,1) is not a solution to the equation 3x+2y=6 . Therefore the point (4,1) is not on the line.

This is an example of the saying,” A picture is worth a thousand words.” The line shows you all the solutions to the 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 3x+2y=6.

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 below.

Graph of the linear function y = 2x - 3, illustrating its positive slope and y-intercept at -3.

For each ordered pair decide
  1. ⓐ Is the ordered pair a solution to the equation?
  2. ⓑ Is the point on the line?
  1. (a) (0,–3)
  2. (b) (3,3)
  3. (c) (2,−3)
  4. (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.

ⓐ
This image shows a series of ordered pairs tested with the equation to determine if said pair is a solution to the equation.

ⓑ Plot the points A: (0,−3) B: (3,3) C: (2,−3) and D: (−1,−5).
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.

A graph in which a line runs through points (–1, –5), (0, –3), and (3, 3). The point (2, –3) is also plotted on the graph but does not intercept the line.

The points which are solutions to y=2x−3 are on the line, but the point which is not a solution is not on the line.

The graph of y=3x−1 is shown.

For each ordered pair, decide

  1. ⓐ is the ordered pair a solution to the equation?
  2. ⓑ is the point on the line?
A two-dimensional graph displays the linear function y = 3x - 1. The line ascends from left to right, intersecting the y-axis at -1.
  1. (0,−1)
  2. (2,2)
  3. (3,−1)
  4. (−1,−4)
Solution
  1. ⓐ yes ⓑ yes
  2. ⓐ no ⓑ no
  3. ⓐ no ⓑ no
  4. ⓐ yes ⓑ yes

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The method we used at the start of this section to graph is called plotting points, or the Point-Plotting Method.

Let’s graph the equation y=2x+1 by plotting points.

We start by finding three points that are solutions to the equation. We can choose any value for x or y, and then solve for the other variable.

Since y is isolated on the left side of the equation, it is easier to choose values for x. We will use 0,1, and -2 for x for this example. We substitute each value of x into the equation and solve for y.

The figure shows three algebraic substitutions into an equation. The first substitution is for x = -2, with -2 shown in blue. The next line is y = 2 x + 1. The next line is y = 2 open parentheses -2, shown in blue, closed parentheses, + 1. The next line is y = - 4 + 1. The next line is y = -3. The last line is “ordered pair -2, -3”. The second  substitution is for x = 0, with 0 shown in blue. The next line is y = 2 x + 1. The next line is y = 2 open parentheses 0, shown in blue, closed parentheses, + 1. The next line is y = 0 + 1. The next line is y = 1. The last line is “ordered pair 0, 2”. The third substitution is for x = 1, with 1 shown in blue. The next line is y = 2 x + 1. The next line is y = 2 open parentheses 1, shown in blue, closed parentheses, + 1. The next line is y = 2 + 1. The next line is y = 3. The last line is “ordered pair -1, 3”.

We can organize the solutions in a table. See Table 1.

y=2x+1
x y (x,y)
0 1 (0,1)
1 3 (1,3)
−2 −3 (−2,−3)

Now we plot the points on a rectangular coordinate system. Check that the points line up. If they did not line up, it would mean we made a mistake and should double-check all our work. See Figure 2.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. Three labeled points are shown, “ordered pair -2, -3”, “ordered pair 0, 1”, and ordered pair 1, 3”.

Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. The line is the graph of y=2x+1.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair -2, -3”, “ordered pair 0, 1”, and ordered pair 1, 3”.

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 on a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
  3. Draw the line through the 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 plot only 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. See Figure 4.

There are two figures. Figure a shows three points that are all contained on a straight line. There is a line with arrows that passed through the three points. Figure b shows 3 points that are not all arranged in a straight line.
Look at the difference between (a) and (b). All three points in (a) line up so we can draw one line through them. The three points in (b) do not line up. We cannot draw a single straight line through all three points.

Graph the equation y=−3x.

Solution

Solution

Find three points that are solutions to the equation. It’s easier to choose values for x, and solve for y. Do you see why?

The figure shows three algebraic substitutions into an equation. The first substitution is for x = 0, with 0 shown in blue. The next line is y = -3 x. The next line is y = -3 open parentheses 0, shown in blue, closed parentheses. The next line is y = 0. The last line is “ordered pair 0, 0 “. The second substitution is for x = 1, with 0 shown in blue. The next line is y = -3 x. The next line is y = -3 open parentheses 1, shown in blue, closed parentheses. The next line is y = -3. The last line is “ordered pair 1, -3”. The third substitution is for x = -2, with -2 shown in blue. The next line is y = -3 x. The next line is y = -3 open parentheses -2, shown in blue, closed parentheses. The next line is y = 6. The last line is “ordered pair -2, 6 “.

List the points in a table.

y=−3x
x y (x,y)
0 0 (0,0)
1 3 (1,−3)
−2 6 (−2,6)

Plot the points, check that they line up, and draw the line as shown.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair -2, 6”, “ordered pair 0, 0”, and ordered pair 1, -3”. The line is labeled y = -3 x.

Graph the equation by plotting points: y=−4x.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, 0” and “ordered pair 4, -4”.

Graph the equation by plotting points: y=x.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, 0” and “ordered pair 1, -4”.

When an equation includes a fraction as the coefficient of x, we can substitute any numbers for x. But the math is easier if we make ‘good’ choices for the values of x. This way we will avoid fraction answers, which are hard to graph precisely.

Graph the equation y=12x+3.

Solution

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.

The figure shows three algebraic substitutions into an equation. The first substitution is for x = 0, with 0 shown in blue. The next line is y = 1 over 2 x + 3. The next line is y = 1 over 2 open parentheses 0, shown in blue, closed parentheses, + 3.  The next line is y = 3. The last line is “ordered pair 0, 3”. The second substitution is for x = 2, with 2 shown in blue. The next line is y = 1 over 2 x + 3. The next line is y = 1 over 2 open parentheses 2, shown in blue, closed parentheses, + 3.  The next line is y = 4. The last line is “ordered pair 2, 4”. The third substitution is for x = 4, with 4 shown in blue. The next line is y = 1 over 2 x + 3. The next line is y = 1 over 2 open parentheses 4, shown in blue, closed parentheses, + 3.  The next line is y = 5. The last line is “ordered pair 4, 5”.

The points are shown in the table.

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 as shown.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair 0, 3”, “ordered pair 2, 4”, and ordered pair 4, 5”. The line is labeled y = 1 over 2 x + 3.

Graph the equation: y=13x−1.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, -1” and “ordered pair 3, 0”.

Graph the equation: y=14x+2.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, 2” and “ordered pair -12, 0”.

So far, all the equations we graphed had y given in terms of x. Now we’ll graph an equation with x and y on the same side.

Graph the equation x+y=5.

Solution

Solution

Find three points that are solutions to the equation. Remember, you can start with any value of x or y.

The figure shows three algebraic substitutions into an equation. The first substitution is for x = 0, with 0 shown in blue. The next line is x + y = 5. The next line is 0, shown in blue + y = 5. The next line is y = 5. The last line is “ordered pair 0, 5”. The second substitution is for x = 1, with 1 shown in blue. The next line is x + y = 5. The next line is 1, shown in blue + y = 5. The next line is y = 4. The last line is “ordered pair 1, 4”. The third substitution is for x = 4, with 4 shown in blue. The next line is x + y = 5. The next line is 4, shown in blue + y = 5. The next line is y = 1. The last line is “ordered pair 4, 1”.

We list the points in a table.

x+y=5
x y (x,y)
0 5 (0,5)
1 4 (1,4)
4 1 (4,1)

Then plot the points, check that they line up, and draw the line.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair 0, 5”, “ordered pair 1, 4”, and ordered pair 4, 1”. The line is labeled x + y = 5.

Graph the equation: x+y=−2.

Solution


This answer graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  The equation x plus y equals -2 is  shown. A line passes through the intercepts with coordinates 0, –2 and –2, 0.

Graph the equation: x−y=6.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through the points “ordered pair 6, 0” and “ordered pair 0, -6”.

In the previous example, the three points we found were easy to graph. But this is not always the case. Let’s see what happens in the equation 2x+y=3. If y is 0, what is the value of x?

This figure shows an algebraic substitution. The first line is 2 x + y = 3. The second line is 2 x + 0, with 0 shown in red. The third line is 2 x = 3. The last line is x = 3 over 2.

The solution is the point (32,0). This point has a fraction for the x-coordinate. While we could graph this point, it is hard to be precise graphing fractions. Remember in the example y=12x+3, we carefully chose values for x so as not to graph fractions at all. If we solve the equation 2x+y=3 for y, it will be easier to find three solutions to the equation.

2x+y=3
y=−2x+3

Now we can choose values for x that will give coordinates that are integers. The solutions for x=0,x=1, and x=−1 are shown.

y=−2x+3
x y (x,y)
0 3 (0,3)
1 1 (1,1)
−1 5 (-1,5)
The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair -1, 5”, “ordered pair 0, 3”, and ordered pair 1, 1”. The line is labeled 2 x + y = 3.

Graph the equation 3x+y=−1.

Solution

Solution

Find three points that are solutions to the equation.

First, solve the equation for y.

3x+y=−1 y=−3x−1

We’ll let x be 0,1, and −1 to find three points. The ordered pairs are shown in the table. Plot the points, check that they line up, and draw the line.

y=−3x−1
x y (x,y)
0 −1 (0,−1)
1 −4 (1,−4)
−1 2 (−1,2)
The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair -1, 2”, “ordered pair 0, -1”, and ordered pair 1, -4”. The line is labeled 3 x + y = -1.

If you can choose any three points to graph a line, how will you know if your graph matches the one shown in the answers in the book? If the points where the graphs cross the x- and y-axes are the same, the graphs match.

Graph each equation: 2x+y=2.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair -1, 2”, “ordered pair 0, -1”, and ordered pair 1, -4”. The line is labeled 3 x + y = -1.

Graph each equation: 4x+y=−3.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair -1,  4” and “ordered pair 0, -3”.

Graph Vertical and Horizontal Lines

Can we graph an equation with only one variable? Just x and no y, or just y without an x? How will we make a table of values to get the points to plot?

Let’s consider the equation x=−3. The equation says that x is always equal to −3, so its value does not depend on y. No matter what y is, the value of x is always −3.

To make a table of solutions, we write −3 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 size of our coordinate graph, we’ll use 1,2, and 3 for the y-coordinates as shown in the table.

x=−3
x y (x,y)
−3 1 (−3,1)
−3 2 (−3,2)
−3 3 (−3,3)

Then plot the points and connect them with a straight line. Notice in Figure 5 that the graph is a vertical line.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A vertical line passes through three labeled points, “ordered pair -3, 3”, “ordered pair -3, 2”, and ordered pair -3, 1”. The line is labeled x = -3.

Vertical Line

A vertical line is the graph of an equation that can be written in the form x=a.

The line passes through the x-axis at (a,0).

Graph the equation x=2. What type of line does it form?

Solution

Solution

The equation has only variable, x, and x is always equal to 2. We make a table where x is always 2 and we put in any values for y.

x=2
x y (x,y)
2 1 (2,1)
2 2 (2,2)
2 3 (2,3)

Plot the points and connect them as shown.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A vertical line passes through three labeled points, “ordered pair 2, 3”, “ordered pair 2, 2”, and ordered pair 2, 1”. The line is labeled x = 2.

The graph is a vertical line passing through the x-axis at 2.

Graph the equation: x=5.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A vertical line passes through the points “ordered pair 5,  0” and “ordered pair 5, 1”.

Graph the equation: x=−2.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A vertical line passes through the points “ordered pair 5,  0” and “ordered pair 5, 1”.

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.

To make a table of solutions, write 4 for all the y values and then choose any values for x.

We’ll use 0,2, and 4 for the x-values.

y=4
x y (x,y)
0 4 (0,4)
2 4 (2,4)
4 4 (4,4)

Plot the points and connect them, as shown in Figure 6. This graph is a horizontal line passing through the y-axis at 4.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A horizontal  line passes through three labeled points, “ordered pair 0, 4”, “ordered pair 2, 4”, and ordered pair 4, 4”. The line is labeled y = 4.

Horizontal Line

A horizontal line is the graph of an equation that can be written in the form y=b.

The line passes through the y-axis at (0,b).

Graph the equation y=−1.

Solution

Solution

The equation y=−1 has only variable, y. The value of y is constant. All the ordered pairs in the table have the same y-coordinate, −1. We choose 0,3, and −3 as values for x.

y=−1
x y (x,y)
−3 −1 (−3,−1)
0 −1 (0,−1)
3 −1 (3,−1)

The graph is a horizontal line passing through the y-axis at –1 as shown.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A horizontal  line passes through three labeled points, “ordered pair -3, -1”, “ordered pair 0, -1”, and ordered pair 3, -1”. The line is labeled y = -1.

Graph the equation: y=−4.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A horizontal  line passes through the points “ordered pair 0,  -4” and “ordered pair 1, -4”.

Graph the equation: y=3.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A horizontal  line passes through the points “ordered pair 0,  3” and “ordered pair 1, 3”.

The equations for vertical and horizontal lines look very similar to equations like y=4x. 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. The y-coordinate changes according to the value of x.

The equation y=4 has only one variable. The value of y is constant. The y-coordinate is always 4. It does not depend on the value of x.

There are two tables. This first table is titled y = 4 x, which is shown in blue. It has 4 rows and 3 columns. The first row is a header row and it labels each column “x”, “y”, and  “ordered pair x, y”. Under the column “x” are the values  0, 1, and 2. Under the column “y” are the values  0, 4, and 8. Under the column “ordered pair x, y” are the values “ordered pair 0, 0”, “ordered pair 1, 4”, and “ordered pair 2, 8”. This second table is titled y = 4 , which is shown in red. It has 4 rows and 3 columns. The first row is a header row and it labels each column “x”, “y”, and  “ordered pair x, y”. Under the column “x” are the values  0, 1, and 2. Under the column “y” are the values  4, 4, and 4. Under the column “ordered pair x, y” are the values “ordered pair 0, 4”, “ordered pair 1, 4”, and “ordered pair 2, 4”.

The graph shows both equations.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A horizontal line passes through “ordered pair 0, 4” and “ordered pair 1, 4” and is labeled y = 4. A second line passes through “ordered pair 0, 0” and “ordered pair 1, 4” and is labeled y = 4 x. The two lines intersect at “ordered pair 1, 4”.

Notice that the equation y=4x gives a slanted line whereas y=4 gives a horizontal line.

Graph y=−3x and y=−3 in the same rectangular coordinate system.

Solution

Solution

Find three solutions for each equation. Notice that the first equation has the variable x, while the second does not. Solutions for both equations are listed.
There are two tables. This first table is titled y = -3 x, which is shown in red. It has 4 rows and 3 columns. The first row is a header row and it labels each column “x”, “y”, and  “ordered pair x, y”. Under the column “x” are the values  0, 1, and 2. Under the column “y” are the values  0, -3, and -6. Under the column “ordered pair x, y” are the values “ordered pair 0, 0”, “ordered pair 1, -3”, and “ordered pair 2, -6”. This second table is titled y = -3 , which is shown in red. It has 4 rows and 3 columns. The first row is a header row and it labels each column “x”, “y”, and  “ordered pair x, y”. Under the column “x” are the values  0, 1, and 2. Under the column “y” are the values  -3, -3, and -3. Under the column “ordered pair x, y” are the values “ordered pair 0, -3”, “ordered pair 1, -3”, and “ordered pair 2, -3”.

The graph shows both equations.
The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A horizontal line passes through “ordered pair 0, -3” and “ordered pair 1, -3” and is labeled y = -3. A second line passes through “ordered pair 0, 0” and “ordered pair 1, -3” and is labeled y = -3 x. The two lines intersect at “ordered pair 1, -3”.

Graph the equations in the same rectangular coordinate system: y=−4x and y=−4.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A horizontal line passes through “ordered pair 0, -4” and “ordered pair 1, -4” . A second line passes through “ordered pair 0, 0” and “ordered pair 1, -4” . The two lines intersect at “ordered pair 1, -4”.

Graph the equations in the same rectangular coordinate system: y=3 and y=3x.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A horizontal line passes through “ordered pair 0, 3” and “ordered pair 1, 3” . A second line passes through “ordered pair 0, 0” and “ordered pair 1, 3” . The two lines intersect at “ordered pair 1, 3”.

ACCESS ADDITIONAL ONLINE RESOURCES

  • Use a Table of Values
  • Graph a Linear Equation Involving Fractions
  • Graph Horizontal and Vertical Lines

Key Concepts

  • 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 on a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
    3. Draw the line through the points. Extend the line to fill the grid and put arrows on both ends of the line.
  • 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.

Practice Makes Perfect

Recognize the Relation Between the Solutions of an Equation and its Graph

In each of the following exercises, an equation and its graph is shown. For each ordered pair, decide
  1. ⓐ is the ordered pair a solution to the equation?
  2. ⓑ is the point on the line?

y=x+2
A graph with both x- and y-axis ranging from –8 to 8. A line runs upwards, intercepting the x-axis at (–2, 0) and the y-axis at (0, 2).

  1. (0,2)
  2. (1,2)
  3. (−1,1)
  4. (−3,1)
Solution
  1. ⓐ yes ⓑ yes
  2. ⓐ no ⓑ no
  3. ⓐ yes ⓑ yes
  4. ⓐ no ⓑ no

y=x−4
A graph with both x- and y-axis running from –8 to 8. A line is drawn that intercepts the y-axis at (0, –4) and the x-axis at (4, 0)

  1. (0,−4)
  2. (3,−1)
  3. (2,2)
  4. (1,−5)

y=12x−3
A graph with both x- and y-axis running from –8 to 8. A line is drawn that intercepts the y-axis at (0, –3) and the x-axis at (6, 0)

  1. (0,−3)
  2. (2,−2)
  3. (−2,−4)
  4. (4,1)
Solution
  1. ⓐ yes ⓑ yes
  2. ⓐ yes ⓑ yes
  3. ⓐ yes ⓑ yes
  4. ⓐ no ⓑ no

y=13x+2
A linear graph on a grid, showing a line with a positive slope, passing through y-intercept (0, 2) and x-intercept (-7, 0). The x and y axes range from -8 to 8.

  1. (0,2)
  2. (3,3)
  3. (−3,2)
  4. (−6,0)

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

y=3x−1

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  -1” and “ordered pair 1/3, 0”.

y=2x+3

y=−2x+2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  4” and “ordered pair 1, 0”.

y=−3x+1

y=x+2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  2” and “ordered pair 2, 4”.

y=x−3

y=−x−3

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  -3” and “ordered pair 1, -4”.

y=−x−2

y=2x

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  0” and “ordered pair 2, 4”.

y=3x

y=−4x

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  0” and “ordered pair 1, -4”.

y=−2x

y=12x+2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  2” and “ordered pair 4, 4”.

y=13x−1

y=43x−5

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  5” and “ordered pair 4, 2”.

y=32x−3

y=−25x+1

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  1” and “ordered pair 8, -2”.

y=−45x−1

y=−32x+2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  2” and “ordered pair 4, -4”.

y=−53x+4

x+y=6

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  6” and “ordered pair 6, 0”.

x+y=4

x+y=−3

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  -3” and “ordered pair -3, 0”.

x+y=−2

x−y=2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  -2” and “ordered pair 2, 0”.

x−y=1

x−y=−1

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  1” and “ordered pair 8, 8”.

x−y=−3

−x+y=4

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair -4,  0” and “ordered pair 0, 4”.

−x+y=3

−x−y=5

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair -5,  0” and “ordered pair 0, -5”.

−x−y=1

3x+y=7

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 4,  -4” and “ordered pair 0, 8”.

5x+y=6

2x+y=−3

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  -3” and “ordered pair 2, -7”.

4x+y=−5

2x+3y=12

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  4” and “ordered pair 6, 0”.

3x−4y=12

13x+y=2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0,  2” and “ordered pair 6, 0”.

12x+y=3

Graph Vertical and Horizontal lines

In the following exercises, graph the vertical and horizontal lines.

x=4

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A vertical line passes through the points “ordered pair 4,  0” and “ordered pair 4, 1”.

x=3

x=−2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A vertical line passes through the points “ordered pair -2,  0” and “ordered pair -2, 1”.

x=−5

y=3

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A horizontal line passes through the points “ordered pair 0,  3” and “ordered pair 1, 3”.

y=1

y=−5

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A horizontal line passes through the points “ordered pair 0,  -5” and “ordered pair 1, -5”.

y=−2

x=73

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A vertical line passes through the points “ordered pair 7 over 3,  0” and “ordered pair 7 over 3 , 1”.

x=54

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

y=−12x and y=−12

Solution


A Cartesian coordinate system displays a dark blue line that passes through the origin (0,0). The line has a negative slope, intersecting points like (-4, 2) and (4, -2).

y=−13x and y=−13

y=2x and y=2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A horizontal line passes through “ordered pair 0, 2” and “ordered pair 1, 2” . A second line passes through “ordered pair 0, 0” and “ordered pair 1, 2” . The two lines intersect at “ordered pair 1, 2”.

y=5x and y=5

Mixed Practice

In the following exercises, graph each equation.

y=4x

Solution


A coordinate plane shows a straight line passing through the origin (0,0) and the points (1,4) and (-1,-4), indicating a positive slope of 4.

y=2x

y=−12x+3

Solution


A graph displays a blue line with a negative slope on a Cartesian coordinate plane. The line intersects the y-axis at approximately 3.5 and the x-axis at approximately 7.

y=14x−2

y=−x

Solution

A line graph on a Cartesian plane showing a straight line passing through the origin (0,0) with a negative slope. The x and y axes range from -8 to 8.

y=x

x−y=3

Solution

A graph showing a straight line on a Cartesian coordinate plane. The line has a positive slope, intersecting the y-axis at -3 and the x-axis at 4.5. It represents a linear function.

x+y=−5

4x+y=2

Solution

A graph shows a linear function, represented by a solid downward-sloping line, plotted on a Cartesian coordinate system. The line intersects the y-axis at (0, 2) and the x-axis at (1, 0), extending infinitely in both directions as indicated by arrows.

2x+y=6

y=−1

Solution

A Cartesian coordinate system with a grid. The x and y axes are labeled from -8 to 8. A horizontal double-headed arrow is drawn along the x-axis, indicating the entire range.

y=5

2x+6y=12

Solution

A graph on a coordinate plane shows a downward-sloping line. The line intersects the y-axis at (0, 2) and the x-axis at (6, 0), extending infinitely in both directions.

5x+2y=10

x=3

Solution

A graph with x- and y- axis both ranging from –8 to 8. A line is drawn to demonstrate x = 3.

x=−4

Everyday Math

Motor home cost The Robinsons rented a motor home for one week to go on vacation. It cost them $594 plus $0.32 per mile to rent the motor home, so the linear equation y=594+0.32x gives the cost, y, for driving x miles. Calculate the rental cost for driving 400,800,and1,200 miles, and then graph the line.

Solution


The graph shows the x y-coordinate plane. The x-axis runs from 0 to 1000. The y-axis runs from 0 to 1200. A line passes through the points “ordered pair 0,  594” and “ordered pair 800, 850”.
$722, $850, $978

Weekly earning At the art gallery where he works, Salvador gets paid $200 per week plus 15% of the sales he makes, so the equation y=200+0.15x gives the amount y he earns for selling x dollars of artwork. Calculate the amount Salvador earns for selling $900, $1,600,and$2,000, and then graph the line.

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?

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, asking students to rate their ability to graph linear equations by plotting points, and graph vertical and horizontal lines.

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

horizontal line
A horizontal line is the graph of an equation that can be written in the form y=b. The line passes through the y-axis at (0,b).
vertical line
A vertical line is the graph of an equation that can be written in the form x=a. The line passes through the x-axis at (a,0).

Graphing with Intercepts

Learning Objectives

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

  • Identify the intercepts on a graph
  • Find the intercepts from an equation of a line
  • Graph a line using the intercepts
  • Choose the most convenient method to graph a line

Before you get started, take this readiness quiz.

Solve: 3x+4y=−12 for x when y=0.
If you missed this problem, review Example 6 in Solve a Formula for a Specific Variable.

Solution

−4

Is the point (0,−5) on the x-axis or y-axis?
If you missed this problem, review Example 5 in Use the Rectangular Coordinate System.

Solution

y-axis

Which ordered pairs are solutions to the equation 2x−y=6?
ⓐ (6,0)ⓑ (0,−6)ⓒ (4,−2).
If you missed this problem, review Example 8 in Use the Rectangular Coordinate System.

Solution

b

Identify the Intercepts on a Graph

Every linear equation has a unique line that represents all the solutions of the equation. When graphing a line by plotting points, each person who graphs the line can choose any three points, so two people graphing the line might use different sets of points.

At first glance, their two lines might appear different since they would have different points labeled. But if all the work was done correctly, the lines will be exactly the same line. One way to recognize that they are indeed the same line is to focus on where the line crosses the axes. Each of these points is called an intercept of the line.

Intercepts of a Line

Each of the points at which a line crosses the x-axis and the y-axis is called an intercept of the line.

Let’s look at the graph of the lines shown in Figure 1.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through two labeled points, “ordered pair 0, 6” and ordered pair 3, 0”.

First, notice where each of these lines crosses the x- axis:

Figure: The line crosses the x-axis at: Ordered pair of this point
42 3 (3,0)
43 4 (4,0)
44 5 (5,0)
45 0 (0,0)

Do you see a pattern?

For each row, 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.

Now, let's look at the points where these lines cross the y-axis.

Figure: The line crosses the y-axis at: Ordered pair for this point
42 6 (0,6)
43 -3 (0,-3)
44 -5 (0,-5)
45 0 (0,0)

x- intercept and y- intercept of a line

The x-intercept is the point, (a, 0), where the graph crosses the x-axis. The x-intercept occurs when y is zero.

The y-intercept is the point, (0, b), where the graph crosses the y-axis.

The y-intercept occurs when x is zero.

Find the x- andy-intercepts of each line:

This table displays linear equations and their respective graphical representations on an x-y coordinate plane, demonstrating how algebraic expressions translate visually.
ⓐ x+2y=4 The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through the points “ordered pair 0, 2” and “ordered pair 4, 0”.
ⓑ 3x−y=6 The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through the points “ordered pair 0, -6” and “ordered pair 2, 0”.
ⓒ x+y=−5 The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through the points “ordered pair 0, -5” and “ordered pair -5, 0”.
Solution

Solution

This table illustrates the relationship between a graph's axis crossings and their corresponding mathematical x and y-intercept terms.
ⓐ
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).
This table illustrates the relationship between the visual description of graph intercepts and their formal definitions and coordinates.
ⓑ
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).
Examples showing how to derive x- and y-intercepts from descriptions of a graph crossing the axes.
ⓒ
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 of the graph: x−y=2.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through the points “ordered pair 0, 2” and “ordered pair 2, 0”.
Solution

x-intercept (2,0): y-intercept (0,−2)

Find the x- and y-intercepts of the graph: 2x+3y=6.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through the points “ordered pair 0, 2” and “ordered pair 3, 0”.
Solution

x-intercept (3,0); y-intercept (0,2)

Find the Intercepts from an Equation of a Line

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 from the Equation of a Line

Use the equation 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.
x y
0
0

Find the intercepts of 2x+y=6

Solution

We'll fill in Figure 2.

A table illustrates how to find the x-intercept and y-intercept for the equation 2x + y = 6, by substituting 0 for y to find the x-intercept, and 0 for x to find the y-intercept.

To find the x- intercept, let y=0:

Step-by-step procedure for finding the x-intercept of an equation.
A mathematical equation is shown in black text on a white background, reading '2x + y = 6'.
Substitute 0 for y. A mathematical equation '2x + 0 = 6' is displayed, with the number '0' highlighted in red to emphasize its role in the expression.
Add. A simple algebraic equation showing '2x = 6' on a white background. This equation can be solved to find the value of x.
Divide by 2. The image displays the mathematical equation 'x = 3' in black characters against a white background.
The x-intercept is (3, 0).

To find the y- intercept, let x=0:

Steps demonstrating how to find the y-intercept by substituting x=0 and solving the equation.
A linear equation is displayed, which reads '2x + y = 6' in black font against a white background.
Substitute 0 for x. A mathematical equation shows '2 multiplied by 0 plus y equals 6'. The number 0 is highlighted in red, suggesting it might be a specific value being substituted or emphasized in the equation.
Multiply. A mathematical equation is displayed, showing '0 + y = 6' in black text against a white background.
Add. The mathematical equation y = 6 is displayed on a white background, representing a horizontal line in a coordinate system.
The y-intercept is (0, 6).
A two column, four row table. The first row has an equation while the following rows show ordered pairs (3, 0) and (0, 6).

The intercepts are the points (3,0) and (0,6).

Find the intercepts: 3x+y=12

Solution

(4,0) and (0,12)

Find the intercepts: x+4y=8

Solution

(8,0) and (0,2)

Find the intercepts of 4x−3y=12.

Solution

Solution

To find the x-intercept, let y=0.

Steps to find the x-intercept for the equation 4x - 3y = 12, demonstrating how to solve for x when y=0.
4x−3y=12
Substitute 0 for y. 4x−3·0=12
Multiply. 4x−0=12
Subtract. 4x=12
Divide by 4. x=3

The x-intercept is (3, 0).

To find the y-intercept, let x=0.

Demonstrates steps to solve the linear equation 4x - 3y = 12 for y when x = 0, finding the y-intercept.
4x−3y=12
Substitute 0 for x. 4·0−3y=12
Multiply. 0−3y=12
Simplify. −3y=12
Divide by −3. y=−4

The y-intercept is (0,−4).

The intercepts are the points (−3,0) and (0,−4).

4x−3y=12
x y
3 0
0 −4

Find the intercepts of the line: 3x−4y=12.

Solution

x-intercept (4,0); y-intercept: (0,−3)

Find the intercepts of the line: 2x−4y=8.

Solution

x-intercept (4,0); y-intercept: (0,−2)

Graph a Line Using the Intercepts

To graph a linear equation by plotting points, you can use the intercepts as two of your three points. Find the two intercepts, and then a third point to ensure accuracy, and draw the line. This method is often the quickest way to graph a line.

Graph −x+2y=6 using intercepts.

Solution

Solution

First, find the x-intercept. Let y=0,

−x+2y=6−x+2(0)=6−x=6x=−6

The x-intercept is (–6, 0).

Now find the y-intercept. Let x=0.

−x+2y=6−0+2y=62y=6y=3

The y-intercept is (0, 3).

Find a third point. We’ll use x=2,

−x+2y=6−2+2y=62y=8y=4

A third solution to the equation is (2, 4).

Summarize the three points in a table and then plot them on a graph.

−x+2y=6
x y (x,y)
−6 0 (−6,0)
0 3 (0,3)
2 4 (2,4)

The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10. Three labeled points are shown at “ordered pair -6, 0”, “ordered pair 0, 3” and “ordered pair 2, 4”.

Do the points line up? Yes, so draw line through the points.
The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10. Three labeled points are shown at “ordered pair -6, 0”, “ordered pair 0, 3” and “ordered pair 2, 4”.  A line passes through the three labeled points.

Graph the line using the intercepts: x−2y=4.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, -2” and “ordered pair 4, 0”.

Graph the line using the intercepts: −x+3y=6.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, 2” and “ordered pair -6, 0”.

Graph a line 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 then check that they line up.
  4. Draw the line.

Graph 4x−3y=12 using intercepts.

Solution

Solution

Find the intercepts and a third point.

The figure shows 3 solutions to the equation 4 x - 3 y = 12. The first is titled “x-intercept, let y = 0”. The first line is 4 x - 3 y = 12. The second line shows 0 in red substituted for y, reading 4 x - 3 open parentheses 0 closed parentheses = 12. The third line is 4 x = 12. The last line is x = 3. The second solution is titled “y-intercept, let x = 0”. The first line is 4 x - 3 y = 12. The second line shows 0 in red substituted for x, reading 4 open parentheses 0 closed parentheses - 3 y = 12. The third line is -3 y = 12. The last line is y = -4. The third solution is titled “third point, let y = 4”. The first line is 4 x - 3 y = 12. The second line shows 4 in red substituted for y, reading 4 x - 3 open parentheses 4 closed parentheses = 12. The third line is 4 x - 12 = 12. The last line is x = 6.

We list the points and show the graph.

4x−3y=12
x y (x,y)
3 0 (3,0)
0 −4 (0,−4)
6 4 (6,4)
The graph shows the x y-coordinate plane. Both axes run from -7 to 7. Three unlabeled points are drawn at  “ordered pair 0, -4”, “ordered pair 3, 0” and “ordered pair  6, 4”.  A line passes through the points.

Graph the line using the intercepts: 5x−2y=10.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7.  A line passes through the points “ordered pair 0, -5” and “ordered pair 2, 0”.

Graph the line using the intercepts: 3x−4y=12.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7.  A line passes through the points “ordered pair 0, -3” and “ordered pair 4, 0”.

Graph y=5x using the intercepts.

Solution

Solution

The figure shows 2 solutions to y = 5 x. The first solution is titled “x-intercept; Let y = 0.” The first line is y = 5 x. The second line is 0, shown in red, = 5 x. The third line is 0 = x. The fourth line is x = 0. The last line is “The x-intercept is “ordered pair 0, 0”. The second  solution is titled “y-intercept; Let x = 0.” The first line is y = 5 x. The second line is  y = 5 open parentheses 0, shown in red, closed parentheses. The third line is y = 0. The last line is “The y-intercept is “ordered pair 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 intercepts are the same point, we need two more points to graph the line. As always, we can choose any values for x, so we’ll let x be 1 and −1.
The figure shows two substitutions in the equation y = 5 x. In the first substitution,  the first line is y = 5 x. The second line is y = 5 open parentheses 1, shown in red, closed parentheses. The third line is y =5. The last line is “ordered pair 1, 5”.  In the second substitution,  the first line is y = 5 x. The second line is y = 5 open parentheses -1, shown in red, closed parentheses. The third line is y = -5. The last line is “ordered pair -1, -5”.

Organize the points in a 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 graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through three labeled points, “ordered pair -1, -5”, “ordered pair 0, 0”, and ordered pair 1, 5”.

Graph using the intercepts: y=4x.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, 0” and “ordered pair 1, 3”.

Graph using the intercepts: y=−x.

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, 0” and “ordered pair 1, -1”.

Choose the Most Convenient Method to Graph a Line

While we could graph any linear equation by plotting points, it may not always be the most convenient method. This table shows six of equations we’ve graphed in this chapter, and the methods we used to graph them.

Equation Method
#1 y=2x+1 Plotting points
#2 y=12x+3 Plotting points
#3 x=−7 Vertical line
#4 y=4 Horizontal line
#5 2x+y=6 Intercepts
#6 4x−3y=12 Intercepts

What is it about the form of equation that can help us choose the most convenient method to graph its line?

Notice that in equations #1 and #2, y is isolated on one side of the equation, and its coefficient is 1. We found points by substituting values for x on the right side of the equation and then simplifying to get the corresponding y- values.

Equations #3 and #4 each have just one variable. Remember, in this kind of equation 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 #5 and #6, 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 and x=0 to find the x- and y- intercepts, and then found a third point by choosing a value for x or y.

This leads to the following strategy for choosing the most convenient method to graph a line.

Choose the most convenient method to graph a line.

  1. If the equation has only 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.
  2. If y is isolated on one side of the equation. Graph by plotting points.
    • Choose any three values for x and then solve for the corresponding y- values.
  3. If the equation is of the form Ax+By=C, find the intercepts.
    • Find the x- and y- intercepts and then a third point.
Identify the most convenient method to graph each line:
  1. ⓐ y=−3
  2. ⓑ 4x−6y=12
  3. ⓒ x=2
  4. ⓓ y=25x−1
Solution

Solution

ⓐ y=−3

This equation has only one variable, y. Its graph is a horizontal line crossing the y-axis at −3.

ⓑ 4x−6y=12

This equation is of the form Ax+By=C. Find the intercepts and one more point.

ⓒ x=2

There is only one variable, x. The graph is a vertical line crossing the x-axis at 2.

ⓓ y=25x−1

Since y is isolated on the left side of the equation, it will be easiest to graph this line by plotting three points.

Identify the most convenient method to graph each line:

  1. ⓐ 3x+2y=12
  2. ⓑ y=4
  3. ⓒ y=15x−4
  4. ⓓ x=−7
Solution
  1. ⓐ intercepts
  2. ⓑ horizontal line
  3. ⓒ plotting points
  4. ⓓ vertical line

Identify the most convenient method to graph each line:

  1. ⓐ x=6
  2. ⓑ y=−34x+1
  3. ⓒ y=−8
  4. ⓓ 4x−3y=−1
Solution
  1. ⓐ vertical line
  2. ⓑ plotting points
  3. ⓒ horizontal line
  4. ⓓ intercepts

ACCESS ADDITIONAL ONLINE RESOURCES

  • Graph by Finding Intercepts
  • Use Intercepts to Graph
  • State the Intercepts from a Graph

Key Concepts

  • Intercepts
    • The x-intercept is the point, (a,0), where the graph crosses the x-axis. The x-intercept occurs when y is zero.
    • The y-intercept is the point, (0,b), where the graph crosses the y-axis. The y-intercept occurs when x is zero.
    • The x-intercept occurs when y is zero.
    • The y-intercept occurs when x is zero.
  • Find the x and y intercepts from the equation of a line
    • To find the x-intercept of the line, let y=0 and solve for x.
    • To find the y-intercept of the line, let x=0 and solve for y.
      x y
      0
      0
  • Graph a line 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 then check that they line up.
    4. Draw the line.
  • Choose the most convenient method to graph a line
    1. Determine if the equation has only one variable. Then 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.
    2. Determine if y is isolated on one side of the equation. The graph by plotting points.
      Choose any three values for x and then solve for the corresponding y- values.
    3. Determine if the equation is of the form Ax+By=C, find the intercepts.
      Find the x- and y- intercepts and then a third point.

Practice Makes Perfect

Identify the Intercepts on a Graph

In the following exercises, find the x- and y- intercepts.

The graph shows the x y-coordinate plane. The axes run from -10 to 10.  A line passes through the points “ordered pair 0, 3” and “ordered pair 3, 0”.
Solution

(3,0),(0,3)

The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, 2” and “ordered pair 2, 0”.
The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, -5” and “ordered pair 5, 0”.
Solution

(5,0),(0,−5)

The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, -1” and “ordered pair 1, 0”.
The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, -2” and “ordered pair -2, 0”.
Solution

(−2,0),(0,−2)

The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, -3” and “ordered pair -3, 0”.
The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, 1” and “ordered pair -1, 0”.
Solution

(−1,0),(0,1)

The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, 5” and “ordered pair -5, 0”.
The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7.  A line passes through the points “ordered pair 0, 0” and “ordered pair 4, 2”.
Solution

(0,0)

The graph shows the x y-coordinate plane. The x and y-axis each run from -10 to 10.  A line passes through the points “ordered pair 0, 0” and “ordered pair 1, 1”.

Find the x and y Intercepts from an Equation of a Line

In the following exercises, find the intercepts.

x+y=4

Solution

(4,0),(0,4)

x+y=3

x+y=−2

Solution

(−2,0),(0,−2)

x+y=−5

x−y=5

Solution

(5,0),(0,−5)

x−y=1

x−y=−3

Solution

(−3,0),(0,3)

x−y=−4

x+2y=8

Solution

(8,0),(0,4)

x+2y=10

3x+y=6

Solution

(2,0),(0,6)

3x+y=9

x−3y=12

Solution

(12,0),(0,−4)

x−2y=8

4x−y=8

Solution

(2,0),(0,−8)

5x−y=5

2x+5y=10

Solution

(5,0),(0,2)

2x+3y=6

3x−2y=12

Solution

(4,0),(0,−6)

3x−5y=30

y=13x−1

Solution

(3,0),(0,−1)

y=14x−1

y=15x+2

Solution

(−10,0),(0,2)

y=13x+4

y=3x

Solution

(0,0)

y=−2x

y=−4x

Solution

(0,0)

y=5x

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

−x+5y=10

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, 2” and “ordered pair -10, 0”.

−x+4y=8

x+2y=4

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, 2” and “ordered pair 4, 0”.

x+2y=6

x+y=2

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, 2” and “ordered pair 2, 0”.

x+y=5

x+y=3

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, 3” and “ordered pair 3, 0”.

x+y=−1

x−y=1

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0, -1” and “ordered pair 1, 0”.

x−y=2

x−y=−4

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  4” and “ordered pair -4, 0”.

x−y=−3

4x+y=4

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  4” and “ordered pair 1, 0”.

3x+y=3

3x−y=−6

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  6” and “ordered pair -2, 0”.

2x−y=−8

2x+4y=12

Solution
The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  3” and “ordered pair 6, 0”.

3x+2y=12

3x−2y=6

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  -3” and “ordered pair 2, 0”.

5x−2y=10

2x−5y=−20

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  4” and “ordered pair -10, 0”.

3x−4y=−12

y=−2x

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  0” and “ordered pair 2, -4”.

y=−4x

y=x

Solution


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12.  A line passes through the points “ordered pair 0,  0” and “ordered pair 2, 2”.

y=3x

Choose the Most Convenient Method to Graph a Line

In the following exercises, identify the most convenient method to graph each line.

x=2

Solution

vertical line

y=4

y=5

Solution

horizontal line

x=−3

y=−3x+4

Solution

plotting points

y=−5x+2

x−y=5

Solution

intercepts

x−y=1

y=23x−1

Solution

plotting points

y=45x−3

y=−3

Solution

horizontal line

y=−1

3x−2y=−12

Solution

intercepts

2x−5y=−10

y=−14x+3

Solution

plotting points

y=−13x+5

Everyday Math

Road trip Damien is driving from Chicago to Denver, a distance of 1,000 miles. The x-axis on the graph below shows the time in hours since Damien left Chicago. The y-axis represents the distance he has left to drive.

The graph shows the x y-coordinate plane. The x and y-axis each run from - to .  A line passes through the labeled points “ordered pair 0, 1000” and “ordered pair 15, 0”.

ⓐ Find the x- and y- intercepts

ⓑ Explain what the x- and y- intercepts mean for Damien.

Solution

ⓐ (0,1,000),(15,0). ⓑ At (0,1,000) he left Chicago 0 hours ago and has 1,000 miles left to drive. At (15,0) he left Chicago 15 hours ago and has 0 miles left to drive.

Road trip Ozzie filled up the gas tank of his truck and went on a road trip. The x-axis on the graph shows the number of miles Ozzie drove since filling up. The y-axis represents the number of gallons of gas in the truck’s gas tank.

The graph shows the x y-coordinate plane. The x and y-axis each run from - to .  A line passes through labeled points “ordered pair 0, 16” and “ordered pair 300, 0”.

ⓐ Find the x- and y- intercepts.

ⓑ Explain what the x- and y- intercepts mean for Ozzie.

Writing Exercises

How do you find the x-intercept of the graph of 3x−2y=6?

Solution

Answers will vary.

How do you find the y-intercept of the graph of 5x−y=10?

Do you prefer to graph the equation 4x+y=−4 by plotting points or intercepts? Why?

Solution

Answers will vary.

Do you prefer to graph the equation y=23x−2 by plotting points or intercepts? Why?

Self Check

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

A student self-assessment rubric on identifying, finding, and graphing line intercepts and selecting appropriate graphing methods.

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

intercepts of a line
Each of the points at which a line crosses the x-axis and the y-axis is called an intercept of the line.

Understand Slope of a Line

Learning Objectives

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

  • Use geoboards to model slope
  • Find the slope of a line from its graph
  • Find the slope of horizontal and vertical lines
  • Use the slope formula to find the slope of a line between two points
  • Graph a line given a point and the slope Solve slope applications

Before you get started, take this readiness quiz.

Simplify: 1−48−2.
If you missed this problem, review Example 13 in Multiply and Divide Mixed Numbers and Complex Fractions.

Solution

−12

Divide: 04,40.
If you missed this problem, review Example 5 in Properties of Identity, Inverses, and Zero.

Solution

0, undefined

Simplify: 15−3,−153,−15−3.
If you missed this problem, review Example 11 in Multiply and Divide Mixed Numbers and Complex Fractions.

Solution

−5,−5,5

As we’ve been graphing linear equations, we’ve seen that some lines slant up as they go from left to right and some lines slant down. Some lines are very steep and some lines are flatter. What determines whether a line slants up or down, and if its slant is steep or flat?

The steepness of the slant of a line is called the slope of the line. The concept of slope has many applications in the real world. The pitch of a roof and the grade of a highway or wheelchair ramp are just some examples in which you literally see slopes. And when you ride a bicycle, you feel the slope as you pump uphill or coast downhill.

Use Geoboards to Model Slope

In this section, we will explore the concepts of slope.

Using rubber bands on a geoboard gives a concrete way to model lines on a coordinate grid. By stretching a rubber band between two pegs on a geoboard, we can discover how to find the slope of a line.

Doing the Manipulative Mathematics activity "Exploring Slope" will help you develop a better understanding of the slope of a line.

We’ll start by stretching a rubber band between two pegs to make a line as shown in Figure 1.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 4 row 2.

Does it look like a line?

Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle as shown in Figure 2. We carefully make a 90° angle around the third peg, so that one side is vertical and the other is horizontal.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 2, column 1 row 4,and column 4 row 2.

To find the slope of the line, we measure the distance along the vertical and horizontal legs of the triangle. The vertical distance is called the rise and the horizontal distance is called the run, as shown in Figure 3.

This figure shows two arrows. The first arrow is vertical and is labeled “rise”. The second arrow begins at the end of the first arrow extending to the right and is labeled “run”.

To help remember the terms, it may help to think of the images shown in Figure 4.

The figure shows an image of a hot air balloon signifying rise as the balloon rises straight up, similar to a y-axis. The second image is of a person jogging, signifying run as the person runs as if they are on an x-axis

On our geoboard, the rise is 2 units because the rubber band goes up 2 spaces on the vertical leg. See Figure 5.

What is the run? Be sure to count the spaces between the pegs rather than the pegs themselves! The rubber band goes across 3 spaces on the horizontal leg, so the run is 3 units.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 2, column 1 row 4, and column 4 row 2. The triangle has a rise of 2 units and a run of 3 units.

The slope of a line is the ratio of the rise to the run. So the slope of our line is 23. In mathematics, the slope is always represented by the letter m.

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.

What is the slope of the line on the geoboard in Figure 5?

m=riserun
m=23
The line has slope23.

When we work with geoboards, it is a good idea to get in the habit of starting at a peg on the left and connecting to a peg to the right. Then we stretch the rubber band to form a right triangle.

If we start by going up the rise is positive, and if we stretch it down the rise is negative. We will count the run from left to right, just like you read this paragraph, so the run will be positive.

Since the slope formula has rise over run, it may be easier to always count out the rise first and then the run.

What is the slope of the line on the geoboard shown?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 5 row 2.
Solution

Solution

Use the definition of slope.

m=riserun

Start at the left peg and make a right triangle by stretching the rubber band up and to the right to reach the second peg.

Count the rise and the run as shown.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 2, column 1 row 5,and column 5 row 2.

The rise is3units.m=3runThe run is4units.m=34The slope is34.

What is the slope of the line on the geoboard shown?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 4 row 1.
Solution

43

What is the slope of the line on the geoboard shown?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 5 row 3.
Solution

14

What is the slope of the line on the geoboard shown?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 3 and the point in column 4 row 4.
Solution

Solution

Use the definition of slope.

m=riserun

Start at the left peg and make a right triangle by stretching the rubber band to the peg on the right. This time we need to stretch the rubber band down to make the vertical leg, so the rise is negative.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 3, column 1 row 4,and column 4 row 4.

The rise is−1.m=−1runThe run is3.m=−13m=−13The slope is−13.

What is the slope of the line on the geoboard?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 4 row 4.
Solution

−23

What is the slope of the line on the geoboard?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 1 and the point in column 4 row 5.
Solution

−43

Notice that in the first example, the slope is positive and in the second example the slope is negative. Do you notice any difference in the two lines shown in Figure 6.

The figure shows two grids of evenly spaced dots. There are 5 rows and 5 columns in each. In the left grid A, a rubberband loops to connect the point in column 1, row 1 and the point in column 5, row 4. In the right grid B, a rubber band loops to connect the point in column 1, row 4 and the point in column 4, row 2.

As you read from left to right, the line in Figure A, is going up; it has positive slope. The line Figure B is going down; it has negative slope.

This image shows two arrows: the left arrow is labeled positive slope and points upward towards the right. The right arrow is labeled negative slope and points downward towards the right.

Use a geoboard to model a line with slope 12.

Solution

Solution

To model a line with a specific slope on a geoboard, we need to know the rise and the run.

This table illustrates the steps and corresponding mathematical expressions for using and manipulating the slope formula.
Use the slope formula. m=riserun
Replace m with 12. 12=riserun

So, the rise is 1 unit and the run is 2 units.

Start at a peg in the lower left of the geoboard. Stretch the rubber band up 1 unit, and then right 2 units.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 3, column 1 row 4,and column 3 row 3.

The hypotenuse of the right triangle formed by the rubber band represents a line with a slope of 12.

Use a geoboard to model a line with the given slope: m=13.

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 2 row 3, column 2 row 4,and column 5 row 3.

Use a geoboard to model a line with the given slope: m=32.

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 1, column 1 row 4,and column 3 row 1.

Use a geoboard to model a line with slope −14,

Solution

Solution

This table illustrates the steps involved in using the slope formula and replacing the slope variable with a specific numerical value.
Use the slope formula. m=riserun
Replace m with −14. −14=riserun

So, the rise is −1 and the run is 4.

Since the rise is negative, we choose a starting peg on the upper left that will give us room to count down. We stretch the rubber band down 1 unit, then to the right 4 units.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 2, column 1 row 3,and column 5 row 3.

The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is −14.

Use a geoboard to model a line with the given slope: m=−21.

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 2 row 3, column 2 row 5,and column 3 row 5.

Use a geoboard to model a line with the given slope: m=−13.

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 1, column 1 row 2,and column 4 row 2.

Find the Slope of a Line from its Graph

Now we’ll look at some graphs on a coordinate grid to find their slopes. The method will be very similar to what we just modeled on our geoboards.

Doing the Manipulative Mathematics activity "Slope of Lines Between Two Points" will help you develop a better understanding of how to find the slope of a line from its graph.

To find the slope, we must count out the rise and the run. But where do we start?

We locate any two points on the line. We try to choose points with coordinates that are integers to make our calculations easier. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.

Find the slope of the line shown:

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 6. The y-axis runs from -4 to 2. A line passes through the points “ordered pair 5,  1” and “ordered pair 0, -3”.
Solution

Solution

Locate two points on the graph, choosing points whose coordinates are integers. We will use (0,−3) and (5,1).

Starting with the point on the left, (0,−3), sketch a right triangle, going from the first point to the second point, (5,1).

Steps and calculations to determine the slope of a line using the rise over run method, with an illustrative graph.
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 6. The y-axis runs from -4 to 2. A line passes through the points “ordered pair 5,  1” and “ordered pair 0, -3”. Two line segments form a triangle with the line. A horizontal line connects “ordered pair 0, 1” and “ordered pair 5,1 ”. A vertical line segment connects “ordered pair 0, -3” and “ordered pair 0, 1”.
Count the rise on the vertical leg of the triangle. The rise is 4 units.
Count the run on the horizontal leg. The run is 5 units.
Use the slope formula. m=riserun
Substitute the values of the rise and run. m=45
The slope of the line is 45.

Notice that the slope is positive since the line slants upward from left to right.

Find the slope of the line:

The graph shows the x y-coordinate plane. The x-axis runs from -8 to 1. The y-axis runs from -1 to 4. A line passes through the points “ordered pair -8,  1” and “ordered pair 0, 3”.
Solution

25

Find the slope of the line:

The graph shows the x y-coordinate plane. The x-axis runs from -2 to 6. The y-axis runs from -2 to 4. A line passes through the points “ordered pair 4,  2” and “ordered pair 0, -1”.
Solution

34

Find the slope from a graph.

  1. Locate two points on the line whose coordinates are integers.
  2. Starting with the point on the left, sketch a right triangle, with the hypotenuse 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:

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 9. The y-axis runs from -1 to 7. A line passes through the points “ordered pair 4,  2” and “ordered pair 3, 3”.
Solution

Solution

Locate two points on the graph. Look for points with coordinates that are integers. We can choose any points, but we will use (0, 5) and (3, 3). Starting with the point on the left, sketch a right triangle, with the hypotenuse going from the first point to the second point.

Illustrates the step-by-step calculation of a line's slope using rise and run, featuring a graph, the formula, and numerical example.
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 9. The y-axis runs from -1 to 7. A line passes through the points “ordered pair 0,  5” and “ordered pair 3, 3”. Two line segments form a triangle with the line. A horizontal line connects “ordered pair 0, 3” and “ordered pair 3, 3 ”. A vertical line segment connects “ordered pair 0, 3” and “ordered pair 0, 5”. It is labeled “rise”.
Count the rise – 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.

Notice that the slope is negative since the line slants downward from left to right.

What if we had chosen different points? Let’s find the slope of the line again, this time using different points. We will use the points (−3,7) and (6,1).
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 9. The y-axis runs from -1 to 7. A line passes through the points “ordered pair 0, 5” and  “ordered pair 3, 3”.  .

Starting at (−3,7), sketch a right triangle to (6,1).

Step-by-step calculation of the slope of a line using the rise and run method, illustrated by a graph.
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 9. The y-axis runs from -1 to 7. A line passes through the points “ordered pair 0, 5” and  “ordered pair 3, 3”. Two line segments form a triangle with the line. A vertical line connects “ordered pair 0, 3” and “ordered pair 3, 3 ”.  A vertical line segment connects “ordered pair 0, 3” and “ordered pair 0, 5”.
Count the rise. The rise is −6.
Count the run. The run is 9.
Use the slope formula. m=riserun
Substitute the values of the rise and run. m=−69
Simplify the fraction. m=−23
The slope of the line is −23.

It does not matter which points you use—the slope of the line is always the same. The slope of a line is constant!

Find the slope of the line:

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 5. The y-axis runs from -6 to 1. A line passes through the points “ordered pair 3,  -6” and “ordered pair 0, -2”.
Solution

−43

Find the slope of the line:

The graph shows the x y-coordinate plane. The x-axis runs from -3 to 6. The y-axis runs from -3 to 2. A line passes through the points “ordered pair 5,  -2” and “ordered pair 0, 1”.
Solution

−35

The lines in the previous examples had y-intercepts with integer values, so it was convenient to use the y-intercept as one of the points we used to find the slope. In the next example, the y-intercept is a fraction. The calculations are easier if we use two points with integer coordinates.

Find the slope of the line shown: The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 8. A line passes through the points “ordered pair 2, 3” and “ordered pair 7, 6”.

Solution

Solution

This table outlines instructions and provides examples for working with points and sketching on a graph.
Locate two points on the graph whose coordinates are integers. (2,3) and (7,6)
Which point is on the left? (2,3)
Starting at (2,3), sketch a right angle to (7,6) as shown below.

Steps to calculate the slope of a line from a graph using the rise and run method, including an illustrative image and formula.
The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 8. Two unlabeled points are drawn at  “ordered pair 2, 3” and  “ordered pair 7, 6”.  A line passes through the points. Two line segments form a triangle with the line. A vertical line connects “ordered pair 2, 3” and “ordered pair 2, 6 ”.  It is labeled “rise”. A horizontal line segment connects “ordered pair 2, 6” and “ordered pair 7, 6”. It is labeled “run”.
Count the rise. The rise is 3.
Count the run. The run is 5.
Use the slope formula. m=riserun
Substitute the values of the rise and run. m=35
The slope of the line is 35.

Find the slope of the line:

The graph shows the x y-coordinate plane. The x-axis runs from -4 to 2. The y-axis runs from -5 to 2. A line passes through the points “ordered pair -3, -4” and “ordered pair 1, 1”.
Solution

54

Find the slope of the line:

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 4. The y-axis runs from -2 to 3. A line passes through the points “ordered pair 3, 2” and “ordered pair 1, -1”.
Solution

32

Find the Slope of Horizontal and Vertical Lines

Do you remember what was special about horizontal and vertical lines? Their equations had just one variable.

  • horizontal line y=b; all the y-coordinates are the same.
  • vertical line x=a; all the x-coordinates are the same.

So how do we find the slope of the horizontal line y=4? One approach would be to graph the horizontal line, find two points on it, and count the rise and the run. Let’s see what happens in Figure 8. We’ll use the two points (0,4) and (3,4) to count the rise and run.

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 5. The y-axis runs from -1 to 7. A horizontal line passes through the labeled points “ordered pair 0, 4” and “ordered pair 3, 4”.
This table illustrates the calculation of slope, demonstrating that a zero rise over a given run results in a zero slope.
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=4 is 0.

All horizontal lines have slope 0. When the y-coordinates are the same, the rise is 0.

Slope of a Horizontal Line

The slope of a horizontal line, y=b, is 0.

Now we’ll consider a vertical line, such as the line x=3, shown in Figure 9. We’ll use the two points (3,0) and (3,2) to count the rise and run.

The graph shows the x y-coordinate plane. Both axes run from -5 to 5. A vertical line passes through the labeled points “ordered pair 3, 2” and “ordered pair 3, 0”.
Slope calculation example defining rise and run, demonstrating an undefined slope with a run of zero.
What is the rise? The rise is 2.
What is the run? The run is 0.
What is the slope? m=riserun
m=20

But we can’t divide by 0. Division by 0 is undefined. So we say that the slope of the vertical line x=3 is undefined. The slope of all vertical lines is undefined, because the run is 0.

Slope of a Vertical Line

The slope of a vertical line, x=a, is undefined.

Find the slope of each line:
  1. ⓐ x=8
  2. ⓑ y=−5
Solution

Solution

ⓐ x=8

This is a vertical line, so its slope is undefined.

ⓑ y=−5

This is a horizontal line, so its slope is 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 figure shows 4 arrows. The first rises from left to right with the arrow point upwards. It is labeled “positive”. The second goes down from left to right with the arrow pointing downwards. It is labeled “negative”. The third is horizontal with arrow heads on both ends. It is labeled “zero”. The last is vertical with arrow heads on both ends. It is labeled “undefined.”

Use the Slope Formula to find the Slope of a Line between Two Points

Sometimes we need to find the slope of a line between two points and we might not 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 there is a way to find the slope without graphing.

Before we get to it, we need to introduce some new 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 between the points. A subscript is a small number written to the right of, and a little lower than, a variable.

  • (x1,y1)readxsub1,ysub1
  • (x2,y2)readxsub2,ysub2

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.

To see how the rise and run relate to the coordinates of the two points, let’s take another look at the slope of the line between the points (2,3) and (7,6) in Figure 10.

The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 7. A line runs through the labeled points 2, 3 and 7, 6. A line segment runs from the point 2, 3 to the unlabeled point 2, 6. It is labeled y sub 2 minus y sub 1, 6 minus 3, 3. A line segment runs from the point 7, 6 to the unlabeled point 2, 6.  It os labeled x sub 2 minus x sub 1, 7 minus 2, 5.

Since we have two points, we will use subscript notation.

(2,3)x1,y1(7,6)x2,y2

On the graph, we counted the rise of 3. The rise can also be found by subtracting the y-coordinates of the points.

y2−y16−33

We counted a run of 5. The run can also be found by subtracting the x-coordinates.

x2−x17−25
Step-by-step derivation of the slope formula, progressing from the basic rise/run definition to the coordinate-based formula.
We know m=riserun
So m=35
We rewrite the rise and run by putting in the coordinates. m=6−37−2
But 6 is the y-coordinate of the second point, y2
and 3 is the y-coordinate of the first point y1.
So we can rewrite the rise using subscript notation.
m=y2−y17−2
Also 7 is the x-coordinate of the second point, x2
and 2 is the x-coordinate of the first point x2.
So we rewrite the run 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 Formula

The slope of the line between two points (x1,y1) and (x2,y2) is

m=y2−y1x2−x1

Say the formula to yourself to help you remember it:

Slope isyof the second point minusyof the first point
over
xof the second point minusxof the first point.
Doing the Manipulative Mathematics activity “Slope of Lines Between Two Points” will help you develop a better understanding of how to find the slope of a line between two points.

Find the slope of the line between the points (1,2) and (4,5).

Solution

Solution

This table illustrates the step-by-step process of calculating the slope between two points, (1,2) and (4,5), using the slope formula.
We’ll call (1,2) point #1 and (4,5)point #2. (1,2)x1,y1and(4,5)x2,y2
Use the slope formula. m=y2−y1x2−x1
Substitute the values in the slope formula:
y of the second point minus y of the first point m=5−2x2−x1
x of the second point minus x of the first point m=5−24−1
Simplify the numerator and the denominator. m=33
m=1

Let’s confirm this by counting out the slope on the graph.
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 7. The y-axis runs from -1 to 7. Two labeled points are drawn at  “ordered pair 1, 2” and  “ordered pair 4, 5”.  A line passes through the points. Two line segments form a triangle with the line. A vertical line connects “ordered pair 1, 2” and “ordered pair 1, 5 ”.  It is labeled “rise”. A horizontal line segment connects “ordered pair 1, 5” and “ordered pair 4, 5”. It is labeled “run”.

The rise is 3 and the run is 3, so

m=riserunm=33m=1

Find the slope of the line through the given points: (8,5) and (6,3).

Solution

1

Find the slope of the line through the given points: (1,5) and (5,9).

Solution

1

How do we know which point to call #1 and which to call #2? Let’s find the slope again, this time switching the names of the points to see what happens. Since we will now be counting the run from right to left, it will be negative.

Step-by-step calculation of the slope between two points (4,5) and (1,2) using the slope formula.
We’ll call (4,5) point #1 and (1,2) point #2. (4,5)x1,y1and(1,2)x2,y2
Use the slope formula. m=y2−y1x2−x1
Substitute the values in the slope formula:
y of the second point minus y of the first point m=2−5x2−x1
x of the second point minus x of the first point m=2−51−4
Simplify the numerator and the denominator. m=−3−3
m=1

The slope is the same no matter which order we use the points.

Find the slope of the line through the points (−2,−3) and (−7,4).

Solution

Solution

Step-by-step calculation of the slope between two points, (-2,-3) and (-7,4), illustrating the application of the slope formula.
We’ll call (−2,−3) point #1 and (−7,4) point #2. (−2,−3)x1,y1and(−7,4)x2,y2
Use the slope formula. m=y2−y1x2−x1
Substitute the values
y of the second point minus y of the first point m=4−(−3)x2−x1
x of the second point minus x of the first point m=4−(−3)−7−(−2)
Simplify. m=7−5
m=−75

Let’s confirm this on the graph shown.
The graph shows the x y-coordinate plane. The x-axis runs from -8 to 2. The y-axis runs from -6 to 5. Two unlabeled points are drawn at  “ordered pair -7, 4” and  “ordered pair -2, -3”.  A line passes through the points. Two line segments form a triangle with the line. A vertical line connects “ordered pair -7, 4” and “ordered pair -7, -3 ”.  It is labeled “rise”. A horizontal line segment connects “ordered pair -7, -3” and “ordered pair -2, -3”. It is labeled “run”.

m=riserunm=−75m=−75

Find the slope of the line through the pair of points: (−3,4) and (2,−1).

Solution

−1

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

In this chapter, we graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.

Another method we can use to graph lines is the point-slope method. Sometimes, we will be given one point and the slope of the line, instead of its equation. When this happens, we use the definition of slope to draw the graph of the line.

Graph the line passing through the point (1,−1) whose slope is m=34.

Solution

Solution

Plot the given point, (1,−1).
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 7. The y-axis runs from -3 to 4. A labeled point is drawn at “ordered pair 1, -1”.

Use the slope formula m=riserun to identify the rise and the run.

m=34riserun=34rise=3run=4

Starting at the point we plotted, count out the rise and run to mark the second point. We count 3 units up and 4 units right.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. Two line segments are drawn. A vertical line segment connects the points “ordered pair 1, -1” and “order pair “1, 2”. It is labeled “3”. A horizontal line segment starts at the top of the vertical line segment and goes to the right, connecting the points “ordered pair 1, 2” and “ordered pair 5, 2”. It is labeled “4”.

Then we connect the points with a line and draw arrows at the ends to show it continues.
The graph shows the x y-coordinate plane. The x-axis runs from -3 to 5. The y-axis runs from -1 to 7. Two unlabeled points are drawn at  “ordered pair 1, -1” and  “ordered pair 5, 2”.  A line passes through the points. Two line segments form a triangle with the line. A vertical line connects “ordered pair 1, -1” and “ordered pair 1, 2 ”.  A horizontal line segment connects “ordered pair 1, 2” and “ordered pair 5, 2”.

We can check our line by starting at any point and counting up 3 and to the right 4. We should get to another point on the line.

Graph the line passing through the point with the given slope:

(2,−2),m=43

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from -12 to 12. A line passes through the points “ordered pair -2, 3” and “ordered pair 8, 6”.

Graph the line passing through the point with the given slope:

(−2,3),m=14

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from -12 to 12. A line passes through the points “ordered pair -2, 3” and “ordered pair 2, 4”.

Graph a line given a point and a slope.

  1. Plot the given point.
  2. Use the slope formula 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 the line with y-intercept (0,2) and slope m=−23.

Solution

Solution

Plot the given point, the y-intercept (0,2).
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 4. The y-axis runs from -1 to 3. The point “ordered pair 0, 2” is labeled.

Use the slope formula m=riserun to identify the rise and the run.

m=−23riserun=−23rise=–2run=3

Starting at (0,2), count the rise and the run and mark the second point.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. A vertical line segment connects points at “ordered pair 0, 2” and “ordered pair 0, 0” and is labeled “down 2”. A horizontal line segment connects “ordered pair 0, 0” and “ordered pair 0, 3” and is labeled “right 3”.

Connect the points with a line.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. Two labeled points are drawn at  “ordered pair 0, 2” and  “ordered pair 3, 0”.  A line passes through the points. Two line segments form a triangle with the line. A vertical line connects “ordered pair 0, 2” and “ordered pair 0, 0 ”.  A horizontal line segment connects “ordered pair 0, 0” and “ordered pair 3, 0”.

Graph the line with the given intercept and slope:

y-intercept 4,m=−52

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from -12 to 12. A line passes through the points “ordered pair 0, 4” and “ordered pair 4, -6”.

Graph the line with the given intercept and slope:

x-intercept −3,m=−34

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from -12 to 12. A line passes through the points “ordered pair -3, 0” and “ordered pair 8, -8”.

Graph the line passing through the point (−1,−3) whose slope is m=4.

Solution

Solution

Plot the given point.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. The point “ordered pair -1, -3” is labeled.

This table illustrates the steps to identify the rise and run from a given slope value, showing the mathematical representation at each stage.
Identify the rise and the run. m=4
Write 4 as a fraction. riserun=41
rise=4run=1

Count the rise and run.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. The y-axis runs from -4 to 2. A vertical line segment connects points at “ordered pair -1,  -3” and “ordered pair -1, 1” and is labeled “up 4”. A horizontal line segment connects “ordered pair -1, 1” and “ordered pair 0, 1” and is labeled “over 1”.

Mark the second point. Connect the two points with a line.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. Two labeled points are drawn at  “ordered pair -1, -3” and  “ordered pair -1, 1”.  A line passes through the points. Two line segments form a triangle with the line. A vertical line connects “ordered pair -1, -3” and “ordered pair -1, 1 ”. It is labeled “up 4” A horizontal line segment connects “ordered pair -1, 1” and “ordered pair 0, 1”. It is labeled “over 1”

Graph the line passing through the point (−2,1) and with slope m=3.

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -7 to 7. The y-axis runs from -7 to 7. A line passes through the points “ordered pair -2, 1” and “ordered pair 0, 7”.

Graph the line passing through the point (4,−2) and with slope m=−2.

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -7 to 7. The y-axis runs from -7 to 7. A line passes through the points “ordered pair 0, 6” and “ordered pair 0, 3”.

Solve Slope Applications

At the beginning of this section, we said there are many applications of slope in the real world. Let’s look at a few now.

The pitch of a building’s roof is the slope of the roof. Knowing the pitch is important in climates where there is heavy snowfall. If the roof is too flat, the weight of the snow may cause it to collapse. What is the slope of the roof shown?

This figure shows a house with a sloped roof. The roof on one half of the building is labeled “pitch of the roof”. There is a line segment with arrows at each end measuring the vertical length of the roof and is labeled “rise = 9 feet”. There is a line segment with arrows at each end measuring the horizontal length of the root and is labeled “run = 18 feet”.
Solution

Solution

Illustrates the step-by-step calculation of a roof's slope using the rise and run formula.
Use the slope formula. m=riserun
Substitute the values for rise and run. m=9 ft18 ft
Simplify. m=12
The slope of the roof is 12.

Find the slope given rise and run: A roof with a rise =14 and run =24.

Solution

712

Find the slope given rise and run: A roof with a rise =15 and run =36.

Solution

512

Have you ever thought about the sewage pipes going from your house to the street? Their slope is an important factor in how they take waste away from your house.

Sewage pipes must slope down 14 inch per foot in order to drain properly. What is the required slope?

This figure shows a  right triangle. The short leg is vertical and is labeled “1 over 4 inch”. The long leg labeled “1 foot”.
Solution

Solution

Step-by-step calculation of a pipe's slope, illustrating the application of the slope formula, unit conversion, and final simplification to determine the result.
Use the slope formula. m=riserun
m=−14in.1ft
m=−14in.1ft
Convert 1 foot to 12 inches. m=−14in.12in.
Simplify. m=−148
The slope of the pipe is −148.

Find the slope of the pipe: The pipe slopes down 13 inch per foot.

Solution

−136

Find the slope of the pipe: The pipe slopes down 34 inch per yard.

Solution

−148

ACCESS ADDITIONAL ONLINE RESOURCES

  • Determine Positive slope from a Graph
  • Determine Negative slope from a Graph
  • Determine Slope from Two Points

Key Concepts

  • Find the slope from a graph
    1. Locate two points on the line whose coordinates are integers.
    2. Starting with the point on the left, sketch a right triangle, with the hypotenuse 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 Horizontal Line
    • The slope of a horizontal line, y=b, is 0.
  • Slope of a Vertical Line
    • The slope of a vertical line, x=a, is undefined.
  • Slope Formula
    • The slope of the line between two points (x1,y1) and (x2,y2) is m=y2-y1x2-x1
  • Graph a line given a point and a slope.
    1. Plot the given point.
    2. Use the slope formula 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.

Section Exercises

Practice Makes Perfect

Use Geoboards to Model Slope

In the following exercises, find the slope modeled on each geoboard.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 3 and the point in column 5 row 2.
Solution

14

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 4 and the point in column 5 row 2.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 1 and the point in column 4 row 4.
Solution

−32

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 1 and the point in column 4 row 4.

In the following exercises, model each slope. Draw a picture to show your results.

23

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 5 and the point in column 5 row 3.

34

14

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 5 row 3.

43

−12

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 3 row 5.

−34

−23

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 4 row 4.

−32

Find the Slope of a Line from its Graph

In the following exercises, find the slope of each line shown.

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, -4” and “ordered pair 10, 0”.
Solution

25

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, -5” and “ordered pair 3, 0”.
The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from -12 to 12. A line passes through the points “ordered pair 0, -1” and “ordered pair 1, 0”.
Solution

54

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 3” and “ordered pair 6, 0”.
The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 2” and “ordered pair 6, 0”.
Solution

−13

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 2” and “ordered pair 6, 0”.
The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 6” and “ordered pair 8, 0”.
Solution

−34

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair -1,  0” and “ordered pair 0, -1”.
The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair -4,  0” and “ordered pair -4, 6”.
Solution

34

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair -2,  0” and “ordered pair 4, 4”.
The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 4” and “ordered pair 4, -6”.
Solution

−52

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair -8, 8” and “ordered pair 8, -4”.
The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 1,  4” and “ordered pair 7, 0”.
Solution

−23

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 0,  3” and “ordered pair 7, 0”.
The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 2, 0” and “ordered pair 10, 4”.
Solution

14

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 6,  2” and “ordered pair 0, -3”.

Find the Slope of Horizontal and Vertical Lines

In the following exercises, find the slope of each line.

y=3

Solution

0

y=1

x=4

Solution

undefined

x=2

y=−2

Solution

0

y=−3

x=−5

Solution

undefined

x=−4

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.

(1,4),(3,9)

Solution

52

(2,3),(5,7)

(0,3),(4,6)

Solution

34

(0,1),(5,4)

(2,5),(4,0)

Solution

−52

(3,6),(8,0)

(−3,3),(2,−5)

Solution

−85

(−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 the line given a point and the slope.

(1,−2);m=34

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 5,  1” and “ordered pair 1, -2”

(1,−1);m=12

(2,5);m=−13

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 2, 5” and “ordered pair 5, 4”.

(1,4);m=−12

(−3,4);m=−32

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair -3, 4” and “ordered pair -1, 1”.

(−2,5);m=−54

(−1,−4);m=43

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 2,0” and “ordered pair -1, -4”.

(−3,−5);m=32

(0,3);m=−25

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 0, 3” and “ordered pair 5, 1”.

(0,5);m=−43

(−2,0);m=34

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair -2,0” and “ordered pair 2, 3”.

(−1,0);m=15

(−3,3);m=2

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair -3, 3” and “ordered pair -2, 5”.

(−4,2);m=4

(1,5);m=−3

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 1, 5” and “ordered pair 2, 2”.

(2,3);m=−1

Solve Slope Applications

In the following exercises, solve these slope applications.

Slope of a roof A fairly easy way to determine the slope is to take a 12-inch level and set it on one end on the roof surface. Then take a tape measure or ruler, and measure from the other end of the level down to the roof surface. You can use these measurements to calculate the slope of the roof. What is the slope of the roof in this picture?

The figure shows a wood board at a diagonal representing a side-view slice of a pitched roof. A vertical line segment with arrows on both ends measures the vertical change in height of the roof and is labeled “4 inches”. A level tool is in a horizontal position above the board and above it is a line segment with arrows on both ends labeled “12 inches”.
Solution

13

What is the slope of the roof shown?

The figure shows a  diagonal side-view slice of a pitched roof. A ruler in vertical position is at the bottom of the roof segment and shows unit labels 1 through 8 and extends one further unit. A second ruler starts at the “7” label of the vertical ruler and extends horizontally until it hits the rising roof. The horizontal ruler has unit labels 1 through 11 and extends one further unit.

Road grade A local road has a grade of 6%. The grade of a road is its slope expressed as a percent.

  1. ⓐ Find the slope of the road as a fraction and then simplify the fraction.
  2. ⓑ What rise and run would reflect this slope or grade?
Solution

ⓐ 350ⓑ rise=3;run=50

Highway grade A local road rises 2 feet for every 50 feet of highway.

  1. ⓐ What is the slope of the highway?
  2. ⓑ The grade of a highway is its slope expressed as a percent. What is the grade of this highway?

Everyday Math

Wheelchair ramp The rules for wheelchair ramps require a maximum 1 inch rise for a 12 inch run.

  1. ⓐ What run must the ramp have to accommodate a 24-inch rise to the door?
  2. ⓑ Draw a model of this ramp.
Solution
  1. ⓐ 288 inches (24 feet)
  2. ⓑ Models will vary.

Wheelchair ramp A 1-inch rise for a 16-inch run makes it easier for the wheelchair rider to ascend the ramp.

  1. ⓐ What run must the ramp have to easily accommodate a 24-inch rise to the door?
  2. ⓑ Draw a model of this ramp.

Writing Exercises

What does the sign of the slope tell you about a line?

Solution

Answers will vary.

How does the graph of a line with slope m=12 differ from the graph of a line with slope m=2?

Why is the slope of a vertical line undefined?

Solution

Answers will vary.

Explain how you can graph a line given a point and its slope.

Self Check

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

A self-assessment checklist for students to rate their understanding and confidence in various skills related to finding, graphing, and applying slope, with options: 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?

Chapter Review Exercises

Use the Rectangular Coordinate System

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system.

(1,3),(3,1)

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -6 to 6. The y-axis runs from 6 to -6. The points “ordered pair 1,3” and “ordered pair 3,1” are plotted.

(2,5),(5,2)

In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.

  1. ⓐ (−1,−5)
  2. ⓑ (−3,4)
  3. ⓒ (2,−3)
  4. ⓓ (1,52)
Solution
  1. ⓐ III
  2. ⓑ II
  3. ⓒ IV
  4. ⓐ I


A Cartesian coordinate plane, also known as an xy-grid, displays four quadrants (I, II, III, IV) and four distinct points labeled a, b, c, and d, plotted within these quadrants.

  1. ⓐ (3,−2)
  2. ⓑ (−4,−1)
  3. ⓒ (−5,4)
  4. ⓓ (2,103)

Identify Points on a Graph

In the following exercises, name the ordered pair of each point shown in the rectangular coordinate system.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at 5, 3, “b” at 2, -1, “c” at -3,-2, and “d” at -1,4.
Solution
  1. ⓐ (5,3)
  2. ⓑ (2,−1)
  3. ⓒ (−3,−2)
  4. ⓓ (−1,4)
The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at -2, 2, “b” at 3, 5, “c” at 4,-1, and “d” at -1,3.
The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at 2, 0, “b” at 0, -5, “c” at -4,0, and “d” at 0,3.
Solution
  1. ⓐ (2,0)
  2. ⓑ (0,−5)
  3. ⓒ (−4,0)
  4. ⓓ (0,3)
The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at 0, 4, “b” at 5, 0, “c” at 0,-1, and “d” at -3,0.

Verify Solutions to an Equation in Two Variables

In the following exercises, find the ordered pairs that are solutions to the given equation.

5x+y=10

  1. ⓐ (5,1)
  2. ⓑ (2,0)
  3. ⓒ (4,−10)
Solution

ⓑ (2, 0),ⓒ (4, –10)

y=6x−2

  1. ⓐ (1,4)
  2. ⓑ (13,0)
  3. ⓒ (6,−2)

Complete a Table of Solutions to a Linear Equation in Two Variables

In the following exercises, complete the table to find solutions to each linear equation.

y=4x−1

x y (x,y)
0
1
−2
Solution
x y (x,y)
0 −1 (0,−1)
1 3 (1,3)
−2 −9 (−2,−9)

y=−12x+3

x y (x,y)
0
1
−2

x+2y=5

x y (x,y)
0
1
−1
Solution
x y (x,y)
5 0 (5,0)
1 2 (1,2)
−1 3 (−1,3)

3x−2y=6

x y (x,y)
0
0
−2

Find Solutions to a Linear Equation in Two Variables

In the following exercises, find three solutions to each linear equation.

x+y=3

Solution

Answers will vary.

x+y=−4

y=3x+1

Solution

Answers will vary.

y=−x−1

Graphing Linear Equations

Recognize the Relation Between the Solutions of an Equation and its Graph

In each of the following exercises, an equation and its graph is shown. For each ordered pair, decide

  1. ⓐ if the ordered pair is a solution to the equation.
  2. ⓑ if the point is on the line.

y=−x+4
The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0,  4” and “ordered pair 4, 0”.

  1. (0,4)
  2. (−1,3)
  3. (2,2)
  4. (−2,6)
Solution
  1. ⓐ yes ⓑ yes
  2. ⓐ no ⓑ no
  3. ⓐ yes ⓑ yes
  4. ⓐ yes ⓑ yes

y=23x−1
The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0,  -1” and “ordered pair 3, 1”.

  1. (0,−1)
  2. (3,1)
  3. (−3,−3)
  4. (6,4)

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

y=4x−3

Solution


The graph shows the x y-coordinate plane. Each axis runs from -6 to 6. A line passes through the points “ordered pair 1,  1” and “ordered pair 0, -3”.

y=−3x

2x+y=7

Solution


The graph shows the x y-coordinate plane. Each axis runs from -6 to 6.  A line passes through the points “ordered pair 1,  5” and “ordered pair 0, 7”.

Graph Vertical and Horizontal lines

In the following exercises, graph the vertical or horizontal lines.

y=−2

x=3

Solution


The graph shows the x y-coordinate plane. Each axis runs from -6 to 6. A vertical line passes through the point “ordered pair 0, 3”.

Graphing with Intercepts

Identify the Intercepts on a Graph

In the following exercises, find the x- and y-intercepts.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0,  4” and “ordered pair -4, 0”.
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 6. The y-axis runs from -4 to 2. A line passes through the points “ordered pair 5,  1” and “ordered pair 0, -3”.
Solution

(0,3) (3,0)

Find the Intercepts from an Equation of a Line

In the following exercises, find the intercepts.

x+y=5

x−y=−1

Solution

(−1,0) (0,1)

y=34x−12

y=3x

Solution

(0,0)

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

−x+3y=3

x+y=−2

Solution


This answer graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6.  The equation x plus y equals -2 is  shown. A line passes through the intercepts with coordinates 0, –2 and –2, 0.

Choose the Most Convenient Method to Graph a Line

In the following exercises, identify the most convenient method to graph each line.

x=5

y=−3

Solution

horizontal line

2x+y=5

x−y=2

Solution

intercepts

y=12x+2

y=34x−1

Solution

plotting points

Understand Slope of a Line

Use Geoboards to Model Slope

In the following exercises, find the slope modeled on each geoboard.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 4 row 2.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 4 row 1.
Solution

43

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 3 and the point in column 4 row 4.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 4 row 4.
Solution

−23

In the following exercises, model each slope. Draw a picture to show your results.

13

32

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 3 row 2.

−23

−12

Solution


The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 3 row 3.

Find the Slope of a Line from its Graph

In the following exercises, find the slope of each line shown.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0,  0” and “ordered pair 2, -6”.
The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0,  4” and “ordered pair -4, 0”.
Solution

1

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair -4,  -4” and “ordered pair 5, -1”.
The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair -3,  6” and “ordered pair 5, 2”.
Solution

−12

Find the Slope of Horizontal and Vertical Lines

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.

(2,1),(4,5)

(−1,−1),(0,−5)

Solution

−4

(3,5),(4,−1)

(−5,−2),(3,2)

Solution

12

Graph a Line Given a Point and the Slope

In the following exercises, graph the line given a point and the slope.

(2,−2);m=52

(−3,4);m=−13

Solution


The graph shows the x y-coordinate plane. The x-axis runs from -6 to 6. The y-axis runs from -4 to 2. A line passes through the points “ordered pair -3,  4” and “ordered pair 1, 3”.

Solve Slope Applications

In the following exercise, solve the slope application.

A roof has rise 10 feet and run 15 feet. What is its slope?


Chapter Practice Test

Plot and label these points:

  1. ⓐ (2,5)
  2. ⓑ (−1,−3)
  3. ⓒ (−4,0)
  4. ⓓ (3,−5)
  5. ⓔ (−2,1)
Solution


The graph shows the x y-coordinate plane. The axes extend from -6 to 6. a is plotted at 2, 5, b at -1, -3, c at -4, 0, d at 3, -5, and e at -2,1.

Name the ordered pair for each point shown.

The graph shows the x y-coordinate plane. The axes extend from -7 to 7. A is plotted at -4, 1, B at 3, 2, C at 0, -2, D at -1, -4, and E at 4,-3.

Find the x-intercept and y-intercept on the line shown.

 The graph shows the x y-coordinate plane. The x-axis runs from -7 to 7. The y-axis runs from -7 to 7. A line passes through the points “ordered pair 4,  0” and “ordered pair 0, -2”.
Solution

(4,0), (0,−2)

Find the x-intercept and y-intercept of the equation 3x−y=6.

Is (1,3) a solution to the equation x+4y=12? How do you know?

Solution

no; 1 + 4 · 3 ≠ 12

Complete the table to find four solutions to the equation y=−x+1.

x y (x,y)
0
1
3
−2

Complete the table to find three solutions to the equation 4x+y=8

x y (x,y)
0
0
3
Solution
x y (x,y)
0 8 (0,8)
2 0 (2,0)
3 −4 (3,−4)

In the following exercises, find three solutions to each equation and then graph each line.

y=−3x

2x+3y=−6

Solution


 The graph shows the x y-coordinate plane. The x-axis runs from -6 to 6. The y-axis runs from -6 to 6. A line passes through the points “ordered pair 0,  -2” and “ordered pair -3, 0”.

In the following exercises, find the slope of each line.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. The y-axis runs from -5 to -4. A line passes through the points “ordered pair 6,  4” and “ordered pair 0, -3”.
The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 3,  0” and “ordered pair 1, 5”.
Solution

−52

Use the slope formula to find the slope of the line between (0,−4) and (5,2).

Find the slope of the line y=2.

Solution

0

Graph the line passing through (1,1) with slope m=32.

A bicycle route climbs 20 feet for 1,000 feet of horizontal distance. What is the slope of the route?

Solution

150

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.

Cumulative Review

Note: Answers to the Cumulative Review can be found in the Supplemental Resources available to instructors. Please visit https://openstax.org/ to view an updated list of the Learning Resources for this title and how to access them.

Chapter 1 Whole Numbers

No exercises.

Chapter 2 The Language of Algebra

Simplify:

1.5(3+2·6)−82

Solve:

2.17=y−13

3.p+14=23

Translate into an algebraic expression.

4.11 less than the product of 7 and x.

Translate into an algebraic equation and solve.

5.The difference of y and 7 gives 84.

6.Find all the factors of 72.

7.Find the prime factorization of 132.

8.Find the least common multiple of 12 and 20.

Chapter 3 Integers

Simplify:

9.|8−9|−|3−8|

10.−2+4(−3+7)

11.27−(−4−7)

12.28÷(−4)−7

Translate into an algebraic expression or equation.

13.The sum of −5 and 13, increased by 11.

14.The product of −11and8.

15.The quotient of 7 and the sum of −4andm.

16.The product of −3 and y is −51.

Solve:

17.−6r=24

Chapter 4 Fractions

18.Locate the numbers on a number line. 78,53,314,5.

Simplify:

19.21p57q

20.37·(−2845)

21.−634÷92

22.−335÷6

23.−423(−67)

24.−214−38

25.7·8+4(7−12)9·6−2·9

26.−2336+1720

27.12+1334−13

28.358−212

29.−23r=24

Chapter 5 Decimals

Simplify:

30.24.76−7.28

31.12.9+15.633

32.(−5.6)(0.25)

33.$6.29÷12

34.34(13.44−9.6)

35.64+225

36.121x2y2

37.Write in order from smallest to largest: 58,0.75,815

Solve:

38.−8.6x=34.4

39.Using 3.14 as the estimate for pi, approximate the (a) circumference and (b) area of a circle whose radius is 8 inches.

40.Find the mean of the numbers, 18,16,20,12

41.Find the median of the numbers, 24,29,27,28,30

42.Identify the mode of the numbers, 6,4,4,5,6,6,4,4,4,3,5

43.Find the unit price of one t-shirt if they are sold at 3 for $28.97.

Chapter 6 Percents

44.Convert 14.7% to (a) a fraction and (b) a decimal.

Translate and solve.

45.63 is 35% of what number?

46.The nutrition label on a package of granola bars says that each granola bar has 180 calories, and 81 calories are from fat. What percent of the total calories is from fat?

47.Elliot received $510 commission when he sold a $3,400 painting at the art gallery where he works. What was the rate of commission?

48.Nandita bought a set of towels on sale for $67.50. The original price of the towels was $90. What was the discount rate?

49.Alan invested $23,000 in a friend’s business. In 5 years the friend paid him the $23,000 plus $9,200 simple interest. What was the rate of simple interest per year?

Solve:

50.9p=−614

Chapter 7 The Properties of Real Numbers

51.List the (a) whole numbers, (b) integers, (c) rational numbers, (d) irrational numbers,

(e) real numbers −5,−214,−4,0.25¯,135,4

Simplify:

52.(815+47)+37

53.3(y+3)−8(y−4)

54.817·49·178

55.A playground is 55 feet wide. Convert the width to yards.

56.Every day last week Amit recorded the number of minutes he spent reading. The recorded number of minutes he read each day was 48,26,81,54,43,62,106. How many hours did Amit spend reading last week?

57.June walked 2.8 kilometers. Convert this length to miles knowing 1 mile is 1.61 kilometer.

Chapter 8 Solve Linear Equations

Solve:

58.y+13=−8

59.p+25=85

60.48=23x

61.4(a−3)−6a=−18

62.7q+14=−35

63.4v−27=7v

64.78y−6=38y−8

65.26−4(z−2)=6

66.34x−23=12x−56

67.0.7y+4.8=0.84y−5.0

Translate and solve.

68.Four less than n is 13.

Chapter 9 Math Models and Geometry

69.One number is 8 less than another. Their sum is negative twenty-two. Find the numbers.

70.The sum of two consecutive integers is −95. Find the numbers.

71.Wilma has $3.65 in dimes and quarters. The number of dimes is 2 less than the number of quarters. How many of each coin does she have?

72.Two angles are supplementary. The larger angle is 24° more than the smaller angle. Find the measurements of both angles.

73.One angle of a triangle is 20° more than the smallest angle. The largest angle is the sum of the other angles. Find the measurements of all three angles.

74.Erik needs to attach a wire to hold the antenna to the roof of his house, as shown in the figure. The antenna is 12 feet tall and Erik has 15 feet of wire. How far from the base of the antenna can he attach the wire?

A Yagi antenna is mounted on a 12-foot pole, secured by a 15-foot guy wire, forming a right triangle. This common setup demonstrates stable antenna installation for reception.

75.The width of a rectangle is 4 less than the length. The perimeter is 96 inches. Find the length and the width.

76.Find the (a) volume and (b) surface area of a rectangular carton with length 24 inches, width 18 inches, and height 6 inches.

Chapter 10 Polynomials

Simplify:

77.(8m2+12m−5)−(2m2−7m−1)

78.p3·p10

79.(y4)3

80.(3a5)3

81.(x3)5(x2)3

82.(23m3n6)(16m4n4)

83.(y−4)(y+12)

84.(3c+1)(9c−4)

85.(x−1)(x2−3x−2)

86.(8x)0

87.(x3)5(x2)4

88.32a7b212a3b6

89.(ab−3)(a−3b6)

90.Write in scientific notation: (a)4,800,000(b)0.00637

Factor the greatest common factor from the polynomial.

91.3x4−6x3−18x2

Chapter 11 Graphs

Graph:

92.y=4x−3

93.y=−3x

94.y=12x+3

95.x−y=6

96.y=−2

97.Find the intercepts. 2x+3y=12

Graph using the intercepts.

98.2x−4y=8

99.Find the slope of the line shown.

A graph with x-axis ranging from –6 to 6 and a y-axis ranging from –6 to 6. The line runs straight sloping upwards and running through the points (–1, –3) and (4, –1).

100.Use the slope formula to find the slope of the line between the points (−5,−2),(3,2).

101.Graph the line passing through the point (−3,4) and with slope m=−13.

Powers and Roots Tables

n n2 n n3 n3
1 1 1 1 1
2 4 1.414214 8 1.259921
3 9 1.732051 27 1.442250
4 16 2 64 1.587401
5 25 2.236068 125 1.709976
6 36 2.449490 216 1.817121
7 49 2.645751 343 1.912931
8 64 2.828427 512 2
9 81 3 729 2.080084
10 100 3.162278 1,000 2.154435
11 121 3.316625 1,331 2.223980
12 144 3.464102 1,728 2.289428
13 169 3.605551 2,197 2.351335
14 196 3.741657 2,744 2.410142
15 225 3.872983 3,375 2.466212
16 256 4 4,096 2.519842
17 289 4.123106 4,913 2.571282
18 324 4.242641 5,832 2.620741
19 361 4.358899 6,859 2.668402
20 400 4.472136 8,000 2.714418
21 441 4.582576 9,261 2.758924
22 484 4.690416 10,648 2.802039
23 529 4.795832 12,167 2.843867
24 576 4.898979 13,824 2.884499
25 625 5 15,625 2.924018
26 676 5.099020 17,576 2.962496
27 729 5.196152 19,683 3
28 784 5.291503 21,952 3.036589
29 841 5.385165 24,389 3.072317
30 900 5.477226 27,000 3.107233
31 961 5.567764 29,791 3.141381
32 1,024 5.656854 32,768 3.174802
33 1,089 5.744563 35,937 3.207534
34 1,156 5.830952 39,304 3.239612
35 1,225 5.916080 42,875 3.271066
36 1,296 6 46,656 3.301927
37 1,369 6.082763 50653 3.332222
38 1,444 6.164414 54,872 3.361975
39 1,521 6.244998 59,319 3.391211
40 1,600 6.324555 64,000 3.419952
41 1,681 6.403124 68,921 3.448217
42 1,764 6.480741 74,088 3.476027
43 1,849 6.557439 79,507 3.503398
44 1,936 6.633250 85,184 3.530348
45 2,025 6.708204 91,125 3.556893
46 2,116 6.782330 97,336 3.583048
47 2,209 6.855655 103,823 3.608826
48 2,304 6.928203 110,592 3.634241
49 2,401 7 117,649 3.659306
50 2,500 7.071068 125,000 3.684031
51 2,601 7.141428 132,651 3.708430
52 2,704 7.211103 140,608 3.732511
53 2,809 7.280110 148,877 3.756286
54 2,916 7.348469 157,464 3.779763
55 3,025 7.416198 166,375 3.802952
56 3,136 7.483315 175,616 3.825862
57 3,249 7.549834 185,193 3.848501
58 3,364 7.615773 195,112 3.870877
59 3,481 7.681146 205,379 3.892996
60 3,600 7.745967 216,000 3.914868
61 3,721 7.810250 226,981 3.936497
62 3,844 7.874008 238,328 3.957892
63 3,969 7.937254 250,047 3.979057
64 4,096 8 262,144 4
65 4,225 8.062258 274,625 4.020726
66 4,356 8.124038 287,496 4.041240
67 4,489 8.185353 300,763 4.061548
68 4,624 8.246211 314,432 4.081655
69 4,761 8.306624 328,509 4.101566
70 4,900 8.366600 343,000 4.121285
71 5,041 8.426150 357,911 4.140818
72 5,184 8.485281 373,248 4.160168
73 5,329 8.544004 389,017 4.179339
74 5,476 8.602325 405,224 4.198336
75 5,625 8.660254 421,875 4.217163
76 5,776 8.717798 438,976 4.235824
77 5,929 8.774964 456,533 4.254321
78 6,084 8.831761 474,552 4.272659
79 6,241 8.888194 493,039 4.290840
80 6,400 8.944272 512,000 4.308869
81 6,561 9 531,441 4.326749
82 6,724 9.055385 551,368 4.344481
83 6,889 9.110434 571,787 4.362071
84 7,056 9.165151 592,704 4.379519
85 7,225 9.219544 614,125 4.396830
86 7,396 9.273618 636,056 4.414005
87 7,569 9.327379 658,503 4.431048
88 7,744 9.380832 681,472 4.447960
89 7,921 9.433981 704,969 4.464745
90 8,100 9.486833 729,000 4.481405
91 8,281 9.539392 753,571 4.497941
92 8,464 9.591663 778,688 4.514357
93 8,649 9.643651 804,357 4.530655
94 8,836 9.695360 830,584 4.546836
95 9,025 9.746794 857,375 4.562903
96 9,216 9.797959 884,736 4.578857
97 9,409 9.848858 912,673 4.594701
98 9,604 9.899495 941,192 4.610436
99 9,801 9.949874 970,299 4.626065
100 10,000 10 1,000,000 4.641589

Geometric Formulas

2 Dimensions
Name Shape Formulas
Rectangle A rectangular box with the right angle indicated. the length is labeled l and the width is labeled w. Perimeter:P=2l+2wArea:A=lw
Square A square with side length 's' and a right angle indicated in one corner, showcasing a fundamental geometric shape. Perimeter:P=4sArea:A=s2
Triangle A triangle with a dotted line (labeled h) running from the base to the top corner (angle B), creating a 90 degree angle. The legs of the triangle are labeled a, b, and c. The angles of the triangle are labeled B, C, and A. Perimeter:P=a+b+cArea:A=12bhSum of Angles:A+B+C=180°
Right Triangle A right-angled triangle with sides labeled 'a' (vertical), 'b' (horizontal), and 'c' (hypotenuse). A square indicates the right angle at the bottom right corner. Pythagorean Theorem:a2+b2=c2Area:A=12ab
Circle A circle with its center marked 'd'. A dashed line from the center to the circumference indicates the radius 'r'. A horizontal dashed line represents the diameter passing through 'd'. C=2πrCircumference:orC=πdArea:A=πr2
Parellelogram A parallelogram with its base labeled b and its width labeled a. A dotted line (labeled h) runs from the base to the top left corner of the parallelogram, creating a 90 degree angle. Perimeter:P=2a+2bArea:A=bh
Trapezoid A trapezoid with labels a, b, c, and B (for the base). A dotted line (labeled h) runs from the base to the top left corner, creating a 90 degree angle. Perimeter:P=a+b+c+BArea:A=12(B+b)h
3 Dimensions
Name Shape Formulas
Rectangular Solid A 3D rectangular solid with its length, width, and height labeled l, w, and h, respectively. Volume:V=lwhSurface Area:SA=2lw+2wh+2hl
Cube A cube with its height labeled s. Volume:V=s3Surface Area:SA=6s2
Cone An illustration of a right circular cone with its height 'h' and base radius 'r' clearly marked. A dashed line indicates the height from the apex to the center of the circular base, forming a right angle. Volume:V=13πr2hSurface Area:SA=πr2+πrh2+r2
Sphere A diagram of a sphere, displaying its central point and a radius 'r' extending to the edge. A dashed arc indicates the equator, highlighting its three-dimensional structure. Volume:V=43πr3Surface Area:SA=4πr2
Right Circular Cylinder A cylinder with its base radius labeled as r and the height of the cylinder labeled as h. Volume:V=πr2hSurface Area:SA=2πr2+2πrh